JP2684551C - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION (Field of application of the invention)   The present invention divides the object into a plurality of regions, and performs photometry for each of the plurality of regions.
A photo camera or a photo camera that gives proper exposure to the main part of the screen using photometric values
Regarding improvement of camera systems such as video cameras, camera bodies and interchangeable lenses
It is. (Background of the Invention)   Conventionally, an object scene is divided into a plurality of parts, photometry is performed for each area, and the plurality of photometric values are used.
A photometric device which gives a proper exposure to a photographing screen by using a light source is disclosed, for example, in Jpn.
71, JP-A-54-123030 and the like.   The proposed device measures the brightness of each divided area of the object scene,
The condition for obtaining the proper exposure of the screen is selected and the calculation based on the condition is performed. Yo
Therefore, in such an apparatus, it is first necessary to accurately measure the luminance of each area.
However, in the case of cameras with interchangeable shooting lenses, the peripheral image surface
The maximum difference is about two steps when compared with the open F-number. Therefore
In the case of aperture metering with some lenses, accurate metering is not performed, and errors in the calculation result
Problem. (Object of the invention)   SUMMARY OF THE INVENTION An object of the present invention is to solve the above-described problem and to provide correction value information corresponding to image plane light quantity drop.
By correcting the photometric output using the camera, the brightness of the subject can be measured accurately.
Camera system. (Features of the invention)   In order to achieve the above object, the present invention provides an interchangeable lens and the detachable interchangeable lens.
In a camera system comprising a functional camera body,
A correction value output unit that can output correction value information corresponding to the image surface light quantity drop at a regular time
At least the brightness of the central area of the screen and the brightness of the peripheral area
Light-receiving means for independently measuring the degree, the luminance measured by the light-receiving means and the replacement lens
The brightness of each area based on the correction value information obtained from the correction value output means.
The scene is determined using the corrected brightness of each area, and based on the scene determination. Computing means for computing a photometric value at the time of the luminance measurement.
. (Example of the invention)   Hereinafter, the present invention will be described in detail based on illustrated embodiments.   FIG. 1 shows an optical arrangement of a photometric system for realizing the present invention with a single-lens reflex camera.
FIG. In the figure, 1 is a photographing lens, 2 is a quick return mirror, 3 is a reticle, 4
Denotes a pentaprism, 5 denotes an imaging lens, and 6 has a light receiving surface divided into six regions.
A light receiving element, 7 is an eyepiece, and 8 is a film. In FIG.
The subject image formed on the focusing screen 3 is guided by the imaging lens 5 onto the light receiving element 6 and
An image is formed and photometry is performed. FIG. 2 explains the light receiving surface of the light receiving element 6 shown in FIG.
FIG. 6A shows an area for measuring the luminance at the center of the object scene, and FIG.
An area for measuring the luminance, and 6C is an area for measuring the luminance around the area.
Area 6C has four small areas 6C₁~ 6C_FourIt is composed of   The above photometry system performs scene discrimination by multi-segment photometry and automatically performs exposure compensation.
This is a so-called evaluation photometry system. For example, with a conventional camera, one finder screen
Because the photometer was used to measure the light with a light-receiving element with
In a so-called backlight scene, the photometric value is dominated by a bright background, and the main subject is exposed.
There was a drawback that it would be under, but multi-segment metering like this metering system
In the case of, since the photometry can be performed by separating the central part and the peripheral part, the brightness difference between the central part and the peripheral part
Is detected, it can be determined that the scene is a backlight scene, and depending on the degree of backlight,
Appropriate exposure correction can be performed. In addition, a normal light scene or highlight scene
Scenes and shadow scenes can be properly exposed by scene discrimination.
It is.   However, a single-lens reflex camera can be replaced with a standard lens.
If you change the lens even if you calculate the photometric value for each area so that
An error occurs in the exposure value. The reason for this depends on the lens attached as described above.
This is because the degree of decrease in the amount of image surface light in the peripheral portion of the screen is different. That is, the image surface light quantity
Light receiving parts 6A, 6B, 6C due to drop₁~ 6C_FourThe metering value of
This is because an error occurs in the result. FIG. 3 is a graph showing the decrease in the amount of light on the image surface of the photographing lens 1. It shows an example of the degree.   FIG. 4 is a schematic block diagram of the present embodiment showing a signal processing process for correcting the photometric error.
In the figure, reference numeral 101 denotes an image surface light amount drop correction value V of the lens._vgOutput
A correction value output circuit (the circuit is incorporated for each interchangeable lens);
Each light receiving section 6A, 6B, 6C of the light receiving element 6₁~ 6C_Four(In Fig. 4, each area is independent.
Voltage equivalent to the luminance value measured by
V_A, V_B, V_C1~ V_C4, And 108-112 are the correction values.
Correction value V from output circuit 101_vgBrightness value V corrected according to_XB, V_XC1~ V_XC4To
A luminance correction circuit for output, 113 is a photometry value calculation circuit, and 114 is an exposure control circuit
You. By configuring such a signal processing circuit, input is made when the lens is replaced.
Correction value V of the lens_vgBrightness value (photometry for each area) corrected according to the information
Value), and a photometric error due to lens replacement can be prevented.   Next, specific circuit diagrams for realizing the circuit configuration shown in FIG.
It is shown in FIG. Reference numerals 115 to 120 denote light receiving portions 6A, 6B, and 6C of the light receiving element 6, respectively.₁~ 6C_Four
Current iA, iB, iC generated in₁~ IC_FourLogarithmic compression circuit
, Each photocurrent is now logarithmically compressed and the voltage V_A, V_B, V_C1~ V_C4Is output as
. Here, the voltage V_A, V_B, V_C1~ V_C4Is a constant a₁~ A₆(≧ 0), b (≧ 0) and
And photocurrent i_a, I_b, I_C1~ I_C4Can be expressed as follows.     V_A = A₁+ Blni_A     V_B = A_Two+ Blni_B     V_C1= A_Three+ Blni_C1     V_C2= A_Four+ Blni_C2     V_C3= A_Five+ Blni_C3     V_C4= A₆+ Blni_C4 Where a₁~ A₆Is V when the brightness of each light receiving section is equal._A= V_B= V_C1= V_C2= V
_C3= V_C4Is set in advance in the logarithmic compression circuits 115 to 120 so that
And A correction value output circuit disposed on the lens side (in the lens barrel) described above.
Reference numeral 101 denotes two types of reference voltages V corresponding to correction values of the lens._vg1, V_{vg Two} Is output through the contacts 101a and 101b.   121 to 140 are resistors having the same resistance value, and 141 to 145 are operational amplifiers.
, Resistors 121-124 and operational amplifier 141, resistors 125-128 and operational amplifier
143, resistors 133 to 136 and operational amplifier 144, resistors 137 to 140 and op pair
Each of the subtraction circuits is constituted by the amplifiers 145._B
~ V_vg1, V_C1-V_vg2, V_C2-V_vg2, V_C3-V_vg2, V_C4-V_vg2Voltage
Is forced. In the following description, V_B-V_vg1Value is V_bIs defined.   Numeral 20 is output from the logarithmic compression circuits 117 to 120, and is arranged at each subsequent stage.
Corrected voltage V output through the subtraction circuit_C1-V_vg2, V_C2-V_vg2,
V_C3-V_vg2, V_C4-V_vg2Are input terminals IN₁~ IN_FourMore input, the outermost of the scene
Surrounding area 6C₁~ 6C_FourIs calculated, and the voltage V is output from the output terminal OUT._COutput
Peripheral luminance value calculation circuit.   FIG. 6 shows the configuration of the peripheral luminance value calculation circuit 20. In FIG. 6, 32,
Reference numerals 33, 34, 35, and 36 denote magnitude comparison circuits, respectively.₁, I_TwoTo
Of the two input voltage values, the larger one is output from the output terminal OH, and the smaller one is
It is output from the output terminal OL. Specific examples of the magnitude comparison circuits 32 to 35 are described below.
It is shown in FIG. In FIG. 7, reference numeral 66 denotes a positive-phase and negative-phase input terminal and a positive-phase and negative-phase output terminal.
The relationship between the positive-phase input terminal voltage V + and the negative-phase input terminal voltage V− is V + ≧
In the case of V-, a voltage of H (meaning high level) is applied from the positive-phase input terminal to a negative-phase voltage.
An L-level (meaning low-level) voltage is output from the output terminal. 67-70
An analog switch that conducts when the voltage applied to the control terminal is H level
State, and becomes open when it is at the L level. Input terminal I₁Input voltage to V_l
1, I_TwoThe voltage input to the input terminal is V_l2Then V_I1≧ V_I2Comparing when
Each input of the data 66 becomes the positive-phase input terminal voltage V + ≧ the negative-phase input terminal voltage V−. At this time,
The positive-phase output terminal voltage of the comparator 66 becomes H level, and the negative-phase output terminal voltage becomes L level.
The control voltage of the analog switches 67 and 70 is H level,
Since the control voltages of the switches 68 and 69 are at the L level,
7 and 70 are conducting, and the analog switches 68 and 69 are open. Yo The output terminal OH has a voltage V through an analog switch 67._I1Is output and the output terminal
OL has a voltage V through an analog switch 70._I2Is output. Similarly, V_I1<
V_I2, Each input of the comparator 66 is a positive-phase input terminal voltage V + <a negative-phase input terminal
The voltage at the positive-phase output terminal of the comparator 66 is L level,
The pressure goes to the H level. Therefore, the control voltage of the analog switches 67 and 70
Is at L level, the analog switches 67 and 70 are open, and
Since the control voltages of the switching switches 68 and 69 are at the H level,
The switches 68 and 69 become conductive. At this time, an analog switch is connected to the output terminal OH.
Voltage V through 69_I2Is output through the analog switch 68 to the output terminal OL.
Voltage V_I1Is output. Thus, the voltage V is applied to the output terminal OH._I1, V_I2Horse
The larger one is output to the output terminal OL, and the smaller one is output to the output terminal OL. Fig. 6 shows the size comparison circuit.
And the voltage V_C1, V_C2, V_C3, V_C4The largest one is V_H1When
Then, the second largest one is V_H2And the third largest is V_H3As the most
V is also small_H4Output. The size comparison circuit 32
Input terminal I₁Has a voltage (V_C1-V_vg2) Is input and the input terminal I_TwoHas a voltage (V_C2-V_v
g2) Is input, the larger one is output from the output terminal OH, and the smaller one is output to OL.
Output from the power end. Similarly, the input terminal I of the magnitude comparison circuit 33₁Has a voltage (V_C3
-V_vg2) Is input, the larger one is output from the output terminal OH, and the smaller one is output.
The signal is output from the output terminal OL. In the size comparison circuit 34, the size comparison circuits 32, 33
The voltage output from the output terminal OH of the₁, I_TwoTo compare large and small,
Is output from the output terminal OH and the smaller one is output from the output terminal OL.
Here, the voltage output from the output terminal OH is V_C1-V_vg2, V_C2-V_vg2, V_C3-V_vg
Two, V_C4-V_vg2Is the largest voltage. Similarly, in the magnitude comparison circuit 35,
The voltages output from the output terminals OL of the magnitude comparison circuits 32 and 33 are applied to the input terminals 11 and 12.
Output and compare the size, the larger one from the output terminal OH and the smaller one from the output terminal OL
Output from the output terminal OL._C1-V_vg2
, V_C2-V_vg2, V_C3-V_vg2, V_C4-V_vg2Is the smallest voltage. Big and small
The comparison circuit 36 is configured to output the output voltage from the output terminal OL of the magnitude comparison circuit 34 and the magnitude comparison circuit.
35 and the output voltage from the output terminal OH₁, I_TwoInput to the input end And compare the size. In the magnitude comparison circuit 36, V_C1-V_vg2, V_C2-V_vg2, V
_C3-V_vg2, V_C4-V_vg2The two intermediate voltage values excluding the maximum and minimum
From the output terminal OH of the magnitude comparison circuit 36,_C1-V_vg2, V_C2-V_vg2
, V_C3-V_vg2, V_C4-V_vg2, The second largest voltage is output, and the output terminal OL
Output the third largest voltage. Thus, V_C1-V_vg2, V_C
Two-V_vg2, V_C3-V_vg2, V_C4-V_vg2, V_H1≧ V_H2≧ V_H3≧ V_H4To
It is rearranged to become.   Returning to FIG._C1, V_C2, V_C3, V_C4The same for averaging
And the output voltage is V_Cm= (V_C1+ V_C2+ V_C3+ V_C4)
/ 4-V_vg2Becomes 41 is a reference voltage generating circuit for four reference voltages V_r1, V_r2, V_r
Three, V_r4Is the output end₁~ R_FourAnd this reference voltage V_r1, V_r2, V_r3,
V_r4The relationship between V and V is_r1> V_r2> V_r3> V_r4It is. 42 to 45 are comparators
And the voltage V_CmAnd reference voltage V_r1, V_r2, V_r3, V_r4, And V_Cm≧
V_r1, V_Cm≧ V_r2, V_Cm≧ V_r3, V_Cm≧ V_r4At the time of each H level voltage
And outputs V_Cm<V_r1, V_Cm<V_r2, V_Cm<V_r3, V_Cm<V_r4Each time
And outputs an L level voltage. 46, 47, 48, and 49 are inverters;
Reference numerals 0, 51 and 52 are AND gates. The inverter 46 has a comparator 42
The output of the inverter 47 is the output of the comparator 43, and the output of the inverter 48 is
The output of the comparator 44 is the output of the comparator 45, and the output of the comparator 45 is the inverter 49.
Has been entered. The output of the inverter 46 is connected to one input terminal of the AND gate 50.
However, the other input terminal receives the output of the comparator 43 and the one input terminal of the AND gate 51.
The output of the inverter 47 is provided at the input end, and the output of the comparator 44 is provided at the other input end.
, And one input terminal of the AND gate 52 receives the output of the inverter 48 and the other input terminal thereof.
Are supplied with the output of the comparator 45, respectively. 53-56 is Oage
It is. The output of the comparator 42 and the AND gate
The output of the gate 50 and the output of the AND gate 51 are input. Or gate
The output of the comparator 42, the output of the AND gate 50, and the
The output of the gate 51 and the output of the AND gate 52 are input. Or
The output of the AND gate 50 and the AND gate 5 1, the output of the AND gate 52, and the output of the inverter 49 are input respectively.
ing. The output of AND gate 51 and AND gate
The output of the inverter 52 and the output of the inverter 49 are input. 57-60
Analog switches, controlled by OR gates 53 to 56, respectively
Have been. 61 to 64 are resistors having the same resistance value. The resistors 61 to 64
V output from one of the analog switches 57 to 60 that is in a conductive state._H1, V_H2, V_H3,
V_H4Two or three or four in descending order or two in descending order
Or for averaging three voltages. An operational amplifier 65 has an output
The terminal and the negative-phase input terminal are connected and used as a voltage follower. Operational amplifier
65 output terminal equals the positive-phase input terminal voltage regardless of the circuit state after the output terminal.
Output voltage. The output voltage of the operational amplifier 65 is V_COutput terminal OUT
Output from.   Next, V in FIG._r1, V_r2, V_r3, V_r3And V_CmFrom the relationship of size to the relationship
Of the scene and the voltage V at that time_CIs described. (1) V_r1≤V_Cm..... Analog switch 57, 58 conduction   In this case, since a scene with a considerably bright background is assumed, 6C₁, 6C_Two,
6C_Three, 6C_FourIgnoring the brightness of the relatively low brightness area of each area of_C= (
V_H1+ V_H2) / 2. (2) V_r2≤V_Cm<V_r1... Analog switches 57, 58, 59 conduction   At this time, a scene with a slightly bright background, such as outdoors in fine weather, is assumed.
In such scenes, the brightness of a background that has a low brightness value, such as shaded ground,
And inappropriate elements are likely to appear. Therefore, in this case, the luminance of the lowest luminance area
Ignoring and V_C= (V_H1+ V_H2+ V_H3) / 3. (3) V_r3≤V_Cm<V_r2…… Analog switches 57, 58, 59, 60 conduction   At this time, a scene with a standard brightness background is assumed.
Cannot ignore the relatively bright and relatively dark parts of the scene. So this
In such a case, V_C= (V_H1+ V_H2+ V_H3+ V_H4) / 4. (4) V_r4≤V_Cm<V_r3..... Analog switches 58, 59, 60 conduction   At this time, a scene with a slightly dark background is assumed, so ignore the area with the highest brightness. Then V_C= (V_H2+ V_H3+ V_H4) / 3. (5) V_Cm<V_r4...... Analog switches 59 and 60 conducting   In this case, a scene with a considerably dark background such as a night view is assumed.
In the scene, for example, it is inappropriate as the brightness of the background that locally shows high brightness such as electric light
Elements tend to appear. Therefore, in this case, ignoring the relatively high brightness area,
V_C= (V_H3+ V_H4) / 2.   As described above, the peripheral luminance calculation circuit 20 responds to the luminance state of the outermost peripheral part of the scene.
By the way, V_H1, V_H2, V_H3, V_H42 or 3 or 4 from the larger of
Or the average of two or three of the smaller_COutput terminal OU
Output from T.   5 is the output voltage V of the logarithmic compression circuits 115 and 116._A, V_BAnd the circumference
Output voltage V of edge luminance calculation circuit 20_CAre input terminals IN₁, IN_Two, IN_ThreeEnter
, A selection circuit that determines which one of a plurality of arithmetic expressions to be described later is to be selected.
is there. In FIG. 5, reference numerals 22, 23, and 24 denote resistors having the same resistance value.
5 is an analog switch. Resistors 22, 23, 24 and analog switch 25
Constitutes an average value circuit, and the output terminal A / P of the selection circuit 21 outputs an H level signal.
Is output and input to the control terminal of the analog switch 25,
The analog switch 25 is turned on, and the output voltage V₁Is (V_A+
V_b+ V_C) / 3. On the other hand, an L-level voltage from the output terminal A / P of the selection circuit 21
Is output and input to the control terminal of the analog switch 25.
Switch 25 is opened, and the output voltage V₁Is (V_A+ V_b) /
It becomes 2. Reference numeral 26 denotes an operational amplifier.
-The output voltage V of the average circuit is connected to the positive-phase input terminal.₁But
Has been entered. The output terminal voltage of the operational amplifier 26 is changed to the circuit state after the output terminal.
V₁It is. 27, 28, 29 and 30 are resistors having the same resistance value
, 31 are operational amplifiers. For resistors 27, 28, 29, 30 and operational amplifier 31
This constitutes a subtraction circuit. The voltage from the output terminal OUT of the selection circuit 21 is V_TwoWhen
Then, the output voltage of the subtraction circuit becomes V₁-V_TwoIt is. This voltage V₁-V_TwoBut the real
In the embodiment, photometry determined by a plurality of arithmetic expressions described later Expresses a value.   FIG. 8 is a circuit diagram showing a specific example of the selection circuit 21 in FIG. 71,7
2, 73, 74 are resistors having the same resistance value, and 75 is an operational amplifier,
It constitutes an arithmetic circuit. Similarly, 76, 77, 78 and 79 have the same resistance value
The resistor 80 is an operational amplifier and constitutes a second subtraction circuit. First subtraction
Voltage V_AAnd V_bAnd the output voltage is V_b-V_AIt is. Second subtraction
Circuit has voltage V_bAnd V_CAnd the output voltage is V_C-V_bIt is. 81 is the base
Reference voltage V in reference voltage generation circuit_r5Has occurred. 82 is a comparator whose positive
Voltage V at the phase input terminal_CIs input, and the reference voltage V_r5Is input and V_C
≧ V_r5Output H-level voltage when_C<V_r5At the time of L level voltage
Output. Reference numeral 85 denotes a reference voltage generation circuit._Pa, V_Pb, V_Qa, V_QbGenerate
You. The reference voltage generation circuit 85 has a control terminal B / D, and the comparator 8
2 is input as the control voltage. H level at terminal B / D
The reference voltage when the control voltage of_Pa= V_P1, V_Pb= V_P2, V_Q
a= V_Q1, V_Qb= V_Q2On the other hand, an L level control voltage is input to the terminal B / D.
The reference voltage when the_Pa= V_P3, V_Pb= V_P4, V_Qa= V_Q3, V_Qb= V_Q4Too
Good. Reference voltage V_P1, V_P2Or V_P3, V_P4, V_Q1, V_Q2Or V_Q3, V_Q4The sign of
V_P1, V_P3, V_Q1, V_Q3Is positive, V_P2, V_P4, V_Q2, V_Q4Is negative. 86,87
Is a comparator, and 88 and 89 are inverters. Comparator 86, 87 positive
The output voltage V of the operational amplifier 75 of the first subtraction circuit is connected to the phase input terminal._b-V_AIs entered
You. The negative-phase input terminal of the comparator 86 has a control of the reference voltage generation circuit 85.
The reference voltage is set according to the H level and L level of the control voltage input to the terminal B / D.
Pressure V_P1Or V_P3Is entered. Similarly, the negative-phase input terminal of the comparator 87
Reference voltage V_P2Or V_P4Is entered. 90, 91 are comparators, 92, 9
3 is an inverter. The second subtraction is provided to the positive phase input terminals of the comparators 90 and 91.
The output voltage V of the operational amplifier 80 of the circuit_C-V_bIs entered. Reverse of comparator 91
A phase input terminal is connected to a control terminal B / D of the reference voltage generation circuit 85.
The reference voltage V depends on the H level and L level of the control voltage._Q1Or V_Q3Is input
Is done. Similarly, the negative-phase input terminal of the comparator 90 has a reference Voltage V_Q2Or V_Q4Is entered. Output of comparators 86, 87, 90, 91
Depending on the magnitude of the positive-phase input terminal voltage V + and the negative-phase input terminal voltage V−, V + ≧ V−
A high-level voltage is generated when V + <V-, and a low-level voltage is generated when V + <V−
I do. 94-99, 200-202 are AND gates, and comparators 86,
Combination of H level and L level output from each output terminal of 87, 90, 91
One of the AND gates outputs an H-level voltage, and the other
The gate outputs an L level voltage. V_b-V_AAnd V_PaAnd V_Pb, V_C-V_bAnd V
_QaAnd V_QbWhich of AND gates 94-99 and 200-202
Whether the AND gate outputs the H level is described below. 1) V_b-V_A≧ V_Pa     i) V_C-V_b≧ V_Qa                  … And gate 94     ii) V_Qa> V_C-V_b≧ V_Qb            … And gate 95    iii) V_Qb> V_C-V_b                  … And gate 96 2) V_Pa> V_b-V_A≧ V_Pb     i) V_C-V_b≧ V_Qa                  … And gate 97     ii) V_Qa> V_C-V_b≧ V_Qb            … And gate 98    iii) V_Qb> V_C-V_b                  … And gate 99 3) V_Pb> V_b-V_A     i) V_C-V_b≧ V_Qa                  … And gate 200     ii) V_Qa> V_C-V_b≧ V_Qb            … And gate 201    iii) V_Qb> V_C-V_b                  … And gate 202   An inverter 203 inverts the output state of the comparator 82.
208 and 209 are OR gates, 210 and 211 are AND gates, 212 is an OR gate
It is. AND gates 98, 99, and 2 are connected to the four inputs of OR gate 208.
01, 202, and AND gates 98, 99, 201, 21
When any one of the AND gates outputs an H level voltage, the OR gate 20
The output voltage of No. 8 becomes H level. Similarly, the six inputs of OR gate 209
Outputs of AND gates 96, 98, 99, 200 to 202 are input, and
Any one of the gates 96, 98, 99, 200 to 202 has an H level. When the bell voltage is output, the output voltage of the OR gate 209 becomes H level. Ann
The output of the OR gate 208 is connected to one input terminal of the
Are the outputs of the comparator 82, respectively. When the comparator 82 is
When the bell is being output, the output of the AND gate 210 is
Equal to output state. At this time, one of the two input terminals of the AND gate 211 is
The output of the inverter 203 (the inverted output of the comparator 82) is input.
Therefore, its output is at L level. Also, one of the two input terminals of the AND gate 211
The output of the OR gate 209 is input to the other end, and the output of the inverter 203 is input to the other end.
Output is input. When the comparator 82 outputs an L-level voltage
Of the AND gate 210, an L level voltage is input to one input terminal of the AND gate 210.
The output of the gate 210 is an L level voltage. At this time, the inverter
The output voltage of AND gate 211 is at H level, and the output of AND gate 211 is OR gate 2
09 output state. Each output of AND gates 210 and 211 is OR
Input to the gate 212 and at least one of the AND gates 210 and 211
Outputs the H level voltage, the output of the OR gate 212 is at the H level.
It becomes. The output terminal of the OR gate 212 is an output terminal A / P. 213 ~
217, 219 to 223 are AND gates, 218, 224 are OR gates, 225
Is a NOR gate. Reference numeral 226 denotes a reference voltage generation circuit,_r8, V_r9, V_r1
0, V_r11, -V_r12, -V_r13, -V_r14, -V_r15Occurs. 227-235
It is an analog switch. One input terminal of two-input AND gates 213 to 217
, The output of the comparator 82 is input, and the two-input AND gates 219 to 223
The output of the inverter 203 is input to one of the input terminals. In contrast,
The output of the AND gate 94 is connected to the other input terminal of the
The other input terminal of the AND 214 receives the output of the AND gate 95 and the other input of the AND gate 215.
An AND gate 96 is connected to one input terminal, and an AND gate 96 is connected to the other input terminal of the AND gate 216.
The output of the AND gate 99 is connected to the other input terminal of the AND gate 217.
The output of 202 is input. Also, the other input of AND gate 219
The output terminal of the AND gate 94 is connected to the input terminal, and the output terminal of the AND gate 220 is connected to the other input terminal.
The output of the gate 97 is connected to the other input terminal of the AND gate 221. The output of the gate 200 is connected to the other input terminal of the AND gate 222 by the AND gate 2.
01 is output from the AND gate 202 to the other input terminal of the AND gate 223.
Forces are each entered. Therefore, the comparator 82 outputs the H level voltage.
Is output, the H-level voltage is applied to one of the input terminals of the AND gates 213 to 217.
, 219 to 223, an L level voltage is input via the inverter 203.
Input, the AND gate 213 directly outputs the output from the AND gate 94.
The gate 214 receives the output from the AND gate 95 as it is, and
5 is the output from the AND gate 96, and the AND gate 216 is the AND gate.
The output from the AND gate 202 is output from the AND gate
The outputs are output as they are, while the output voltages of the AND gates 219 to 223 are output.
Becomes L level. Conversely, the comparator 82 sets the L level and the inverter 203
When each voltage of H level is output, the output voltage of AND gates 213 to 217 becomes
The output from the AND gate 94 to the L level.
The AND gate 220 outputs the output from the AND gate 97 as it is
The gate 221 receives the output from the AND gate 200 as it is, and the AND gate 222
And the output from the gate 201 as it is, and the AND gate 223
02 are output as they are. Of two inputs of OR gate 218
The output of the AND gate 214 is output to one of them, and the output of the AND gate 215 is output to the other.
Each force is entered. The output of the OR gate 218 is
, 216 are at the H level when at least one of them is at the H level.
You. One of the two input terminals of the OR gate 224 receives the output of the AND gate 221.
The other end receives the output of the AND gate 222, respectively. OR gate 2
24 is at least one of the output voltages of the AND gates 221 and 222.
It becomes H level when it is at H level. The 8-input NOR gate 225 has
The output of the gate 213, the output of the OR gate 218, the output of the AND gate 216,
The output of the AND gate 217 and the output of the AND gate 219 are
The output of 0, the output of the OR gate 224, the output of the AND gate 223,
And the outputs of the AND gates 213 to 217 and 219 to 223 are all
When the voltage is at the L level, the output voltage of the NOR gate 225 becomes the H level, Otherwise, it goes to L level.   227 to 230 and 232 to 235 are analog switches.
The above-described reference voltage is applied from the reference voltage generation circuit 226. Analog switch
Switch 227 has a reference voltage V_r8, The input terminal of the analog switch 228
Reference voltage V_r9, The input terminal of the analog switch 229 has a reference voltage V_r10, Analogs
The input terminal of the switch 230 has a reference voltage V_r11, To the input terminal of the analog switch 232
Is the reference voltage -V_r12, A reference voltage −V is applied to the input terminal of the analog switch 233._r13,
A reference voltage −V is applied to the input terminal of the analog switch 234._r14, Analog switch 235
Reference voltage -V_r15Has been added. Turn on analog switch 231
The force end is at 0V. The output terminals of the analog switches 227 to 235 are connected to each other.
, Its output voltage V_TwoAre output from the output terminal OUT. Of the analog switch 227
The control terminal receives the output of the AND gate 213 and the output of the analog switch 228.
The output of the OR gate 218 is connected to the control terminal of the analog switch 229.
The output of the AND gate 216 is connected to the control terminal of the analog switch 230.
The output of the AND gate 217 is connected to the control terminal of the analog switch 213.
The output of the NOR gate 225 is connected to the control terminal of the analog switch 232.
The output of the AND gate 219 is connected to the control terminal of the analog switch 233.
The output of the AND gate 220 is connected to the control terminal of the analog switch 234 at the control terminal.
The output of the OR gate 224 is connected to the control terminal of the analog switch 235.
The output of the AND gate 223 is input to the roll terminal. The choice
Three input voltages V input to the circuit 21_A, V_b, V_CVoltage level, depending on the magnitude relationship
And AND gate 213, OR gate 218, AND gate 216, AND gate
Gate 217, NOR gate 225, AND gate 219, AND gate 220,
One of the output terminals of the gate 224 and the AND gate 223 becomes H level,
Since the output terminals of the other gates become L level, the analog switches 227 to 23
Out of 5, only one analog switch to which the H level control voltage is applied
It becomes conductive and the other analog switches are open. AND Gate 2
13 is at H level, the analog switch 227 is turned on.
Voltage V_r8However, when the output of the OR gate 218 is at the H level, The switch 228 is turned on and the voltage V_r9Is the H level output of the AND gate 216
The analog switch 229 is turned on and the voltage V_r10But Ann
When the output of the gate 217 is at the H level, the analog switch 230 is conductive.
Voltage V_r11However, when the NOR gate 225 is at the H level, the analog switch
Switch 232 becomes conductive, and in this case, 0 V is applied, and the output of the AND gate 219 becomes
At the time of the H level, the analog switch 232 is turned on, and the voltage −V_r12
However, when the output of the AND gate 220 is at the H level voltage, the analog switch 2
33 becomes conductive and the voltage −V_r13However, the output of the OR gate 224 is at the H level.
Sometimes, the analog switch 234 is turned on and the voltage −V_r14But andge
When the output of the port 223 is at the H level, the analog switch 235 is turned on.
Become voltage -V_r15Is output from the output terminal OUT. Where V_r8, V_r9, V_r
Ten, V_r11> 0, -V_r12, -V_r13, -V_r14, -V_r15<0.   Next, the circuit operation of FIGS. 5 and 8 will be described with reference to FIGS. 9 to 11.
You. (1) The outermost region 6C (6C) of the light receiving element 6 shown in FIG.₁~ 6C_Four) Get more
Voltage (hereinafter referred to as a luminance signal) V corresponding to the luminance value to be obtained_CIs the reference voltage V_r5Greater than
In other words, V_C> V_r5When it is determined that the scene is outside (
If it is determined that a bright subject such as the sky is in the background and the periphery of the screen is bright),
When the luminance signal difference V_b-V_A(Hereinafter abbreviated as △ BA) and luminance signal difference V_C-V_b(
Depending on the value of △ CB, the reference voltage V_P1, V_P2, V_Q1, V_Q2
(The relation of the reference voltage is V_P2<0 <V_P1, V_Q2<0 <V_Q1) For the next operation
More photometric value V₁-V_TwoAsk for. (1-1) At this time, physically, as shown in FIG. 9 (a), the central part of the object field, that is, the area 6
A and the luminance signal difference ΔBA measured by the surrounding area 6B become smaller,
On the other hand, the luminance signal difference ΔCB between the region 6B and the outermost region 6 is a predetermined value V on the + side._Q1Greater than At this time, the main subject exists at both the positions of the area 6A and the area 6B.
Can often be determined. Therefore, the photometric value V₁-V_TwoIs a moderate exposure for the main subject
, The luminance signal V of the area 6A and the area 6B_A, V_BAnd the correction value is 0
From the following equation (1).     V₁-V_Two= (V_A+ V_b) / 2 ... (1)-   Then, based on the circuit operation of FIGS. 5 and 8, first, the selection circuit 21 of FIG.
To explain, the outermost area 6C (6C₁~ 6C_Four) Luminance signal (voltage) V_CBut
Reference voltage V_r5Therefore, the comparator 82 outputs the H level. Therefore
An H-level signal is supplied to the control terminal B / D of the reference voltage generation circuit 85,
The reference voltage of the circuit 85 is V_Pa= V_P1, V_Pb= V_P2, V_Qa= V_Q1, V_Qb= V_Q2Tona
You. On the other hand, the output signal (voltage) V of the operational amplifier 75_b-V_AIs V_P2<△ BA <V_P1
Therefore, the output of the comparator 86 is at L level, and the output of the comparator 87 is
The force becomes H level, and the output signal (voltage) V of the operational amplifier 80_C-V_bIs V_Q1
<△ CB, so that the outputs of the comparators 90 and 91 are both at H level.
Become. Therefore, only the output of the AND gate 97 becomes H level,
97 and H-level outputs and AND gates 94-96, 98, 99, 200-20
2, the output of the OR gate 212 is set to L level by the output terminal A /
P is set to L level, only the NOR gate 225 is set to H level,
The gates 213 to 223 are set to L level, and the voltage V from the output terminal OUT is_TwoTo 0V
You. Accordingly, the operational amplifier 26 shown in FIG._A+ V_b) / 2 is output,
Since 0 V is supplied to the negative-phase input terminal of the operational amplifier 31, the operational amplifier 31
Output V₁-V_TwoIs the voltage (V_A+ V_b) / 2. (1-2) In this case, specifically, as shown in FIG. 9B, the luminance signal difference 領域 between the area 6A and the area 6B is obtained.
BA is the predetermined value V on the + side_P1The luminance signal difference ΔC between the area 6B and the area 6C.
B is also a predetermined value V on the + side._Q1In this case, the main subject is the entire area 6A
It can be determined that there are many cases where the data exists in a part of the area 6B. In this case, the area 6A Luminance signal V_APhotometry value V₁-V_TwoYou can ask for the shine of the background
Considering the degree signal to some extent results in empirically good exposure. Therefore,
Photometric value V₁-V_TwoIs the main subject V_A, V_BAnd the correction value V_r8The next performance using
Equation (1) —determined from:     V₁-V_Two= (V_A+ V_b) / 2-V_r8                    ... (1)-   The circuit operation in this case will be described. The comparators 86, 87, 90, 91 and
The comparators 82 all output the H level, and only the AND gate 94 outputs the H level.
And the other AND gates 95 to 99 and 200 to 202 output L level.
You. Therefore, the output of the OR gate 212 is set to L level, and the output of the AND gate 213 is output.
When the force is set to H level (the analog switch 227 is turned on),
The output of the operational amplifier 26 shown in FIG._A+ V_b) / 2, and the operational amplifier 3
Output V₁-V_TwoIs (V_A+ V_b) / 2-V_r8Becomes (1-3) Specifically, as shown in FIG. 9 (C), the brightness of the area 6A and the area 6B
The signal difference ΔBA is a predetermined value V on the + side._P1While the area 6B and the area 6C
Since the luminance signal difference ΔCB is small, in this case, the main subject exists in the entire area 6A.
It can be determined that there are many cases where the main subject is particularly small. In this case the area
The photometric value V for the main subject within 6A₁-V_TwoIs good, but the background
The luminance signal V of the corresponding area 6B_BTo give a moderate exposure to the main subject
The photometric value V₁-V_TwoRepresents the correction values V for the areas 6A and 6B._r9(V_r9> V_r8
) Is obtained from the following equation (1).     V₁-V_Two= (V_A+ V_b) / 2-V_r9                    ... (1)-   The circuit operation in this case will be described. The comparators 86, 87 and 90 are at the H level.
And only the AND gate 95 outputs the H level, and the other AND gate 9
4, 96 to 99, 200 to 202 output the L level, and the comparator 82 outputs the H level.
Output a bell. Therefore, the output of the OR gate 212 is set to L level, and
The output of the switch 218 is at the H level (the analog switch 228 is turned on). The output of the operational amplifier 26 shown in FIG._A+ V_b) / 2
Output V of the pair amplifier 31₁-V_TwoIs (V_A+ V_b) / 2-V_r9Becomes (1-4) Specifically, as shown in FIG. 9 (d), the brightness of the region 6A and the region 6B
The signal difference ΔBA is a predetermined value V on the + side._P1While the area 6B and the area 6C
Is a predetermined value V on the negative side._Q2The main subject at this time
The size is about the same as that described in the above (1-3) and considerably high brightness in the region 6B.
Region (for example, the sun, sea surface reflection, etc.)
It can be determined that a subject having a considerably high luminance is located in B. In this case, the data
Photometric values for all the luminances of the areas 6A, 6B and 6C, with the correction value set to 0
V₁-V_Two[V₁-V_Two= (V_A+ V_b+ V_C/ 3) also obtained good results
The inventor has recognized that this method is effective, but in this embodiment, another good effect is obtained.
In other words, as in the case of the above (1-3), the brightness of the regions 6A and 6B
Correction value V_r9Is obtained from the following equation (1).     V₁-V_Two= (V_A+ V_b) / 2-V_r9                    ... (1)-   Explaining the circuit operation in this case, the comparators 86 and 87 output H level.
While the comparators 90 and 91 output L level,
Outputs an H level, and the other AND gates 94, 95, 97 to 99, 200 to
202 outputs the L level, and the comparator 82 outputs the H level. Therefore,
The output of the OR gate 212 is set to L level, and the output of the AND gate 218 is set to H level.
(This causes the analog switch 228 to be in a conductive state), as shown in FIG.
The output of the amplifier 26 is a voltage (V_A+ V_b) / 2, the output V of the operational amplifier 31₁−
V_TwoIs (V_A+ V_b) / 2-V_r9Becomes (1-5) Specifically, as shown in FIG. 9 (e), the brightness of the area 6A and the area 6B The signal difference ΔBA is small, and the luminance signal difference ΔCB between the area 6B and the area 6C is negative and V_Q2
Since the absolute value is larger, in this case, the main subject exists in both the areas 6A and 6B.
It can be determined that the image is large and white. In this case, the area 6A
Although the photometric value may be obtained for the luminance of the area 6B, the peripheral part of the screen is located closer to the center than the central part.
In the case of low luminance, the light metering value is calculated in consideration of the luminance of the area 2C at the periphery of the screen to some extent.
The person who sought to make sure that the whitish main subject appears white (highlight control).
It is known that a good exposure value can be obtained in terms of data. Therefore
In this case, in order to give an exposure that highlights the main subject,
Signal V_A, V_b, V_CAnd the photometric value V₁-V_TwoIs the correction value V_r10Using,
It is obtained from the following equation (1).     V₁-V_Two= (V_A+ V_b+ V_C) / 3-V_r10              ... (1)-   The circuit operation in this case will be described.
On the other hand, comparators 86, 90 and 91 output L level, and AND gate 99 outputs
H level is output, and the other AND gates 94 to 98 and 200 to 202 output L level.
And the comparator 82 outputs the H level.   Therefore, the output of the OR gate 212 is set to the H level, and the AND gate 216 is set to the H level.
The bell (which causes the analog switch 229 to be conductive) as shown in FIG.
The output of the operational amplifier 26 is a voltage (V_A+ V_b+ V_C) / 3, and the operational amplifier 31
Output V₁-V_TwoIs (V_A+ V_b+ V_C) / 3-V_r10Becomes (1-6) Specifically, as shown in FIG. 9 (f), the brightness of the area 6A and the area 6B
The signal difference ΔBA is a negative value and a predetermined value V_P2The absolute value is larger than that of the area 6B and the area 6C.
The luminance signal difference ΔCB is also a negative value and a predetermined value V._Q2Since the absolute value is larger,
When the subject is in the middle of the entire area 6A and a part of the area 6B and is white.
It can be determined that this is the case of the subject. In this case, the case of the above (1-5)
In order to give an exposure that highlights the main subject in the same manner as
6B and 6C, the photometric value V₁-V_TwoIs the correction value V_r11(V_r11<V _r10 ) Is obtained from the following equation (1).     V₁-V_Two= (V_A+ V_b+ V_C) / 3-V_r10              ... (1)-   The circuit operation in this case will be described.
And the AND gate 202 outputs the H level.
99, 100 to 101 output L level, and comparator 82 outputs H level
I do. Therefore, the output of the OR gate 212 is set to the H level, and the output of the AND gate 217 is
Set the output to the H level (the analog switch 230 becomes conductive).
The output of the operational amplifier 26 shown in FIG._A+ V_b+ V_C) / 3
Output V of amplifier 31₁-V_TwoIs (V_A+ V_b+ V_C) / 3-V_r11Becomes (1-7) Specifically, as shown in FIG. 9 (g), the brightness of the region 6A and the region 6B
The signal difference ΔBA is a negative value and a predetermined value V_P2The absolute value is larger than that of the area 6B and the area 6C.
Since the luminance signal difference ΔCB is small, the main subject exists in the entire area 6A at this time.
Or especially when the main subject is small and the main subject is whitish.
It can be determined that the object is a copy. In this case as well, the main subject is
In order to obtain the exposure as described on the site, the regions 6A, 6B and 6C are targeted and supplemented.
Assuming that the positive value is 0, it is obtained from the following equation (1)-.     V₁-V_Two= (V_A+ V_b+ V_C) / 3 ... (1)-   To explain the circuit operation in this case, all the comparators 90 output H level.
On the other hand, comparators 86, 87 and 91 output L level, and AND gate 2
02 outputs the H level, and the other AND gates 94 to 99, 200, 202
The L level is output, and the comparator 82 outputs the H level.   Therefore, the output of the OR gate 212 is set to the H level, and the output of the NOR gate 225 is
The H level (which causes the analog switch 230 to be in the conductive state),
The output of the operational amplifier 26 shown in FIG._A+ V_b+ V_C) / 3, and the operational amplifier 3
Output V of 1₁-V_TwoIs (V_A+ V_b+ V_C) / 3. (1-8) Specifically, as shown in FIG. 9 (h), the luminance signal between the area 6A and the area 6B is
The signal difference ΔBA is a negative value and a predetermined value V_P2Absolute value is larger, brightness of area 6B and area 6C
Degree signal difference ΔCB is a predetermined value V_Q1In this case, the main subject is
The same size as the case described in 1-1), and there is a difference in brightness in the main subject portion,
When the area 6A has a slightly high luminance, or when the area 6A is a low
It can be determined that the object is occupied by a bright subject. In this case, data
And the correction value is set to 0 for all the luminances of the areas 6A, 6B, and 6C.
₁-V_Two[V₁-V_Two= (V_A+ V_b+ V_C) / 3] also gives good results
As a method, the inventor has recognized, but in this embodiment, another good effect can be obtained.
The method, that is, the luminance of the area 6A and the area 6B as in the case of the above (1-1).
And the correction value is set to 0, and is calculated from the following equation (1).     V₁-V_Two= (V_A+ V_b) / 2 ... (1)-   The circuit operation in this case will be described. Comparators 90 and 91 output H level.
While the comparators 86 and 87 output the L level, and the AND gate 200
H level is output, and the other AND gates 94 to 99, 201, 202 are at L level.
And the comparator 82 outputs the H level. Therefore, the OR gate 21
2, the output of NOR gate 225 is set to H level, and the output of NOR gate 225 is set to H level.
The output of the operational amplifier 26 is a voltage (V_A+ V_b) / 2, the output V of the operational amplifier 31
₁-V_TwoIs (V_A+ V_b) / 2. (1-9) Specifically, as shown in FIG. 9 (i), the luminance signal between the area 6A and the area 6B is
The signal difference ΔBA is small, and the luminance signal difference ΔCB between the region 6B and the region 6C is also small.
In this case, when the main subject occupies the entire object scene,
It is determined that there is no intention to set the main subject. In this case, the area 6A
, 6B and 6C, the correction value is set to 0, and the following equation (1)- Request.     V₁-V_Two= (V_A+ V_b+ V_C) / 3 ... (1)-   Explaining the circuit operation in this case, the comparators 87 and 90 output H level.
While the comparators 86 and 91 output the L level, and the AND gate 98 outputs
H level is output, and the other AND gates 94-97, 99, 200-202
The L level is output, and the comparator 82 outputs the H level. Therefore,
The output of the gate 212 is set to the H level, and the output of the NOR gate 225 is set to the H level.
The output of the operational amplifier 26 is a voltage (V_A+ V_b+ V_C) / 3 and the output of the operational amplifier 31
Force V₁-V_TwoIs (V_A+ V_b+ V_C) / 3.   FIG. 10 (a) is a graph corresponding to each of the above-described luminance states (1-1) to (1-9).
It shows the relation of each of the arithmetic expressions performed. (2) The outermost region 6C (6C) of the light receiving element 6 shown in FIG.₁~ 6C_Four) Get more
Luminance signal V_CIs the reference voltage V_r5Less than, ie V_C<V_r5As tall
When it is determined that the room is such that the indoor wall is located in the scenery,
The constant V is determined by the values of the luminance signal difference ΔBA and the luminance signal difference ΔCB in the same manner as (1)._P3
, V_P4, V_Q3, V_Q4(V_P4<0 <V_P3, V_Q4<0 <V_Q3) For the next operation
More photometric value V₁-V_TwoAsk for. (2-1) Specifically, as shown in FIG. 11A, the luminance signal difference between the area 6A and the area 6B
ΔBA becomes small, and the luminance signal difference ΔCB between the area 6B and the area 6C becomes a predetermined value V_Q3Than
In this case, the size of the main subject exists in both the area 6A and the area 6B.
It can be determined that the subject is dark and the subject is dark.   In this case, the brightness of the areas 6A and 6B is simply measured and the correction value is set to 0.
The light value may be obtained, but the dark subject is surely taken black (shadow control
In this embodiment, the main subject portion is shadow
In order to provide an exposure as shown in FIG.₁
-V_TwoIs the negative correction value -V_r13Is calculated from the following equation (2).     V₁-V_Two= (V_A+ V_b) / 2 + V_r13                   ... (2)-   The circuit operation in this case will be described. The outermost area 6C (6C₁~ 6C_Four) Shine
Degree signal (voltage) V_CIs the reference voltage V_r5Therefore, the comparator 82 sets the L level.
Output to the control terminal B / D of the reference voltage generating circuit 85.
Is supplied, and the reference voltage of the circuit 85 is V_Pa= V_P3, V_Pb= V_P4, V_Qa= V
_Q3, V_Qb= V_Q4Becomes On the other hand, the comparator 86 outputs L level,
Parators 87, 90 and 91 output H level. And AND gate 97 is H
Output level, and other AND gates 94-96, 98, 99, 200-20
2 outputs the L level, and the inverter 203 outputs the H level. So, or
The output of the gate 212 is set to L level, and the output of the AND gate 220 is set to H level (this
As a result, the analog switch 233 is turned on).
The output of the loop 26 is a voltage (V_A+ V_b) / 2, the output V of the operational amplifier 31₁-V_TwoIs
(V_A+ V_b) / 2 + V_r13Becomes (2-2) Specifically, as shown in FIG. 11 (b), the luminance signal difference between the area 6A and the area 6B
ΔBA is a predetermined value V_P3The luminance signal difference ΔCB between the region 6B and the region 6C is larger than the predetermined value.
Value V_Q3At this time, the main subject is located in the entire area 6A and a part of the area 6B.
It can be determined that the object exists and that the object is blackish. In that case
Is an exposure such that a shadow is drawn on the main subject in the same manner as in the above (2-1).
, The brightness of the areas 6A and 6B is targeted, and the correction value V₁-V_TwoIs the complement of
Positive value -V_r12To (| V_r12| <| V_r13|), The following equation (2)
Ask.     V₁-V_Two= (V_A+ V_b) / 2 + V_r12                   ... (2)-   The circuit operation in this case will be described.
And the AND gate 94 outputs an H level, and the other AND gates 95 to 9
9, 200 to 202 output L level, and inverter 203 outputs H level.
You. Therefore, the output of the OR gate 212 is set to L level, and the output of the AND gate 220 is output. When the output is set to the H level, the output of the operational amplifier 26 shown in FIG._A+ V_b) / 2
Output V of the operational amplifier 31₁-V_TwoIs (V_A+ V_b) / 2 + V_r12Becomes (2-3) In this case, specifically, as shown in FIG. 11 (c), the luminance signal between the area 6A and the area 6B is
Signal difference ΔBA is a predetermined value V_P3The luminance signal difference Δ between the area 6B and the area 6C.
Since CB is small, at this time, the main subject exists in the entire area 6A, or
It can be determined that the main subject is small and the subject is blackish
. In this case, in order to give the exposure that the main subject
The correction value V is set for the luminance of the areas 6A and 6B.₁-V_TwoSets the correction value to 0 and
Equation (2) −Calculates     V₁-V_Two= (V_A+ V_b) / 2… (2) −   The circuit operation in this case will be described. The comparator 91 outputs L level,
The comparators 86 to 90 output H level, and the AND gate 95 outputs H level.
And the other AND gates 94, 96 to 99, 200 to 202 output L level
Then, inverter 203 outputs an H level. Therefore, the OR gate 212 is set to L level.
5, the NOR gate 225 is set to the H level, and the output of the operational amplifier 26 in FIG.
Is the voltage (V_A+ V_b) / 2, the output V of the operational amplifier 31₁-V_TwoAlso (V_A+ V_b)
/ 2. (2-4) Specifically, as shown in FIG. 11 (d), the luminance signal between the area 6A and the area 6B is
Signal difference ΔBA is a predetermined value V_P3The luminance signal difference ΔCB between the area 6B and the area 6C is larger.
Predetermined value V with negative value_Q4Since the absolute value is larger, the main subject is in the case of (2-3) described above.
And a high-luminance object (eg,
(Light). High brightness in area 6B indoors as in this case When there is a subject of the degree, compared with the case where the outdoor sun etc. are located in the area 6B
It can be said that the influence of the high luminance is small in terms of data.₁-V_TwoIs the area
For the luminances of 6A, 6B and 6C, the correction value is set to 0 and the following equation (2)-
Find more.     V₁-V_Two= (V_A+ V_b+ V_C) / 3 ... (2)-   Explaining the circuit operation in this case, the comparators 90 and 91 output L level.
While the comparators 86 and 87 output the H level, and the AND gate 96 outputs the H level.
Output level, and other AND gates 94, 95, 97-99, 200-20
2 outputs an L level, and the inverter 203 outputs an H level. So, or
The gate 212 goes low, the NOR gate 225 goes high, and the
The output of the loop 26 is a voltage (V_A+ V_b+ V_C) / 3, the output V of the operational amplifier 31₁−
V_TwoAlso (V_A+ V_b+ V_C) / 3. (2-5) Specifically, as shown in FIG. 11 (e), the brightness of the region 6A and the region 6B
The degree signal difference ΔBA is small, and the luminance signal difference ΔCB between the area 6B and the area 6C is a negative value.
Constant value V_Q4Since the absolute value is larger, the main subject exists in both areas 6A and 6B.
And only the areas 6A and 6B are illuminated by lights or the like.
It can be determined that there is. In that case, the luminance is measured only for the areas 6A and 6B.
Although there is a concept of obtaining a light value, in this embodiment, a dark peripheral area of the object scene, that is,
Considering the area 6C to some extent, the photometric value V₁-V_TwoRepresents the brightness of the areas 6A, 6B and 6C.
With the correction value set to 0, it is determined by the following equation (2)-.     V₁-V_Two= (V_A+ V_b+ V_C) / 3 ... (2)-   The circuit operation in this case will be described.
The comparator 87 outputs the H level, and the AND gate 99 outputs the H level.
Output level, and other AND gates 94-98, 200-202 are at L level
And the inverter 203 outputs the H level.   Therefore, the OR gate 212 becomes H level, and the NOR gate 225 becomes H level. The output of the operational amplifier 26 shown in FIG._A+ V_b+ V_C) / 3
Output V of the pair amplifier 31₁-V_TwoAlso (V_A+ V_b+ V_C) / 3. (2-6) Specifically, as shown in FIG. 11 (f), the luminance signal between the area 6A and the area 6B is
The signal difference ΔBA is a negative value and a predetermined value V_P4The absolute value is larger than that of the area 6B and the area 6C.
The luminance signal difference ΔCB is also a negative value and a predetermined value V._Q4Since the absolute value is larger, the main subject is 6
A and a part of the area 6B, and all of the area 6A and the area
It can be determined that a part of 6B is illuminated by a light or the like. in this case
Unlike the case of (2-5) above, the correction with the correction value
It is necessary to give a proper exposure, and it is necessary to target the brightness of the areas 6A, 6B and 6C.
, Photometric value V₁-V_TwoIs the negative correction value -V_r15And the following equation (2)-
Confuse.     V₁-V_Two= (V_A+ V_b+ V_C) / 3 + V_r15              ... (2)-   The circuit operation in this case will be described.
And the AND gate 202 outputs an H level, and the other AND gates 94 to 94
99, 200 and 201 output L level, and inverter 203 outputs H level.
I do. Therefore, the OR gate 212 goes high and the AND gate 223 goes high.
As an output level, the output of the operational amplifier 26 in FIG._A+ V_b+ V_C) / 3
Output V of the operational amplifier 31₁-V_TwoIs (V_A+ V_b+ V_C) / 3 + V_r15Becomes (2-7) Specifically, as shown in FIG. 11 (g), the brightness of the area 6A and the area 6B
The signal difference ΔBA is a negative value and a predetermined value V_P4The absolute value is larger than that of the area 6B and the area 6C.
Since the luminance signal difference ΔCB is small, in this case, the main subject exists in the entire area 6A.
Alternatively, it can be determined that the main subject is small and exists in a part of the area 6A. In this case, the areas 6A, 6B and 6C
Of the photometric value V₁-V_TwoIs the negative correction value -V_r14(| V_r14| <| V_r1
Five|), And is calculated by the following equation (2)-.     V₁-V_Two= (V_A+ V_b+ V_C) / 3 + V_r14              ... (2)-   The circuit operation in this case will be described. The comparators 86, 87 and 91 are at L level.
And the comparator 90 outputs the H level, and the AND gate 202 outputs
H level is output, and the other AND gates 94 to 99 and 200 to 202 output L level.
And the inverter 203 outputs the H level. Therefore, the OR gate 21
2 outputs the H level, and the OR gate 224 outputs the H level (the analog switch
Switch 234 is turned on), and the output of the operational amplifier 26 in FIG.
Pressure (V_A+ V_b+ V_C) / 3, the output V of the operational amplifier 31₁-V_TwoIs (V_A+ V_b
+ V_C) / 3 + V_r14Becomes (2-8) Specifically, as shown in FIG. 11 (h), the brightness of the area 6A and the area 6B
The signal difference ΔBA is a negative value and a predetermined value V_P4The absolute value is larger than that of the area 6B and the area 6C.
The luminance signal difference ΔCB is a predetermined value V_Q3In this case, the main subject is
The size of the area 6A is almost the same as that of the case 2-1), and there is a contrast in the main subject.
When the subject is slightly bright but the overall subject is blackish,
It can be determined that the region 6B is occupied by an object having a considerably low luminance.
In this case, all luminances of the areas 6A, 6B and 6C are targeted as data, and
Photometric value V with positive value as 0₁-V_Two[V₁-V_Two= (V_A+ V_b+ V_C) / 3]
We also know that good results can be obtained depending on the conditions of the scene,
In the example, a method that gives good overall results, that is, the regions 6A, 6B and 6C
Correction value V for all luminances₁-V_TwoIs the negative correction value -V_r14Using the following
It is calculated by the equation (2)-.     V₁-V_Two= (V_A+ V_b+ V_C) / 3 + V_r14              ... (2)-   Explaining the circuit operation in this case, the comparators 86 and 87 output the L level. While the comparators 90 and 91 output an H level signal and the AND gate 200
Outputs an H level, and the other AND gates 94 to 99, 201, and 202 output an L level.
And the inverter 203 outputs an H level. Therefore, OR gate 2
12 outputs an H level, and the OR gate 224 outputs an H level.
The output of the operational amplifier 26 is a voltage (V_A+ V_b+ V_C) / 3, and the operational amplifier 31
Output V₁-V_TwoIs (V_A+ V_b+ V_C) / 3 + V_r14Becomes (2-9) Specifically, as shown in FIG. 11 (i), the brightness of the region 6A and the region 6B
The signal difference ΔBA is small, and the luminance signal difference ΔCB between the region 6B and the region 6C is also small.
So, in this case, if the main subject occupies the whole scene, or
It can be determined that the main subject is not specifically set as shown in FIG. This place
In order to provide a proper exposure to the entire area 6A, 6B and 6C, the area 6A
６6C, the photometric value V₁-V_TwoIs the following equation with the correction value set to 0
(2) Determined by-.     V₁-V_Two= (V_A+ V_b+ V_C) / 3 ... (2)-   Explaining the circuit operation in this case, the comparators 86 and 91 output L level.
While the comparators 87 and 90 output the H level, and the AND gate 98 outputs
H level output, and the other AND gates 94-97, 99, 200-202
The L level is output, and the inverter 203 outputs the H level. Therefore,
The gate 212 outputs an H level, and the NOR gate 225 outputs an H level.
The output of the amplifier 26 is a voltage (V_A+ V_b+ V_C) / 3, the output of the operational amplifier 31
V₁-V_TwoAlso (V_A+ V_b+ V_C) / 3.   FIG. 10 (b) corresponds to each of the luminance states (2-1) to (2-9) described above.
It shows the relation of each of the arithmetic expressions performed.   The characteristic feature of the photometric device of the embodiment described above is that the main subject is
Assuming that it is located at the center of the body region (region 6A), the size of the main subject is
It is possible to make a determination and make a correction according to the size of the main subject.   In FIG. 5, if the peripheral image drop information is not transmitted from the lens side,
, The reference voltage V output from the correction value output circuit 101_vg1, V_vg2Light receiving element 6
Each area 6B, 6C₁~ 6C_FourThis causes an error in the luminance value. That is, adjacent
Scenes are determined at the level of the luminance difference between the
And the calculated photometric value V₁~ V_TwoCauses an error. In this embodiment,
When the lens is replaced, the correction value output circuit 101 drops the image plane light amount suitable for that lens.
Is used to input the reference voltage (correction value) corresponding to
Because the correction of the drop in image plane light intensity is added to the value, that is, the brightness of the
Since the measurement can be performed accurately regardless of the surface position, photometric errors due to lens replacement are eliminated.
The main subject in the center can be properly exposed.
Becomes   In the present embodiment, each small area 6C of the outermost area 6C of the light receiving element 6 is used.₁~
6C_FourWhen the average value of the obtained luminance signal is on the high luminance side, the small area 6C₁
~ 6C_FourFor example, by ignoring the output of a small area that results in a low luminance signal
If there is an inappropriate element for background brightness such as shaded ground outdoors in fine weather,
Can be removed. In addition, each of the small areas 6C₁~ 6C_FourThe average value of the luminance signal obtained from
Side, the small area 6C₁~ 6C_FourOf a small area that is a high-luminance signal
By ignoring the power, the brightness of the background, such as electric light in a night view,
Eliminate inappropriate elements.   Further, in the photometric device of the present embodiment, it is determined that the subject is white or black.
When detected, subjects that are whitish are consciously white, and subjects that are dark are black.
Highlight description (highlight control) or shadow description (shadow control)
Control), and this correction is performed for the size of the main subject.
Extremely effective photometric control because the amount of correction is changed by judging the height
Is possible.   Note that, in the explanatory diagrams of FIGS. 10 and 11 used in the description of the above-described embodiment,
The value of the brightness level of each of the areas 6A to 6C is determined when the brightness difference between adjacent areas is small.
, But at the same level, of course, this is a slight difference in actual metering
(The predetermined value compared, for example, V_P1Meaning that the difference in luminance is smaller) FIGS. 10 and 11 are illustrations for facilitating understanding in the present embodiment.
It only plays an eye. In this embodiment, the photometric value V₁-V_TwoOperation of
There are four methods (V_C1~ V_C4)
However, the photometric value may be obtained by dividing the area more finely.
Around the surface (V_C) May be measured as one area. Also, the photometric value V₁-V_TwoPerformance
The calculation method is to set the target metering area to V_AAnd V_BOnly if and full screen
V_A, V_B, V_CIs divided into two cases.
Calculate the photometric value in the same way as described above, with part or all of the
You may make it. The field is divided into three or more concentric regions, and
The luminance signal difference between the adjacent area and the adjacent area may be used.   Further, in the present embodiment, the selection circuit 21 is constituted by a logic circuit.
Naturally, processing using software using a computer would not be an embodiment of the present invention.
It is a theory. (Correspondence between invention and embodiment)   In the present embodiment, the correction value output circuit 101 outputs the light to the correction value output means of the present invention.
The element 6 serves as a light receiving means, and includes luminance value correction circuits 108 to 112 and a photometric value calculation circuit 113.
Correspond to the calculation means. (The invention's effect)   As described above, according to the present invention, the brightness of the central region and the peripheral region of the screen,
And the decrease in image plane light quantity during luminance measurement obtained from the correction value output means of the interchangeable lens.
The brightness of each area is corrected based on the corresponding correction value information, and the brightness of each corrected area is corrected.
The scene is determined using the degree, and the photometric value at the time of the luminance measurement is determined based on the scene.
Is calculated, so that photometric output is
By correcting the force, the luminance of the subject can be accurately measured.

BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is an optical arrangement diagram of a photometric system when the present invention is applied to a single-lens reflex camera, FIG. 2 is a diagram for explaining a light receiving surface of a light receiving element shown in FIG. FIG. 3 is a diagram showing an example of a decrease in the amount of image surface light in the taking lens, FIG. 4 is a schematic block diagram showing an example of a signal processing system for implementing the present invention, and FIG. FIG. 6 is a circuit diagram showing a configuration example of the peripheral luminance calculation circuit shown in FIG. 5, FIG. 7 is a circuit diagram showing a configuration example of the magnitude comparison circuit shown in FIG. 6, and FIG. 9 is a circuit diagram showing a configuration example of the selection circuit shown in FIG. 5, and FIGS. 9 to 11 are diagrams for explaining arithmetic expressions for obtaining a photometric value in each luminance state of the object scene. DESCRIPTION OF SYMBOLS 6 ... Light receiving element, 6A-6C ... Area | region, 20 ... Peripheral luminance calculation circuit, 21 ... Selection circuit, 22-24 ... Resistance, 25 ... Analog switch, 26 ... Operational amplifier, 28-30
... resistance, 31 ... operational amplifier, 101 ... correction value output circuit, 115 to 120 ... logarithmic compression circuits, 121 to 140 ... resistor, 141 - 145 ... operational amplifier, V _A ~V _C ... voltage, V _vg1 ~V _vg2 ... reference voltage , V ₁ -V ₂ ... Photometric values.

Claims (1)

(1) In a camera system including an interchangeable lens and a camera body to which the interchangeable lens is detachable, correction value information corresponding to a decrease in image surface light amount at the time of luminance measurement is stored in the interchangeable lens. A correction value output means for enabling output; a light receiving means for independently measuring at least the luminance of the central area of the screen and the luminance of the peripheral area of the camera body; and the luminance measured by the light receiving means and the exchange. The brightness of each area is corrected based on the correction value information obtained from the correction value output means of the lens, scene determination is performed using the corrected brightness of each area, and the brightness is measured based on the scene determination. And a calculating means for calculating the photometric value in the camera. (2) In a camera body to which an interchangeable lens having a correction value output unit capable of outputting correction value information corresponding to a decrease in image surface light amount at the time of luminance measurement is detachable, at least the luminance and the peripheral part in the central area of the screen A light-receiving means for independently measuring the brightness of the area, and correcting the brightness of each area based on the brightness measured by the light-receiving means and the correction value information obtained from the correction value output means of the interchangeable lens; A camera body that performs a scene determination using the brightness of each of the determined areas and calculates a photometric value at the time of the brightness measurement based on the scene determination. (3) A light receiving means for independently measuring at least the luminance of the central area of the screen and the luminance of the peripheral area, and the luminance measured by the light receiving means and the correction value information obtained from the correction value output means of the interchangeable lens. And a calculating means for calculating a photometric value at the time of measuring the luminance based on the scene determination. A possible interchangeable lens, characterized in that correction value output means for outputting correction value information corresponding to a decrease in the amount of image surface light during luminance measurement is provided to the calculation means of the camera body. interchangeable lens.