JP2684551B2 - Camera system, camera body and interchangeable lens - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION (Field of application of the invention)   The present invention divides the object scene into a plurality of regions,
It is suitable for the main part of the screen that uses each photometric value
Photo camera or video camera designed to give a positive exposure
Of camera system such as camera, camera body and interchangeable lens
It is about improvement. (Background of the Invention)   Conventionally, the field is divided into multiple areas, and photometry is performed for each area.
Providing proper exposure on the shooting screen using multiple photometric values
Such a photometric device is disclosed, for example, in Japanese Utility Model Publication No. Sho 51-9271,
It is disclosed by 4-123030 etc.   The proposed device calculates the brightness of each divided area of the object scene.
Conditions for metering and obtaining proper exposure of the screen from multiple outputs
Is selected and an operation based on it is performed. Yottteko
In such a device, first measure the brightness of each area accurately
There is a need to. However, the shooting lens is replaceable.
In the case of Mera, the amount of light loss on the peripheral image plane is larger due to the lens.
In contrast, when compared with the open F number, the maximum difference is about two steps.
You. Therefore, depending on the lens, it is accurate in the case of aperture open metering.
The problem that accurate photometry is not performed, resulting in an error in the calculation result
Had a point. (Object of the invention)   The object of the present invention is to solve the above-mentioned problems and to reduce the amount of light on the image plane.
The photometric output can be corrected using the correction value information corresponding to
By using, you can accurately measure the subject brightness.
It is to provide a camera system. (Features of the invention)   In order to achieve the above object, the present invention provides an interchangeable lens
And a camera body with the interchangeable lens detachable.
When measuring the brightness of the interchangeable lens in the Mela system
It is possible to output correction value information corresponding to the image surface light falloff at
By providing a correction value output means for
Also independently measures the brightness of the center and peripheral areas of the screen.
Light receiving means, brightness measured by the light receiving means, and the exchange
The correction value information obtained from the correction value output means of the lens
It is characterized by the provision of a calculation means for calculating the photometric value based on
Sign. (Example of the invention)   Hereinafter, the present invention will be described in detail based on illustrated embodiments.
You.   FIG. 1 shows the present invention implemented by a single-lens reflex camera.
FIG. 3 is an optical arrangement diagram of a photometry system for 1 in the figure is a shooting lens
, 2 is a quick return mirror, 3 is a focusing screen, 4 is a pen
Center prism, 5 is an imaging lens, 6 is divided into 6 areas
A light-receiving element having a light-receiving surface, 7 is an eyepiece lens, and 8 is a lens.
It is the film. In FIG. 1, the focusing lens 3 is fixed by the taking lens 1.
The image of the object formed on the
The light is guided and imaged on the top to measure the light. Figure 2 is Figure 1
FIG. 6A is a diagram for explaining the light-receiving surface of the light-receiving element 6 shown in FIG.
Is the area for measuring the brightness in the center of the scene, 6B is the area around it
6C is the area where the brightness of
The area 6C is defined by four small areas 6C.₁~ 6C_FourYo
It is configured.   The above photometry system performs scene discrimination by multi-segment photometry,
This is a so-called evaluative metering system that automatically performs exposure compensation. An example
For example, in a conventional camera, the entire viewfinder screen is
Since the photometer was using a light receiving element with a
In a so-called backlit scene like a person
The photometry value is dominated by the scenery, and the main subject is underexposed.
There were some drawbacks when it came to light, but like the photometric system
In the case of multi-segment photometry, the central part and the peripheral part are separated
Therefore, it is possible to detect that there is a brightness difference between the central part and the peripheral part.
It is possible to determine that the scene is backlit and
Therefore, it becomes possible to perform proper exposure compensation. Other order
For light scenes, highlight scenes, shadow scenes
Even if it is, it is possible to obtain the proper exposure by scene discrimination.
is there.   However, for single-lens reflex cameras, lens replacement
Since it is possible to adjust the exposure value with a standard lens,
Even if the metering value is calculated for each range, the exposure value can be obtained by changing the lens.
Error will occur. The reason is that it is installed as described above.
By the lens that is
This is because they are different. That is, it is received due to the decrease in the amount of light on the image plane.
Optical part 6A, 6B, 6C₁~ 6C_FourThe metering value of the
This is because there is an error in the result. Figure 3 is a shooting lens
3 shows an example of the degree of image plane light amount drop of No. 1).   FIG. 4 shows a signal processing process for correcting the above photometric error.
FIG. 1 is a schematic block diagram of this embodiment, in which 101 is the
Correction value V_vgOutput correction value
Force circuit (the circuit is built into each interchangeable lens), 10
2 to 107 are light receiving portions 6A, 6B, 6C of the light receiving element 6₁~ 6C_Four(No.
In Fig. 4, each area is equivalently shown as an independent light receiving element.
The voltage V corresponding to the brightness value measured in_A, V_B, V
_C1~ V_C4A luminance value output circuit for outputting
Correction value V from the positive value output circuit 101_vgLuminance corrected according to
Value V_XB, V_XC1~ V_XC4Brightness value correction circuit that outputs
A light value calculation circuit, 114 is an exposure control circuit. Such a belief
By configuring the signal processing circuit,
Input correction value V of the lens_vgComplementary according to information
It is possible to obtain the brightness value (photometric value) for each corrected area.
And prevent photometric errors due to lens replacement.
You.   Next, a tool for realizing the circuit configuration shown in FIG.
The embodied circuit diagrams are shown in FIGS. 115 to 120 are photo detectors
Each light receiving portion 6A, 6B, 6C of the child 6₁~ 6C_FourPhotocurrent iA, iB,
I c₁~I c_FourIs a logarithmic compression circuit that logarithmically compresses each photocurrent
Is logarithmically compressed here and the voltage V_A, V_B, V_C1~ V_C4Output as
Is done. Where the voltage V_A, V_B, V_C1~ V_C4Is the constant a₁~ A₆
(≧ 0), b (≧ 0) and photocurrent i_a, i_b, i_c1~ I_c4Using
It can 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 the same as V_A
= V_B= V_C1= V_C2= V_C3= V_C4The logarithmic compression circuit
It shall be preset within 115-120. In addition,
Output correction value to be placed on the side of the lens (inside the lens barrel)
The circuit 101 has two types of correction values corresponding to the correction values of the lens.
Reference voltage V_vg1, V_vg2Is output through contacts 101a and 101b.
Road.   121 to 140 are resistors having the same resistance value, and 141 to 145 are operational resistors.
It is an amplifier, resistors 121-124, operational amplifier 141, resistor 125
~ 128 and operational amplifier 142, resistors 129 to 132 and operational amplifier 143,
Resistors 133-136 and Op Amp 144, Resistors 137-140 and Operation
Each of the subtraction circuits is composed of
V from the subtraction circuit_B~ V_vg1, V_C1−V_vg2, V_C2−V_vg2, V_C3−V
_vg2, V_C4−V_vg2Is output. The following theory
At Ming, V_B−V_vg1Becomes V_bIs defined.   20 is output from the logarithmic compression circuits 117 to 120, respectively.
Compensation output through the subtraction circuit placed in the subsequent stage
Voltage V_C1−V_vg2, V_C2−V_vg2, V_C3−V_vg2, V_C4−V_vg2
Input end IN₁~ IN_FourEnter from the outermost area of the scene
6C₁~ 6C_FourBrightness value is calculated and the voltage V_C
Is a peripheral luminance value calculation circuit that outputs   FIG. 6 shows the configuration of the peripheral luminance value calculation circuit 20. No.
In FIG. 6, reference numerals 32, 33, 34, 35, and 36 are magnitude comparison circuits, respectively.
Input terminal I₁, I_TwoOf the two voltage values input to
Of these, the larger one is output from the output end OH, and the smaller one is output
It is output from the end OL. This size comparison circuit 32-35
A specific example of is shown in FIG. In FIG. 7, 66 is a positive phase,
Comparator with negative phase input terminal and positive and negative phase output terminals
Then, the relationship between the positive phase input terminal voltage V + and the negative phase input terminal voltage V-
When V + ≧ V-, H from the positive phase output terminal (high level means
Taste level voltage from the reverse phase output end to L (low level)
Level) is output. 67-70 is Ana
With the log switch, the voltage applied to the control terminal
When H level, it becomes conductive, and when L level, it opens.
State. Input I₁Input voltage to V_I1, I_TwoAt the input end
Input voltage is V_I2Then, V_I1≧ V_I2When
Each input of the palletizer 66 is a positive phase input terminal voltage V + ≥ negative phase input terminal
The voltage becomes V-. At this time, the positive phase output of the comparator 66
The terminal voltage becomes H level, the negative phase output terminal voltage becomes L level,
The control voltage of analog switches 67 and 70 is H level.
Control voltage of analog switches 68 and 69 is L level.
Since it is a bell, the analog switches 67 and 70 are in the conductive state,
The analog switches 68 and 69 are open. Out
Voltage V is applied to the power terminal OH through the analog switch 67._I1Is output
The voltage V is applied to the output terminal OL through the analog switch 70.
_I2Is output. Similarly, V_I1<V_I2When the
Each input of the oscillator 66 is positive phase input terminal voltage V + <negative phase input terminal voltage.
The voltage becomes V-, and the positive phase output end voltage of the comparator 66 is L level.
Bell, the negative phase output terminal voltage becomes H level. Yottte, Ana
The control voltage of the log switches 67 and 70 is L level.
The analog switches 67 and 70 are open,
The control voltage of the analog switches 68 and 69 is H level
Therefore, the analog switches 68 and 69 are in a conductive state.
At this time, output terminal OH is powered through analog switch 69.
Pressure V_I2Is output, and the analog switch 68 is connected to the output end OL.
Through voltage V_I1Is output. In this way, the output end O
H has voltage V_I1, V_I2The larger one is smaller for the output end OL
One is output. In Fig. 6, five magnitude comparison circuits are used.
Voltage V_C1, V_C2, V_C3, V_C4The largest of them is V
_H1And the second largest is V_H2As a third
Big one V_H3The smallest one is V_H4Out as
It is structured so that it will work. Input terminal of the size comparison circuit 32
I₁Is the voltage (V_C1−V_vg2) Is input and input end I_TwoHas
Pressure (V_C2−V_vg2) Is input and the larger one is output
Output from the end OH, and output the smaller one from the OL output end.
You. Similarly, the input terminal I of the magnitude comparison circuit 33₁Is the voltage (V_C3
−V_vg2) Is input, and whichever is larger is output from OH.
It outputs, and the smaller one is output from the output end OL. Large / small ratio
The comparison circuit 34 outputs from the output terminal OH of the size comparison circuits 32 and 33.
Input voltage I₁, I_TwoType in to compare big and small,
The output side OH outputs the threshold value and the output end OL outputs the smaller value
It is outputting, but here the voltage output from the output terminal OH
Is V_C1−V_vg2, V_C2−V_vg2, V_C3−V_vg2, V_C4−V_vg2Out of
It is the highest voltage. Similarly, in the size comparison circuit 35,
Input the voltage output from the output terminal OL of the magnitude comparison circuits 32 and 33.
Input to the power terminals 11 and 12 to compare the magnitude, and the larger one to the output terminal O
From H, the smaller one is output from the output end OL respectively.
However, the voltage output from the output terminal OL is V_C1−V_vg2, V
_C2−V_vg2, V_C3−V_vg2, V_C4−V_vg2With the smallest voltage of
is there. Whether the size comparison circuit 36 is the output end OL of the size comparison circuit 34
And the output voltage from the output terminal OH of the magnitude comparison circuit 35.
Pressure and I₁, I_TwoInput to the input terminal and compare the magnitude. Big
In the small comparison circuit 36, V_C1−V_vg2, V_C2−V_vg2, V_C3−V_vg2, V
_C4−V_vg2Ratio of the two intermediate voltage values excluding the maximum and minimum of
Output from the output terminal OH of the size comparison circuit 36.
Is V_C1−V_vg2, V_C2−V_vg2, V_C3−V_vg2, V_C4−V_vg2Out of
The second largest voltage is output and the third from the output end OL.
A large voltage is output to the eyes. In this way, V_C1−V
_vg2, V_C2−V_vg2, V_C3−V_vg2, V_C4−V_vg2, In descending order of V_H1
≧ V_H2≧ V_H3≧ V_H4Are rearranged so that   Returning to FIG. 6, 37-40 are V_C1, V_C2, V_C3, V_C4Average
Resistor with the same resistance value for its output voltage
Is V_cm= (V_C1+ V_C2+ V_C3+ V_C4) / 4-V_vg2Becomes 41 is
4 reference voltages V in the reference voltage generation circuit_r1, V_r2, V_r3, V_r4To
Output end r₁~ R_FourGenerated by this reference voltage V_r1, V_r2,
V_r3, V_r4The magnitude relationship of V_r1> V_r2> V_r3> V_r4It is. 42 ~ 4
5 is a comparator, the voltage V_cmAnd reference voltage V_r1, V
_r2, V_r3, V_r4Compare with, V_cm≧ V_r1, V_cm≧ V_r2, V_cm≧ V_r3, V
_cm≧ V_r4At the time of, the H level voltage is output respectively, and V
_cm<V_r1, V_cm<V_r2, V_cm<V_r3, V_cm<V_r4That's when
Outputs an L level voltage. 46,47,48,49 are invar
50 and 51 and 52 are AND gates. Inverter
The output of comparator 42 is input to 46, and the output of inverter 47 is output to inverter 47.
The output of the parator 43 is fed to the inverter 48 by the comparator 44.
Output of the comparator 45 is output to the inverter 49.
Each has been entered. One input end of AND gate 50
The output of the inverter 46 is connected to the
The output of the data 43 is input to one input terminal of the AND gate 51.
The output of the converter 47 is connected to the comparator 44 at the other input.
Output of the AND gate 52
The output of the comparator 48 and the output of the comparator 45 on the other input.
Have been entered respectively. 53-56 are OR gates
You. The output of the comparator 42 is connected to the three input terminals of the OR gate 53.
The output of AND gate 50 and the output of AND gate 51.
Each has been entered. At the 4 input ends of the Oagade 54,
Comparator 42 output, AND gate 50 output, AND
Output of gate 51 and output of AND gate 52 are input
Have been. The AND gate is connected to the four inputs of the OR gate 55.
Output of AND 50, output of AND gate 51, output of AND gate 52
The output and the output of the inverter 49 are respectively input.
The output of the AND gate 51 is connected to the 3 inputs of the OR gate 56,
The output of the AND gate 52 and the output of the inverter 49 are respectively
Has been entered. 57 to 60 are analog switches,
Controlled by OR gates 53-56 respectively
You. 61 to 64 are resistors having the same resistance value. Resistance 61 ~
64 is the analog switch 57 to 60 that is in the conductive state.
Force V_H1, V_H2, V_H3, V_H42 or
3 or 4 or 2 or 3 in ascending order
It is for averaging the voltage. 65 is an operational amplifier
The output end and the negative-phase input end are connected, and the voltage
It is used as a. From the output terminal of the operational amplifier 65,
Regardless of the circuit condition after the output terminal of the
Output voltage. The output voltage of this operational amplifier 65 is V_C
And output from the output terminal OUT.   Next, V in FIG._r1, V_r2, V_r3, V_r4And V_cmIs it big or small
The characteristics of the object field in that relationship and the voltage V at that time._C
The value of is described. (1) V_r1≤V_cm...... Analog switch 57, 58 conduction   At this time, a scene with a fairly bright background is assumed.
So 6C₁, 6C_Two, 6C_Three, 6C_FourRelatively low brightness in each area of
Ignore the brightness of the area_C= (V_H1+ V_H2) / 2. (2) V_r2≤V_cm<V_r1...... Analog switch 57,58,59
Continuity   At this time, the background is slightly bright, like outdoors in sunny weather.
A scene is assumed, and in such a scene, for example, shade
Not suitable as background brightness, which shows low brightness values, like the ground of
Appropriate elements are likely to appear. Therefore, in this case, the lowest
Ignoring the brightness of the brightness area, V_C= (V_H1+ V_H2+ V_H3) /
Set to 3. (3) V_r3≤V_cm<V_r2...... Analog switch 57,58,59,
60 continuity   In this case, a background scene with standard brightness is assumed.
In such a scene, the relatively bright part of the scene
The relatively dark part cannot be ignored. Therefore like this
In case V_C= (V_H1+ V_H2+ V_H3+ V_H4) / 4. (4) V_r4≤V_cm<V_r3...... Analog switch 58,59,60
Continuity   At this time, a scene with a slightly dark background is assumed, so
Ignore the brightest areas and use V_C= (V_H2+ V_H3+ V_H4) /
Set to 3. (5) V_cm<V_r4...... Analog switch 59, 60 conduction   At this time, a scene with a fairly dark background such as a night view
Is assumed, and in such a scene, for example,
Inappropriate elements as background brightness that shows high brightness locally
Is likely to appear. Therefore, in this case,
Ignore the region, V_C= (V_H3+ V_H4) / 2.   As described above, the peripheral luminance calculation circuit 20 is
V depending on the brightness of the surrounding area_H1, V_H2, V_H3, V_H4Large of
2 or 3 or 4 from your choice, or
The average value of 2 or 3 from the smaller one is V_CAs output
Output from the end OUT.   21 in FIG. 5 is the output voltage V of the logarithmic compression circuits 115 and 116._A, V_B
And the output voltage V of the peripheral brightness calculation circuit 20_CEach input end
IN₁, IN_Two, IN_ThreeEnter in the
A selection circuit that determines which arithmetic expression is selected.
In Fig. 5, 22,23,24 are resistors having the same resistance value.
Yes, 25 is an analog switch. Resistors 22, 23, 24 and
The analog switch 25 constitutes an average value circuit.
The output terminal A / P of the selection circuit 21 outputs an H level voltage.
Input to the control terminal of analog switch 25.
The analog switch 25 becomes conductive and the average value
Circuit output voltage V₁Is (V_A+ V_b+ V_C) / 3. on the other hand,
The L level voltage is output from the output terminal A / P of the selection circuit 21.
Input to the control terminal of analog switch 25.
Then, the analog switch 25 opens and the average value
Output voltage V₁Is (V_A+ V_b) / 2. 26 is an operational amplifier
And connect its output and negative-phase input to voltage
It is used as a follower, and the average value times
Output voltage V₁Is entered. Output of operational amplifier 26
The terminal voltage is V regardless of the circuit condition after the output terminal.₁In
You. 27, 28, 29, 30 are resistors having the same resistance value, 31
Is an operational amplifier. Resistors 27, 28, 29, 30 and operational amplifier 3
1 constitutes a subtraction circuit. Output terminal O of selection circuit 21
The voltage from UT is V_TwoThen, the output voltage of this subtraction circuit is
V₁−V_TwoIt is. This voltage V₁−V_TwoHowever, in this embodiment,
Represents the photometric value determined by multiple arithmetic expressions described later
doing.   FIG. 8 is a circuit showing a concrete example of the selection circuit 21 in FIG.
It is a road map. 71, 72, 73, 74 are resistors with the same resistance value, 7
Reference numeral 5 is an operational amplifier, which constitutes the first subtraction circuit.
You. Similarly, 76, 77, 78, 79 are resistors with the same resistance value, 8
0 is an operational amplifier, which constitutes the second subtraction circuit
You. The voltage V is applied to the first subtraction circuit._AAnd V_bIs entered and its
Output voltage is V_b−V_AIt is. The voltage V is applied to the second subtraction circuit._bPassing
And V_CIs input and its output voltage is V_C−V_bIt is. 81 is the base
Reference voltage V with quasi-voltage generator_r5Is occurring. 82 is
The voltage V at the positive phase input terminal of the comparator._CIs entered,
Reference voltage V at the negative phase input terminal_r5Is entered, V_C≧ V_r5When
H level voltage is output to V_C<V_r5When is L level
Output voltage. 85 is a reference voltage generation circuit
_Pa, V_Pb, V_Qa, V_QbOccurs. The reference voltage generator circuit 85
Output voltage of comparator 82 with control terminal B / D
Is input as the control voltage. To terminal B / D
Reference voltage when H level control voltage is input
Pressure V_Pa= V_P1, V_Pb= V_P2, V_Qa= V_Q1, V_Qb= V_Q2, While the end
When the L level control voltage is input to the child B / D
Reference voltage of V_Pa= V_P3, V_Pb= V_P4, V_Qa= V_Q3, V_Qb= V_Q4When
deep. Reference voltage V_P1, V_P2Or V_P3, V_P4, V_Q1, V_Q2Or V_Q3, V
_Q4The sign of V_P1, V_P2, V_Q1, V_Q3Is positive, V_P2, V_P4, V_Q2, V_Q4But
Is negative. 86,87 is a comparator, 88,89 is an inverter
is there. The first pull-out terminal is connected to the positive phase input terminals of the comparators 86 and 87.
Output voltage V of the operational amplifier 75 of the arithmetic circuit_b−V_AIs entered.
The reference voltage generating circuit 85 is connected to the negative-phase input terminal of the comparator 86.
Control voltage input to the control terminal B / D of
Reference voltage V depending on the H level and L level_P1Or V_P3
Is entered. The same applies to the negative phase input terminal of the comparator 87.
Reference voltage V_P2Or V_P4Is entered. 90, 91 are con
Parators 92 and 93 are inverters. Comparator 90,
The positive-phase input terminal of 91 is the operational amplifier of the second subtraction circuit 80
Output voltage V_C−V_bIs entered. Reverse phase input of comparator 91
The control terminal B / D of the reference voltage generation circuit 85
To the H level and L level of the control voltage input to
Therefore, the reference voltage V_Q1Or V_Q3Is entered. Compare
Similarly, the reference voltage V_Q2There
Is V_Q4Is entered. Output terminal of comparator 86,87,90,91
Is the magnitude of the positive phase input end voltage V + and the negative phase input end voltage V-.
Thus, when V + ≧ V−, the H level voltage becomes V + <
When the voltage is V-, L level voltage is generated. 94 ~
99,200 to 202 are AND gates, and comparators 86,8
H level and L level output from each output terminal of 7,90,91
Depending on how to combine, one of the AND gates
It outputs H level voltage and other AND gates are L level.
Outputs the bell voltage. V_b−V_AAnd V_PaAnd V_Pb, V_C−V_bAnd V
_QaAnd V_QbAnd gate 94-99,200-202
Which AND gate outputs H level
As described in. 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   203 is an inverter, the output state of the comparator 82
Has been reversed. 208,209 are oaged, 210,211 are
AND gate, 212 is an OR gate. Orgate 208
The outputs of AND gates 98,99,201,202 are at the four input terminals.
Has been entered and is one of AND gates 98,99,201,202
When one AND gate outputs H level voltage,
The output voltage of the agate 208 becomes H level. Similarly,
The AND gates 96, 98, 99, are provided at the six input terminals of the agate 209.
The output of 200 to 202 is input, AND gate 96, 98,
One of the 99,200-202 AND gates has an H level
When the voltage is output, the output voltage of OAGADE 209 becomes H level.
Become One of the inputs of the AND gate 210 has an oage
The output of the comparator 208 is connected to the other input terminal of the comparator 82.
Each output is input. The comparator 82
Output of AND gate 210 when outputting bell
Is equal to the output state of OR gate 208. At this time, Ann
The inverter 203 is provided at one of the two input terminals of the gate 211.
Output (the inverted output of the comparator 82) is input
Therefore, its output is at L level. Also AND gate
One of the two inputs of 211 has the output of OR gate 209
Is being input, and the output of the inverter 203 is being input to the other
Have been. The comparator 82 outputs the L level voltage
The AND gate 210 has one input terminal L
Output of AND gate 210 because level voltage is input
Is an L level voltage. At this time, the inverter
The output voltage of 203 is H level, and the output of AND gate 211 is
The force is equal to the output state of OR gate 209. And gate 2
The outputs of 10,211 are respectively input to the OR gate 212,
AND gate 210, 211 Output of at least one of
When the H level voltage is output, the OR gate 212
Output becomes H level. The output end of the OR gate 212 is
It is connected to the output terminal A / P. 213-217,219-223 is Ande
218, 224 are OR gates, and 225 are NOR gates.
226 is a reference voltage generation circuit, which has a reference voltage V_r8, V_r9, V_r10,
V_r11, -V_r12, -V_r13, -V_r14, -V_r15Occurs. 227 ~ 2
Reference numeral 35 is an analog switch. Two-input AND gate 21
The output of the comparator 82 is input to one of the input terminals of 3 to 217.
And one of the two-input AND gates 219 to 223
Is output from the inverter 203. Against this
And AND gate 9 is connected to the other input terminal of AND gate 213.
The output of 4 goes to the other input of AND gate 214.
The output of gate 95 goes to the other input of AND gate 215
AND gate 96 is the other input terminal of AND gate 216
Has the output of AND gate 99 and the other of AND gate 217.
The output of the AND gate 202 is input to the input end of
Have been. Also, the other input terminal of the AND gate 219
The output of AND gate 94 is the other input of AND gate 220.
The output of the AND gate 97 is at the output end of the AND gate 221.
The output of the AND gate 200 is connected to the other input terminal.
The output of the AND gate 201 is at the other input end of the gate 222,
The other input end of the AND gate 223 is connected to the AND gate 202.
Each output is input. Yotsutte
When an H level voltage is output from the
H-level voltage at one input terminal of 213 to 217,
The L level voltage is applied to one input terminal via the inverter 203.
Since it is input, AND gate 213 is AND gate 94
The output from the AND gate 214 is AND gate 95
And gate 215 is AND gate
AND gate 216 AND gate
The output from gate 99 remains unchanged and AND gate 217 outputs
The output from the NAND gate 202 is output as it is.
On the other hand, the output voltage of the AND gates 219 to 223 becomes L level.
Become. Conversely, L level from the comparator 82, inverter
When the H level voltage is output from 203, the AND gate
The output voltage of the switches 213 to 217 becomes L level, while
The gate 219 outputs the output from the AND gate 94 as it is.
The gate 220 outputs the output from the AND gate 97 as it is.
AND gate 221 outputs the output from AND gate 200 as it is.
The AND gate 222 outputs the output from the AND gate 201.
As is, AND gate 223 is the output from AND gate 202
Are output as they are. 2 pieces of OR gate 218
The output of the AND gate 214 is connected to one of the power
The output of the AND gate 215 is input to each
You. The output of OR gate 218 is the output of AND gates 215 and 216.
H when at least one of the input voltage is H level
Level. One of the two input terminals of the OR gate 224
Is the output of AND gate 221, and the other is AND gate 2
Each of the 22 outputs is input. Or gate 224
The output is at least the output voltage of the AND gates 221,222.
When either one is H level, it becomes H level. 8 pieces
The output of the AND gate 213, Oage
The output of the gate 218 is the output of the AND gate 216
The output of the gate 217 is the output of the AND gate 219.
The output of the gate 220 is the output of the OR gate 224.
The output of the
The outputs of 213 to 217 and 219 to 223 are all at the L level voltage.
When the output voltage of the NOR gate 225 becomes H level,
In all other cases, it becomes L level.   227-230 and 232-235 are analog switches.
The input terminal receives the reference voltage from the reference voltage generation circuit 226.
Has been added. At the input end of analog switch 227
Reference voltage V_r8, A reference voltage at the input end of the analog switch 228.
Pressure V_r9, A reference voltage is applied to the input terminal of analog switch 229.
V_r10, A reference voltage is applied to the input terminal of the analog switch 230.
V_r11, Reference voltage −V at the input end of analog switch 232
_r12, The reference voltage −
V_r13, The reference voltage −V is applied to the input terminal of the analog switch 234.
_r14, Reference voltage −V at the input end of analog switch 235_r15
Has been added. The input end of analog switch 231 is OV
It is. Connect the output terminals of analog switches 227-235 to each other.
And its output voltage V_TwoIs output from the output terminal OUT.
An analog switch 227 is
The output of the switch 213 is controlled by the analog switch 228.
The output of the OR gate 218 is connected to the analog switch 229
The output of the AND gate 216 is connected to the control terminal of
The control terminal of the analog switch 230 has an AND gate
Output of the switch 217 is the control end of the analog switch 213.
The output of NOR gate 225 is on the child
The output of the AND gate 219 is
AND gate 2 on the control terminal of log switch 233
20 outputs are analog switch 234 control terminals
The output of the OR gate 224 is connected to the analog switch 235.
The output of the AND gate 223 is
Has been entered. Three inputs to the selection circuit 21
Voltage V_A, V_b, V_CVoltage level, regardless of magnitude
Gate 213, OR Gate 218, AND Gate 216, AND
Gate 217, NOR Gate 225, AND Gate 219, AND
Any of gate 220, or gate 224, and gate 223
One output terminal goes to H level and the other gate outputs
Since the end is at the L level, the analog switches 227 to 235
One of them is the one to which the H level control voltage is applied.
Only the log switch becomes conductive, and other analog
The switch is open. The output of AND gate 213 is
When it is at H level, the analog switch 227 is conductive.
Voltage V_r8However, the output of the OR gate 218 is at the H level
At times, the analog switch 228 becomes conductive and
Pressure V_r9When the output of AND gate 216 is level,
When the analog switch 229 becomes conductive, the voltage V_r10But
When the output of the NAND gate 217 is H level, the analog switch
Switch 230 becomes conductive and voltage V_r11But Noah Gate 2
When 25 is H level, analog switch 232 is conductive
In this case OV is the output of AND gate 219
Is at the H level, the analog switch 232 is in the conductive state.
Voltage -V_r12However, the output of the AND gate 220 is H
At the level voltage, the analog switch 233 is in the conductive state.
Voltage -V_r13However, the output of the OR gate 224 is H level.
When the bell is on, the analog switch 234 becomes conductive.
Voltage −V_r14However, the output of the AND gate 223 is at the H level.
At times, the analog switch 235 becomes conductive and
Pressure-V_r15Is output from the output terminal OUT. However, V_r8, V
_r9, V_r10, V_r11> 0, -V_r12, -V_r13, -V_r14, -V_r15<0
Have a relationship.   Next, referring to FIGS. 9 to 11, FIG. 5 and FIG.
The circuit operation of the figure will be described. (1) Outermost region 6C of the light receiving element 6 shown in FIG.
(6C₁~ 6C_Four) Corresponding to the brightness value obtained from
Luminance signal) V_CIs the reference voltage V_r5If it is larger,
Nachi, V_C> V_r5Is determined to be outdoors
When a bright subject such as the sky enters the background and the periphery of the screen
If it is judged to be bright), then the luminance signal difference
V_b−V_A(Abbreviated as △ BA below) and luminance signal difference V_C−V_b(Less than
Based on the value of ΔCB), the reference voltage V_P1,
V_P2, V_Q1, V_Q2(The relationship of the reference voltage is V_P2<0 <V_P1, V_Q2<
0 <V_Q1), The photometric value V₁−V_TwoSeeking
Confuse. , Specifically, as shown in FIG. 9 (a),
By the central part of the area, that is, the area 6A and the surrounding area 6B
The measured luminance signal difference ΔBA becomes smaller, while the area 6B
And the luminance signal difference ΔCB between the outermost area 6 and the predetermined value V on the positive side
_Q1It becomes larger, and the main subject is the area 6A and
It can be judged that it often exists in both positions of area 6B.
You. Therefore, the photometric value V₁−V_TwoProvides moderate exposure to the main subject
In order to give, the luminance signal V of area 6A and area 6B_A, V_BTargeting
Then, assuming that the correction value is 0, it is calculated from the following arithmetic expression (1)-
You. V₁−V_Two= (V_A+ V_b) / 2… (1)-   Then, regarding the circuit operation of FIG. 5 and FIG.
Explaining the selection circuit 21 of FIG. 8, the outermost area 6C
(6C₁~ 6C_Four) Luminance signal (voltage) V_CIs the reference voltage V_r5Than
Since it is large, the comparator 82 outputs the H level. Yo
Then, set the control terminal B / D of the reference voltage generation circuit 85 to H level.
Bell signal is supplied and the reference voltage of the circuit 85 is V_Pa=
V_P1, V_Pb= V_P2, V_Qa= V_Q1, V_Qb= V_Q2Becomes Meanwhile, au pair
Output signal (voltage) V of pump 75_b−V_AIs V_P2<ΔBA <V_P1of
As the condition is met, the output of the comparator 86 is L level,
The output of the comparator 87 becomes H level, and the operational amplifier
80 output signal (voltage) V_C−V_bIs V_Q1<For the condition of ΔCB
Therefore, the outputs of comparators 90 and 91 are both at the H level.
Become. Therefore, the output of only AND gate 97 becomes H level.
And, this AND gate 97 H level output and
With the L level output of 94 to 96,98,99,200 to 202,
The output of the agate 212 is set to L level and the output terminal A / P is set to L level.
And set only Noah gate 225 to H level,
The other AND gates 213 to 223 are set to L level, and the output terminal OUT
Voltage of V_TwoIs OV. Therefore, the operational amplifier of FIG.
From 26 the voltage (V_A+ V_b) / 2 is output, and the operational amplifier 31
Since OV is supplied to the negative phase input terminal,
Output V₁−V_TwoIs the voltage (V_A+ V_b) / 2. At this time, specifically, as shown in FIG. 9 (b), the area 6A and the area
Brightness signal difference ΔBA from 6B is + side predetermined value V_P1Greater,
Further, the brightness signal difference ΔCB between the areas 6B and 6C is also the predetermined value V on the positive side.
_Q1Since it is larger, the main subject at this time is the entire area 6A.
It can be judged that it is often present in the part and part of the region 6B.
In this case, the luminance signal V of area 6A_APhotometry for only
Value V₁−V_TwoYou may ask for the background luminance signal
It is empirical that a better exposure value can be obtained by considering the degree.
You. Therefore, the photometric value V₁−V_TwoIs the main subject V_A, V_BTargeting
Correction value V_r8Is calculated from the following arithmetic expression (1)-
Confuse. V₁−V_Two= (V_A+ V_b) / 2-V_r8             … (1)-   To explain the circuit operation in this case, the comparators 86 and 8
7, 90, 91 and comparator 82 all output H level,
Only AND gate 94 outputs H level,
The gates 95 to 99 and 200 to 202 output the L level. Yotsu
The output of the OR gate 212 to the L level, and
The output of the switch 213 to the H level (the analog switch 2
27 becomes conductive), the operational amplifier 26 of FIG.
Output is voltage (V_A+ V_b) / 2, and the output V of the operational amplifier 31_I
−V_TwoIs (V_A+ V_b) / 2-V_r8Becomes , Specifically, as shown in FIG. 9 (c),
The luminance signal difference ΔBA between 6A and the area 6B is a predetermined value V on the positive side._P1Than
On the other hand, the luminance signal difference ΔCB between area 6B and area 6C increases
Is small, so the main subject will cover the entire area 6A at this time.
It is determined that there are many cases where the main subject exists or exists.
it can. In this case, target the main subject in area 6A
Photometric value V₁−V_TwoYou can ask for the
Luminance signal V in area 6B_BConsidering the
Metering value V to give exposure₁−V_TwoTargets areas 6A and 6B
Correction value V_r9(V_r9> V_r8) Using
(1) -determined from V₁−V_Two= (V_A+ V_b) / 2-V_r9             … (1)-   To explain the circuit operation in this case, the comparators 86 and 8
7 and 90 output H level, only AND gate 95 outputs H level
Output level and other AND gate 94,96 ~ 99,200 ~ 2
02 outputs L level, and comparator 82 outputs H level.
Power. Therefore, the output of the OR gate 212 is set to the L level.
The output of the AND gate 218 to H level (
As shown in FIG. 5, the analog switch 228 becomes conductive.
The output of the operational amplifier 26 of the_A+ V_b) / 2, the operation
Output V of amplifier 31₁−V_TwoIs (V_A+ V_b) / 2-V_r9Becomes At this time, specifically, as shown in FIG.
The luminance signal difference ΔBA between 6A and the area 6B is a predetermined value V on the positive side._P1Than
On the other hand, the luminance signal difference ΔCB between area 6B and area 6C increases
Is the negative value V_Q2Since it is smaller, the main cover
The size of the image is about the same as in (1-3) above.
In addition, a subject having a considerably high brightness in the area 6B (for example, the sun,
Sea surface reflection, etc.), or in area 6B for landscape photography, etc.
It can be determined that the subject with considerably high brightness is located. This
In the case of, all the brightness of areas 6A, 6B and 6C in terms of data
, The photometric value V₁−V_TwoAsk for
〔V₁−V_Two= (V_A+ V_b+ V_C) / 3] can also get good results
The inventor recognized as a method, but another one is used in this embodiment.
A method that yields good results in, that is, (1-
As in the case of 3), the brightness of the area 6A and the area 6B is targeted.
Correction value V_r9Is calculated from the following arithmetic expression (1)-
Confuse. V₁−V_Two= (V_A+ V_b) / 2-V_r9             … (1)-   The circuit operation in this case will be explained.
And 87 output H level, while comparators 90 and 91
Outputs L level, and AND gate 96 outputs H level
And other AND gates 94,95,97 to 99,200 to 202 are L
Outputs the level, and the comparator 82 outputs the H level.
You. Therefore, the output of the OR gate 212 is set to L level and
The output of the NAND gate 218 to the H level (the analog
Switch 228 becomes conductive), the operation of FIG.
The output of the amplifier 26 is the voltage (V_A+ V_b) / 2, the operational amplifier
31 outputs V₁−V_TwoIs (V_A+ V_b) / 2-V_r9Becomes , Specifically, as shown in FIG. 9 (e),
The brightness signal difference ΔBA between 6A and 6B is small, and
6C luminance signal difference ΔCB is a negative value V_Q2Greater absolute value
Therefore, at this time, the main subject exists in both areas 6A and 6B.
It can be judged that it is white and white. This
In the case of, the photometric value is for the brightness of area 6A and area 6B.
However, if the peripheral area of the screen has lower brightness than the central area,
In this case, consider the brightness of area 2C at the periphery of the screen to some extent.
To obtain a photometric value, it is possible to
You can take a picture (highlight control).
It is known that a good exposure value can be obtained in terms of data.
Therefore, at this time, the main subject part is highlighted.
Brightness signal V_A, V_b, V_CAll of
Elephant, photometric value V₁−V_TwoIs the correction value V_r10And the following operation
It is calculated from the equation (1)-. V₁−V_Two= (V_A+ V_b+ V_C) / 3-V_r10        … (1)-   To explain the circuit operation in this case, the comparator 87
Outputs H level, while comparators 86, 90 and 91
Outputs L level and AND gate 99 outputs H level
And other AND gates 94-98, 200-202 are at L level
Is output, and the comparator 82 outputs the H level.   Therefore, the output of the OR gate 212 is set to H level, and
The gate 216 to the H level (the analog switch 2
29 becomes conductive), the operational amplifier 26 of FIG.
Output is voltage (V_A+ V_b+ V_C) / 3, and the output of operational amplifier 31
Force V₁−V_TwoIs (V_A+ V_b+ V_C) / 3-V_r10Becomes At this time, specifically, as shown in FIG. 9 (f), the area 6A
And the brightness signal difference ΔBA between the area 6B and the area 6B are negative values and the predetermined value V_P2Than
The absolute value is large, and the brightness signal difference ΔCB between area 6B and area 6C is also negative.
Value of V_Q2Since the absolute value is larger than this,
Main subject is in all of area 6A and part of 6B Medium
It can be judged that it is the case of
Wear. In this case, as in the case of (1-5) described above,
To give the main subject an exposure that will be highlighted
, The brightness of the areas 6A, 6B, and 6C, and the photometric value V₁−V_Two
Is the correction value V_r11(V_r11<V_r10) Using
(1) -determined from V₁−V_Two= (V_A+ V_b+ V_C) / 3-V_r11        … (1)-   The circuit operation in this case will be described.
91 outputs all L level, and AND gate 202 outputs H level.
The other AND gates 94 to 99,100 to 101
The L level is output, and the comparator 82 outputs the H level.
You. Therefore, the output of the OR gate 212 is set to H level,
The output of the NAND gate 217 to the H level (this
Switch 230 becomes conductive), the operation of FIG.
The output of the amplifier 26 is the voltage (V_A+ V_b+ V_C) / 3, and opea
Output V of pump 31₁−V_TwoIs (V_A+ V_b+ V_C) / 3-V_r11Becomes At this time, specifically, as shown in FIG. 9 (g), the area 6A
And the brightness signal difference ΔBA between the area 6B and the area 6B are negative values and the predetermined value V_P2Than
The absolute value is large, and the brightness signal difference ΔCB between area 6B and area 6C is small.
Therefore, at this time, the main subject exists in the entire area 6A.
Or, especially when the main subject is small,
It can be determined that the main subject is a white subject.
In this case as well, the main subject is highlighted as in the above case.
Regions 6A, 6B and 6C are paired to obtain the exposure as shown.
Assuming that the correction value is 0, the following equation (1)-
Ask. V₁−V_Two= (V_A+ V_b+ V_C) / 3 (1)-   Explaining the circuit operation in this case, the comparator 90
All outputs H level, while comparators 86, 87 and 9
1 outputs L level and AND gate 202 outputs H level
And other AND gates 94 to 99,200,202 are at 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 NOR gate
The output of the gate 225 goes high (the analog switch
Switch 230 becomes conductive), the operation of FIG.
The output of_A+ V_b+ V_C) / 3, the operational amplifier
31 outputs V₁−V_TwoIs (V_A+ V_b+ V_C) / 3. At this time, specifically, as shown in FIG. 9 (h), the area 6A
And the brightness signal difference ΔBA between the area 6B and the area 6B are negative values and the predetermined value V_P2Than
The absolute value is large, and the brightness signal ΔCB of area 6B and area 6C is
Value V_Q1At this time, the main subject is
The size of the main subject is about the same as the case described in (1-1).
There is a difference in light and shade in the body part, and the area 6A becomes slightly brighter.
Area, or a landscape photograph, etc.
It can be judged that the case is occupied by the body. in this case
In terms of data, all brightness in areas 6A, 6B, and 6C is targeted.
And set the correction value to 0 and the photometric value V₁−V_Two[V₁−V_Two=
(V_A+ V_b+ V_C) / 3] can also be a way to get good results
The inventor recognized that, in this embodiment, another good
The method by which the result is obtained, that is, in the case of (1-1) above
Similarly, the correction value is for the brightness of the areas 6A and 6B.
0 is obtained from the following arithmetic expression (1)-. V₁−V_Two= (V_A+ V_b) / 2… (1)-   The circuit operation in this case will be explained.
And 91 output H level, while comparators 86 and 87 output L level.
The level is output, and AND gate 200 outputs the H level
However, other AND gates 94 to 99,201,202 are set to L level.
Then, the comparator 82 outputs the H level. Yotsu
The output of OR gate 212 to L level, and NOR gate 2
The output of the operational amplifier 26 of FIG.
Force is voltage (V_A+ V_b) / 2, and the output V of the operational amplifier 31₁−
V_TwoIs (V_A+ V_b) / 2. At this time, specifically, as shown in FIG. 9 (i), the area 6A
And the area 6B have a small luminance signal difference ΔBA, and the area 6B and
Since the luminance signal difference ΔCB with the range 6C is also small, the main object at this time is
If the object occupies the entire scene, or if the scene
It can be judged that there is no intention to set the main subject
You. In this case, the brightness of the areas 6A, 6B and 6C is targeted.
Then, the correction value is set to 0, and it is calculated from the following arithmetic expression (1)-
You. V₁−V_Two= (V_A+ V_b+ V_C) / 3 (1)-   The circuit operation in this case is explained below.
And 90 output H level, while comparators 86 and 91
Outputs L level and AND gate 98 outputs H level
However, other AND gates 94 to 97,99,200 to 202 have L level.
And the comparator 82 outputs the H level. Yo
Then, set the output of the OR gate 212 to H level and
The output of the operational amplifier 26 is
Voltage (V_A+ V_b+ V_C) / 3, and the output V of the operational amplifier 31₁−
V_TwoIs (V_A+ V_b+ V_C) / 3.   FIG. 10 (a) shows the above (1-1) to (1-9).
The relation of each of the arithmetic expressions performed according to each luminance state in
It represents a person in charge. (2) Outermost region 6C of the light receiving element 6 shown in FIG.
(6C₁~ 6C_Four) Luminance signal V obtained from_CIs the reference voltage V_r5Yo
Less than V, that is, V_C<V_r5Indoor wall as background
When it is judged that it is a room where
Also in the same manner as (1) above, the luminance signal difference ΔBA and the luminance signal are
Depending on the value of the signal difference ΔCB, a constant V_P3, V_P4, V_Q3, V_Q4(V_P4<0 <
V_P3, V_Q4<0 <V_Q3), The photometric value V₁
−V_TwoAsk for.At this time, specifically, as shown in FIG.
The brightness signal difference ΔBA in area 6B becomes smaller, and
Brightness signal difference ΔCB is a predetermined value V_Q3It ’s bigger, so at this time
Is large if the main subject is in both areas 6A and 6B
Therefore, it can be determined that the subject is a blackish subject.   In this case, simply targeting the brightness of areas 6A and 6B,
The photometric value may be calculated with the correction value set to 0, but
Make sure the subject is black (shear control)
Is more preferable, and in this embodiment, the main subject portion is
Area to give the exposure that the minutes are described as shadow
For 6A and 6B brightness, the photometric value V₁−V_TwoIs the negative correction value
−V_r13Is calculated from the following arithmetic expression (2)-. V₁−V_Two= (V_A+ V_b) / 2 + V_r13            (2)-   To explain the circuit operation in this case, the outermost area 6C
(6C₁~ 6C_Four) Luminance signal (voltage) V_CIs the reference voltage V_r5Than
Since it is large, the comparator 82 outputs the L level,
The control terminal B / D of the reference voltage generating circuit 85 to the L level.
Signal is supplied, 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
Data output from the L86, while comparators 87, 90 and 9
1 outputs to H level. And gate 97 is H level
And AND gates 94 to 96,98,99,200 ...
202 outputs L level and inverter 203 outputs H level.
Power. Therefore, the output of the OR gate 212 is set to the L level.
The output of the AND gate 220 to H level (
As shown in FIG.
The output of the operational amplifier 26 of the_A+ V_b) / 2, the operation
Output V of amplifier 31₁−V_TwoIs (V_A+ V_b) / 2 + V_r13Becomes At this time, specifically, as shown in FIG. 11 (b), the area 6A and the area
6B brightness signal difference ΔBA is a predetermined value V_P3Larger, area 6B
And area 6C luminance signal difference ΔCB is a predetermined value V_Q3Greater than
Then, at this time, the main subject is in the entire area 6A and part of the area 6B.
If it exists, it is judged to be a blackish subject
Wear. In that case, the above (2-1) is added to the main subject part.
In order to give the exposure that is described as a shadow in the same way as
, The brightness of the areas 6A and 6B, and the photometric value V₁−V_TwoIs the-side
Correction value of −V_r12To (| V_r12| <| V_r13|)
Calculated from the formula (2)-. V₁−V_Two= (V_A+ V_b) / 2 + V_r12            (2)-   The circuit operation in this case will be described.
91 outputs all H level, and AND gate 94 outputs H level
And other AND gates 95 to 99, 200 to 202 are L
Outputs the level, and the inverter 203 outputs the H level.
You. Therefore, the output of the OR gate 212 is set to L level and
The output of the NAND gate 220 is set to H level, and the operation of FIG.
The output of the amplifier 26 is the voltage (V_A+ V_b) / 2, the operational amplifier
31 outputs V₁−V_TwoIs (V_A+ V_b) / 2 + V_r12Becomes At this time, specifically, as shown in FIG.
The brightness signal difference ΔBA from the area 6B is a predetermined value V_P3Greater, one
Since the brightness signal difference ΔCB between the two areas 6B and 6C is small, this
, The main subject exists in the entire area 6A, or
When the main subject is small, and
It can be judged that it is the case. In this case, the main subject part
Area 6 to give the exposure as shown in shadow
For A and 6B brightness, the photometric value V₁−V_TwoSets the correction value to 0
Then, it is obtained from the following arithmetic expression (2)-. V₁−V_Two= (V_A+ V_b) / 2… (2)-   To explain the circuit operation in this case, the comparator 91
L level is output, and comparators 86-90 output H level.
The AND gate 95 outputs H level,
AND gates 94,96 to 99,200 to 202 output L level, and
The inverter 203 outputs the H level. Yotsuge
Set 212 to L level and NOR gate 225 to H level
The output of the operational amplifier 26 in FIG._A+ V_b) / 2 and
And the output V of the operational amplifier 31₁−V_TwoAlso (V_A+ V_b) / 2
You. At this time, specifically, as shown in FIG. 11 (d), the area 6A and
The brightness signal difference ΔBA from the area 6B is a predetermined value V_P3Greater than
The difference in brightness signal between area 6B and area 6C ΔCB is a negative value and the specified value V_Q4Yo
Since the absolute value is large, the main subject is in the case of (2-3) above.
It is a normal subject of the same size as
It is known that there is a high-brightness subject (for example, a light).
Can be turned off. As in this case, in the room 6B, the high brightness
When the body is present, the outdoor sun etc. is located in area 6B
Compared to the case, data that this high brightness has less effect
Photometric value V₁−V_TwoIs the brightness of areas 6A, 6B and 6C
The following calculation formula (2)-
Request. V₁−V_Two= (V_A+ V_b+ V_C) / 3 (2)-   The circuit operation in this case will be explained.
And 91 output L level, while the comparators 86 and 87 output H level.
And the AND gate 96 outputs H level,
Other AND gates 94,95,97 to 99,200 to 202 are L bells
And the inverter 203 outputs H level. Yotsu
OR gate 212 goes to H level, and NOR gate 225
As the H level, the output of the operational amplifier 26 is the voltage (V_A+ V_b+
V_C) / 3, and the output V of the operational amplifier 31₁−V_TwoAlso (V_A+ V_b+
V_C) / 3. , Specifically, as shown in FIG. 11 (e), the area
The brightness signal difference ΔBA between 6A and area 6B is small, and
6C luminance signal difference ΔCB is a negative value and the specified value V_Q4More absolute value
Because it is large, if the main subject exists in both areas 6A and 6B,
And only area 6A and area 6B are
It can be determined that it is illuminated. In that case
Determines the photometric value for the brightness of areas 6A and 6B only.
There is also a way of thinking, but in the present embodiment, the peripheral area where the object scene is dark
The photometric value V₁−V_Two
Is for the brightness of areas 6A, 6B and 6C, and the correction value is 0
And calculated by the following arithmetic expression (2)-. V₁−V_Two= (V_A+ V_b+ V_C) / 3 (2)-   To explain the circuit operation in this case, the comparators 86 and 9
0 and 91 output L level, while comparator 87 outputs H level.
And the AND gate 99 outputs H level,
Other AND gates 94-98, 200-202 output L level
Then, the inverter 203 outputs the H level.   Therefore, the OR gate 212 becomes H level, and
Output of operational amplifier 26 in FIG.
Is the voltage (V_A+ V_b+ V_C) / 3, and the output V of the operational amplifier 31₁
−V_TwoAlso (V_A+ V_b+ V_C) / 3. At this time, specifically, as shown in FIG. 11 (f), the area 6A and
The brightness signal difference ΔBA from the area 6B is a negative value and the predetermined value V_P4More perfect
The logarithmic value is large and the luminance signal difference ΔCB between area 6B and area 6C is also negative.
Value of V_Q4Since the absolute value is larger, the main subject is 6
If it exists in all of A and part of region 6B, and
The entire area 6A and part of the area 6B are illuminated by a light, etc.
It can be determined that it is. In this case
Unlike the case of (2-5), it can be corrected with the correction value.
It is necessary to give a proper exposure to the main subject part,
For the brightness of the areas 6A, 6B and 6C, the photometric value V₁−V_TwoIs the-side
Correction value of −V_r15Is calculated by the following arithmetic expression (2)-
You. V₁−V_Two= (V_A+ V_b+ V_C) / 3 + V_r15        (2)-   The circuit operation in this case will be described.
91 outputs all L level, and AND gate 202 outputs H level.
And AND gates 94 to 99,200,201 are L
The inverter 203 outputs the H level.
You. Therefore, the OR gate 212 becomes H level, and
With the gate 223 at the H level, the operational amplifier 26 of FIG.
Output is voltage (V_A+ V_b+ V_C) / 3, and the output of operational amplifier 31
Force V₁−V_TwoIs (V_A+ V_b+ V_C) / 3 + V_r15Becomes At this time, specifically, as shown in FIG. 11 (g), the area 6A
And the brightness signal difference ΔBA between the area 6B and the area 6B are negative values and the predetermined value V_P4Than
The absolute value is large, and the brightness signal difference ΔCB between area 6B and area 6C is small.
Therefore, at this time, the main subject exists in the entire area 6A.
Or the main subject is small and exists in part of area 6A
Can be determined. In this case, it is suitable for the main subject part
The brightness of areas 6A, 6B and 6C
And photometric value V₁−V_TwoIs the minus correction value_r14(| V_r14|> | V
_r15|) Is used to obtain the following equation (2)-. V₁−V_Two= (V_A+ V_b+ V_C) / 3 + V_r14        (2)-   To explain the circuit operation in this case, the comparators 86 and 8
7 and 91 output L level, while comparator 90 outputs H level.
The level is output, and AND gate 202 outputs the H level
However, other AND gates 94 to 99, 200 to 202 are at L level
And the inverter 203 outputs H level. Yotsu
OR gate 212 outputs H level, and OR gate 224
Is at the H level (which causes the analog switch 234 to become conductive).
The output of the operational amplifier 26 in FIG. 5 is
Voltage (V_A+ V_b+ V_C) / 3, and the output V of the operational amplifier 31₁−
V_TwoIs (V_A+ V_b+ V_C) / 3 + V_r14Becomes At this time, specifically, as shown in FIG. 11 (h), the area 6A
And the brightness signal difference ΔBA between the area 6B and the area 6B are negative values and the predetermined value V_P4Than
The absolute value is large, and the luminance signal difference ΔCB between area 6B and area 6C is
Fixed value V_Q3Since it is larger, the main subject is above
The same size as the case of (2-1) and the main subject part
There is a difference in light and shade, and area 6A is slightly brighter, but
For subjects that are black as a body or for landscape photography, etc.
This is a case where a subject of considerably low brightness occupies the area 6B.
Can be determined. In this case, data 6A, 6B and
And all the brightness of 6C, the photometric value with the correction value as 0
V₁−V_Two[V₁−V_Two= (V_A+ V_b+ V_C) / 3]
We also found that good results can be obtained depending on the conditions of the field.
However, in the example, those who obtained good overall results
Method, i.e., for all intensities in regions 6A, 6B and 6C
And photometric value V₁−V_TwoIs the minus correction value -V_r14Using
It is calculated by the formula (2)-. V₁−V_Two= (V_A+ V_b+ V_C) / 3 + V_r14        (2)-   The circuit operation in this case will be explained.
And 87 output L level, while comparators 90 and 91
H level is output, and AND gate 200 outputs H level
However, other AND gates 94 to 99,201,202 are at L level.
Then, the inverter 203 outputs the H level. Yotsu
OR gate 212 outputs H level, and OR gate 224
Outputs H level, the output of operational amplifier 26 in FIG.
Force is voltage (V_A+ V_b+ V_C) / 3, the output of the operational amplifier 31
V₁−V_TwoIs (V_A+ V_b+ V_C) / 3 + V_r14Becomes At this time, specifically, as shown in FIG. 11 (i), the area 6A
And the area 6B have a small brightness signal difference ΔBA, and the area 6B and
Since the brightness signal difference ΔCB with the area 6C is also small, the main object at this time is
If the object occupies the entire scene, or a landscape photo
This is the case when the main subject is not set, such as
You can judge that. In this case, the areas 6A, 6B and 6C
The entire area 6A-6C should be
For brightness, the photometric value V₁−V_TwoSets the correction value to 0 and
It is calculated by the formula (2)-. V₁−V_Two= (V_A+ V_b+ V_C) / 3 (2)-   The circuit operation in this case will be explained.
And 91 output L level, while comparators 87 and 90 output
H level is output, and AND gate 98 outputs H level
However, other AND gates 94 to 97,99,200 to 202 have L level.
And the inverter 203 outputs H level. Yo
Then, the OR gate 212 outputs the H level, and the NOR gate 2
Since 25 outputs H level, the output of operational amplifier 26 is voltage
(V_A+ V_b+ V_C) / 3, and the output V of the operational amplifier 31₁−V_TwoAlso
(V_A+ V_b+ V_C) / 3.   FIG. 10 (b) shows the above (2-1) to (2-9).
The relation of each of the arithmetic expressions performed according to each luminance state in
It represents a person in charge.   Above, the characteristic features of the photometric device of the embodiment described
Means that the main subject is in the center of the object area (area 6A).
Assuming that the size of the main subject can be judged
And make corrections according to the size of the main subject.
it can.   In Fig. 5, the peripheral image plane drop information is transmitted from the lens side.
If not reached, output from the correction value output circuit 101
Reference voltage V_vg1, V_vg2Each area 6B of the light receiving element 6
6C₁~ 6C_FourThere is an error in the brightness value. That is,
Perform scene discrimination based on the level of brightness difference between adjacent areas
Therefore, the wrong scene will be identified and calculated.
Photometric value V₁−V_TwoError. In this embodiment, the lens
Fits the lens from the correction value output circuit 101 due to replacement of the lens
Input the reference voltage (correction value) that corresponds to the decrease in the amount of light on the image plane.
By using the fact that the
Since the correction of the drop is added, that is, the field
The brightness of can be accurately measured regardless of the positions on both sides.
Therefore, it is possible to eliminate the photometric error due to lens replacement,
Providing proper exposure to the main subject in the center
It becomes possible.   Further, in the present embodiment, the outermost region of the light receiving element 6
6C each small area 6C₁~ 6C_FourThe average value of the obtained luminance signal is
If it is on the high brightness side, the small area 6C₁~ 6C_FourLower of
By ignoring the small area output, which is the luminance signal,
For example, on a background like a shaded ground outdoors on a sunny day
It is possible to remove elements that are inappropriate for brightness. Also, each of the above small areas
6C₁~ 6C_FourThe average value of the luminance signal obtained from
If the small area 6C₁~ 6C_FourWith a high brightness signal
By ignoring the output of small areas
Inappropriate factors as background brightness such as lighting
Exclude.   Furthermore, in the photometric device of this embodiment, the subject is white
Or, when it detects that it is black, it is consciously white.
The subject should be white, and the black subject should be black.
Illumination (highlight control) or
Correct the shadow description (shadow control)
In addition, this correction is based on the determination of the size of the main subject.
Since the correction amount is changed, the
It is possible to control.   It should be noted that FIGS. 10 and 11 used in the description of the above embodiment
In the illustration of the figure, the value of the brightness level of each area 6A to 6C
Is the same level when the brightness difference between adjacent areas is small.
This is, of course, a little in the case of actual photometry.
There is a difference (a predetermined value compared, for example V_P1Than
Means that the brightness difference is small.)
Role of explanatory diagram for easy understanding in this embodiment
It is the only thing that can be achieved. In addition, in this embodiment
, The photometric value V₁−V_TwoThe calculation method of is the brightness level around the screen.
There are 4 ways by bell (V_C1~ V_C4), But
Even if you obtain a photometric value by dividing it into smaller areas
Good, and conversely the screen peripheral (V_C) As one area
You may. Also, the photometric value V₁−V_TwoThe calculation method of
Set the metering area to V in the center and middle_AAnd V_BOnly and the whole picture
Face V_A, V_B, V_CAlthough it was divided into two cases, the case of multiple division
Always cover some or all of the surrounding areas
In this way, the photometric value is obtained in the same manner as described above.
May be. Also divide the field into three or more concentric regions
And use the difference in luminance signal between adjacent areas.
You may.   Further, in this embodiment, the selection circuit 21 is composed of a logic circuit.
However, it is processed by software using a microcomputer.
Of course, it goes without saying that the present invention is implemented.
You. (Correspondence between invention and embodiment)   In the present embodiment, the correction value output circuit 101 is an auxiliary circuit of the present invention.
Corrects the brightness value by the positive value output means and the light receiving element 6 by the light receiving means.
The circuits 108 to 112 and the photometric value calculation circuit 113 serve as calculation means.
Equivalent to each. (The invention's effect)   As described above, according to the present invention,
Obtained from the peripheral brightness and the correction value output means of the interchangeable lens.
Correction value information corresponding to the image surface light loss during the measured luminance
Since the photometric value is calculated based on the
Correct the photometric output using the correction value information that corresponds to the dropout.
By doing so, the subject brightness can be measured accurately.
You.

BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is an optical layout diagram of a photometric system when the present invention is applied to a single-lens reflex camera, and FIG. 2 is a diagram for explaining a light receiving surface of a light receiving element shown in FIG. FIG. 4 is a diagram showing an example of an image plane light amount drop in a photographing lens, FIG. 4 is a schematic block diagram showing an example of a signal processing system for carrying out the present invention, and FIG. 5 is a concrete block diagram of each circuit shown in FIG. Circuit diagram, Figure 6 is the fifth
7 is a circuit diagram showing a configuration example of a peripheral brightness calculation circuit shown in FIG.
6 is a circuit diagram showing a configuration example of the magnitude comparison circuit shown in FIG.
FIG. 8 is a circuit diagram showing a configuration example of the selection circuit shown in FIG.
9 to 11 are diagrams for explaining an arithmetic expression for obtaining a photometric value in each luminance state of the object scene. 6 ... Light receiving element, 6A to 6C area, 20 ... peripheral brightness calculation circuit, 21 ... selection circuit, 22-24 ... resistance, 25 ... analog switch, 26 ... operational amplifier, 28-30 ... Resistance, 31 ...
… Op Amp, 101 …… Correction value output circuit, 115-120 ……
Logarithmic compression circuit, 121 to 140 ... resistance, 141 to 145 ... operational amplifier, V _{A to} V _C ... voltage, V _vg1 , V _vg2 ... reference voltage, V ₁ -V
₂ …… Photometric value.

   ────────────────────────────────────────────────── ─── Continuation of front page    (72) Inventor Shuichi Kiyohara               770 Shimonoge, Takatsu-ku, Kawasaki-shi Canon               Tamagawa Works Co., Ltd.                (56) References JP-A-56-74226 (JP, A)

Claims (1)

(57) [Claims] In a camera system including an interchangeable lens and a camera body to which the interchangeable lens is attachable / detachable, correction value output means capable of outputting correction value information corresponding to a drop in light amount on an image plane during luminance measurement to the interchangeable lens. The camera body is provided with at least a light receiving unit for independently measuring the brightness of the central portion and the peripheral portion of the screen, and the brightness measured by the light receiving unit and the correction value output unit of the interchangeable lens. A camera system comprising: a calculation unit that calculates a photometric value based on correction value information. 2. In a camera body with an interchangeable lens that has a correction value output unit capable of outputting correction value information corresponding to the decrease in light amount on the image plane during brightness measurement, at least the brightness at the center of the screen and the brightness at the periphery are independent. And a calculation means for calculating a photometric value based on the brightness measured by the light reception means and the correction value information obtained from the correction value output means of the interchangeable lens. The camera body to do. 3. At least the photometric value based on the correction value information obtained from the light receiving means for independently measuring the brightness of the central part and the peripheral part of the screen and the brightness measured by the light receiving means and the correction value output means of the interchangeable lens. An interchangeable lens that can be attached to and detached from a camera body having a computing means for computing, and is capable of outputting, to the computing means of the camera body, correction value information corresponding to an image plane light amount drop during luminance measurement. An interchangeable lens having a value output means.