JPS636820B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a light receiving circuit that operates in response to light.

Conventionally, various light receiving circuits have been proposed, but they cannot avoid the influence of the dark current of the light receiving element, and when the temperature rises, the dark current value also increases, causing circuit malfunction. The defect was that it could not detect very small amounts of light due to dark current.

The present invention provides a light receiving circuit in which the influence of dark current does not appear on the output, and eliminates the above-mentioned conventional drawbacks. That is,
The present invention is realized by providing a light-receiving element that does not receive light as a dark current canceling element, and configuring a current mirror circuit between the circuit of the non-light-receiving element and the circuit of the light-receiving element that actually receives light. .

Embodiments of the present invention will be described below with reference to the drawings.

FIG. 1 is an example of a light receiving circuit for obtaining on/off output.

Transistors Tr _Z1 , Tr _Z2 , Tr _Z3 and resistor Rz constitute a constant voltage circuit, for example,
Even if any voltage of 8V or higher is applied, a constant voltage V _Z of about 6.5V is always output at the emitter of the transistor Tr _Z3 . light emitting diode
The LED serves as a light source for appropriately irradiating light to a light-receiving element, which will be described later, depending on the operation of a shield or the like.

The phototransistor PT ₁ is provided as a dark current canceling transistor, and its surface is covered with aluminum to prevent light from entering, so that only dark current is output. The other phototransistor PT ₂ actually receives light, but the area, diffusion depth, etc. of phototransistors PT ₁ and PT ₂ are made under the same conditions, so that their dark currents are equal. has been done. The collector-base connected transistor Tr ₁ has a phototransistor PT ₁ connected in series to form a current sink circuit.
In addition, in this current sink circuit, the resistor R _I1 and the transistors Tr _I1 , Tr _I2 , and Tr _I3 connected in parallel with the phototransistor PT ₁ obtain a comparison reference current, and the current I ₁ is I ₁ = V _Z −V _BE ×4/R _I1 ...... However, V _BE is the transistor Tr _I1 , Tr _I2 , Tr _I3 ,
It is expressed as the base-emitter voltage of Tr ₁ . Now I _PT1 phototransistor _PT1
Assuming that the output current is the dark current, the emitter current I _D of the transistor Tr ₁ is I _D = I ₁ + I _PT1 , so according to the formula, I _D = V _Z −4V _BE /R _I1 + I _PT1 .

Transistor Tr ₂ operates with the same base-emitter voltage as transistor Tr ₁ , and in the illustrated circuit example, at least these transistors
Tr ₁ and Tr ₂ are included in the integrated circuit of the same chip, and the transistors Tr ₁ and Tr 2 are respectively
Assuming that the base and emitter areas of Tr ₂ are the same,
The sinkable current (emitter current) I ₃ of the transistor Tr ₂ is the same as the emitter current _ID of the transistor Tr ₁ , I ₃ =I _D . This, in other words, constitutes a current mirror circuit. A phototransistor PT ₂ that actually receives light is connected in series to the transistor Tr ₂ of this current mirror circuit.

When light enters the phototransistor _PT2 , an output current _IPT2 proportional to the incident light flows through the phototransistor.
The base current I _b of the amplifier transistor Tr ₃ flowing through PT ₂ is I _b = I _PT2 − I ₃ ...... From the formula, I _b = I _PT2 − V _Z −4V _BE /R _I1 − I _PT1 = ( I _PT2 −I _PT1 )−V _Z −4V _BE /R _I1 …. As mentioned above, I _PT1 is the dark current of the phototransistor PT ₁ , and I _PT2 is _the dark current flowing through the phototransistor PT ₂ that can receive light. photocurrent produced
I _PT2 ′) + (dark current I _PT2 ″) Moreover, the dark current of phototransistors PT ₁ and PT ₂
I _PT1 , I _PT2 ″ are set so that I _PT1 = I _PT2 ″,
In the end, the equation is expressed as I _b = I _PT2 ′−V _Z −4V _BE /R _I1 ………′. As is clear from this equation, I _b is only related to the photocurrent I _PT2 ' caused by the incident light of the phototransistor PT ₂ , and the output can be obtained without being affected by dark current.

By the way, in the example of the light receiving circuit shown in Fig. 1, when the incident light on the phototransistor PT ₂ is small and I _PT2 ′≦V _Z −4V _BE /R _I1 , the value of I _b in the equation ′ is negative, that is, I _b =
0, the output transistor Tr ₄ is off, and the output voltage Vout becomes the power supply voltage Vcc. When the incident light on the phototransistor PT ₂ increases and I _PT2 ′>V _Z −4V _BE /R _I1 , the transistor Tr ₄ turns on with I _b > 0,
Obtain a voltage approximately at ground (GND) level as the output voltage Vout. The inversion boundary value of the output voltage Vout can be determined by varying the resistor R _I1 and arbitrarily setting the comparison reference current I ₁ according to the equation '.

The usefulness of this circuit will be explained by citing an example in which the above light receiving circuit is applied to a tachometer.

FIG. 2 is a diagram of the principle of the tachometer, and FIGS. 3a, b, and c are time charts showing the signal waveforms of the main parts of the light receiving circuit of FIG. 1 when used as a tachometer. As shown in FIG. 2, when a light emitting diode LED and a phototransistor PT ₂ are placed opposite to each other with a rotary plate R having slits S, S, _. . .
The output current I _PT2 and output voltage Vout of PT ₂ are as shown in Figure 3 a, b, c, respectively, and the output voltage Vout
The number of rotations of the rotating plate R can be detected by counting the number of pulses.

In such a tachometer, when the temperature near the device becomes high, the dark current component of the phototransistor _PT2 increases. For example, in some cases, the dark current at 100°C is approximately 2000 times higher than at 25°C. In a conventional light receiving circuit (a light receiving circuit that does not include a so-called dark current cancel circuit), when the temperature reaches a high temperature, the minimum value I _b (Min) becomes the inversion operating current value I _b (b) of the output transistor Tr ₄ .
This may cause the output voltage Vout to remain almost at the GND level. However, when a light receiving circuit as shown in FIG. 1 is used, the influence of dark current is canceled and normal operation can be performed even at high temperatures.

FIG. 4 shows an example of a light receiving circuit for obtaining an analog output proportional to incident light. Components having the same functions as those in FIG. 1 are designated by the same reference numerals as in FIG. 1. In terms of circuitry, compared to the one in Figure 1, there is no circuit to obtain the comparison reference current _I1 ,
The configuration is almost the same as that of FIG. 1 except that the final output stage is an amplifier transistor Tr ₃ .

In the example of the light receiving circuit shown in Fig. 4, I _D = I _PT1 ...... I ₃ = I _D ......, and the base current I _b of transistor Tr ₃ is expressed as I _b = I _PT2 − I ₃ ...... Therefore, from the formula, I _b = I _PT2 − I _D = I _PT2 − I _PT1 ………. Here, I _PT2 is I _PT2 = I _PT2 ′ (photocurrent generated by incident light) + I _PT2 ″ (dark current), and since I _PT1 = I _PT2 ″, the formula is I _b = I _PT2 ′+I _PT2 ″−I _PT1 = I _PT2 ′ ...... Then, assuming the current amplification factor of transistor Tr ₃ and β, the collector current Io is Io = βI _b = βI _PT2 ′, and therefore the output voltage Vout is the load resistance
By R _L , Vout=Vcc−βI _PT2 ′×R _L is obtained.

In this way, in this circuit example, dark current can be ignored, and operation at high temperatures can be performed without any problem. Furthermore, there is no effect of dark current even at room temperature, and this circuit example can detect minute amounts of light that could not be detected because they were hidden by dark current in conventional light receiving circuits.

In the above embodiments, a phototransistor was used as the light receiving element, but other elements such as a photodiode may be used in the same manner. In addition, as an application example, we described the case where it is used in a tachometer.
It goes without saying that this method can be applied to all cases when it is desired to detect light.

In this way, in the present invention, a light-receiving element that does not receive light is provided as a dark current canceling element, and a current mirror circuit is configured between the circuit of this non-light-receiving element and the circuit of the light-receiving element that actually receives light. We have created a useful light receiving circuit that eliminates the effects of dark current with a simple configuration, can operate without any problems even at high temperatures, and can detect minute amounts of light that are hidden by dark current even at room temperature. provide.

[Brief explanation of the drawing]

FIG. 1 is a light receiving circuit showing an embodiment of the present invention, FIG. 2 is a principle diagram explaining an application example of the light receiving circuit of FIG. 1, and FIG. 3 is an outline of FIG. FIG. 4 is a time chart showing partial signal waveforms, and FIG. 4 is a light receiving circuit showing another embodiment of the present invention. PT ₁ , PT ₂ ... Phototransistor, Tr ₁ , Tr ₂
...Transistor.

Claims (1)

[Claims] 1. A light-receiving circuit comprising a light-receiving element that does not receive light as a dark current canceling element, and a current mirror circuit configured between the circuit of the non-light-receiving element and the circuit of the light-receiving element that receives light.