JPS6022857B2 - Photoelectric conversion circuit - Google Patents

Description

[Detailed Description of the Invention] [Industrial Application Field] The present invention relates to a photoelectric conversion circuit, and particularly to a photoelectric conversion circuit suitable for direct current transmission type and medium-to-long distance digital optical transmission of several MD/s. It is.

[Conventional technology]

FIG. 1 is a circuit diagram of a main part of this type of photoelectric conversion circuit, which has been conventionally known.

This circuit has a feedback resistor Rf added to an amplifier consisting of an emitter-grounded transistor Q and a collector-grounded transistor Q2, and the incident light is passed through a photodiode PD.
It is converted into electricity, amplified and output to the output side of transistor Q2.

[Problems to be Solved by the Invention] However, this follower circuit has two drawbacks.

The first drawback is that the amplification characteristics deteriorate in the high frequency region due to the collector-base capacitance Cc of the transistor Q. The second drawback is that when there is no optical signal input,
Output bias voltage V.

ut (emitter voltage of transistor Q2) is given by formula [1}, but as is well known, VBE,, VBE2
,B. etc. have temperature characteristics, so V. ut also has temperature characteristics, and if DC transmission is required, temperature fluctuations in the output bias will be magnified in subsequent stage DC amplifiers (not shown), and the operation of the final comparator will become extremely unstable. It is. v skin = v skin + zumo (v hiro - v job)….

) In other words, since the photodiode PD can be ignored in Fig. 1, it can be considered as shown in Fig. 6, and since the base potential of transistor Q is VB8,
The base current lb, of , is ,b,=V.

Since ut-VB8,Rf, the collector current lc of the transistor Q is lc,=8,XV.

ut-vBE,Rf On the other hand, the emitter current le2 of transistor Q2 is ~Rf
＞＞If Re2, then V is connected, so the base current 1 is the current IL flowing through the load resistance ∊c of the R mother transistor Q, ，
(The collector potential of transistor Q2 is VBE2
), IL. -VCQ2guV side+VB82) RC, Also, lc,=1L, 10lb2, so evening, V skin -V
BE, -Vcc-(V.

ut10VBE20R? 1 Rc, V.

However, when we solve this for Buddha ut, Rf/8, Rc, and Rf/8, (10P2
) The R solution term is usually negligible compared to 1.

Therefore, the formula [1} is derived.

Here, VBE, , VBE2: transistors Q, , Q2, respectively
The base of the emitter problem is pressure drop.

81: Transistor Q. Emitter ground current amplification factor.

VQ; Power supply voltage.

Rc・, Rr, Re2; resistances Rc, , Rf, R, respectively
Resistance value of e2.

In this case, if the reproducibility and linearity of the temperature characteristics of Vout are good, it is possible to provide a temperature compensation circuit after FIG.
Since V88·, VB82, and 8 are related to out, and there are large variations in 8, there is a problem that the temperature characteristics to be compensated for are different for each circuit, making it unsuitable for mass production.

[Object of the Invention] An object of the present invention is to provide a photoelectric conversion circuit which eliminates the above-mentioned drawbacks of the prior art, has high temperature stability, and is suitable for mass production.

[Means for solving problems]

In amplifiers using transistors, bias fluctuations based on the temperature characteristics of the current amplification factor and base-emitter voltage effect are unavoidable. In order to ensure reproducibility and to avoid variations between elements, the output bias voltage when there is no signal using the photodiode output current as input reduces the base-emitter voltage effect to VB.
8, a feedback amplifier made of transistors whose voltage is the sum of 2VBE offset voltages and circuit means for generating a voltage of 2VBE are provided, and the output of the feedback amplifier is inputted to one input side of the comparator. , the other input side is characterized in that a voltage of 2VBE from the circuit means is inputted as a reference voltage.

〔Example〕

Hereinafter, the present invention will be explained in detail using the embodiment shown in FIG. 4, and FIGS. 2, 3, 5, 7, and 8.

First, the main points of the present invention will be explained using FIG. 2.

The differences between FIG. 2 and FIG. 1 are, firstly, that an emitter follower Q is added, and secondly, that the collector impedance of the transistor Q4 is used as the emitter load of the emitter follower Q3. The role of the emitter follower Q3 is to separate the feedback resistor Rf from the collector/capacitance Cc of the transistor Q to improve high frequency characteristics.
Therefore, the emitter follower Q3 is designed to have a kind of buffer effect. In this case, in principle, a resistor Rb indicated by a broken line may be used instead of the transistor Q4, but if this is done, the output bias voltage V of the amplifier shown in FIG.

The mountain is given by the formula ■. V. Mountain VBE3 10 {・10 (.
10 heat expansion} V group... [21 That is, as in the case of formula '1}, since the photodiode PD can be ignored in Fig. 2, it can be considered as shown in Fig. 7, and therefore the transistors Q,, Q3, The base potentials of Q2 are VB8, , VBE2 and VB83, V, respectively.

It becomes ut+VBE2. The emitter current le3 of the emitter follower Q at the base of the transistor Q is: ・Anchor V881/Rbl+lbl...(1) However, 1 puma is the base current 1 of the emitter follower Q3.
Because it is 10B times 3.

I={VOut-(VBEI+VBE3)}/Rfl
Spot = VOut-NBE, 10VBE3) (1183) ...
(□) Rr Also, at the collector of transistor Q, (ignoring the base current of transistor Q2) 8, lb, 2 {V - (VOut + VBE2)} / RCI ... (m
)(0), (m) into (1) and get VOut・(
VBE, 10VBE3) (1181) Rf-Wani Jushin (VO
ut+VBE2) 8, Rc, this is V.

Solving for mountains is A10B+C V.

Mountain = “Blade Re” A = (1133) (VBE.10VBE3)
/RfB=VBE, /Rb,C Ni (VKoichi VBE2).
/BIRCI D=(1183)/Rf E=1/P, Rc, However, since C is usually very small compared to A and B and E is very small compared to D and can be ignored, =VBE30 {
・Jusawashi; }VBE. and [Equation 21 can be derived.

Here, VBs, , VB83; each transistor Q
,,Base of Emitsuta Follower Q3. Emitter voltage drop.

B3; Current amplification factor of emitter follower Q.

Rb, is the resistance value of the resistor Rb. ■In the formula,
'1} Similarly, VBE, , VB83, 83, etc. are V.

It contributes to the temperature characteristics of ut. Moreover, the contribution of 83 in the curly brackets is quite large, and the element characteristics of 83
Due to the large variation in V. The temperature characteristics of ut cannot be easily compensated for. However, in formula ■, R
b, →, then VOut=VBE, 10VBE3 22VBE. Therefore, the output bias voltage is V.

ut is now determined only by the voltage drop between the base and emitter, and the temperature coefficient of the voltage drop between the base and emitter is -2 hV/°C, which has small variations depending on the element, making it easy to compensate for the temperature characteristics. Therefore, in FIG. 2, the resistor Rb is replaced with the collector impedance of the transistor Q. In the circuit shown in Figure 2, the output bias voltage V is VOut=VBE30V
Skin Ten Magazines 3/Agriculture. (VBB-VB84)...(3}i.e.
[Similar to equations 11 and (2), since the photodiode PD can be ignored in FIG. 2, it can be considered as shown in FIG. 8. Therefore, the emitter current le3 of the transistor Q, lbl
...(W) On the other hand, in the emitter follower Q3, the emitter current le3 is the base current lb3 (lb3={V.

ut-(VBE, 10VBE3)}/Rf)
) times, so ・Fin = (10B3)V. Mountain-WBE, 10V
BE di... (V) In the Rf transistor Q, the collector current 1 is the base current lb4 (lb4 = (V88 - V
Since it is 84 times BE4)/Rb4), ・B41M=84 {(VBB-VB84)/RM}...(M) At the collector of transistor Q, (transistor Q2
(ignoring the base current lb2), the collector current lc,
is equal to 8,lb, which is the base current lb, multiplied by 8,
and load current {Vcc (V.

Since it is equal to ut ten VB82)}/Rc, 8,Ibl
i {V is - (VOut + VBE2)} /RCI/
, lbl 2 {V-(VOut 1 VBE2)}/
81RC1... (skin) (V), (W), (skin) formula (
W) Substitute into formula (1183)VOut-(VB81
10VBE2Rf284 VBBnaVBE40VCC(V
Out 10VB82) Rb4 8, Rc, this is V.

Solving for ut, V skin ni (F+G+H)/(J+K
)
4(VBB-V884)/R Thou H=(VQ-VBE2)
/B. RC, J = (1163)/Rf K = 1/8, RC, However, for H', F and D, K is usually very small and can be ignored with respect to J, so VOut
=(F+G)/J = {(10B3XV88, 10VBB3)Rf 184VBB・VBE4}/{118} Rb4 :VBE30V leg gap・Agriculture・(VBB−VBE4) and {3−Eq. can be derived.

Here, 34; current amplification factor of transistor Q.

RM: resistance value of resistor Rb4. VBB; power supply voltage VBE4: Base of transistor Q4.

Emitter voltage drop.

And VBB=VBE4...
If {41}, the voltage drop between the base and emitter of any transistor is generally approximately equal, so the formula '3} is V.

Stop=VB83 10VBBI=2VB82
”-(51 becomes “V. The mountain becomes a shape that is easy to compensate.
Also, V.

The fact that the peak does not depend on the power supply voltage V ratio means that power supply voltage fluctuations are strong, which is advantageous. Next, what kind of circuit should actually be used for the circuit shown in FIG. 2 will be explained using the embodiments shown in FIGS. 3 and 4.

When light is incident on the circuit shown in FIG. 2, a current lp flows through the photodiode PD. By similar calculation, the glue formula becomes Rf V heart V skin WBE 30 sheep - string - (VBB - VB84) - lpRr, and compared to when there is no optical signal, lpRf (=V
p) only.

If the power supply voltage VBB is connected to the base of Q as shown in Fig. 3, it becomes VBB-VBE4ni0, so "V.

Mountain = VB81 10VBE3 -Vp 2V88, 1Vp (Here, VBE., VB82, VB83, ...VBE
Since n, etc. are generally approximately equal, these are designated as V88.

) Therefore, if no light is input to the photodiode PD, Vp=0, and V side t=2V88.

Since VBB in FIG. 2 must satisfy equation (2), the emitter-grounded transistor Q. Figure 3 uses the base of .

In the case of the circuit shown in Figure 3, the output point (emitter fluoro Q3) when no optical signal is input from the photodiode PD
The DC bias of the emitter) is 2VBE from formula '5'.
It becomes.

When an optical signal is input, the potential at this point is Vp (=lpRf
) to 2VBE-Vp. That is, the emitter potential V of the emitter follower Q depends on whether the optical signal is "OJ" (off) or "1" (on). The mountain is 2VBE or 2
VB8-Vp. The change in V skin depending on the presence or absence of light input is Vp, and this Vp is V. Since it is very small compared to ut, it must be amplified. Therefore, in the circuit shown in and after the circuit shown in FIG. 3, it is necessary to generate a sufficiently large voltage change depending on the presence or absence of optical input.

To do this, input the output of the circuit shown in Fig. 3 to a comparator, and use the reference voltage as E0(2VBE-Vp).
} = Cho - Mamoru ---'61, but it is a little troublesome to create a reference voltage of 2VBE - Vp/2.

Therefore, the output of the circuit of FIG. 3 and the comparator reference voltage when the optical signal is "0" or "1" are all made to be the sum of Vp/2. In other words, the output of the circuit in Figure 3 is 2vBE depending on whether the optical signal is "0" or "1".
+ Morimata'ma, 2VBE - Mori, and the comparator reference voltage was 2VBE.

Even if the DC bias is shifted by Vp/2 in this way,
This has no effect on the comparator operation (as this is well known, detailed explanation will be omitted). Now, as a means for giving an offset (deviation) of Vp/2 to the output of the circuit shown in FIG. 3 (the emitter of the emitter follower Q), as shown in FIG. may be grounded to the offset power supply VoF (=Vp/2). Further, in order to create a reference voltage of 2VB8, as shown in FIG. 4, by using a resistor Rc6 and diode-connected transistors Q and Q7, a voltage of 2VB8 can be obtained between the base of the transistor Q and the ground.

Note that CM in FIG. 4 is a comparator. As mentioned above,
The output voltage of the circuit in Figure 3 with a voltage offset called VoP added to one input of the comparator CM (2VBE+Vp/2 when there is no optical signal, 2VBE- when the optical signal is input)
Vp/2), and also set the comparator as a reference voltage of M by twice the base-emitter voltage drop, that is, 2V.
If B8 is supplied, even if the bias of the input signal to comparator CM changes due to a temperature change, the reference voltage and 2VBE will change at exactly the same ratio in conjunction with this, so comparator CM
It does not affect the operation of the device and provides stable operation. Next, the operation of comparator CM will be explained using the time chart shown in FIG.

In FIG. 5, 10 of a is the transistor Q of FIG.
This is the output voltage of the feedback amplifier consisting of Q4, and in the "0" state where no output current is input from the photodiode PD, Vout=VoF12 VBE=Vp/20WB8
However, in the state of "1" when the output current is input, it decreases by the amplitude Vp, and it becomes 2VB8-V.
It becomes p/2.

By the way, since VR=2VB8 (VR is the reference voltage), for example, even if the output bias voltage V of the feedback amplifier decreases as shown by 13 due to temperature change,
Following this, the reference voltage VR also decreases, and the V side - VR = 1
/2Vp, the comparator CM performs the correct comparison operation and outputs 12 as shown in FIG. 5b.
is obtained. As described above, the embodiments of the present invention have extremely high temperature stability, are less affected by temperature changes, and have a simple circuit, so they can be miniaturized and integrated circuits.
It is also suitable for mass production. It is also resistant to fluctuations in power supply voltage. In addition, in the above explanation, the offset voltage V
Although 1/2Vp is used as oP, it is not limited to this value, and can be any value from 0 to Vp.

However, it is most preferable to set it to 1/2Vp.

[Effects of the Invention] As described above, the present invention has the advantage of having high temperature stability, being able to be miniaturized, and being suitable for mass production.

[Brief explanation of drawings]

Figure 1 is a circuit diagram of the main parts of a conventional photoelectric conversion circuit, Figures 2 and 3 are circuit diagrams for explaining the main points of the present invention, and Figure 4 is an example of the location of the photoelectric conversion circuit of the present invention. Circuit showing, fifth
The figure is a time chart for explaining the effect of the circuit in Figure 4, and Figure 6 is a time chart for explaining the effect of the circuit in Figure 4.
Each figure is a circuit diagram when there is no optical input to the photodiode in FIG. 2. PD...Phototransistor, Q, ~Q7...
...・Transistor, Rc,, Rf, Re2, Rb4, R
c6...Resistance, CM...Comparator. Sai'zuzai 2 galley illustrations 4 illustrations S drawings from S drawings 7 figures complete 8 figures

Claims (1)

[Claims] 1. An amplifier consisting of a first emitter follower, an emitter grounding circuit, and a second emitter follower is provided with offset adjustment means for the output voltage of the amplifier, and a feedback resistor is provided between the second emitter follower and the first emitter follower. a feedback amplifier, which is provided with a collector impedance of a transistor connected as a load of the first emitter follower, and whose output is the emitter of the second emitter follower, and whose output is the output of the feedback amplifier as one input. , a comparator whose other input is the output of a circuit means that generates a voltage twice the base-emitter voltage drop as a reference voltage; and a photodiode connected to the base of the first emitter follower of the amplifier. The photoelectric conversion circuit is characterized in that when there is no optical signal input to the photodiode, the output bias voltage of the amplifier is the sum of twice the base-emitter voltage drop, the offset, and the voltage. photoelectric conversion circuit.