JPH03289203A - optical receiver - Google Patents

Abstract

(57) [Summary] This bulletin contains application data before electronic filing, so abstract data is not recorded.

Description

DETAILED DESCRIPTION OF THE INVENTION Field of the Invention The present invention relates to an optical receiver for converting digital optical signals into electrical signals, and more particularly to an optical receiver in which circuits are integrated on a semiconductor substrate.

2. Description of the Related Art In recent years, optical receivers have come into widespread use in optical fiber communications, optical disc signal reproduction, and the like.

Conventionally, the optical receiver shown in FIG. 2 has been used. In the figure, 1 is a photodiode, 22 is an amplifier circuit, 23 is a comparator for waveform shaping the output of the amplifier circuit, 24 is a reference voltage circuit that generates the same voltage as the DC voltage of the amplifier circuit when there is no signal, and 25 is a This is a threshold controller that generates a threshold voltage to be applied to a comparator.

Generally, the reference voltage circuit 24 has the same configuration as the amplifier circuit and is integrated on the same semiconductor substrate as the amplifier circuit 22 in order to have the same temperature fluctuations and power supply voltage fluctuations as the amplifier circuit 22. Furthermore, the reference voltage circuit 24
In order to follow the variation in the output level of the amplifier circuit caused by the dark current, a photodiode 26 for generating dark current whose light receiving surface is covered with an opaque material is connected to an amplifier circuit 27 having the same configuration as the amplifier circuit 22. The configuration is as shown in Figure 3. As such a conventional example, for example, Japanese Patent Application Laid-Open No. 57-1
There is one published in Publication No. 41160.

Problems to be Solved by the Invention However, the conventional configuration described above requires a photodiode 26 of approximately the same size as the photodiode 1, which poses a problem of increased cost. Similarly, when the photodiode and amplifier circuit are integrated,
There is a problem in that the chip area becomes larger due to the photodiode 26, and the cost increases accordingly. This problem becomes more pronounced as the area of the photodiode increases.

The present invention solves the above-mentioned conventional problems, and an object of the present invention is to provide an optical receiver in which a dark current generating photodiode is smaller than a light receiving photodiode.

Means for Solving the Problems To achieve this object, the optical receiver of the invention comprises a reference voltage circuit having a larger transimpedance than the amplifier circuit.

Function: With this configuration, a large reference voltage fluctuation can be generated with a small dark current, so the area of the dark current generating photodiode can be reduced and the cost can be reduced.

EXAMPLE Hereinafter, an example of the present invention will be described with reference to the drawings.

FIG. 1 shows a circuit diagram of an optical receiver in an embodiment of the present invention. In the figure, 1 is a photodiode as a light receiving element for receiving optical signals, 2 is an amplifier circuit, 3 is a comparator for waveform shaping the output of the amplifier circuit, 4 is a reference voltage circuit, and 5 is a voltage applied to the comparator. This is a threshold controller that generates a threshold voltage. A dark current generating photodiode 6 is connected to the reference voltage circuit 4. The light-receiving surface of the photodiode is covered with an opaque material so that it does not receive optical signals.

The amplifier circuit 2 consists of two transistors 7.8 and three resistors R.
It consists of 8I, 'Rel, and Rfl. The base-emitter voltage of transistor 7 is VBEI
When the base current is IBI, the DC output level V1 when there is no signal is given by v, "VBEI + IBI" Rfl, ignoring the dark current of the photodiode 1.

The reference voltage circuit 4 also includes two transistors 9 and 10 and three resistors Rc2, Re2, and Rf2.

However, Re2= Rel, Re2= 2 X Rf
1, Re2-2 x Rcl, and the emitter area of transistor 9 is half that of transistor 7. When the moose-emitter voltage of the transistor 9 is VBE2 and the moose current is ra2, the reference voltage v2 is given by V2 = VBE2 + rB2@Rr2, ignoring the dark current of the photodiode 6. Here, since Re2 is twice Rc I, the collector current of transistor 9 is half the collector current of transistor 7, but the emitter area is also half, so VBE2 '' VBE. Also, IB2 = Ia + / 2. Yes, but Rr==2
- Since it is Rfl, IB2@Rf1: ■B10Rf2. Therefore, v2 = V, and the amplifier output and the reference voltage output when there is no signal are equal.

In a circuit configured like amplifier circuit 2, the input current amplitude is il, and the output voltage amplitude is V. Then, ■o=iI −Rfl holds true. That is, the transformer impedance R1
"Vo/l i is equal to the feedback resistance Rf+. Therefore, the transformer impedance R2 of the reference voltage circuit 4 is equal to Rf2 and twice R1.

In the optical receiver configured as described above, if the dark current of photodiode 1 is Idl and the dark current of photodiode 6 is Id2, the dark current causes V! v2 increases across ΔvI=Id1・Rfl ΔV2”Id2・Rf, respectively. In order to have ΔV,=ΔV2, Re2
= 2 x R41, so Id2-L+/2. In other words, Id2 can be half of Tdl, so
The area of the photodiode 6 may be half that of the photodiode 1.

In addition, in this example, in order to make the explanation easier to understand,
Although the transformer impedance of the reference voltage circuit is twice the transformer impedance of the amplifier circuit, the present invention is not limited to this. The ratio of the transformer impedance may be any real number n (>1), and in this case, the area of the photodiode 6 may be 1/n of the area of the photodiode 1.

As described above, according to this embodiment, the area of the dark current generating photodiode can be reduced by making the transformer impedance of the reference voltage circuit larger than the transformer impedance of the amplifier circuit. for example,
In order to receive an optical signal from a plastic fiber with a diameter of 1 m, it is assumed that the diameter of the photodiode for light reception is 1 m, Rf+=100 is set to Ω, and Rrx=200 is set to Ω. Compared to the conventional method, the area of Rf2 is 0.01+++
Although it only increases by about n2, the area of the dark current generating photodiode decreases by 0.4 m2.

Effects of the Invention According to the present invention, by making the transformer impedance of the reference voltage circuit larger than the transformer impedance of the amplifier circuit, the dark current generating photodiode can be made smaller, and therefore an inexpensive optical receiver can be realized.

[Brief explanation of the drawing]

FIG. 1 is a circuit diagram of an optical receiver in an embodiment of the present invention;
FIG. 2 is a block diagram of a conventional optical receiver, and FIG. 3 is a block diagram of a reference voltage circuit of the conventional optical receiver. 1.6...Photodiode, 2...
Amplifier circuit, 3... Comparator, 4...
-Reference voltage circuit, 5... Threshold controller, 7, 8.9.10... Transistor.

Claims (1)

[Claims] An optical receiver comprising an amplifier circuit for amplifying an output signal of a light receiving element, a threshold controller, and a reference voltage circuit having the same configuration as the amplifier circuit for generating a reference voltage to be applied to the threshold controller. , an optical receiver in which the transformer impedance of the reference voltage circuit is larger than the transformer impedance of the amplifier circuit.