# Amplifying photodiode current with BJT transistor

I'm trying to develop a kind of light detector. To do that, I'm going to use a photodiode and a bjt transistor to amplify its current. I found a circuit in this tutorial:

which explains how the phototransistors work. It is pretty simple, but I'm not able to understand two things:

1. It is said that Rb is optional. But the photodiode needs bias to work in that way. How on Earth it will be biased? Maybe due to base-emitter junction, but how exactly does it work?
2. What if the Rl resistor would be connected to collector? I tried such circuit with this photodiode and this transistor. The Rb and Rl was both 100K. I expected Rb voltage to be about 1,2V because of the 12uA current produced by the diode, but actually it was about 0.7V and the circuit was extremely unstable - it turns off and on for unexplainable reason.

I guess I'm missing something obvious, but just can't get, what exactly.

• You might have better luck with a MOSFET or JFET which has a high impedance gate. Photodiodes aren't so good with low impedance inputs. – DKNguyen Feb 1 at 19:42
• You will want a circuit with hysteresis. Just FYI. – jonk Feb 2 at 5:40

1. A photodiode doesn’t have to be biased; it will try and force a photo current into whatever circuit it is connected to. In your emitter follower circuit if you removed Rb, there will be a path for photo current to flow through the base emitter junction, through the emitter resistor and back to the negative terminal on the supply. The base emitter junction behaves like a forward biased diode and drops around 0.7 volts in a typical circuit but, this can range from a couple of hundred milli volts to around 1 volt (depending on the circuit).
2. You can put a resistor in the collector and short emitter to ground and you get a very high gain amplifier. Because base and emitter conduct at typically around 0.7 volts (as per (1) explanation), you won’t ever see 1.2 volts. Yes it can be unstable and be influenced by temperature and light changes (sometimes) and, that is why it tends not to get used like that in engineered circuits.
• Well to actually get very high gain, since $\beta$ may vary sensibly depending on collector current, such collector resistor should be chosen carefully to make $I_c$ fall in range where $\beta$ is high. Clearly depends on what BJT will be used. – edmz Feb 1 at 12:24
• @edmz I’m unsure of the relevance of your comment. For the OP, your comment might serve to confuse. If you think there’s relevance, please do make an answer to this question. – Andy aka Feb 1 at 12:30
• The OP says this circuit will be actually developed: I do agree with your answer, but I just wanted to point out that you can't just put the transistor and expect a high gain (though he didn't tell us how much amplifications they're looking for exactly), but rather depends on the technology and the biasing chosen. – edmz Feb 1 at 12:35
• Now it makes sense!!! – Andy aka Feb 1 at 12:46
• @Andy aka Thank you! I missed the fact that the Rb voltage will be equal to base-emitter voltage drop. Please, could you tell me more about the biasing? Why do we need it at all - only to make diode work faster? According to wikipedia, biased and unbiased diodes works in different modes - I thought that it means completely different physical processes. – msmirnov91 Feb 1 at 12:54

It is said that Rb is optional. But the photodiode needs bias to work in that way. How on Earth it will be biased? Maybe due to base-emitter junction, but how exactly does it work?

Without Rb the diode gets its bias through the Base-Emitter junction. This provides maximum sensitivity because the transistor gets all the photodiode current. But sometimes you don't want maximum sensitivity. Rb shunts small photo currents away from the Base so the transistor won't respond to low light levels.

When the transistor is connected in Emitter Follower mode, as in this circuit, it acts as a buffer which reproduces the voltage on Rb - Vbe. So the voltage across RL becomes proportional to light intensity once the voltage across Rb reaches the threshold set by Vbe (~0.6 V).

What if the Rl resistor would be connected to collector? I tried such circuit with this photodiode and this transistor.

Unlike Common Collector configuration which has a voltage gain of 1, in Common Emitter configuration the voltage gain is very high (typically ~100 times). So once the voltage across Rb reaches ~0.6 V the output voltage drops rapidly with increasing light level.

I expected Rb voltage to be about 1,2V because of the 12uA current produced by the diode, but actually it was about 0.7V and the circuit was extremely unstable - it turns off and on for unexplainable reason.

Rb cannot be 1.2 V because the transistor's Base-Emitter junction draws exponentially increasing current as Vbe rises above ~0.6 V (this is the reason for the high voltage gain in Common Emitter configuration). The circuit didn't turn on for an 'unexplainable' reason, it just became very sensitive to light intensity variations above a threshold level.

This is useful when you want a circuit to turn on and off when the light level goes above or below a certain intensity, for example in a nightlight daylight sensor. However since Vbe is temperature sensitive and ambient light levels may vary, it is more commonly done with a linear light sensor followed by a comparator with adjustable threshold to set the switching point.

• Thank you for such a verbose answer! I'm a bit confused: Andy aka in his answer wrote that photodiode doesn't have to be biased at all. Does it mean that it is not biased in this particular circuit, or it just doesn't need bias in common case? And one more question about base-emitter junction: doesn't in need to be biased itself to make photodiode biased? – msmirnov91 Feb 2 at 16:39
• A photodiode can be modelled as a current source (proportional to light intensity) in parallel with a diode. In photovoltaic mode (with no external bias) the photocurrent itself forward-biases the diode and all the current goes though it (unless an external circuit draws some off). Due to the diode I/V curve the output voltage is nonlinear and doesn't go above ~0.7V. In photoconductive mode a negative voltage is applied which reverse-biases the diode, and all the photocurrent is 'sucked out' into the external circuit. This provides a linear response (if the external circuit is linear). – Bruce Abbott Feb 2 at 22:34
• You could wire a photodiode directly across the transistor's Base-Emitter and it would work (in photovoltaic mode), but the photocurrent would split between the transistor b-e junction and the photodiode's internal diode. The result would be non-linear response and lower sensitivity (dependent on the current splitting ratio). It can be useful if you don't have a suitable bias voltage available, for example in 'PhotoMOS' relays, which have several photodiodes in series producing enough voltage to turn on a MOSFET. – Bruce Abbott Feb 2 at 22:50
• Thank you, you have helped to make it more clear for me! – msmirnov91 Feb 4 at 21:20

The transistor chosen is a PNP type (2N2905 equivalent). The example circuit requires a NPN transistor, something like a 2N2222. You can certainly get a PNP transistor to amplify, but you must use a different circuit:

simulate this circuit – Schematic created using CircuitLab

With Rc=100k, sensitivity to light may be excessive, so that even with room light, transistor Q1 is saturated...its collector voltage will be close to BAT1+. In that case, use a lower-value resistor for Rc.
Rb can be removed, if you want to be very sensitive to low light.

I've quite successfully used a phototransistor, with negative-DC-feedback to stabilize the collector voltage. This also removes some/most of the response to incandescent bulb photon flux. The DC feedback also removes SUNLIGHT variation.

• It would be nice if you shared the circuit. – Mattman944 Feb 2 at 16:58