# Laser Diode driver circuit help

I'd like to understand the functions of Q2 and Q4 in the driver below. Also, why there's a shunt RC branch at Q5 base? Is Q2 simply a buffer or does something else? Also, in the diagram, it says "0.2mA" at the photodiode branch, with that current Q2 would be in cutoff even when no signal at Q3, so the laser would be off - that makes no sense...

• Q4 is likely for SOA fold-back to limit the laser drive current to (1Vbe/10) or approximately 60 mA. Q1 & Q2 together form an error-amplifier attempting to maintain 1.2 V at node X. Commented Mar 3, 2021 at 2:20
• Could you cite the source that this came from, as a link if it's on the web? Does it not describe the circuit operation? Commented Mar 3, 2021 at 2:43
• @TimWescott it's from The Art of Electronics by Horowitz. But the circuit explanation is really brief, not much details provided. Commented Mar 3, 2021 at 3:36
• @strange_bakery Probably the most important thing to know in explaining the circuit is the expected leakage in the photodiode when the laser is off. This relates to R4. The rest is easy. RC and CC just limit how quickly Q5's base can be pulled up. It's a good idea. Q1 and Q2 are just a diff-amp. I'd probably need to read the LM385 datasheet, as I hope it works well in hundreds of microamps. Also, I'm curious how it fails if the current is still less. But the basic idea seems sound without having all the details at hand.
– jonk
Commented Mar 3, 2021 at 7:16

This looks like a RAP or Winfred Hill design that was made to work well in the lab, but not optimized for production tolerances without a pot. It was designed for binary ASK. (low/hi level current lasing)

In addition to the other fine explanations. I'll add my two bits below.

R3 and R2/R1 control the desired binary current levels with some smaller subtracted current from the PD and use a pot for R2 to bring up current slowly at first try. Q1,Q2 have roughly the same Vbe values of 1.2V, so Q1 may turn off as the pair act as a differential.

Q2 does not turn off as the current from R3 exceeds the PD (photodiode) feedback with two selected base currents for an optical AM transmitter hi/lo, not on/off.

All the parts in yellow with the PD feedback affect the luminous intensity levels with very low effective CTR (current transfer ratio) for the necessity of optical feedback. These come included with the laser diode.

The input RC filter gives some edge boost while the output RC filter is a snubber to prevent the laser from having a lethal orgasm with rapid NTC effects on power levels and tendencies towards thermal runaway. So the slew rate is reduced for stability.

R4 controls the low current level.

There's a lot of parts that affect current levels in the laser. So one would start with a pot in R2 and bring the current up slowly with the 10 Ohm < 60 mA safety net on Q4 limiter.

• Thanks for the explanation! But still, I don't get the 0.2mA - like, if Q3 is on then the voltage drop at R3 is 6800*0.2/1000+Vcbq3 ~= 1.66V which is higher than 1.2V - shouldn't Q2 emitter-base be reverse biased in this case? Commented Mar 4, 2021 at 22:29
• No Vb is around 1.2V from regulated other side so without diode current is 1.2V / 29k= < 400uA and 200uA modulates this. Commented Mar 4, 2021 at 22:48

Laser diodes are notoriously unstable and easy to blow up. (It's not if, but when if you are designing drivers for them. I speak from bitter experience. It is best to develop and learn on cheap ones. Most suppliers of the high end ones will sell a driver with the diode, you buy a sealed unit.)

The photo diode and Q2 are negative feedback to reduce laser current if it gets too bright. Reducing Q2's current will reduce diode drive, so the photodiode has to decrease its voltage with more light.

Q4 is current limiting, it will start to shunt current away from the drive transistor at about 50mA.

One issue with laser diodes can be that absolutely any ringing in the leads going to them will create transients which will kill the diode after a few million pulses. It's often a good idea to limit switching slew rates to reduce this. That may well be what the CR network is about. Their time constant is about 150ns, so this circuit can probably pulse the diode down to a us or so.

EDIT - One comment about this circuit. Because of the horrible vulnerability of laser diodes to any transients (causing damage by overvoltage), it is not generally recommended to drive them with a transistor and a resistor as here. It might be OK if the diode is on the driver PCB (thus extremely short traces) but I ran into problems with leaded diodes where the leads were only about 4 or 5cm with such an approach, with lasers failing after many millions of pulses. We then went to another approach which was to drive the laser with a current source, always on, and then put a "soft switch" transistor in parallel with the laser, shunting current away from it to turn it off. By turning the shunt transistor on and off softly (say 200ns) we avoided diode failures. I can't say whether that is a commonly used approach, but it worked quite well for us.

I think the book does a decent job explaining the circuit:

Just compare the monitor photodiode current with a threshold, and adjust the drive current accordingly. It’s a good idea to include a current limit, so you don’t blow out the diode; and you’ll need to compensate the feedback loop to make it stable. Figure 12.67 shows a “good-enough” driver circuit: it converts the monitor current to a proportional voltage (switchable between two levels by Q3), compared with a diodecompensated 1.2V reference voltage. Q5 drives the laser, with current limiting via Q4. CC and RC compensate the loop, with final “cut-and-try” values as shown.

Source: Horowitz, Hill: The Art of Electronics 3rd edition

So the use of Q4 for current limiting the laser current should be clear. A higher laser current leads to higher voltage drop at $$\R_{CL}\$$ which leads to Q4 pulling the base of Q5 lower.

The photodiode current is converted to a voltage either with $$\R_3\$$ and $$\R_4\$$ in series if the input signal is low, or with $$\R_3\$$ and via Q3 to ground if the input signal is high. The circuit of Q1 and Q2 build the mentioned comparator. You could also think of it as a differential amplifier with one input being about 1.2V if that helps.

Also, in the diagram, it says "0.2mA" at the photodiode branch, with that current Q2 would be in cutoff even when no signal at Q3, so the laser would be off - that makes no sense...

When the input signal at Q3 is low, Q2 is supposed to be in cutoff. But when the input signal at Q3 is high, the voltage at the base of Q2 is lower so that it is not in cutoff region.

The overall idea of the circuit is to monitor a part of the laser power using the photodiode and use the measured power feedback for biasing to keep the average laser power relatively constant. Otherwise the optical output power of the laser diode would vary a lot with temperature and aging.

$$\C_C\$$ and $$\R_C\$$ make sure that this feedback doesn't cause oscillations.