This is an example circuit providing a constant current for a laser diode:
What does the two diodes in the circuit do?
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7 Answers
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At a 'moderate' current (set by R1) a single Si diode drops ~ 0.6V, two in series drop 1.2V. Hence the voltage at the top of the potentiometer is ~ 1.2V, more or less independent of the battery voltage.
The rest of the circuit serves to get a constant current through the laser diode: the OpAmp will vary its output to get the same voltage on its two inputs, hence the voltage across the resistor R4 will equal the voltage set by the potentiometer. If the voltage across R4 is fixed so is the current, which is the same as the current through the laser (minus the much smaller base current for Q1).
answered Jan 4, 2013 at 22:35
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They're generating a stable reference voltage. In this case, the designer relied on two diode drops, which would give a reference voltage of 1.2-1.4V. The potentiometer can then be used to adjust the voltage at the non-inverting pin of the op amp, and control the intensity of the laser diode.
The use of a dedicated reference diode would guarantee the required reference voltage, with much tighter manufacturing tolerances.
answered Jan 4, 2013 at 22:31
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The provide a reasonably accurate voltage reference (twice the forward voltage of an 1N4148) that is then divided by the pot and fed into the opamp. There it is compared to the voltage drop over the shunt resistor, which is proportional to the current through the laser diode and shunt, thus a constant current (adjustable with the pot) is maintained.
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The 2 diodes provide a reasonably stable low voltage reference which is picked off by R2 and input to pin 3 of U1. U1 will keep the voltage on its pin 2 the same as pin 3. The voltage at pin 3 when divided by R4 yields a current that will be very close to the current in the laser since the collector current of Q1 will be very close to its emitter current. Q1 provides current gain so that U1 does not have to provide the laser current. Thus the overall circuit acts to provide a constant, stable current to the laser.
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It looks like they are acting like the bottom part of a voltage divider. This post talks about some of this: What are the advantages and disadvantages of using signal diodes instead of resistors?
answered Jan 4, 2013 at 22:31
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The Multisim analysis [before someone vandalised the question to delete that part] is wrong; perhaps the supply voltage is about 3V in which case it would be about right for entirely the wrong reason.
The two diodes will ensure the voltage across them is two diode drops, regardless of the supply voltage. As to why, there is probably a note that accompanies the schematic. The diodes and current sense resistor R4 are marked with an asterisk; the note presumably explains why.
It might explain that the diodes have to be mounted in close thermal contact with the laser diode. Then as the laser heats up, the diode voltage reduces, and so controls the laser current.
answered Jan 4, 2013 at 22:30
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1\$\begingroup\$ The temperature compensation aspect of it is probably key to this application. \$\endgroup\$ Jan 4, 2013 at 22:37
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\$\begingroup\$ Thanks to everyone. Actually if freaking bothered to read the description it told me exactly how it works Stupid! RTFD (read the freaking description) Anyway here's the bit I really don't understand what the heck is a common mode input? the description said "must have common mode input all the way to ground" . will this work with a TLC272 since that all I have but that particular opamp I may be able to source from evil maplin. \$\endgroup\$– AgeisJan 5, 2013 at 1:20
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\$\begingroup\$ I just noticed the current between the pot wiper and the non inverting input is either negative,positive or zero depending on how high I turn the pot why? Hmm also what sort of wattage R4 be? I'm thinking of 3A! I like my toys to be beefy size does matter!! \$\endgroup\$– AgeisJan 5, 2013 at 1:46
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\$\begingroup\$ Common mode input: a lot of opamps stop working when the inputs get too close to either supply rail - if the "common mode input range" includes 0V (when the supplies are 0/+V) the opamp keeps working; you need that here. TLC272 datasheet says "Common-Mode Input Voltage Range Extends Below the Negative Rail" so it's good... just remember its pinout is different! \$\endgroup\$ Jan 5, 2013 at 8:58
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The power supply provides the current (through R1) to forward bias the two diodes. Each diode, when forward biased, has a voltage drop of approximately 0.7V. In series, they will drop about 1.4 V. Potentiometer R2 functions as a voltage divider that allows adjustment of the voltage at pin 3 of U1 between ground and ~1.4 V.
