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I came across the following circuit used to set a microcontroller-controlled fixed current in an LED to around 100mA (target current): Op-amp controlled emitter follower with 2 BJTs

As far as I understand, the voltage reference at the IN+ (1.2V) and the negative feedback force the voltage at IN- and consequently across the R1 to be the same 1.2V. R1 is then chosen to limit the current to 100mA. The right-hand side BJT acts as a switch and when a micro outputs a +5V, the transistor is biased into the linear mode and current flows through the LED. If a micro outputs a 0V, the BJT is in cutoff.

This is all very good and I know it works but I don't really see how this topology would be advantageous over a classic emitter follower with a single BJT that would simultaneously act as a current amplifier and a switch. Isn't this an overkill and unnecessary complication? The final device is going to use many multiples of this circuit and having unnecessary components would be very unreasonable.

Classic emitter follower

Unfortunately, I have no chance to ask the designer whether he chose it for some specific reason, so I wonder whether I'm missing some perks of the circuit in question.

Also, I'm not sure how justified the use of C6 is at the emitter of U6.

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The two circuits do the same things, but with different emphasis.

The opamp circuit allows the LED current to be changed by altering Vref. It also allows the LED current to be completely independent of the voltage drop across the LED. This circuit would tend to be used for testing and characterisation where the flexibility and accuracy is needed.

The switch and resistor circuit does neither of these things, but is simpler, and will usually be adequate once the LED type and desired current has been determined.

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  • \$\begingroup\$ Your answer lead me to an idea that turned out to be the case, thank you very much. \$\endgroup\$
    – vasus
    Oct 29, 2015 at 15:13
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The first circuit uses a number of relatively expensive components.
It is over specified but under designed.
It will not work correctly in all cases with production components if either or both of Vcc and U6B base drive are <= 5V.
See below - Problems with op-amp circuit:

If it did work as intended it would provide a potentially more stable and accurate result than the 2nd circuit.
C6 isnot strictly necessary unless there are special circumstances but does no harm.

The second circuit will work at Vcc=5V. It is less accurate and potentially less stable in setting a defined current than the 1st circuit. The set current depends on Vin, Vbe, Vf_LED and R1 accuracy.
I_LED ~= (Vin- Vbe - V_LED)

The second circuit can be improved by moving the LED (but not the resistor ) to between Vcc & U6B collector. This then places Vin - Vbe across R1 and I_LED != (Vin-Vbe) / R1 and does not depend on V_LED changes with current or device.

Dual transistor used - MMDT5551 datasheet


Problems with op-amp circuit:

IF [ [ D1 is blue or white] AND [Vcc is 5V or less] ] THEN ...
the top circuit is incompetently designed and marginal unless D1 is specially chosen with Vf at 100 mA of under 3V.

Because, if Vcc = 5V :
V_R1 = 1.2V, V_DA=3V say,
V_CE_U6B = 0.75 typical

(see datasheet fig 4) & (emitter follower with Vb<=5V)
So "headroom for U6A
= Vcc - V_U6B -V_D1 - V_R1
== 5V - 0.75 - 3 - 1.2 = 0.05V

But, as above, the transistor Vce at 100 mA is typically 0.75V so the transistor typically can not be turned on hard enough to provide required LED current.

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  • \$\begingroup\$ At 100mA it's likely to be a IR LED. \$\endgroup\$ Oct 29, 2015 at 6:09
  • \$\begingroup\$ The previous comment is correct, the target diodes have a forward voltage drop of 1.4-1.6 V so the op-amp circuit does work in practice, I verified it myself. Thanks for the information though! \$\endgroup\$
    – vasus
    Oct 29, 2015 at 15:12
  • \$\begingroup\$ @Vasus It's always nice when people supply all the information needed to allow a good answer to be given so that much effort is not expended answering the wrong question. Do you have an LED part number available? \$\endgroup\$
    – Russell McMahon
    Oct 29, 2015 at 16:22

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