Need help with understanding constant current control circuit

Question

I am an electronics noob struggling to understand a circuit I have reverse-engineered from a circuit board. The circuit will power up a spectral lamp with a current control from 0-25 mA. From what I understand, R6 is the current sense resistor and with the 10k pot I am able to control/limit the current on the load. The opamp circuit is what I yet have to figure out. It looks like something between a differential and non-inverting amplifier but unsure what it actually is (it looks different from typical textbook examples). I am also confused by the additional 1k resistor R3 between the inverting input and R6. So far, I have recreated the circuit and it works as expected.

Any tips on where to start would be highly appreciated.

Disclaimer: This is not a high school or college assignment.

Answer 1

The opamp circuit will drive its output in such a way as to make its '+' and '-' input voltages equal.

The '+' voltage is fixed and controlled by the pot R2.

The opamps output drives the base of Q1 which in turn drives R6. Thus as the output rises, Q1 turns on stronger, and the V at the top of R6 rises. This V is (nearly directly) connected to the '-' input.

When it settles, the V on the '-' will equal the pot V on the '+'. Thus the current in R6 is defined. Since the transistor Q1 has quite high beta, its collector current is quite close (say 98 %) of the emitter current (i.e. R6's current). Thus (as long as the transistor doesn't saturate; i.e. the load R is not too high), the pot V controls the V across R6 which in turn defines the current in R6 and the load.

If Q1 was a MOSFET, then most of the (small) gain error would be gone. There would still be a small error (about 0.7 %) from R3 and R4.

Answer 2

R3 is necessary to provide negative feedback from the sense resistor R6, so for the pot voltages of 0 to 5 volts, the output current will be 0 to 73 mA. R4 and C1 in the feedback path seem to be a low pass filter to stabilize the op-amp and avoid oscillation.

Answer 3

The way to look at this circuit is to realize that what you're really controlling is the emitter current of Q1, and the way that it's being controlled is by controlling the voltage of the hot end of R6.

Take everything inside the red circle and replace it with one theoretically perfect op-amp. When you do this, the voltage across R6 will exactly mirror the voltage at the wiper of R2.

Now you can look at some detail.

The purpose of C1 is to stabilize the circuit. The transistor will add some lag to the circuit; op-amp unity-gain followers are already close to being unstable, so C1 adds some lead -- at the cost of slowing the circuit down.

Note that this circuit has some imperfections: the collector current of Q1 is less than it's emitter current, by as much as 2.5% (look at the \$h_{FE}\$ vs. current plot -- collector current = \$\frac{h_{FE}}{1 + h_{FE}}\$).

Moreover, the designer has decided to limit the gain of the op-amp slightly, by putting R4 in parallel with C1. This sort of "leaky integrator" topology is common; it makes a circuit that recovers better from large transients, particularly if the op-amp in question isn't well behaved when it's operated with its inputs saturated.