Does this linear current source driver deliver constant current?

Question

I was reading on LED driver designs, and I have a question about the linear current source driver. In the circuit below, will D1 ever turn off when there is too much current through R2? In other words, will D1 turn off when Q1 turn on?

Because the base and the drain are driven by the same voltage Vcc, I am assuming Q2 is an N-Channel MOSFET (and not a transistor). And I can see when Q1 is activated, Q2 turning off because the voltage at the node between R1, Q1 and Q2 would be ~0V (GND).

However, there is no mention of this side effect. Ideally, the circuit regulates current in itself without switching off the LED.

Is my assumption false and does this circuit works like a constant current source?

Answer 1

Both Q1 and Q2 are BJT transistors. That is indicated entirely by the shape of the symbol. Furthermore, they're NPN type BJT transistors.

BJT transistors look like this:

MOSFETs look like this:

This circuit limits to a specific current only if the value of R2 is correctly chosen to correspond to that current. You want the voltage across R2 to equal about 0.7V when the current reaches your specified maximum. When the voltage across R2 reaches 0.7V, Q1 will begin conducting across its collector and emitter. That will drive the voltage at the base of Q2 lower, which will effectively reduce the current through Q2. The more Q1 conducts, the less Q2 will conduct. The combination of Q1 and Q2 acts as a negative feedback system that maintains the current through Q2 at a maximum value or less.

So, for example, let's say you wanted to limit the current through the LED to 50mA. Using Ohm's Law, the value of R2 should be: $$R_{2}=\frac{V_{R2}}{I_{R2}} = \frac{0.7V}{0.05A} = 14\Omega$$

The magic value of 0.7V is used because that's approximately the voltage that the base-emitter junction begins to conduct on most BJT transistors. When the base-emitter begins conducting, it allows proportionately more current through the collector-emitter.

Answer 2

Q1 sucks enough of the base current (coming through R1) from Q2 to keep the emitter at around 0.6V higher than ground, because when the transistor is in the active region, that's Vbe. More voltage means the transistor diverts more current from the base of Q2, less voltage and it diverts less.

There are limits- if the value of R1 is chosen too high, then Q2 won't have enough base current to drive the LED fully on. If the value of R2 is chosen too low, then Q1 won't be able to divert the base current as well, and it won't regulate very well.

If you assume Vbe of Q1 is 0.6V-0.7V (not a great approximation) and ignore the base current of Q2 in relation to the LED current (a good approximation), the LED current will be a constant current of ~0.65V/R2

It's easy to simulate this with the simulator here:

^{simulate this circuit – Schematic created using CircuitLab}

As you can see, once the supply voltage exceeds 2.5 or 3V, the current is pretty constant. Similar plots could be made for LED voltage drop and for temperature and you'd see that the current is "fairly" constant, not by instrumentation standards, but visually (LED brightness) it's pretty decent, especially for such a simple circuit, and the current is close to the predicted 0.65V/50R = 13mA.