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?

Linear Current Source / Current Limiter

  • 1
    \$\begingroup\$ Don't think of these transistors as switches. They're not being operated in their extreme on/off regions but in between. \$\endgroup\$ – brhans Oct 28 '14 at 15:35
  • \$\begingroup\$ @brhans in other words, Q2 does not deactivate? \$\endgroup\$ – Adam Lee Oct 28 '14 at 15:38
  • 1
    \$\begingroup\$ No, Q2 wouldn't ever deactivate. Think about what would happen if it did - no current out of its emitter, therefore no current through or voltage across R2 or Q1's base-emitter, therefore Q1 is off, therefore Q2 is turned on by the current from R1. Both transistors are in a partially-on state. This is an example of negative feedback. \$\endgroup\$ – brhans Oct 28 '14 at 15:43
  • 2
    \$\begingroup\$ A MOSFET is a transistor; more specifically, it's a metal–oxide–semiconductor field-effect transistor. \$\endgroup\$ – ntoskrnl Oct 28 '14 at 18:25

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:
enter image description here

MOSFETs look like this:
enter image description here
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.

| improve this answer | |
  • \$\begingroup\$ Great, I was treating these transistors with only on/off states. Was not thinking that the more voltage it is given, the more it conduct. Thanks. \$\endgroup\$ – Adam Lee Oct 28 '14 at 16:21
  • \$\begingroup\$ (little cleanup of the comments...) \$\endgroup\$ – clabacchio Oct 28 '14 at 17:38
  • \$\begingroup\$ <nitpicking mode> Re " ...This circuit works only if the value of R2 is chosen correctly.." -> This circuit always works correctly as ling as the ratings of the parts is not exceeded.However, like any CC cct, it only produces the current you want if you use the value of R2 that the design calls for. I know you know that :-). But, as put, it may confuse some who do not understand it well. \$\endgroup\$ – Russell McMahon Oct 29 '14 at 13:14
  • \$\begingroup\$ @Russell McMahon, sure, I was using the phrase "This circuit works" to mean "This circuit works as intended". But I see the potential ambiguity of the phrase. I'll edit to clarify. \$\endgroup\$ – Dan Laks Oct 29 '14 at 16:24

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

enter image description here

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.

| improve this answer | |
  • 1
    \$\begingroup\$ This is also a great answer. Too bad I can't pick multiple answers. What's the simulation tool you are using? \$\endgroup\$ – Adam Lee Oct 28 '14 at 16:20
  • 1
    \$\begingroup\$ I used the built-in Circuitlab simulator. You can play with it and change values. \$\endgroup\$ – Spehro Pefhany Oct 28 '14 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.