I'm not sure if my design is done correctly. I don't understand transistors that well, and the LED driver adds even more confusion. Basically I'm using an LED driver (TLC5917) that sinks current to drive the base of a PNP BJT. My source voltage is 15.5 V, my resistance is up to 30 ohms, and my target current is 500 mA.

It seems to me that changing the base current (by changing the set resistor on my LED driver) should change the load current but it doesn't.

How do I choose the current I'm after? Accuracy isn't terribly important, I just need it roughly there. I can't find any examples online of using an LED driver to control a BJT.

Edit: I'm using a different BJT than the one in the image. That was just my test one.

LED driver BJT circuit

  • 3
    \$\begingroup\$ The TLC5917 is designed to drive LEDs, not transistors. \$\endgroup\$
    – Finbarr
    Apr 18, 2023 at 17:11
  • \$\begingroup\$ As a beginner, I would recommend using chips the way they are intended to be used. That's complicated enough as it is. \$\endgroup\$
    – Drew
    Apr 18, 2023 at 17:13
  • \$\begingroup\$ The problem here is that the driver can't "see" the current going through the LED, so it has no way of knowing it's correct or keeping it constant. \$\endgroup\$
    – Finbarr
    Apr 18, 2023 at 17:14
  • \$\begingroup\$ Your transistor is rated for 100mA collector current. The graphs don't go any further than that. \$\endgroup\$
    – hobbs
    Apr 18, 2023 at 17:21

2 Answers 2


R_EXT controls the current into each of the outputs of the TLC5917 (intended for direct LED drive). That current is only rather roughly related to the LED current in your circuit (current gain of the S9015 at 5V and 1mA is somewhere between 60 and 1000 times the base current).

Worse, the little S9015 is only capable of absolute maximum 100mA collector current.

In short, this circuit is unsuited for what you are trying to do. If you describe what you are trying to do exactly you may be able to get better advice.

If you used a much more suitable p-channel MOSFET or PNP and a simple series resistor for the LED it might be workable. You'd also probably want a resistor from base to emitter or gate to source.

  • \$\begingroup\$ Thanks. I've got a suitable (800 mA) PNP transistor coming in the mail today, the 9015 was just my test one. I'm trying to set off talon igniters for fireworks. I'm working closely with a professional to make sure that end of things is safe. I've found resistances between 2.5 and 30 ohms but the target current is the same for all of them, so a simple series resistor is not suitable. \$\endgroup\$
    – Steffen
    Apr 18, 2023 at 17:23
  • \$\begingroup\$ You'll need a different circuit to have controlled current. \$\endgroup\$ Apr 18, 2023 at 17:28
  • \$\begingroup\$ Can you share an example of such a circuit? \$\endgroup\$
    – Steffen
    Apr 18, 2023 at 17:30
  • \$\begingroup\$ I could but I think this would better be posed as a new question with full details of your squibs etc. Others may have better insight into any special requirements. \$\endgroup\$ Apr 18, 2023 at 18:15

It is possible to make a current source out of your circuit. The combination of a standard diode (Vf = 0.67 V) and a schottky diode (Vf = 0.21V) provides a relative constant voltage reference (0.88 V). This voltage together with R2 defines the maximum current through Q1. A higher current would rise the voltage across R2, which would reduce Vbe of Q1, which would reduce the current through Q1.

Be careful: All these mentioned voltages have a temperature drift, so this current source is not very exact. You should test it in different outdoor conditions.

Since Circuit Lab seems to favor ancient, obsolete and hard to get components, I had to select this absurd bulky transistor here. There are much better devices, which would need a slightly different value of R2.

The supply voltage of 15.5 V is just a bit too low to allow 0.5 A with 30 ohm load because the minimum voltage drop across Q1 and R2 is around 0.9 V.

The power dissipation in Q1 can be up to 7 W. Either the time the current is enabled is very short or a heat sink is required or you really use 2N2955.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Awesome, thank you so much! I can't figure out what the TLC5917 is doing in that circuit, but making it control the switch makes it work just fine. I tweaked some things including increasing the voltage and I'm getting less than 10% variation between current at 2.5 ohms and at 30, which is perfect. \$\endgroup\$
    – Steffen
    Apr 19, 2023 at 4:14
  • \$\begingroup\$ @Steffen Well, the switch SW1 was added to test the circuit in the simulator, this job is done by the TLC5917 in a real circuit. You may (but don't have to) use a simpler chip as driver, because the special LED driver features of the TLC5917 are not used here. I like MIC5841, STPIC6C595 or NPIC6C595 in such cases. There is a whole group of xxxPIC6x595 chips. \$\endgroup\$
    – Jens
    Apr 19, 2023 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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