I am new to electronics so I'm not too sure about how I should be calculating the base resistor to use in this circuit.

I think the load resistor is calculated correctly (data sheet at bottom for reference), and I've used an online calculator on this site here to get the base resistor value. However, I'm not confident I used the correct values to start with.

Looking for any corrections or comments on this design.

Constant current driver has been suggested to me in another thread, but I think this might be too complicated for me to calculator resistor requirements.

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  • 2
    \$\begingroup\$ That circuit won't work. You've wired the transistor as a voltage follower. The emitter voltage will be 0.7 V below the base so the best you'll get out is about 3.3 - 0.7. = 2.6 V. \$\endgroup\$
    – Transistor
    Oct 28, 2021 at 22:47
  • \$\begingroup\$ You calculated 20mA per LED, but you have 6 strings with 20mA each. Plus the transistor is not used as a switch but emitter follower so the topology is not what you expect and even without base resistor at all the output would not be 5V but only about 2.6 to 2.7V. \$\endgroup\$
    – Justme
    Oct 28, 2021 at 22:50

1 Answer 1


Several things.

  • Put the transistor on the low side, emitter to ground, collector to the LEDs. This will allow it to be driven into saturation.
  • The transistor beta will be more like 100, not 35, given the current you’re working with.
  • Consider placing 3 LEDs in series, in 4 strings instead of the 2 x 6 you have now. Recompute your load resistors using 3.6V forward drop instead of 2.4V. (I calculate about 70 ohms for 20mA, 1.4V drop across the resistor.)

With these changes you’re looking at 80mA drive, instead of 120mA with the 2 x 6 arrangement. Base resistor could be sized to provide about 1mA or so to the base. That’s about 2.4k, or less, to a 3.3V GPIO.

  • \$\begingroup\$ The emitter-follower is wrong, a saturated switch circuit is needed with the base current 1/10th the collector current. The current-limiting resistors must be calculated for the forward voltage of the IR LEDs which might be 1.0V producing a very high current or a forward voltage of 1.5V for less current. \$\endgroup\$
    – Audioguru
    Oct 29, 2021 at 0:22
  • \$\begingroup\$ At Ic=150mA, saturation happens at about 3.0mA for the 2N2222. So between 1.5 and 2mA of base drive ought to be enough. Base resistor of ~1.2k then. 8mA is a bit much for the R.pi to drive, I would keep it below 4. Any more than that I'd be looking to use an n-FET instead. \$\endgroup\$ Oct 29, 2021 at 1:19
  • \$\begingroup\$ The datasheet for an old ON Semi 2N2222 says that its minimum hFE at 150mA is 100 when its Vce is at least 10v or is 50 when its Vce is 1V, it is not saturated. Its maximum Vce saturation voltage is 0.3V when its base current is 15mA. You cannot buy a more sensitive one than most, you get whatever is available. \$\endgroup\$
    – Audioguru
    Oct 29, 2021 at 22:31
  • \$\begingroup\$ There is an answer to that dilemma; use a Darlington configuration, or use a MOSFET. \$\endgroup\$ Oct 30, 2021 at 0:05

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