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I am using a transistor to drive an LED with a GPIO pin of a microcontroller.

Ic is specified to be 25mA

I designed the circuit so that Ic can be as high as 100mA before the transistor itself acts as a current-limiter, and then I decided to use a resistor to limit the current through the LED (just felt safer to me - idk is this dumb?).

Now I am confused on how to do KVL on the output loop to solve for the value of the current-limiting resistor. Do I just assume Vce is very small?

enter image description here

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  • \$\begingroup\$ Ic = 100mA, Ib = Ic/B = 1mA, Rb = (3.3-0.7)/1mA = 2k6 ..... With this you will limit Ic to 100mA max. It's very temperature and Beta dependent, but for Led should be enough if you measure the exact Beta of transistor you use. \$\endgroup\$
    – user208862
    Nov 20, 2021 at 2:18
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    \$\begingroup\$ Normally for saturated operation a beta of 10-25 is used to ensure the transistor is saturated. A Vce of 200mV is a good place to start for calculating the series resistor. Using a resistor as you are doing is the recommended method, not dumb at all. \$\endgroup\$ Nov 20, 2021 at 2:19
  • \$\begingroup\$ Yes, Vce is simply negligible there, 0.2V, once you give enough IB, check the TR datasheet, nothing special. \$\endgroup\$
    – jay
    Nov 20, 2021 at 3:02

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I would not use base resistor for current limiting, as it depends on beta, and this varies a lot depending on temperature and manufacturing tolerances.

You don't care about high speed here, so just choose R1 to definitely saturate your transistor -- if you have 2N2222 for example, it promises 75-375 beta range (datasheet link)... So choose base current to be 2.5mA just to be sure.

Then you can consult the datasheet and look at "Collector −Emitter Saturation Voltage" -- this depends on many factors, so it will be a graph. In the 2N2222 datasheet, it is Figure 4, and it would be less than 0.1V. This is likely less than power supply tolerance, so just treat the transistor as short circuit while calculating R2.

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  • \$\begingroup\$ Haha I was answering EXACTLY like this and you beat me to it. OP, just drop 1k on that base resistor, and use R2 to limit LED current. We use this circuit by the billion (with a "b") in production exactly as you show. 1k base resistor will "always" work when driving a single small LED like that. (Do not apply same "rule of thumb" when targeting motors or other higher-current loads). \$\endgroup\$
    – Kyle B
    Nov 20, 2021 at 2:59

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