Say I have this configuration:


simulate this circuit – Schematic created using CircuitLab

I want to calculate the current going to the base of the NPN transistor.

Specifically for PMBT3904,215 transistor, but I am trying to figure out a general rule for calculating the Ibase.

Because of R2 and the 0.7V voltage drop of the NPN and the 1.2V drop of the LED, the current will be I=(5 - 0.7 - 1.2V)/(100+400[Ω]) =6.2mA right?

I do not know if there will be any "surprises" for using the transistor on the high side.

  • 2
    \$\begingroup\$ I do not know if there will be any "surprises" for using the transistor on the high side. The "surprise" is that the transistor will drop some voltage (around 0.6 V) while if you would use it low side, that could be reduced to 0.1 V. In my opinion, it is rather pointless to use a transistor like this. \$\endgroup\$ Feb 26, 2021 at 9:58
  • \$\begingroup\$ I agree, but the part of the led and R2 is on a different system that i have no access to. And a bunch of other limiting things that made me use the above configuration. \$\endgroup\$ Feb 26, 2021 at 10:03
  • 1
    \$\begingroup\$ Another way to look at this configuration is to consider that R_E (400Ω ish) in this case dominates the situation, and determines R_B ≈ β.R_E . With β=100 or more, the 100Ω is negligible in comparison. You have pretty much a diode-connected transistor. \$\endgroup\$
    – Pete W
    Feb 26, 2021 at 16:18

1 Answer 1


the current will be I=(5 - 0.7 - 1.2V)/(100+400[Ω]) =6.2mA right?

No, that's not quite right. It would be correct if the collector was disconnected from the 5 volt rail but, it isn't.

What we can say (as a fairly reasonable approximation) is that the the BJT emitter will be at around 4.3 volts\$^1\$ and this of course assumes that the base is close to 5 volts. Now, the current into the LED is the voltage across R2 divided by R2 hence, it is 3.1 volts ÷ 400 Ω = 7.75 mA.

And, nearly all of that current comes through the collector to the emitter. The base current will contribute a little bit but maybe only about 1% of the 7.75 mA. This, of course, is due to the current gain of the transistor. We tend to say that when the BJT isn't saturated that the current gain is about 100 hence, we might expect about 77.5 μA through the base.

Looking at the data sheet for the PMBT3904,215 transistor, the current gain will probably be more like 200 typically

So, the volt drop is the 100 Ω resistor (R1) is about 7.75 mV i.e. the base is pretty close to 5 volts thus bolstering our original assumption that assumed the emitter would be around 4.3 volts.

\$^1\$ - the emitter always follows the base but, because the base-emitter junction is a forward biased diode we assume that the voltage at the emitter (for an NPN) is around 0.7 volts lower.


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