How is it that R3 draws 0.6 mA current in the circuit described below?

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The "divider" formed by R2-R3 may be confusing: R3's job is to keep Q3 off when Q2 is off; and when Q2 pulls its collector low, most of its collector current comes from Q3's base (because only ~0.6 mA of the 4.4mA collector current comes from R3 -- make sure you understand why. That is, R3 does not have much effect on Q3's saturation. Another way to say it is that the divider would sit at about +11.6V (rather than +14.4V), were it not for Q3's base-emitter diode, which consequently gets most of Q2's collector current. In any case, the value of R3 is not critical and could be made larger; the tradeoff is slower turn-off of Q3, owing to capacitive effects.

  • \$\begingroup\$ The Photon has given you an excellent answer that directly addresses your question, as I read it. What else are you looking for? \$\endgroup\$ Commented Sep 2, 2023 at 6:46

2 Answers 2


We can assume that in forward active region, the base-emitter junction of Q3 has 0.6 V across it (other authors might assume 0.7 V or some other value, but 0.6 V is a reasonable assumption).

It would be a good idea to verify that the assumption that Q3 is in forward active mode leads to consistent results for the rest of the circuit.

Since the same voltage is across R3, and R3 is 1 kohm, then there must be 0.6 mA through R3.

  • \$\begingroup\$ I thought since first r3 and r2 is in series for the npn the current drawn is 3.4mA for both , after that I applied current division rule since pnp acting as voltage divider will make the current parallel:(1/4.3)*3.4. =0.79,but I guess my method is too crude. Thankyou for proper explanation \$\endgroup\$
    – user232591
    Commented Aug 29, 2023 at 4:40
  • \$\begingroup\$ @user232591 As you grow in sophistication, you will also be able to see how it is (in the right-side circuit) that R1 can be removed from the base (replace with wire) and placed in the Q2 emitter leg (with a different value) to create a current sink. This allows using a wire for R2 (if Q2 can handle R2's earlier dissipation needs.) This lightens the loading on whatever is driving Q2, since Q2 remains in active, not saturated mode, should that be useful. \$\endgroup\$ Commented Aug 29, 2023 at 12:57

R3 is connected across a base-emitter junction. The equivalent circuit is:


simulate this circuit – Schematic created using CircuitLab

There are two extremes:

  1. B-E forward biased: +1V across B-E, so 1mA max through R3.

  2. B-E reverse biased: -6V across R3 at B-E at the point of base-emitter reverse breakdown (approximately), so -6mA max through R3.

So, the current across R3 will be between +1mA and -6mA worst-case. That's no matter what the application is - this subcircuit inherently limits the B-E voltage, and thus the voltage across R3.

In the high-side switch application, the reverse bias won't happen, and the maximum base current is limited by R2 to under 5mA. At +5mA, a typical base will drop about +0.6V, and that's the voltage R3 will "see". Thus 0.6V/1kΩ=0.6mA mentioned in the text.


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