0
\$\begingroup\$

So lets say that we are looking at NPN transistor:

schematic

simulate this circuit – Schematic created using CircuitLab

I know the practical and theoretical basics of transistor but what I don't understand is next:

1.) Collector current flows in and out of transistor if we have load on the collector. How is that possible - in and out at the same time?

2.) Ic=β*Ib, right? So when we are using this equation, which Ic do we use - Ic in or Ic out?

\$\endgroup\$
3
  • \$\begingroup\$ I think you misunderstood something or drew this circuit wrong. In this case, Q1 is on. R is essentially shorted to ground, Current out ~= 0, and Current in ~= R2/V. \$\endgroup\$
    – Jim
    Oct 20, 2016 at 21:57
  • \$\begingroup\$ I don't even know where "out" and "in" refer to. In and out from where? This may sound childish, but sometimes it helps to imagine that electrons are a mob of tiny people who set out marching from the "+" and are trying to find the easiest route to the "-". And the ones who go through R1 are, um, holding a gate open for the others. And you can't have people going in opposite directions along the same path. \$\endgroup\$
    – Ian Bland
    Oct 20, 2016 at 23:05
  • \$\begingroup\$ Current in a circuit is a scalar quantity, so whether it is flowing in or out is just a matter of the sign. It's either one or the other unless it's zero. \$\endgroup\$
    – Spehro Pefhany
    Oct 21, 2016 at 1:27

2 Answers 2

Reset to default
1
\$\begingroup\$

in and out at the same time?

It doesn't flow in and out at the same time. It flows through R2 and then in to the BJT collector and into the load (R).

which Ic do we use - Ic in or Ic out?

Ic will be flowing in to an NPN transistor if it is operating in forward active or saturated mode.

\$\endgroup\$
0
\$\begingroup\$

In the circuit shown, current flows through R2, and then flows EITHER into Q1 collector OR load R, or divides with some flowing into each. There is no current out of Q1 collector.

Ic = β*Ib, right.

So if R1 is infinite, Ib = 0, Ic = 0, and all the current from R2 flows through R. I(out) = V/(R + R2), and Vout = Iout * R.

If R1 is low enough, Ib is large, turning the transistor fully on (saturating it), so all the available current lows into the collector.

Now Vout = 0 (actually, saturation voltage is typically below 0.2V), Ic = V/R2, and Iout = 0 (actually < 0.2V * R). Now saturation is also commonly defined as the point where β falls to 10 or less, so you can take Ib as Ic/10, and work out a suitable R1 from that.

For an intermediate value of Ib, current divides at the collector between Ic (= β * Ib) and Iout. Databooks give approximate values of β at different values of Vce the collector-emitter voltage (which is also Vout in this circuit).

\$\endgroup\$

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.