I have been given this problem as Homework and I can't seem to figure this out. It seems that the only thing I have is emitter current Ie. However, it is known that BJT is in saturation mode, so I can't think of a way to find Vb, Ve, Vc, Ic, Ib. Assuming Vce = 0.1V for saturation and Vbe = 0.7V

Edited couple values that I previously entered incorrectly.

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  • \$\begingroup\$ Well, at least you know that if you have any one of \$V_E\$, \$V_B\$, or \$V_C\$, that you have the other two, right? And is that current there really a full Amp? Or did you forget to adjust it? \$\endgroup\$
    – jonk
    Nov 14, 2017 at 20:52
  • \$\begingroup\$ ANd is that a current source or a current measurement... \$\endgroup\$
    – Trevor_G
    Nov 14, 2017 at 21:04
  • \$\begingroup\$ It was supposed to be a 1mA current,sorry \$\endgroup\$
    – B.James
    Nov 15, 2017 at 5:28
  • \$\begingroup\$ With 1 mA emitter current, there is going to be 20V across R2. Whatever voltage remains from the +/-10V power supply (0V) has to be divided over the BJT and the current source (assuming that the current source is actually a current sink). I think you need to reconsider your design. \$\endgroup\$
    – Bart
    Nov 15, 2017 at 9:36
  • \$\begingroup\$ Why 20V across R2? How did u get that? \$\endgroup\$
    – B.James
    Nov 15, 2017 at 16:44

1 Answer 1


The first thing to note, of course, is that you only need one of \$V_E\$, \$V_C\$, or \$V_B\$ in order to get the other two. And if you have those, you can compute \$I_C\$ and \$I_B\$ (obviously, \$I_E\$ is a given value here.) So the entire problem turns on being able to solve for one of those.

I'll just completely ignore your specific values and treat this more generally. I'm also just going to pick \$V_C\$ at random, base everything on that value, and see where that takes us. (It doesn't really matter which one I pick.) Finally, this is an NPN so I will consider \$I_E\$ as positive if exiting the emitter, \$I_C\$ as positive if entering the collector, and \$I_B\$ as positive if entering the base. This means that I can rely upon the usual \$I_E=I_C+I_B\$.

It's pretty easy to see that \$V_C=V_2-R_2\cdot I_C\$. But \$I_C\$ isn't known and it looks difficult to just guess at. Still, you also know that \$I_C=I_E-I_B\$, so that means we could try out this replacement and find \$V_C=V_2-R_2\cdot \left(I_E-I_B\right)\$. We are given \$I_E\$ and we can easily state that \$I_B=\frac{0\:\textrm{V}-V_B}{R_1}\$. So let's stuff all that together and see what this makes for us:

$$V_C=V_2-R_2\cdot \left(I_E-\frac{0\:\textrm{V}-V_B}{R_1}\right)$$

Hmm. We can replace \$V_B\$ by using \$V_C\$, right? We do know that \$V_B=V_E+700\:\textrm{mV}\$ and that \$V_C=V_E+100\:\textrm{mV}\$. So let's do that:

$$V_C=V_2-R_2\cdot \left(I_E-\frac{0\:\textrm{V}-\left(V_C+600\:\textrm{mV}\right)}{R_1}\right)$$

Think you can solve that for \$V_C\$? If so, do you think that may help solve all of the rest, as well?

  • \$\begingroup\$ First of all, I really appreciate your time answering my question. This has helped A LOT. After looking at your answer, I derived it myself it makes sense now! Also, I might have made a mistake for the picture (that I have already edited) and now V2 is positive. Therefore I used +10V instead of -10V and got all the currents, and voltages. Thanks a million! \$\endgroup\$
    – B.James
    Nov 15, 2017 at 6:34
  • \$\begingroup\$ @B.James Thanks for the kind comments. I adjusted my reply to accommodate the use of \$V_2\$ instead of the actual value. Thanks again. \$\endgroup\$
    – jonk
    Nov 15, 2017 at 7:15

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