enter image description here

I have a common-emitter circuit as above. I need to find the value of Vbe and IC. The picture below is the equivalent large-DC analysis of the circuit.

enter image description here

Applying KVl

  • Ic = (VTH - VBE)/((1/B)RTH+(B+1/B)RE)).
  • Ic = Isexp(Vbe/Vt)

Where (from my calculations) Vth = 5 V and Rth = 29166 khms or 2.9 kohms.

I know I can find Vbe by iteration but I'm really lost as to how. I tried using Excel but the value I obtained was wrong.

Can anyone point out how I can obtain Vbe? I can't move forward to obtain the other small parameters if I can't determined the value of Vbe (I'm doing manual calculation.)

here's an example of my work [The values listed down are different, it's from another problem but basically, this is how I try to solve for Vbe].

enter image description here

  • 1
    \$\begingroup\$ Show us your work. How did you attempt to calculate \$V_{BE}\$? Explain why you don't just run the simulation and find what you need. \$\endgroup\$ Commented Jun 28, 2021 at 17:30
  • \$\begingroup\$ Peace, are you wanting to include the Shockley diode equation as it applies to an active mode BJT? Because that's the only way you get there, directly. \$\endgroup\$
    – jonk
    Commented Jun 28, 2021 at 17:36
  • \$\begingroup\$ Yes, in order to find the Ic @jonk \$\endgroup\$ Commented Jun 28, 2021 at 17:43
  • \$\begingroup\$ It's true that I can use simulation, but I'm trying to practice the manual calculations in order to get a clearer perspective on how amplifiers work @Elliot Alderson. \$\endgroup\$ Commented Jun 28, 2021 at 17:45
  • 1
    \$\begingroup\$ I'm really thankful for the info! Never thought that there's an other equation that can help me find Vbe @jonk \$\endgroup\$ Commented Jun 29, 2021 at 4:53

1 Answer 1


For your circuit, we can try to use iteration.

But first, we need to find: \$R_{TH}\$ and \$V_{TH}\$

\$V_{TH} = V_{CC} \frac{R_2}{R_1 + R_2} = 5V\$ and \$R_{TH} = R_1||R_2 = 29.17k\Omega\$

More about it here:

Calculation of base current and what decides the current through collector-emitter branch

We star the iteration prosec by asuming \$V_{BE}\$ value and calculate the base current:

$$I_B = \frac{V_{TH} - V_{BE}}{R_{TH} + (\beta +1)R_E } = \frac{5V - 0.6V}{29.17k\Omega +121*200\Omega} \approx 82.443 \mu A$$

And solving for \$I_C\$ current

$$I_C = 9.89mA$$ and the collector voltage \$V_C = 12V -9.89mA*1.5k\Omega= -2.83V\$

Since we are getting the negative value, this indicates that the transistor in your circuit is in a saturation region.

Thus, our equations do not hold anymore in the saturation region. In that case, we need to use KCL (\$I_E = I_B+I_C\$) and solve for currents when the transistor is operating in saturation.

$$I_E = I_B + I_C$$

$$\frac{V_E}{R_E}=\frac{V_{TH}-(V_{BE}+ V_E)}{R_{TH}}+\frac{V_{CC} - (V_{CEsat}+V_E)}{R_C}$$

And I solve it for \$V_E\$

$$V_E = \left(\frac{V_{TH} - V_{BE}}{R_{TH}} +\frac{V_{CC} -V_{CEsat}}{R_C}\right)\cdot R_E||R_{TH}||R_C $$

But this time we also need to guess \$V_{BE}\$ value and \$V_{CEsat}\$ as well.

Therefore, the first guess is:

$$ V_{BE} = V_T \ln \left(\frac{I_B}{\frac{I_{S}}{\beta}}+1\right)=0.663V$$


$$V_{CEsat} = 0.2V$$

Thus our first iteration is

\$V_E = 1.40597V\$ and \$I_E \approx 7mA\$

And the base curent is:

$$I_B = \frac{V_{TH} - (V_{BE}+V_E)}{R_{TH}} = 100\mu A$$

And the new (second iteration) Vbe value is

$$ V_{BE}(2) = V_T \ln \left(\frac{I_B}{\frac{I_{S}}{\beta}}+1\right)=0.668V$$

And this would end the process. There is no need to make more accurate calculations because the transistor is saturated.

  • \$\begingroup\$ Thank you very much! I think I understand it a little better now. \$\endgroup\$ Commented Jun 29, 2021 at 4:31

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.