# How can I find the VBE of a common-emitter amplifier?

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.

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].

• 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. Commented Jun 28, 2021 at 17:30
• 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.
– jonk
Commented Jun 28, 2021 at 17:36
• Yes, in order to find the Ic @jonk Commented Jun 28, 2021 at 17:43
• 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. Commented Jun 28, 2021 at 17:45
• I'm really thankful for the info! Never thought that there's an other equation that can help me find Vbe @jonk Commented Jun 29, 2021 at 4:53

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:

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

And

$$V_{CEsat} = 0.2V$$

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.

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