# Using Thevenin Theorem to solve for unknown values in BJT voltage divider circuit

Assuming this is a DC bias of a voltage-divider type BJT circuit:

V_CC = 10V
Beta value = Not Known
V_BE = 0.73V (Base-emitter voltage)


And the Thevenin circuit being:

How do I go about finding the base/collector/emitter currents without knowing the beta value? Can I assume the base resistance as Rth, and the base current to be Vth/Rth? If the latter were true, then I don't think it would be correct if I were to plug it in this equation (KVL across base and emitter to ground, Ibr = Ith = Vth/Rth):

• Try to read this electronics.stackexchange.com/questions/471906/…
– G36
Commented Feb 1, 2020 at 7:19
• You'll need the beta value, unless it's not in active mode.
– jonk
Commented Feb 1, 2020 at 8:10
• Have any details from the original question been omitted?
– Chu
Commented Feb 1, 2020 at 9:47

First, I'd assume Vbe = 0.7 V, and beta = 100, and after one pass see how big an impact that makes. For your specific example, Vth = 10/11 V = 0.909 V, and Rth = 100k/11 = 9.1 kOhms.

0.909 = 9.1k*Ie/100 + 0.7 + Ie*2k

0.209 = Ie*(2k+0.09k) = Ie*2.09k

Ie = 0.1 mA; given those assumptions.

How sensitive would this be to Vbe? Try with 0.75 V, and 0.65 V. Subtract the currents and divide by 100mV and you get a I/V transconductance value - this is an important concept. How sensitive would it be to beta? Say beta were 10.

0.209 = Ie*2.9k = 0.072 mA.

How do I go about finding the base/collector/emitter currents without knowing the beta value?

You could start by finding a symbolic expression for the base/collector/emitter currents in terms of $$\\ \beta_F\$$

One simple way to model a BJT circuit is to say that the "apparent" resistance of $$\\ R_e\$$, when viewed from the base of the transistor, is $$\\ r_{e}=(\beta_F+1) R_e\$$.

Now you can find the voltage on the base of the transistor (compared to gnd) easily, by modelling the circuit as follows:

simulate this circuit – Schematic created using CircuitLab

Let's apply KVL and KCL;

$$\\ V_{b}-V_{be}-r_e I_{re}=0\implies I_{re}=\frac{V_{b}-V_{be}}{r_e}\$$

$$\\ \frac{V_{cc}-V_{b}}{R_1}-\frac{V_{b}}{R_2}-I_{re}=0\implies I_{re}=\frac{V_{cc}-V_{b}}{R_1}-\frac{V_{b}}{R_2}\$$

$$\\ \frac{V_{b}-V_{be}}{r_e}=\frac{V_{cc}-V_{b}}{R_1}-\frac{V_{b}}{R_2}\implies V_b=\frac{R_2 r_e V_{cc} + R_1 R_2 V_{be}}{R_1 r_e + R_2 r_e + R_1 R_2}\$$

$$\\ I_c=\frac{\beta_F}{\beta_F+1}I_{Re}=\frac{\beta_F (V_b-V_{be})}{(\beta_F+1) R_e}\$$

$$\\ V_c=V_{cc}-R_c I_c\$$

Assuming this is a DC bias of a voltage-divider type BJT circuit:

I'm not sure I understand what you mean by that, but if you want to model the circuit as two voltage dividers you can, assuming $$\\ V_{cc}>>V_{be}\$$ you can set $$\\ \frac{V_{be}}{V_{cc}}=0\$$ and get rid of $$\\ V_{be}\$$ from the expressions for $$\\ V_b\$$ and $$\\ V_c\$$. I will leave this for you to practice your algebra..