# Voltage Divider Bias

I am new to biasing calculations and I cannot understand why in a voltage divider bias we can apply the Thevenin Theorem to find the potential in the transistor base. For example, here http://www.learningaboutelectronics.com/Articles/Voltage-divider-bias-of-a-BJT-transistor why is V in the base calculated as if the current though the base is 0?

• ... THAT website strikes again! It really needs a health warning... (1) You can often live with the error caused by assuming Ib=0, provided you know the current in the divider is at least 10*Ib. (2) But you're right, it IS an error and the website ought to point that out.
– user16324
Dec 21, 2015 at 14:14
• I have seen here en.wikipedia.org/wiki/… a correct formula (I think :) ) How it is calculated? How to calculate correctly the potential at the base? Dec 21, 2015 at 14:30

It is an approximation.

You typically work around the approximation by setting the current flow the two bias divider resistors be 10X what the maximum expected base current would be. Some folks like to use a higher ratio than 10X.

• Ok. Thanks! But how would it be correct to calculate the voltage at the base? (without approximating the current to 0). Dec 21, 2015 at 14:18

"But how would it be correct to calculate the voltage at the base?"

Starting with the base voltage VB (against common ground) it is not a problem to calculate both resistors using Ohm`s law:

Top resistor R1=(Vcc-VB)/C*IB and R2=VB/(C-1)*IB.

C=Bias factor (usually, C=10, or larger). The DC base voltage VB is VB=VBE-VE with VE depending on the emitter dc voltage VE=IE*RE.

The voltage V of the Thevenin equivalent circuit is not exactly the voltage applied to the base, otherwise, the voltage at the base terminal, when the transistor is not connected, i.e., the Open Circuit Voltage.
The other important element to consider is the Thevenin equivalent resistance, so that the entire model line connected to the transistor circuit is completed.

This model allows to simplify the analysis of non-linear circuit. The condition for applying Thevenin's theorem is that we must "remove" the transistor from the circuit to calculate Thevenin's voltage.

The Thevenin equivalent resistance is the resistance "that the transistor sees" connected to its input terminals.
Based on the operating point of the transistor, through the base will circulate a certain current, which will produce a voltage drop across the equivalent resistance, applying the base voltage that would be measured in the real circuit.

Keep the formula for Ib on the wiki page in mind. You can calculate it by assuming the Vbe is 0.65 volts and beta (the transistor current gain) is about 100. You can calculate the Thevenin resistance Rt. You know the Thevenin voltage Vt, so the voltage at the base is $$V_b = V_t - R_t I_b$$ Is this close enough? It's easy to determine. Recalculate for Vbe of 0.6 and 0.7, combined with beta of 50 and 150. Using all three values of each variable will give you 9 different results. If the results are close enough together, you can accept the results. If this is not the case, you'll have to find a data sheet for the actual transistor being used, and use this to find, for instance, what beta and Vbe really are. Because both of these numbers are nonlinear and hard to characterize analytically (especially beta), you'll probably need to do an iterative process, where you plug in numbers and see how far off they are from being consistent over the entire circuit, then modify and try again. Computers are a really big help here.