# Confusion with understanding the fixed bias circuit

Below shows a bipolar transistor in fixed current bias configuration. A text says that this topology is independent of β:

What I understand is, since Vcc, Vbe and Rb is constant the base current becomes fixed as:

Ib = (Vcc-Vbe) / Rb

As we see in the above formula since all three variables Vcc Vbe and Rb are constant, Ib is constant and so fixed.

My confusion is the following:

Imagine if we now change the transistor with the same type but with a different β, will the Ic change?

Thought 1: I'm asking because I guess we can say that after the transistor is changed the Vbe will not change(?). And according to the Ebers Moll equation the Ic will not change since Vbe will not change.(Ic is determined by Vbe)

Thought 2: But if we think again after changing the same type transistor with different β which means Vbe will not change and so that means Ib is fixed at the same value as well. But now the new transistor has a different β and Ic = β × Ib. So this tells us that Ic will change.

Which thought above is correct and where am I making the logical flaw?

• Yes, you are right, the Ic will change when a beta value change and Vbe also. In this circuit, only Ib current is fixed (not exactly true). And in real life, you shouldn't use this type of a biasing circuit.
– G36
Sep 9, 2018 at 16:54
• So the root reason that the β changes the change in Vbe by construction? Sep 9, 2018 at 16:55
• But you are saying Ib is fixed, how come Ib is fixed if Vbe changes? For Ib to be fixed Vbe should be constant isnt it? Maybe I have a problem in the meaning of "fixed" here.. Sep 9, 2018 at 16:56
• Every single BJT's will have a different β value. So the Ic will also change. If Vcc >> Vbe we can assume that the Ib ≈ Vcc/Rb is more or less fixed.
– G36
Sep 9, 2018 at 17:00
• Or maybe your book treat the BJT as a current controlled current source (Ic = Ib x β ) with the fixed Vbe value. In this case, only change in β can change the Ic value.
– G36
Sep 9, 2018 at 17:03

The Ebers-Moll equation does account for changes in $\beta$, but they use a parameter $\alpha$ where
$$\alpha = 1 - \frac{1}{\beta}$$
We often say that $V_{BE}$ is fixed in this circuit because the changes in $V_{BE}$ are usually small compared to $V_{CC}$. It follows that $I_B$ is (essentially) fixed since it is determined by Ohm's Law and the voltage drop across RB.
Your understanding is correct then that if $\beta$ (or $\alpha$) changes then $I_C$ will change significantly, causing $V_{CE}$ to change proportionally.