# Finding thevenin voltage in a common emittor circuit

simulate this circuit – Schematic created using CircuitLab

β=100, Vbe(on)=0,7, VA= ∞

Lower corner frequency is what actually wanted. In the solutions of the book, it began with finding thevenin resistance and thevenin voltage, then using them, it found IBQ and ICQ.

I get confused because finding thevenin resistance and thevenin voltage means that we have to analyse a pair of circuit nodes. Which two circuit nodes do i have to analyse here?

Thevenin equivalent is normally used to calculate bias point of the transistor.

To know bias point, you need Vb and from Vb you can obtain Ic. First of all, there is a mistake in your circuit: the 12V power supply must have the "+" to RC and the "-" to the ground, that (other mistake) is not assigned in your schematich. Put it to the "-" of Vi.

To calculate bias point you have to open all capacitors considering them all as open circuit, so your circuit as far as the biasing point is concerned, is only composed by Q1 and the resistors R1, R2, Rc and Re as Cc insulate Vi from the circuit.

Now, you have to calculate Thevenin on the base of the Q1 to the ground, that is the "-" of Vi, and in this case just mean applying the voltage divider formula to obtain Vb:

Vb = 12*(R2/(R1+R2)) = 1,56V

Now you can suppose Q1 in active zone and put Vbe = 0,6, so Ve = 1,56-0,6 = 0,96 and so Ie = 0,96/100 = 9,6mA

As Beta is given and equal to 100, you can now suppose that Ib is negligible compared to Ic, so Ic almost equal to Ie = 9.6mA.

All just let you know that Q1 actually is in active zone, so now you can use the right model to study the full amplifier (that is a standard 10 times single stage amplifier) to check the position of lower cut off frequency.