1
\$\begingroup\$

This is an image from a book showing us how to solve BJT circuits via AC analysis:
enter image description here

So I understand that when doing AC analysis to ground all the DC voltage sources but I don't understand why. Can anyone explain this concept, the book just says to ground it as if it is some simple EE concept that needs no explaining.

\$\endgroup\$
3
\$\begingroup\$

If you look at your top circuit you will notice that there are shorts where the input, emitter and output capacitors were - this is the first step to doing an AC analysis. The caps are assumed to pass AC without hinderance so thay are shorted. Resistors are presumed to to attenuate so these are left in.

Whether a resistor is connected to Vsupply or ground is irrelevant - The power supply is assumed to be one big capacitor and so this is also shorted to ground (note the ground symbols on R1, Rc and Rout1).

You then use the Hfe of the transistor (current gain of the transistor) to calculate the signal-gain as it goes from base to collector.

So, on the base is a fraction of Vin and this fraction is initially determined by the attenuation produced by Rs, R1 and R2. BUT importantly it is mainly dictated by the B-E of the transistor - it can be assumed to be forward-conducting and sometimes it is easier to assume it is a short circuit hence the current into the base is purely Vin / Rs.

This current is amplified by Hfe (current gain) and produces a voltage on the collector that is dictated by the parallel combination of Rc, Rout1 and Rout2.

Thus you can determine the approximate signal gain for AC.

Me, I use a simulator - it's quicker especially if you are trying different values out AND takes all the transistor details into account and can give a fairly accurate frequency response too.

\$\endgroup\$
1
\$\begingroup\$

the DC source and AC source all have a internal resistance, Rs. for a AC signal, a ideal DC source is shorted.it means that when we doing AC analysis, only consider the path of AC current, even if we could not ignore the Rs of DC source in some case, notice the common junction betweeen circuit and Rs of DC source, and then think about the AC current path..

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.