Hi. Can anyone help explain this concept, which usually reads something like this in most textbooks? "the DC voltage supply Vcc is assumed to have a very small internal resistance, typically 0.01 ohms. Thus to all intents the DC supply looks like a short circuit as far as the AC signal is concerned". Well, evidently...

Why does the AC signal at the base of the BJT (assuming voltage divider biasing) see Vcc as an earth? The voltage divider causes a reasonably stiff voltage at the base. The AC input signal comes along and sits on it, modulating the base bias voltage slightly up and down.

If Vcc is constant, presumably if the AC input signal goes slightly positive, the voltage at the bottom of the upper biasing resistor goes up from its quiescent value, meaning there is now less voltage across this resistor, thus reducing the current flowing through it from the Vcc rail towards the base..I cant quite see things from the AC signal's perspective...something's just not making sense in my brain!! Any help would be greatly appreciated.

Mango S

 

The AC voltage produces a change in the current passing through the resistor connected to Vcc, and also produces a change in the current in the resistor connected to ground. The changes in current produced in the two resistors by the AC voltage do not depend on the DC supply voltage. In an extreme case, you can have zero Vcc (i.e. Vcc becomes identical to ground), and the AC still produces the same current changes.

You will really need to understand and accept this; it is a key concept.

 

The Vcc is seen as earth because it usually has a large capacitor placed across it. A large capacitor behaves like a short circuit to AC. Any signal that leaks into the Vcc rail gets shunted to ground.
The AC signal does not modulate the bias voltage up and down. It modulates the bias CURRENT up and down. The voltage across the base-emitter junction is steady (0.6v for silicon and 0.3v for germanium). A BJT is really a current to voltage amplifier.

 

The Vcc is seen as earth because it usually has a large capacitor placed across it. A large capacitor behaves like a short circuit to AC. Any signal that leaks into the Vcc rail gets shunted to ground.
The AC signal does not modulate the bias voltage up and down. It modulates the bias CURRENT up and down. The voltage across the base-emitter junction is steady (0.6v for silicon and 0.3v for germanium). A BJT is really a current to voltage amplifier.

That is not accurate. The base-emitter voltage is indeed modulated. Although the BJT can by considered to be a current-controlled device, its input characteristic is not equivalent to a perfect 0.6V voltage with zero impedance. An accepted method of small-signal analysis for a transistor regards it as being voltage-controlled, with an output conductance controlled by vbe. Please see the following link.

http://en.wikipedia.org/wiki/Hybrid-pi_model

 

If we could regard the base-emitter voltage as fixed, the AC input would produce zero voltage change across either of the bias resistors, and there would therefore be no need to consider their effect on the signal. The transistor's (zero) input impedance would completely dominate the situation.

It is only because the transistor has a finite input impedance that the loading effects of the bias resistors need to be considered.

 

The datasheet for the 2N3904 has a graph that shows its common-emitter input impedance at various collector currents when it does not have an external emitter resistor. The input impedance is no where near zero ohms.
So of course its input voltage and its input current both are varied by an input signal.

 

Thanks for your help!
It helps to see the inclusion of a power supply decoupling capacitor in the picture and how that appears as a short to the AC signal.