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
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