Electronics > Beginners
Simple Transistor Impedance Understanding
spitta:
I'm struggling understanding a fundamental transistor impedance equation.
See pictured: common emitter amp with R1 and R2 divider at the base
When there is a voltage divider at the base of a transistor, they are considered "in parallel".
Furthermore, the input signal sees R1 (110kΩ), R2 (10kΩ), and the impedance looking into the base in parallel. I've accepted this for years, but I don't truly understand it. WHY ARE R1 AND R2 CONSIDERED "IN PARALLEL"? They are not in parallel. In the picture, R1 is connected to 20V rail, while R2 is connected to ground. They only share one connection point. They are not in parallel... right?
billbyrd1945:
The odds of my being able to answer any question on an electronics forum is near zero. However, I think I can answer the part about R1 and R2. They are separated by a node (actually two nodes). I'm pretty sure that that alone would constitute a parallel configuration.
spitta:
Well, series configurations also share a node. Parallel resistors must share both nodes.
I think there is a fundamental assumption with basic transistor configuration equations that I am missing.
IanMacdonald:
R1 and R2 are in series as far as the DC supply is concerned. As far as an AC signal is concerned they are in parallel because R1 returns to ground via the supply decoupling capacitor. The fact that the top end of R1 is not at 0v makes no odds to the AC signal.
magic:
They aren't parallel. But the junction between them behaves exactly the same as if they were two parallel resistors connected to 1.66V.
1.66V is the voltage at the junction if no current flows from the divider to anywhere outside.
If an external load pulls the voltage down by 1mV, the top resistor will source 1mV/110k more current into the junction and the bottom will sink 1mV/10k less. So the junction will source 1mV/110k+1mV/10k into the load. Just like two resistors in parallel with 1mV across them.
See Thevenin's theorem.
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