Please Help - RC filtering action

136

01:43:39

(simulation not needed, I just need to understand logic of how RC filtering works)
Assuming transistor gain=100, then Re(1K) seen from the base = approx 100k (impedance reflection). Therefore total impedance looking into base is approx 10k || 20k || 100k = 6.24k. This can be modelled as the 1uF input capacitor (Cin) in series with 6.24K equivalent resistance. Because of arrangement of Cin and 6.24k, this forms a high-pass filter action. We can find -3dB cut-off by 1/2pi(R)(Cin)=1/2pi(6.24k)(1uF)=25Hz. This I understand and makes sense.

My question:

I was told the output is also acting as a high-pass filter. I don't understand how this is when 1uF Cout capacitor and 1k (Re) resistor are not arranged in high-pass series configuration like at input (which can be visualised easily by drawing equivalent resistance and putting in series with Cin).  I was told -3dB cut-off for output is calculated as 1/2pi(Re)(Cout) = 1/2pi(1k)(1uF).  I don't understand also why 1K Re resistor is used as it's not in series with Cout. Please can someone explain I'm so confused

published 5 years ago

PrathikP

5 years ago

The 1k Re resistor is an emitter degeneration resistor used for negative feedback. It is not a part of the output HPF. Your circuit is actually incomplete, which is why you don't see the output HPF. Connect a load to the output, like a 10k load, and you will see the HPF. Also, connect an input signal and simulate the circuit to better understand how it works.

PrathikP

5 years ago

Also, your circuit is missing a collector resistor

dB7

5 years ago

I understand about connecting the load to then see the hpf.. i understand that now, thank u. But why would 1k be used to calculate cutoff frequency if no load? Also, circuit is not missing a collector resistor, it is meant to be like that. It is a common collector /emitter follower configuration which is providing current amplification only.

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