# How the low pass filter works in this circuit? simulate this circuit – Schematic created using CircuitLab

Vi=input, Vo=output signal. The capacitors and resistors of the output terminal of the above circuit is supposed to act as low pass filter (i.e. attenuate all frequencies of the input signal greater than a particular frequency). The only problem is that I can not understand how a low pass filter is obtained at the output in the above circuit. The problem is that (as in my textbook) 'CU' is calculated as if it is in series with 'RC' and acting as that low pass filter. But 'CU' is parallel to 'RC'! Also 'CO' is calculated as if 'CO' is in series with 'RC' and acting as high pass filter when 'Vo' is considered as input. And thus it becomes very confusing. Please help me understand it.

• Only one of those capacitors can act as part of a low pass filter. Identify it and focus on that. – Brian Drummond Jan 17 '17 at 15:00
• to lead on from Brian's comment, what's the output impedance of a common emitter transistor? We usually assume, and it's generally true, that the supply voltage is an AC ground. Why should that matter? – Neil_UK Jan 17 '17 at 15:02
• What you're looking for is the simplest lowpass filter imaginable: a resistor and a capacitor. Think how you can make a lowpass filter with a resistor and a capacitor. Such a circuit is in there. – Bimpelrekkie Jan 17 '17 at 15:02
• If you know how to make a small signal equivalent circuit then it will become even more obvious. – Bimpelrekkie Jan 17 '17 at 15:03
• You are right - seen from the collector node, the elements RC and CU are in parallel. However, why do you expect a problem? Don´t forget that the transistor acts as a current source and NOT as a low-impedance voltage source. – LvW Jan 17 '17 at 15:24 