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Yes sir I read in my book that it controls the gain and together they are called volume control. But I want to know how the capacitor C4 works physically in that circuit. I Know a capacitor is considered shorted in a.c at the oprating freq and I already know a capacitor is considered open in d.c and it is also used as a filter, but with all these knowledge I still cant figure out the particular work that C4 does in the circuit. Please help me understand it. Thanks.
just from the AC view: Consider the capacitor as zero impedance.
If the pot R10 is adjusted to the bottom then the emiter AC impedance is 130 Ohm + 5000 Ohm = 5130 Ohm. --> Low gain
If the pot R10 is adjusted to half value then the emiter AC impedance is 130 Ohm + 2500 Ohm = 2630 Ohm. --> Medium gain
If the pot R10 is adjusted to the top then the emiter AC impedance is 130 Ohm --> High gain
The higher the emitter impedance the lower the gain.
The keyword is "NEGATIVE FEEDBACK".
Negative feedback alwqays reduces the gain and - at the same time - improves linearity and reduces the influence of tolerance uncertainties.
For stabilizing the DC operating point (against tolerances and temperature) we always use an emitter resitance - however, because of the (perhaps unwanted) gain reduction we place a capacitor across this resistor (or a part of this resistor), thus cancelling (resp. reducing) the feedback for frequencies above the corresponding cut-off frequency.
just from the AC view: Consider the capacitor as zero impedance.
If the pot R10 is adjusted to the bottom then the emiter AC impedance is 130 Ohm + 5000 Ohm = 5130 Ohm. --> Low gain
If the pot R10 is adjusted to half value then the emiter AC impedance is 130 Ohm + 2500 Ohm = 2630 Ohm. --> Medium gain
If the pot R10 is adjusted to the top then the emiter AC impedance is 130 Ohm --> High gain
The higher the emitter impedance the lower the gain.
As you recognized before: C4 is high impedance for DC.
If you now short C4 the path is low impedance for DC. So a change of R10 changes the DC operation point.
As you recognized before: C4 is high impedance for DC.
If you now short C4 the path is low impedance for DC. So a change of R10 changes the DC operation point.
So in conclusion, C4 is placed tbere to control the impedance since by the a.c signal, but not the resistance seen by d.c so that when the pot is adjusted the d.c operating point is not affected. Am I correct?
So in conclusion, C4 is placed tbere to control the impedance since by the a.c signal, but not the resistance seen by d.c so that when the pot is adjusted the d.c operating point is not affected. Am I correct?
For each transistor it is the emitter path that provides negative feedback: A rising emitter current (due to temperature, for example) causes an increase in the emitter node voltage and reduces VBE. Reduction in VBE - in turn - causes again a reduction of the increased emitter current.
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