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Plz tell me the use of resistor R1 and capacitance C(s) and C(e) in the given circuit and also their necessity.

Diagram

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    \$\begingroup\$ Cs is required only if there is a possibility of the DC component in the signal. Also safety. If you do without it, but then connect to something with DC, you'd fry your amp (I have). ReCe is a frequency dependent local negative feedback, as described in the answer below. \$\endgroup\$ – safesphere Sep 1 '17 at 20:39
  • \$\begingroup\$ look up H bias common emitter \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 1 '17 at 21:06
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R1 and R2 form a resistor ladder with a voltage that with no signal from the input will be somewhere between Vcc and 0. This is to bias the transistor so that it is always conducting at least some. The capacitor Cs is a blocking capacitor to keep any DC portion of the input signal out, so we only see the changing part at the base of the transistor. CE is there to shunt any HF part of what the emitter sees to ground. Given no actual values for the components we can only describe what they are doing, not how much of it.

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  • \$\begingroup\$ I got capacitors function.. Bt plz explain the first line only... My doubt why the resistors are so neccessary there.. What if they arent present? \$\endgroup\$ – user167814 Sep 1 '17 at 20:30
  • \$\begingroup\$ One doubt more.. Why we are draining HF part via C(e)? Why it is so necessary here? \$\endgroup\$ – user167814 Sep 1 '17 at 20:33
  • \$\begingroup\$ Suppose Cs is directly connected to the base. For the negative part of the wave, the transistor will be cut off, so you will only see the positive part of the signal. \$\endgroup\$ – zeta-band Sep 1 '17 at 21:47
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To elaborate a bit on the precious answer, an NPN transistor needs about 0.6V on the base in order for it to turn on. R1 is there to pull the base voltage up so that the transistor is on.

Without it, the transistor would be completely off the negative cycle of the AC input, and also for any part of the cycle less than 0.6V.

Cs isolates the bias voltage produced by R1 and R2 from any DC component present in the input signal. It only lets through the AC.

As zeta-band has said, Ce will tend to dump any high frequency components that get through the amplifier, by effectively shorting them to ground. Presumably the designer of this circuit didn't want them. High frequencies could include noise signals that are above any expected input frequency.

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  • \$\begingroup\$ Thank u all... I got it.. \$\endgroup\$ – user167814 Sep 2 '17 at 4:22

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