# What is the effect of capacitor put in parallel with the resistor in emisor pin?

1. ## What is the effect of capacitor put in parallel with the resistor in emisor pin?

What is the efect of the capacitor put in paralell with the resistor in emisor pin.

2. ## Re: CAPACITOR IN BJT

Rise the gain of the amplifier, this happen because , the capacitor goes to short circuit in AC and the emisor pin goes to ground and the amplifier become a emisor comun amplifier who has the higest gain.

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3. ## Re: CAPACITOR IN BJT

Actually we can 2 things by putting capcitor (CE) parallel with resistor(RE) at emitter.

Amplifier without RE has more gain as snaider said. (for AC)

But Amplifier with RE has more stable. (for DC biasing)

If you want more gain with stable circuit put CE.

4. ## CAPACITOR IN BJT

fo small signal analysis, the capacitor is considerd to be shorted in AC right? why RE can reduce the Gain? because of feedback?

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5. ## Re: CAPACITOR IN BJT

gain (Av) = Vout / Vin
Av = (Ic.rL) / (Ib.Beta(re'+RE))
Av = (Beta.Ib.rL) / (Beta.Ib(re'+RE))
Av = rL / (re' + RE) <-- if the RE increased, the gain reduced

is it right?? please correct me if it's wrong.

6. ## CAPACITOR IN BJT

We need to stabilize the Q-point,so,
we include the RE resistor,
which is a feedback resistor.
Let's do the DC analysis of the Common Emitter amplifier:
Without RE,IC=Beta*IB=Beta*[(VBB-0.7)/RB]
whose value depend on Beta,which is a variant parameter;
With RE,IC=Beta*IB
=Beta*[(VBB-0.7)/(RB+Beta*RE)]
=Beta*[(VBB-0.7)/Beta*RE]
=(VBB-0.7)/RE
it's a fixed value.

O.K. we get DC Q-point stabilization now,
Do the ac analysis:
Without RE:
A=-ic*RC=-Beta*ib*RC
=-Beta*[(1/(RB+r-phi)]*RC
=-Beta*[1/RB]*RC
=-Beta*(RC/RB) --->a large value

With RE:
A=-ic*RC=-Beta*ib*RC
=-Beta*[1/(r-phi+Beta*RE)]*RC
=-Beta*[1/Beta*RE]*RC
=-RC/RE
=-1*(RC/RE)---------> a smaller value

yean,
with RE ,we get stabilization but we lose ac gain.

How to improve ?
If we connect a capacitor CE across the RE.
At DC analysis,the CE will be open circuit.
we still get the stabilization.

At AC analysis,the CE will be short circuit and
"short out" RE (Bypass)
we get the ac gain.

7. ## Re: CAPACITOR IN BJT

The purpose of the amplifier is get maximum output signal (a.c.) from the input at the same time we need to make Q-point stable so that it doesn't happen that on side (day +ive) our signal gets amplified but on the other side it gets chopped due to improper location of Q-point. Q-point is dependant upon how the circuit is baised. The resistor at Emitter plays a major role in circuit biasing and Q-point stabilization but this resistor would also reduce the gain of the amplifier so, which isnot desirable. Thus we put a capacitor in parallel with the resistor. Now for DC capacitor virtually acts as an open circuit hence no effect on Q-point stabilization. For AC it essentially acts as a short circuit nullifying the effect of parallel resistor this resulting in maximum gain.

8. ## Re: CAPACITOR IN BJT

hi,
Thanks for those info ed_surfer. How do we determine the value of the capacitor parallel to RE. Is there any formula/equation related to it?? Thanks

9. ## Re: CAPACITOR IN BJT

The reason for this is-
1>for a.c it acts like a short ckt so it behaves as a.c ground
2> for d.c it acts as open circuit.d.c biasing is un disturbed.

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10. ## Re: CAPACITOR IN BJT

It can increase high frequency gain.

best regards

Originally Posted by fab
What is the efect of the capacitor put in paralell with the resistor in emisor pin.

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