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Power Gain Of Amplifer

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shermaine

New Member
How to show that power gain of amplifier can be expressed as Gp = (M2/M1) Ga
where Ga is the available gain, M1 and M2 are the input and output of impedance mismatch factors.
 

Electronworks

New Member
How to show that power gain of amplifier can be expressed as Gp = (M2/M1) Ga
where Ga is the available gain, M1 and M2 are the input and output of impedance mismatch factors.
should that be Ga^2?

Power in = (Vin^2)/Rin
Power out = (Vout^2)/Rout

Power gain = Pout/Pin = [((Vin x Ga)^2)/Rout]/[Vin^2/Rin]

Vin^2 on top and bottom cancel giving:

= [Ga^2/Rout]/[1/Rin]

= Ga^2 [Rin/Rout]

Anyone got any other ideas...?
 

BrownOut

Banned
Ga is just the available gain. If the output power is V^2/Rout, and Rout is the output resistance plus the resistance of the load ( or next amplifier stage ) and the voltage delivered is Vout *Rd/Rout + Rd (where Rd is the load resistance), then

load power over output power is:


(vout * Rd/Rout + Rd)^2 /Rd
----------------------------
(vout^2/Rout + Rd)


= Rd/Rout + Rin

Effective gain is Ga(Rd/Rout + Rd)

In the special case where Rd = Rout, the Effective Gain is: Ga/2, and that is the maximum Effective gain possble.
 
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