Power Gain Of Amplifer

[shermaine]

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]

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]

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.