# Why does power driven to the load has this expression?

#### SebastianOspina

Joined Mar 21, 2021
1
Hello everybody, I have seen this expression in some design books.
According to this books, the power driven to the load (RL) is:
P = Icq2 × RL/8

I have try to deduce this equation and I have got this expression.
Let's assume the current (I) is the current across the load resistor (for example, using a simulator and measuring this signal).
P(rms) = IRMS^2 × RL.
P(rms) = Ipeak^2 × RL / 2.
P(rms) = Ip-p2 ^2× RL / 8.
It's a similar expression, however it depends on the alternal current across the resistor, not the Icq (average DC transistor's polarization current)

How can I relation this expression with Q-point?
Also I have another question, let's say I have a voltage amplifier (CE). It has a gain of 2.

If the input signal is 1, I would have 2 volts in the output. So my RL with have this 2 volts signal across it, and this would generate a power.
Now I have 2 volts in the input, this will generate 4 volts in RL. So the power will increase.

According to this, the power depends on the input, the amplifier gain and the RL, not Q-point, so how does this expression makes valid ?
P(rms) = Icq2 × RL/8 Thank you.