Load power calculation

Hi,

I was looking through the application notes of an attenuator. The image is from the same link:

Attenuator Application Note

I am a bit unclear in the first step where the load power is calculated. According to me, the load power should be the same barring the Rs term. I feel there should not be an Rs term. The way I did it, I just calculated the voltage across the load resistor using voltage division, squared it and then divided it by the load resistor. So, how is Rs appearing in the numerator? Am I doing something wrong. I'd be grateful if someone could throw some light on this.

Thanks.

PS: E is the 1V source shown.

the equation stands for maximum power Rs should be same as Rl
it is called impedance matching.

Why? The resistor is there so it must be used in the calculation.

So that must have contained Rs. Load voltage is E * RL/(RS+RL). Power is that squared divided by RL.

Keith.

Thanks for the replies. Yes Keith, that is exactly what I got as the power. But if you notice in the equation shown, there is an Rs in the numerator as well. That is the Rs I am doubtful about. The one in the denominator in (Rs + RL) is fine.

Ok, I see what you mean now. Their equation is wrong I think. I would make it (RL*E /(RS +RL)) ^2 /RL

Keith

Thanks for the reply Keith. I was a little doubtful because it was mentioned in an application note and I thought I might me missing something there. Thanks again for clearing that up.

At first I thought it was a re-arrangement of the same equation but it isn't. Unfortunately not everything you read is correct - I have a few text books with mistakes in them and it can be very confusing!

Keith

yes...I thought it was a rearrangement too at first..thanks again