# Average current of sine wave

#### tyassin Hi

I have a question regarding the attached page from "RF Power Amplifiers". On the given page which deals with a class D amplifier, the average current or DC through the is calculated with the equation in (4.78).

I dont understand why the result is the peak amplitude divided by pi?
Should it not be (2/pi)*peak?

Thank you

#### Attachments

• 51.6 KB Views: 13

#### Easy peasy for a class D the input current is in lumps - as the D goes from 0% ( or 50% ) pwm to max ( say 98% ) back to 0% ( or 50% depending on how its done ) so the current supplied is less than the ave of a sine wave - which is rms x 1.111, the modulation of the D makes it Ipk / pi

= Irms x sqrt(2) / pi = 0.45 x Irms - quite a saving, the L carries the current when the switch is off ...

#### tyassin for a class D the input current is in lumps - as the D goes from 0% ( or 50% ) pwm to max ( say 98% ) back to 0% ( or 50% depending on how its done ) so the current supplied is less than the ave of a sine wave - which is rms x 1.111, the modulation of the D makes it Ipk / pi

= Irms x sqrt(2) / pi = 0.45 x Irms - quite a saving, the L carries the current when the switch is off ...
I am still not sure I follow. If I consider to calculate the average current(DC) for a sine wave (to get the power supply current), I follow the the equation given in (4.78) in my attached link. Then the DC component is calculated by integrating only the half sinewave. I think this is also what is shown in the equation by changing the integration from 2*pi to 0 too pi to 0. But then it is still averaged over 2*pi. Should this not be just pi?
I can see you arrive at the same result (Ipk/pi), but not sure how to get there?

Maybe I get it now Because the class D when operating at 50% duty-cycle is only supplying current 50% at the time, the integration time from pi to 2*pi is zero and therefore the result is Ipk*pi instead of Ipk*(2/pi). Also according to equation (4.77) in the attachment.

Last edited: