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baab Senior Member Offline Posts: 178 EA	Class B power amplifier. Aug 28^th, 2013, 10:31pm Hi, please help me understand about class B power amplifier. Thank you.
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baab Senior Member Offline Posts: 178 EA	Re: Class B power amplifier. Reply #1 - Aug 28^th, 2013, 10:32pm The second image:
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Class B power amplifier. Reply #2 - Aug 29^th, 2013, 4:02am Average voltage across inductor is zero over the cycle, so please integrate the voltage and make it zero surely you will find the solution.
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summi Community Member Offline Posts: 38	Re: Class B power amplifier. Reply #3 - Aug 30^th, 2013, 7:38am Dear Raja.cedt, I did intergration, no ans. Br, Summi.
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Class B power amplifier. Reply #4 - Sep 1^st, 2013, 9:41am Hello, Rough derivation.... Thanks, Raj.
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baab

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Posts: 178
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Re: Class B power amplifier.
Reply #5 - Sep 3^rd, 2013, 8:21pm

Thank you, Raja and sorry for the late reply.
I still has two questions I am confused.
1. In your answer, the voltage above VDD is Vp/π. I can't figure out why Vp = Ip*Rp* 2√2
Can you tell me why?
2. Can you tell me why the voltage at points X, Y (Drains of M1 and M2) has the shape like that? I think it should be a sinosoid but not. Here is my thought:
Vx = VDD - VL
Where VL is the voltage across half of primary winding.
VL = Ldi/dt
With ID1 has the shape said above, then Vx has to be a sinosoid.
Can you tell me where I am wrong?

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aaron_do

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Re: Class B power amplifier.
Reply #6 - Sep 4^th, 2013, 5:47pm

Hi,

Quote:

2. Can you tell me why the voltage at points X, Y (Drains of M1 and M2) has the shape like that? I think it should be a sinosoid but not.

The author has simplified the load. The transformer is assumed to be ideal, and so the load seen by the transistors is purely real. So the output waveform is just the current waveform multiplied by a constant.

Aaron

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