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Design >> RF Design >> Class B power amplifier.
https://designers-guide.org/forum/YaBB.pl?num=1377754309

Message started by baab on Aug 28^th, 2013, 10:31pm

Title: Class B power amplifier.
Post by baab on Aug 28^th, 2013, 10:31pm

Hi, please help me understand about class B power amplifier. Thank you.

Title: Re: Class B power amplifier.
Post by baab on Aug 28^th, 2013, 10:32pm

The second image:

Title: Re: Class B power amplifier.
Post by raja.cedt on 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.

Title: Re: Class B power amplifier.
Post by summi on Aug 30^th, 2013, 7:38am

Dear Raja.cedt,
I did intergration, no ans.

Br,
Summi.

Title: Re: Class B power amplifier.
Post by raja.cedt on Sep 1^st, 2013, 9:41am

Hello,
Rough derivation....

Thanks,
Raj.

Title: Re: Class B power amplifier.
Post by baab on 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?

Title: Re: Class B power amplifier.
Post by aaron_do on 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