Audio Amplifier. Which class should I use?

[goldsmith]

Sir I am not getting what you have said here?
What is ciss?
What is dissipation across GS junction? I have never heard about it and never calculated.
You are also talking about high value of inrush gate current. As much as I know, FET's have high input impedance therefore they don't have gate current. So what was your talking about.

Hi shayaan. i came back again !
Forget them it is not for this time , i'll illustrate them for you , soon !

Sir you had given me a question here, (post#334)
so can you calculate the dissipation across it's DS with D.C=0.04 ? consider frequency is just 50KHZ . vin is 250 volts and vout is 10 volts and RL is 1 ohms . ok ? show me the result .

Here is how I solved it..

Ploss=(V^2)/(RL)
Ploss=(10^2)/(1)
Ploss=100W

Dissipation across its DS,
Average = amplitude * D.C.
Average = 100 * 0.04
Average = 4W

Why V^2/RL ? the dissipation across a switch ( mosfet , in on state ) will be D.C* Rdson*Id^2 . did you forget it ?
can you calculate Id ? i think you can . did you ask from yourself that why i have asked you to find RDson of that mosfet ? of course for this example .
Retry .

 

[Shayaan_Mustafa]

Hello Welcome back sir.

Id=248.3/50
Id=4.996A

Average=D.C. * Id^2 * Rdson
Average=0.04 * 4.996^2 * 0.4
Average=0.016 * 24.96
Average=0.399W

Sir I am getting this result.
But it is very much low. I think it is wrong again. I made mistake somewhere.

 

[goldsmith]

No shayaan ., it is not wrong ! its is pretty correct ! . now can you calculate the power across the RL ? ( 1 ohms ) ? and then can you calculate the efficiency ? if yes , show me the result .

 

[Shayaan_Mustafa]

Power across the RL:

P=Id^2 / RL
P=4.996^2 / 1
P=24.96W

Efficiency:[\U]
efficiency = Po/(Po+Ploss) * 100
efficiency = 24.96/(24.96+0.399) * 100
efficiency = 98.426%

Oh my ***. It is very good..

I don't believe.. Awesome..

 

[goldsmith]

can you remember my example ? how much is the value of load that i told ? and how much is the out put ? retry your calculations with that ratio . don't forget that each things need average taking in this example .

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Id=10/1=10A 10/1 =10A
ploss= 10^2*0.4*0.04=1.6w
pout= 10^*1=100W =======> efficiency= (100/100+1.6 )*100=98 percent ! ok ?

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Now can you remember the example with that BJT , when vout was 10 volts and vin was 250 volts ? how much was the efficiency ? tell me .

 

[Shayaan_Mustafa]

can you remember my example ? how much is the value of load that i told ? and how much is the out put ? retry your calculations with that ratio . don't forget that each things need average taking in this example .

You mean, I should re-calculate my result of post#345?

Now can you remember the example with that BJT , when vout was 10 volts and vin was 250 volts ? how much was the efficiency ? tell me

Yes sir, I remember.
I had calculated efficiency for this in post#286 and it was 3.78%

 

[goldsmith]

You mean, I should re-calculate my result of post#345?

No i corrected it for you . did you see my latest post accurately ?

Yes sir, I remember.
I had calculated efficiency for this in post#286 and it was 3.78%

Can you compare it with the efficiency of this mosfet ? 3.78 percent instead of 98 percent ? did you understand that what i have tried to do ? can you predict , that what this action called ?

 

[Shayaan_Mustafa]

Id=10/1=10A 10/1 =10A
ploss= 10^2*0.4*0.04=1.6w
pout= 10^*1=100W =======> efficiency= (100/100+1.6 )*100=98 percent ! ok ?

I am still confused with this.
Why did you recalculate Id rather we had it is 5A.
And also I am getting 98.42% and your answer is 98%
So both are equal. This mean whether way I can adopt to calculate efficiency. I calculated 98% and you also calculated 98% so it means both ways are correct. Isn't it?

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Can you compare it with the efficiency of this mosfet ? 3.78 percent instead of 98 percent ? did you understand that what i have tried to do ? can you predict , that what this action called ?

I think you are suggesting here to replace BJT by MOSFET due to high efficiency. Right?
And therefore this action will be called replacement. lol hehehehe... :-D

 

[goldsmith]

I am still confused with this.
Why did you recalculate Id rather we had it is 5A.
And also I am getting 98.42% and your answer is 98%
So both are equal. This mean whether way I can adopt to calculate efficiency. I calculated 98% and you also calculated 98% so it means both ways are correct. Isn't it?

Of course both are equal together , but don't forget that you calculated Id instead of 50 ohms load !

I think you are suggesting here to replace BJT by MOSFET due to high efficiency. Right?
And therefore this action will be called replacement. lol hehehehe...

No no no ! can you remember that bjt example ? the base voltage was a dc voltage . but for this mosfet the GS voltage is a square wave . that bjt was in linear region as a variable resistor but this mosfet is in on / off region . you forgot it ?

 

[Shayaan_Mustafa]

Of course both are equal together , but don't forget that you calculated Id instead of 50 ohms load !

I got it.

No no no ! can you remember that bjt example ? the base voltage was a dc voltage . but for this mosfet the GS voltage is a square wave . that bjt was in linear region as a variable resistor but this mosfet is in on / off region . you forgot it ?

No sir, I didn't forget it. Rather than I didn't notice it because in university we had not so much practical with FETs as we had BJTs that's why it is not in my practice. If it was in my practice then immediately I could realize that.

No I have cleared. So on/off region is the study related to power electronics I think..

 

[goldsmith]

could you understand what this method called ? of course it called Switch mode regulator !!! haha ! could you understand what happened ? :grin: you have learned it's basics here , simply and without any problem !

 

[Shayaan_Mustafa]

Yes sir I have understood and I guessed it should be a switched mode regulator.
Great teaching up til now sir.

 

[goldsmith]

So let me continue . until now , we used mathematics to get average ( just) ! but can you tell me or can you predict what circuit we can use to get average ? ( between mosfet and load ) ?

 

[Shayaan_Mustafa]

Yes sir. In the study of my linear integrated circuits (LIC) I have read about averaging amplifier. I think we can use this. Isn't it?

 

[goldsmith]

No no . averaging amplifier is for low powers ! i'm referring to the passive networks .

 

[Shayaan_Mustafa]

OK. So passive networks, means using network of resistors, inductors or capacitors etc.. Right?

 

[goldsmith]

of course yes . but i think a ........... . another thing came in my mind . keep this average taker issue in your mind . before that , tell me , please , can you write an equation for out put voltage for this filter ? ( network function )


can you write AV and H(s) for these filters ?

 

[Shayaan_Mustafa]

I will try to write output voltage for this filter.

But what is H(s)?
I think Av is the voltage gain . Right?

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I will try to write output voltage for this filter.

But what is H(s)?
I think Av is the voltage gain . Right?

 

[goldsmith]

Of course it is voltage gain . but don't you know what is H(s) ? voltage gain in laplace ! are you familiar with that ?

 

[Shayaan_Mustafa]

No sir. I don't know laplace. I don't know anything about laplace. I didn't do anything using laplace transformation. I don't know. Now what happened :-(