Audio Amplifier. Which class should I use?

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I don't see, that the voltage and current numbers refer to a particular circuit respectively a design problem. So we don't need to worry about real transistor ratings or heatsinks at this point.

I noticed that various circuits have been discussed a few fifty posts above. So if the calculation is referring to any of these, it would be a good idea to tell which it is. The reason for choosing the said voltage and current quantities would be interesting, too.
 

Why 240.7 volts ? vout will be around 9.4 volts . do you know why ? and voltage across the CE junction will be around 250-9.4 if we neglect from vcesat . ok ?
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Really I didn't get anything which you said above sir.

And yes 4.4KW is pretty high loss of power. This implies it must have low efficiency. So if we made efficiency high then loss of power can be reduced.

Formula for efficiency is,
efficiency=Pout/Pin*100

I have one power of them and don't have other. I must have both powers, Pout and Pin, in order to calculate efficiency.
So I am not getting idea to get efficiency sir.
 

Dear FvM
Hi
Shayaan decided to design an amplifier , and we guided it . and now we noticed that he will need a PSU to supply it . and the reason of this discussion is that . i will conduct him to the best PSU , and advantages and disadvantages of some of PSUs .
With My Best Regards
Goldsmith

Hi shayaan
Of course that is formula of efficiency . but here you will have this : efficiency=po/po+ploss you understand ? and about the collector emitter voltage : it means VC-VE . ok ? VE is9.4 volts and VC is 250 volts ok ?
 
Of course that is formula of efficiency . but here you will have this : efficiency=po/po+ploss you understand ?
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OK OK sir so I should this,
Pout=50W (As it is our requirement which we choose in designing this amplifier. Right sir?)
And,
Ploss=4.4K

so,
efficiency=50/(50+4.4K)*100
efficiency=50/4.45K*100
efficiency=0.0112359*100
efficiency=1.12359%

Is it correct sir?

BTW: I didn't understand what FvM said. I can't understand hard English.
 

OH... yes sir you are right..

Pout=(VE)(IE)
Pout=(9.3)(18.6)
Pout=172.98
and
Ploss=4.4K

so
efficiency=Pout/(Pout+Ploss) *100
efficiency=172.98/(172.98*4.4K) *100
efficiency=172.98/(4.572K) *100
efficiency=0.0378 *100
efficiency=3.78 %

It is also not too much.. it is bad..
 

Humm , well done ! congratulations !
So , in the other step , what we can do , to decrease power dissipation ? consider we have to use 250 volts as input and we want 172.98 Watts . what should we do to decrease loss power , and increase efficiency ? think on it , and tell me the result . ok ?
 
Shayaan decided to design an amplifier , and we guided it . and now we noticed that he will need a PSU to supply it . and the reason of this discussion is that . i will conduct him to the best PSU , and advantages and disadvantages of some of PSUs.
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I see. Seriously speaking, I have a problem with threads running out of topic. In my view, edaboard visitors can expect that the discussion in a thread is somhow related to the original question. In the present case, it is only in a broad sense.
 

Dear FvM
Hi
I agree this discussion is a bit long and will be longer than this , but i think readers can solve many of their problems in this thread ( problems regarding audio amplifier design . ) but if you have better , idea , of course i will accept it , because i believe that you have better ways that we can follow , ever .
Best Regards
Goldsmith
 

Certainly yes . but we don't need higher out puts !( for example ! ) . let me help you : 1- consider that we have a switch . when it is open , the voltage across it will be around vin and current will be zero . ok ?
And it means P=V*I =0 W 2- and when switch is close , voltage across the switch is zero and current can be infinite ! ( for example ) . then , again loss power across the switch will be zero . are you agree with me ?
 
Yes I agree with you.

So then we should decrease power loss rather than we increase output power. Right sir?

So in order to decrease power loss we have to decrease emitter current. Isn't it?

power loss is given as,
Ploss=Vce*Ie

If we decrease emitter current then we can decrease Ploss. Similarly we can decrease voltage across transistor too (Vce). But I think we should decrease Ie instead we decrease Vce. Am I right sir?
 

Dear Bruno
Hi and welcome to the EDAboard !

Hi shayaan .
No . my mean wasn't that we change the current . let me show you another way : compare my latest post and the average formula of square wave . and tell me what you can understand from that comparison .
 

Hello sir..
I tried to compare. But seriously I didn't find any result..

your post#291 says either to decrease current or voltage.. it can reduce power..
 

No that will tell you , you can use a switch ( for example a transistor ) to switch the high value of voltage , thus the dissipation across the transistor will be negligible . ( for example switch it with a square wave . thus the emitter voltage will be a square wave but with high value of amplitude ( e.g 250 volts ) if you change the duty cycle , and then take average from it the voltage will change as well but with pretty low dissipation )
You understand ?
 

thanks sir.

yes I understand but little..

- - - Updated - - -

Sir I think you should continue topic. Many things we understand during other things. So may be when we discuss further then I could understand this more clearly..
 

Sir I think you should continue topic
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What do you mean by that ? is that means that you don't like thinking about things ? if i give you the answer simply , your mind never can't be active . it is a standard way . don't you want to trust me ?
 

no no sir. this is not the matter. actually i mean many things we learn by time..

i mean i get you post#296 but i have confusion.. it is being difficult to understand for me sir..
 

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