Audio Amplifier. Which class should I use?

[Shayaan_Mustafa]

1) 0.1*50=5

2) 0.2*50=10

3) 0.5*50=25

4) 0.8*50=40

5) 1*50=50

There are results. But I have question.

I don't use frequency and calculated result. So is it understood that these are the calculated results of 200KHz frequency?

 

[godfreyl]

It is the same result for any frequency.

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What is important is duty cycle.

Duty cycle = Ton / (Ton + Toff).

 

[Shayaan_Mustafa]

Yes you are right sir.

Thanks a lot for helping me. you both are genius.

 

[godfreyl]

No, I am not a genius or an expert.

 

[Shayaan_Mustafa]

Can you tell me why goldsmith used suddenly square wave? What can be advantage of using this here?

 

[goldsmith]

Hi to both of you my friends , shayaan and Godfreyl
so , shayaan , could you understand that what i'm trying to say ? what is my meaning by that example ?

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Can you tell me why goldsmith used suddenly square wave? What can be advantage of using this here?

Humm let me , simplify my aim . but before that , tell me are you familiar with mosfets and modes of operation ?

 

[Shayaan_Mustafa]

Yes you are. of course.

 

[goldsmith]

BTW : suppose that we have a bjt transistor , which has 10 volts in base ( npn ) and 250 volts in collector , and a 0.5 ohms load in emitter . can you tell me how much is the voltage across the 0.5 ohms resistor ? and how much is power loss across the transistor ?

 

[Shayaan_Mustafa]

@goldsmith

Yes sir I have understood. Thanks to godfreyl he helped me in understanding that.

Yes I know operation regarding MOSFETs. Both E-MOSFETs and D-MOSFETs. But FET topics were remain just 1 month in our study and we don't have so much practice on these transistors as I have on BJTs so I will have so trouble with them but I will pickup many things which you will tell me.

 

[goldsmith]

So can you calculate the latest thing that i asked ?

 

[Shayaan_Mustafa]

Sir I will give you answer tomorrow coz its too night. So I have to go. I have noted down your question. It is about BJT

 

[goldsmith]

Well if you want to go and sleep , i wish good dreams for you ! ;-)

 

[Shayaan_Mustafa]

We have given:
\[{ V}_{B }\] = 10V
\[{ V}_{C }\] = 250V
We know,
\[{V }_{CB }\] = \[{V }_{C }\] - \[{V }_{B }\] = 250 - 10 = 240V
\[{ V}_{BE }\] = 0.7V
\[{ C}_{CE }\] = 250 - \[{ V}_{E }\] --------- (1)
\[{V }_{CB }\] = \[{V }_{CE }\] - \[{V }_{BE }\] --------------(2)

Put \[{V }_{CB }\] = 240V and \[{ V}_{ BE}\] = 0.7V and eq (1) in eq (2)

eq(2)=>

240 = 250 - \[{ V}_{E }\] -0.7
So,
\[{ V}_{E }\] = 250 - 0.7 - 240
\[{ V}_{E }\] = 9.3V

I have calculated VE. This is voltage from emitter to ground. So this is the voltage across the resistor load too. Isn't it?

And about power. I know the formula for output power but still I didn't study power dissipation of either BJTs or FETs.

 

[godfreyl]

More simply:

\[{ V}_{B }\] = 10V
\[{ V}_{BE }\] = 0.7V

So \[{ V}_{E }\] = 10V - 0.7V = 9.3V

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So this is the voltage across the resistor load too. Isn't it?

Yes. So now you can calculate the current through the resistor.

 

[Shayaan_Mustafa]

godfreyl thank you sir. I am really duffer.
Thanks a lot.

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Yes sir.

I am getting
IE=VE/RE
IE=9.3/0.5
IE=18.6amp

 

[godfreyl]

Now you know the current through the transistor. You also know the voltage across the transistor. So you can calculate the power dissipated by the transistor.

 

[Shayaan_Mustafa]

So IE=18.6amp is the current across the transistor?

But voltage??

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VE is the voltage across the transistor?

 

[goldsmith]

Hi shayaan , sorry , for two days , i was pretty busy , and i was not in my home ! now i came back .
Why emitter voltage ? the voltage across the transistor ! means Collector-Emitter voltage . ok ? it is enough that you just assume you haven't any transistor ! , consider that you just have a variable resistor . ok ?
Best Wishes
Goldsmith

 

[Shayaan_Mustafa]

Hi shayaan , sorry , for two days , i was pretty busy , and i was not in my home ! now i came back .

It's OK sir. Business is the part of life. I understand.

Why emitter voltage ? the voltage across the transistor ! means Collector-Emitter voltage . ok ? it is enough that you just assume you haven't any transistor ! , consider that you just have a variable resistor . ok ?
Best Wishes
Goldsmith

Thank to tell me voltage across transistor. Because I didn't know that. Yes if I assume it as a variable resistor then it will be easy.

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I have
IE=18.6amp and VCE=240.7V

So,
P=VI
P=(240.7)(18.6)
P=4477.02W

I think this loss is too much.. I must use heat sink. Also I must be sure that I calculated right value.

 

[goldsmith]

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 ?
However the result is pretty near .
it will be around 4.4 KW . is that reasonable , really ? i think you can use it as a stove , isn't it ?
So , what should we do if we need 9.4 volts from 250 volts ? should we tolerate this value of dissipation ?
Think about it and tell me each thing that can go through your mind .
By the way , can you calculate the efficiency ? if yes , tell me it's value in percent .
Best Luck
Goldsmith