Audio Amplifier. Which class should I use?

Status
Not open for further replies.
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?
 

It is the same result for any frequency.

- - - Updated - - -

What is important is duty cycle.

Duty cycle = Ton / (Ton + Toff).
 

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

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 ?

- - - Updated - - -

Can you tell me why goldsmith used suddenly square wave? What can be advantage of using this here?
Click to expand...
Humm let me , simplify my aim . but before that , tell me are you familiar with mosfets and modes of operation ?
 

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 ?
 

@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.
 

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
 

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.
 

More simply:

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

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

- - - Updated - - -

Shayaan_Mustafa said:
So this is the voltage across the resistor load too. Isn't it?
Click to expand...
Yes. So now you can calculate the current through the resistor.
 

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

- - - Updated - - -

Yes sir.

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

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.
 

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

But voltage??

- - - Updated - - -

VE is the voltage across the transistor?
 

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
 

goldsmith said:
Hi shayaan , sorry , for two days , i was pretty busy , and i was not in my home ! now i came back .
Click to expand...

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

goldsmith said:
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
Click to expand...

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.

- - - Updated - - -

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.
 

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
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…