Audio Amplifier. Which class should I use?

Status
Not open for further replies.
so in most of the applications we prefer lowest rise time for our square wave . ok ? and of course lowest fall time .
hence in our example you can select Tr and TF , a number around 1 ns . ok ? and for Td too ( i will tell you what td does mean it is not for this time ) . and PW means pulse width . and Per derived from period , per=1/F and when you want D.C around 50 percent you should select PW around per/2 ok ? now can you show me the result ?
 

V1 is positive half cycle and V2 is negative half cycle , so for V1 you can select 15 volts and for V2 -15 volts
 

Shayaan ? i think you are sleepy ? am i right ? when i told that you short circuit the out put of your source ? my friend i told that use a half wave rectifier with 1n4007 and a 1k ohms resistor as load . and show me the voltage across the load . ;-)
 

Don't mention it ! a mistake is a usual thing ! if you see my mistakes in my projects you will kill yourself ! ha ha !!!! :grin: don't lose your hope .
 



- - - Updated - - -

You are right. We must not lose our hope. In fact, we are human beings the most genius living thing in the universe so we must believe on our self that we can do everything.

- - - Updated - - -

You are right. We must not lose our hope. In fact, we are human beings the most genius living thing in the universe so we must believe on our self that we can do everything.
 

So didn't you ask from yourself that why when i have used a diode in series the out put voltage still is AC ?

- - - Updated - - -

BTW , increase the end time of simulations thus you can see more cycles .
 

Yes sir. You are right. Output is still AC. As we have across the inductor.

It is due to switching.
 

Switching ????????????????????????? we have a diode and resistor just ! . replace the diode with a 1n4148 and show me the result
 

Here is the result of D1n4148


- - - Updated - - -

Yes sir. You are right. I must not use the word switching because diode is not on or off. It is conducting all the time that is why I am getting positive and negative cycle on the output.
 

Hi shayaan
Now can you say why the results are not the same ? can you predict why 1n4148 could rectify that wave but 1n4007 couldn't ?
By the way , a diode is like a switch too . if you give it the 0.6 volts ( for silicon ) it will be on . but the reason , was not that ! ( that AC voltage after the diode )

- - - Updated - - -

By the way , why you just want to see one cycle ? it is not good ! you should be able to see some cycles
 

Now can you say why the results are not the same ? can you predict why 1n4148 could rectify that wave but 1n4007 couldn't ?
Click to expand...
How would I know sir? Even I don't know what is the difference between these two diodes. We have never learned when to use which diode. So really it is approximately impossible for me to say about the difference between two.

By the way , a diode is like a switch too . if you give it the 0.6 volts ( for silicon ) it will be on . but the reason , was not that ! ( that AC voltage after the diode )
Click to expand...
I considered much my attention on this question but I remain unable to get answer to this question. Still I am blank; your were right.
When we use a sine wave at input then we don't receive -ve cycle at output because during -ve cycle of input, diode is turned off. But this is not the case in square wave.

By the way , why you just want to see one cycle ? it is not good ! you should be able to see some cycles
Click to expand...
Yes sir, I have seen some cycles.
 

As I saw output, output is same as input. At output I have +14.3V and -15.7V.

So diode is conducting for both cycles. Why it is not off for the -ve cycle?
 

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…