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 Dj BASS AMP 4th July 2002 01:31 PM

hi all!

I need information and circuit of STK4050V

Thanks!!!...:)

 e96mlo 4th July 2002 02:20 PM

This site

contains the data sheet. It's also a good idea to check the data sheet for STK4048XI as it is contains a bit more information and as they are quite similar.

/Marcus

 Dj BASS AMP 4th July 2002 02:35 PM

thanks!

but in the site not have a circuit of STK4050V

 e96mlo 4th July 2002 02:37 PM

Did you even bother to look in the data sheet?

/Marcus

 Freddie 4th July 2002 02:38 PM

Quote:
 but in the site not have a circuit of STK4050V
STK4050 is there.. You didn't look very carefully did you?

 Dj BASS AMP 4th July 2002 03:00 PM

THANKS!!!

and HOW MACH AMPER THE STK4050V IS NEED?

i now is need 66V but i dont now how mach amper (A) is need??

 e96mlo 4th July 2002 03:13 PM

Ohms law:

U = R*I => I = 66 V/8 Ohm = 8,25 A

16,5 A is needed for a 4 Ohm load. But these are peak currents so 8,25 A/(sqrt(2)) = 5,8 A continous and the double for 4 Ohms.

And add some to count for the losses...

/Marcus

 Dj BASS AMP 4th July 2002 03:35 PM

THANKS U!!!

I build a amp With 2 STK4050V IN 8 Ohm

soo i need a power 66V ~ 16.5A in a 8 Ohm (for two STK)

or 66V ~ 30A in a 4 Ohm ( for two STK)...

yes??

 Dj BASS AMP 4th July 2002 03:47 PM

if .............

66V ~ 8Ohm (IN STK4050V) = 200W R.M.S

soo....

66V ~ 4Ohm (in stk4050V) = W ????

 e96mlo 4th July 2002 03:53 PM

U=R*I
P=U*I

With these formulas you can do most of the calculations you need. But remember that Up-p = (sqrt(2))*Urms and the same for current.

This means that the power doubles when you take half the load.

The transformer you need will be quite large (>1000 VA).

/Marcus

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