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Old 29th December 2012, 07:50 PM   #21
DUG is offline DUG  Canada
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I have a question regarding the definition of CLASS A.

Let's for the moment assume a complimentary/symmetry output stage with active devices in both the top and bottom positions.

Let's assume +/- 16V rails and an 8R load. Let's also assume the device can go to rail just to make it easier.

If my Iq is set to 0.1A and in a positive swing the peak current in the + side goes to 2A and the negative side is minimal. (lets say 0.001A)

That would be class AB.

If I set the Iq to 2.5A and had the same output swing and the bottom device maintained the 2.5A so the top device would have to supply 4.5A (2A o/p + 2.5A Iq from the bottom device)

That would be class A because the bottom device was still in a conduction mode.

What about in between.

I set my Iq to 1A and design the circuitry to maintain that as a minimum.

The positive swing goes to 16V @ 2A output.

The upper device supplies 3A and the lower device still supplies 1A.

This should also be class A...The inactive device (lower) is still maintaining conduction and has not "shut off".

This definition seems to be in accordance with this thread:

What is the definition of Class A?

So if you have designed the output stage not to go below a certain current (never shut "off") then you have class A.

Even on a 400W with only 0.5A Iq.

This should make your class A power supply calculations easier.

Thoughts anyone?
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Old 29th December 2012, 11:31 PM   #22
djk is offline djk
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400W out of four Hexfets?

Have you calculated the junction temperature?

How will it look with a 55C ambient temperature?
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Old 9th July 2013, 04:45 AM   #23
lovemov is offline lovemov  United States
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I doubt the calculation is correct? The calculation is based on 8ohm speaker, so what about 4ohm? For class A bias is fixed, so when change speaker from 8ohm to 4ohm, one gets half the watts output?
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Old 9th July 2013, 11:05 AM   #24
DUG is offline DUG  Canada
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Quote:
Originally Posted by lovemov View Post
I doubt the calculation is correct?...
Which calculation?

Quote:
Originally Posted by lovemov View Post
... The calculation is based on 8ohm speaker, so what about 4ohm? ...
Who's post are your comments directed at?

Help me out here.

Just asking some questions.

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Old 9th July 2013, 09:49 PM   #25
lovemov is offline lovemov  United States
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Many posts referring this: P = I^2*R or P = (I*2)^2*R. So I am wondering, if the bias 'I' (whatever it should be I or I*2) is constant, P will vary with R, which gets lower if R is lower, so 4ohm speaker gets 1/2 power of 8ohm...can't be right?
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Old 10th July 2013, 12:47 AM   #26
DUG is offline DUG  Canada
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P = I^2*R or P = I*I*R

That is what I learned...like what does work...like equivalent heating power.

"P = (I*2)^2*R" came from a peak power calculation.

rhythmsandy:

"...( 3.6 x 2 )^2*8 = 414.72W peak and for continuous ... "

(post #3)

(I don't use that one personally)

The power of 8R vs 4R would double (for 4R) assuming adequate supply of current with the same voltage applied to both 8R and 4R.

The situation discussed was not this case. There was a limit to the current available.

Goto go.

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Old 10th July 2013, 01:15 PM   #27
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lovemov. the formula: (I*2)^2*R is class A power.
and yes. without taking some action, the class A power will be half with half the impedance.
however. in a push-pull design. it will leave class A and continue into class A/B for as long as the voltage allows it to.
then the formula goes: p= V^/R
lets say it can give full rail voltage, just to make it easy. +/-60V rails.
60*60/8ohm=450W peak/225W average.
or. 60*0.707=42.4Vrms
42.4*42.4/8=225W
at 4ohm, this will dobble.
of course the PSU must manage the currents. and hold the voltage up at the same time. thats not likely to happend.

Last edited by AudioSan; 10th July 2013 at 01:19 PM.
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