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Old 4th December 2012, 02:47 PM   #591
BigE is offline BigE  Canada
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Thanks AndrewT.

So, for 40 Vpk loaded, we need 45 Vpk unloaded, or a theoretical 45/1.414 = 31.82, or about 32 VAC for 100 watts RMS into 8R0 when biasing at around 2.6 amps. Using 1.3 instead of 1.414 as the rectification constant, gives us a 35 VAC transformer.

If the rule of thumb for PS droop is about 2 volts/amp/pair of transistors, then for a large number of output transistors, say 4 pairs, each biased at 1 amp, we should get an 8 Volt droop or so.

That means to get 40 Vpk from 4 pairs becomes 48 Vpk, which needs a theoretical 48/1.414= 34 VAC transformer. Using 1.3, gives a 37 VAC transformer. A 36 VAC transformer ought to be close to the the right size for a 100 watt amp using 4 pairs of outputs biased at 1 amp.

Higher bias means more droop, so if the 4 pairs are biased at 1.3 amps, the droop will be closer to 10 volts, which will need 50 Vpk, or between a 36 and 38.5 VAC transformer. That makes a 38 VAC transformer the better choice.

That is much more clear. thank you!

Thanks!

The interesting thing about this is that you can use a lower voltage transformer with fewer pairs, but still acheive the 100 watts output. I'm guessing that fewer pairs means more clarity, while more pairs means higher damping factor and better longevity.

Last edited by BigE; 4th December 2012 at 03:13 PM.
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Old 4th December 2012, 04:49 PM   #592
AndrewT is online now AndrewT  Scotland
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Quote:
Originally Posted by BigE View Post
...............The interesting thing about this is that you can use a lower voltage transformer with fewer pairs, but still acheive the 100 watts output. ..........
That's back to front.
The losses should be lower as the number of output pairs is increased.
Select your idle output bias as 2.6A.
That allows for 5Apk of ClassA output into 8r0.
A single pair would need to be biased to 2.6A. The losses will be the sum of the cable resistance losses + the trace losses + the emitter resistor losses + the return cable losses. While still in ClassA the PSU will suffer virtually no droop from the fully biased voltage.

A dual pair would have each pair biased to 1.3A. The losses through each pair carrying the half current halved. Those wires carrying the full 5Apk would not be any different from the single pair case.

A 3pair would have each pair biased to 867mA. The losses are divided by 3
A 4pair etc.....
A 6pair etc.....

yet another unhelpful post?
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Old 4th December 2012, 06:54 PM   #593
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just started my F5 turbo V3 build. the PCB's i'm using is toecutter's prototype V3 boards. wich i cut in half(output boards that is) to get the heat spread well over my 2x10.080" profiles.
got the PSU layout ready to night. 1000VA 2x40V transformer 2x35A rectifier bridges(on there own sink). and 4x68.000uF caps. softstart will be modified hypex.
this is ofcourse monoblocks
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Last edited by AudioSan; 4th December 2012 at 06:57 PM.
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Old 4th December 2012, 07:36 PM   #594
BigE is offline BigE  Canada
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Quote:
Originally Posted by AndrewT View Post
That's back to front.
The losses should be lower as the number of output pairs is increased.
Select your idle output bias as 2.6A.
That allows for 5Apk of ClassA output into 8r0.
A single pair would need to be biased to 2.6A. The losses will be the sum of the cable resistance losses + the trace losses + the emitter resistor losses + the return cable losses. While still in ClassA the PSU will suffer virtually no droop from the fully biased voltage.

A dual pair would have each pair biased to 1.3A. The losses through each pair carrying the half current halved. Those wires carrying the full 5Apk would not be any different from the single pair case.

A 3pair would have each pair biased to 867mA. The losses are divided by 3
A 4pair etc.....
A 6pair etc.....

yet another unhelpful post?
Actually, it is helpful. 4 pair gets biased to 2.6/4 = .65 ma.

Since the losses are related to the bias current, lower bias => lower losses, in my example, as the bias was not decreased as extra pairs were added, there was a lot of added droop.

But, since the bias goes down with more pairs, as do the losses, then the size of the power transformer will also go down as pairs are added.

Are we suggesting that as the voltage droop is a MAX of 5 volts or so, then the actual droop = 5* 1/number_of_pairs?

eg, With two pairs the droop is halved, three it is a third, 4 it is a quarter? ie, with 4 pairs, the droop is not 5 volts, but only 1.25?

That would mean that a 4 pair amp putting out 100 watts rms into 8R0 would need to make only 41.25 Vpk to cover the looses, which gives about a 32 VAC transformer. (Using 1.3 as the rectification constant.)

With 36 VAC * 1.3 = 46.8 Vpk, or about 45.55 Vrail after losses for 4 pairs yields close to 130 watts RMS Class A power, biased at just 0.65ma.

Do I have it right now? The more pairs, the less losses, the more output power.

I hope this is not like banging your head against a wall -- I'm trying to learn that rule to select the right transformer, without someone just saying use XX vac...

Thanks for helping.
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Old 4th December 2012, 08:01 PM   #595
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the bias don't go down with more pairs. just the bias pr output device.
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Old 4th December 2012, 11:48 PM   #596
BigE is offline BigE  Canada
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I see, the total bias stays the same, but the bias/device changes. So the droop is constant regardless of how many pairs.
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Old 5th December 2012, 12:09 AM   #597
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yes the transformer can influence the drop doh.
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Old 5th December 2012, 01:55 AM   #598
dazed2 is offline dazed2  Canada
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Hmm so this means to get the final 50Vpk loaded, I will need about 55Vpk unloaded
which means 55V = 1.3 x 42Vsec from the transformer.

This will mean that : 50Vpk /1.414 =35.36V and solve P=V^2 /R(8ohm) I should get 156Watts?!
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Old 5th December 2012, 02:17 AM   #599
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or make it alot easyer. 35VAC sec= about 45VDC wich give you aboyt 40Vpk-pk sving and hoooome ruuuuuuuuuuuuuuun
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Old 5th December 2012, 02:34 AM   #600
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I hope you boys have some nice BIG heatsinks...or willing to cook on your amp
Gotta love DIY.
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