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Old 5th May 2005, 02:35 PM   #1
k1jroth is offline k1jroth  Finland
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Default Help needed PSU designing

Hello

I reacently accuired some parts and i going to built linear psu from them.

Parts whict i have:
230V/2x42V (60VDC unveighted) 1000VA toroid
20pcs 63V 12000uF EL-caps
10pcs of MJL4281A (NPN) and 10pcs MJL4302A (PNP) transistors
230W 15A 35Mhz 350V Hfe80-250

I`m trying to design psu with following properties

dual voltage out (+-50V adjustable)
Output current of about 8A
Adjustable current limiting (0-8A) with short circuit protection
Electrical overheat protection (fan or shutdown or both).

I would be greatful if someone could lend a hand give some hints
or maybe some kind schema for design.

Please help beginner.
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Old 5th May 2005, 02:57 PM   #2
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Have a look for 'foldback current limiting'. You will find all the information you need and then tons more good stuff in the 'Art of Electronics' by Horowitz & Hill.
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Old 5th May 2005, 06:39 PM   #3
AndrewT is offline AndrewT  Scotland
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Hi,
John Linsley Hood did a Wireless World design article for an 80w Mosfet stereo amp many years ago (probably mid 80s).
This may be the Williams Hart that is still retailed.
It included the full design of a regulated dual voltage PSU.
The low current side was simple. The hi current side included fold back limiting and DC detection and both pole shutoff and I think there was some other detection built in there.
All the voltages were adjustable within the thermal limits of the pass fets and IMO they should transfer to a BJT amp without difficulty.
By the way he chose not to include the design in his book "The art of linear Electronics"
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Old 5th May 2005, 06:47 PM   #4
dhaen is offline dhaen  Europe
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If it's an analogue supply, give some thought as to how you will limit power dissapation in the "series-pass" transistors.
Consider: Dropping 50v at 8A . That's 400 Watts, which is ourageous.
Continuously variable 0-50v is therefore not an option. A tapped transformer and voltage ranges is, however.
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Old 5th May 2005, 07:04 PM   #5
AndrewT is offline AndrewT  Scotland
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Hi,
fold back limiting solves that.
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Old 5th May 2005, 08:53 PM   #6
dhaen is offline dhaen  Europe
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Quote:
Originally posted by AndrewT
Hi,
fold back limiting solves that.
Andrew,

I think you miss my point.
To supply 50v, the input of the regulator will, in practice have to be at least 55v, so my 50v drop senario is not one of short-circuit (where of course you are right, that foldback solves), but in supplying 5V, not an unusual case. If we consider only 2 amps, the dissapation is still 100 watts.
Pehaps I should have been clearer.
What I am seeking to do, is to make the original poster aware that his wish-list is a difficult one, without taking some extra steps, or reducing his expectations. I know his transformer is a limitation, but there are even ways round this
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Old 5th May 2005, 08:55 PM   #7
mikeks is offline mikeks  United Kingdom
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Regulated supplies for power output stages is just plain bad design....

http://www.diyaudio.com/forums/showt...292#post627292
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Old 5th May 2005, 09:13 PM   #8
AndrewT is offline AndrewT  Scotland
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Hi John,
I don't know if my brain has gone to sleep so please excuse me if I'm wrong but surely if the input voltage is 55v plus half the ripple say another 2 volts, then the average volts drop down to a regulated 50v is (55+0.5*2)-50 = 6v.
The max current is 8A giving power lost = 48w. Manageable.
When an excessive load develops the foldback reduces the output current to maintain a sensible power dissipation in the pass FET.

Mikeks,
I am not advocating Reg output stage, just suggesting a solution.
I personally prefer reg supply to input & drivers and brute force PSU to the output stage.
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Old 5th May 2005, 10:27 PM   #9
dhaen is offline dhaen  Europe
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Hi Andrew,

Hmm, I had assumed that he wanted a bench power supply, perhaps a rash assumption!
Your brain is not asleep Perhaps mine is in overdrive and out of control

Let's wait for some clarification...
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Old 5th May 2005, 10:52 PM   #10
moamps is offline moamps  Croatia
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Quote:
Originally posted by AndrewT
[B]..... if the input voltage is 55v plus half the ripple say another 2 volts, then the average volts drop down to a regulated 50v is (55+0.5*2)-50 = 6v. The max current is 8A giving power lost = 48w.
But, power lost for output voltage 5V, current 8A is
(55+0.5*2)-5= 51V*8A=ca 400W!

Regards,
Milan
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