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Re: [Amps] winding an HV transformer

To: "'Jeff Blaine'" <keepwalking188@yahoo.com>, "'KD7QAE'" <KD7QAE@ARRL.NET>
Subject: Re: [Amps] winding an HV transformer
From: "Alex Eban" <alexeban@gmail.com>
Date: Wed, 1 Apr 2009 20:17:26 +0300
List-post: <amps@contesting.com">mailto:amps@contesting.com>
...not in principle.
I have a friend here in 4Zulu Land, designing switchers at about 20 kW
levels. They work very nicely and quite a few government agencies use them.
The problem seems to be rectifiers and filtering components, more so than
the primary side components or topology.
Besides, they got used to doing it the old way and it works. Why change?
The fact that we would like something more modern, lighter and probably also
cheaper doesn't enter their consideration!
Alex    4Z5KS

-----Original Message-----
From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
Behalf Of Jeff Blaine
Sent: Wednesday, April 01, 2009 9:22 AM
To: KD7QAE
Cc: amps@contesting.com; Paul Decker
Subject: Re: [Amps] winding an HV transformer

Tomm,

Very interesting information here.  I am wondering if any of the 
commercial tube amps are using this kind of circuit for the plate voltage. 

Given a decent 2KW TX alone can set easily set a guy back $400 or so, a 
switching power supply seems the natural evolution for tube amps.  But 
I've not seen a switcher in a tube amp yet. 

Is there is some limitation that prevents easy scaling to those power 
levels and voltages?

73/jeff/ac0c

KD7QAE wrote:
> Paul,
>
> At a 6:1 ratio, assuming the core are supports the flux created by 340V 
> and 6 turns, you will have an output of 6x340 or 2040V.  The transformer 
> takes the 680Vp-p primary square wave and delivers 4080Vp-p to the FWB 
> which 'inverts' the negative pulses and adds them into the gaps between 
> the positive pulses to give you the 2040Vdc output. 
>
> Since this is a square wave, a capacitive filter will suffice to fill in 
> the small gaps due to switch rise and fall times and transformer 
> response.  I would build this filter as a high frequency filter followed 
> by a few hundred uF of energy storage caps. 
>
> If you want to pull the maximum out of the AC line you have to power 
> factor correct the input rectifier so the PS looks resistive to the 
> incoming AC.  This is a simple matter of either building a PFC circuit 
> from PS controller application guides, or, better yet, buying a surplus 
> 2kW computer (server) power supply and reusing the AC front end to get 
> the PFC and even the power MOSFETS.  By the way, this approach will give 
> you line regulation and, if you reuse more of the PSU, load regulation 
> as well.
>
> On a final note, while the 50% square wave chopper is an easy circuit, I 
> see a major drawback in that it has no load or line regulation.  If you 
> were to add a simple PWM controller to this so that the duty cycle of 
> the square wave were variable, and add an output inductor filter, you 
> would then have a regulated HV PSU for not much more trouble than what 
> you are now building.
>
> Tomm, KD7QAE
>
> Paul Decker wrote:
>   
>> I've been holding this question for a couple of days now, I'm sure it is
very simple and perhaps I just need some reassurance on the answer. 
>>
>>
>>
>> If you have been following some of this smps discussion, I've got 100Khz
pulsed DC (0 - 340v) which is generated by directly rectifying and filtering
the 240 V AC mains and providing that into an h-bridge.   The h-bridge dumps
the 340 V 100Khz square wave into the transformer. 
>>
>>
>>
>> As the QST article recommends, I've wound the transformer with five turns
on the primary and had calculated that I need 30 turns on the secondary.
Performing some small signal tests, inputting 3.4v pk-pk square wave from my
signal generator yields about 20.4 volts pk-pk square wave.   I believe this
relationship should be linear and inputting 340 V will yield 2040 volts on
the secondary.    
>>
>>
>>
>> At this point the secondary dumps into a full wave bridge rectifier
followed by a filter capacitor.   This is where I am unclear.    When I
rectify this with the full wave bridge, will I get 2040 * 0.90 or will I get
2040 * 1.414 as the final DC output?     Part of me says I get the 1.414
value of 2885 VDC, however reading through the handbook, I seem to be
reading I'll get 0.90 the output voltage. 
>>
>>
>>
>> thanks, 
>>
>> Paul 
>>
>>
>>
>>   
>>
>>
>>
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>>   
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