Not correct!
With a full bridge rectifier you get the peak to peak voltage out as DC!
All the diodes do is redirect the transformer secondary to the correct
polarity.
What you wrote pertains to double rectification with a pair of diodes and a
center tapped transformer secondary.
Alex 4Z5KS
-----Original Message-----
From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
Behalf Of KD7QAE
Sent: Wednesday, April 01, 2009 8:34 AM
To: Paul Decker
Cc: amps@contesting.com
Subject: Re: [Amps] winding an HV transformer
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 rectifing 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 recomends, 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 yeilds about 20.4 volts pk-pk square wave. I believe this
relationship should be linear and inputting 340 V will yeild 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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> http://lists.contesting.com/mailman/listinfo/amps
>
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