...none!
With full wave rectification of a square wave you get the peak to peak
voltage of the secondary as the DC output.
What scares me after having worked on off line converters for about 6 years
is the primary number of windings: it seems much too low and since you give
no information on the core, I can't verify the number of turns. I would
expect to see 20 turns or so in the primary.
Please give me more info and I shall try to make sure you don't send the
power transistors to the grave.
Alex 4Z5KS
-----Original Message-----
From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
Behalf Of Paul Decker
Sent: Tuesday, March 31, 2009 12:37 AM
To: amps@contesting.com
Subject: [Amps] winding an HV transformer
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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