[Amps] winding an HV transformer

[jeff millar]

Paul...

Since the H bridge generate a square wave voltage on the primary and not 
a sine wave, the output should also have a square ave stepped up by the 
turns ratio.  The 1.4X or 0.9X rules apply to sine waves into a 
capacitive or inductive load. 

For an H bridge fed from a 340V rail, the transform primary will see 340 
Volts, in both polarities, effectively a 680V square wave driving 
voltage.  With your 6:1 turns ratio, this works out to 4080V square wave 
on the output, producing a 4080V DC on the capacitors.

Did I interpret your design correctly?

jeff, wa1hco

Paul Decker wrote:
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> 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. 
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> 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. 
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> 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.    
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> 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. 
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> thanks, 
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> Paul 
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