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:
>
>
> 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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