[Amps] winding an HV transformer

[Paul Decker]

Hi Jeff, 



Thanks for taking the time to look into this.   I think you have interpreted my design correctly.   As I look at it more, I think you might be right, the h-bridge is fed 340 vdc which is switch through the load at alternate polarities (I think) which would mean as you say a square wave of 680 V.   I don't have the h-bridge constructed yet so I can't verify this.   If this is true, then I've made a huge error in the transformer.   I guess this is where I get tripped up.   If the h-bridge produces 680 volt pk-pk square wave, and that is fed to the step up transformer to produce 4080 v pk-pk, when the 4080 volts is fed to the bridge, I should get 4080 vdc with very little glitches in between the transition points of the square wave i.e. the negative portion of the square wave should be "flipped" to the positive side.   The filter cap should just smooth out those little transition (which should be like 100ns in my design).   I think that would mean the filtered DC should be just slightly under 4080 volts.   



Even if I've got the h-bridge voltage wrong, does the above seem correct?     



BTW, I think the other issues I was having making this make sense was because my scope was measuring pk-pk while my dvm was attempting to read rms but having problems at 100 khz . 



thanks, 

Paul 







----- Original Message ----- 
From: " jeff millar " < jeff @wa1hco.net> 
To: "Paul Decker" <kg7hf at comcast.net> 
Cc: amps at contesting.com 
Sent: Monday, March 30, 2009 10:57:59 PM GMT -05:00 US/Canada Eastern 
Subject: Re: [Amps] winding an HV transformer 

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.     
> 
> 
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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. 
> 
> 
> 
> thanks, 
> 
> Paul 
> 
> 
> 
>   
> 
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> 
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