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PUSH-PULL transformer question

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bowman1710

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Hi guys,

I'm looking at a push-pull topology (open loop) with and input of 50V (regulated) and 350V output at 200W. If its open loop (i.e the output will not be regulated) I take it I will need a 1:7 turn ratio on the transformer. I have had issues in the past with the turns ratio being too low causing the magnetising inductance to be to large, so what will be the minimum number of turns in the primary, while keeping the magnetising inductance low? I need to try and keep the design as small as possible (ideally with a planar design if possible).
 

planars are designed with special software usually.
Magnetising inductance should preferably be high to reduce magnetising current, but its not a hard rule.

To get turns ratio, treat it as a buck converter and do vout = D*vin with your max D at min vin.
("vin" is the effective vin as if it were a buck)
Then find what turns ratio you need to give that effective vin at the buck converter...and its simply v(pri)/v(sec) = N(pri)/N(sec).
Remember the leakage means you will need a higher ns/np ratio than calculated....remember to sandwich wind so as to reduce leakage.

I don't like push pull, the primary snubber dissipates leakage flow and magnetising current flow.
 

Treez, I know you would want the magnetising current as high as possible, but doesn't it become a problem when copper losses start taking over? To get enough turns on the primary i may have problems getting the turns out in a planar design. When i work out the turns ratio of say 1:7, is it say 8T primary total or 8T+8T and 56T in the secondary?
 

I know you would want the magnetising current as high as possible
Actually treez said the opposite. But I think you put the cart before the horse anyway.

The primary design criterion for power transforners, both low and high frequency, is core flux. It's limited either by saturation limits or acceptable core losses. Number of turns and everything else is commanded by designed AC flux.

You didn't tell much about your design, we can just guess that it's a high frequency transformer.
 

mag current should be as low as poss.
Regarding turns, its 8t + 8t in pri.....type thing.
But it wont be 56t in secondary......that wouldn't give enough voltage for 350v at the output......remember the output inductor.

Supose yor max duty cycle is 0.7 (choose your max d as you wish)
Then for 350V, vin(eff) = 350v/0.7 = 500v.
now to get 500v from 50v you need 1:10 transformer, or 8t+8t, 80t

.....................
Thanks FvM, you must have written at same time as me
 

The frequency of the design is likely to be 150kHz or 333kHz switching frequency. I gather it was the opposite by the way he mentioned "buck" when that wasn't what I was doing.
 

I understand you are not doing buck.
But to design pushpull.....start as if it were a buck, then get your effective vin...then do the tunrs ratio for it to be pushpull, considering the max D at min vin.
 

How will this effect my design if I am running it with no feedback (open loop) with a 1:10 turns ratio? My initial stage (i.e 50V) is being controlled by feedback and the push-pull is being left open loop. How does duty cycle work in an open loop system?
 
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for a buck Vout = D * VIN.
(its the same for pushpull except remember that the buck vin is what you set your turns ratio for.)

So keep your duty cycle constant at the required value.
if its open loop then presumably you mean fixed duty cycle.

beware on open loop because if you go no-load then your vout will spiral upwards...until vout = effective vin...where effective vin = vin *ns/np
 

How will this effect my design if I am running it with no feedback (open loop) with a 1:10 turns ratio? My initial stage (i.e 50V) is being controlled by feedback and the push-pull is being left open loop. How does duty cycle work in an open loop system?

Other people may be drawing from more experience than I have in this application but from my vantage point more information is needed.

Can you be more concrete about what signal you're putting in (50V square wave? Peak to peak? Something else?), and what you expect to get out (are you expecting 350V rectified?).

I agree that if you're just trying to get stepped up power across a transformer and it's open loop then duty cycle isn't an issue (and in-fact you'd want it to be exactly 50%). But it seems like you haven't provided all the information necessary.
 
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In order to have a push pull converter with a specific output voltage running open loop, i.e. no feedback controlling the converter, then you will have to have a minimum load, or a voltage clamp (e.g. big zeners) or a stop start controller stopping the gate drive when Vout exceeds V desired + 5%. Luckily, also, if you are running at near full duty cycle on the primary side fets (e.g. 49% + 49%) with just enough dead time to give a nice voltage transition on the fets from 25% load to 100% load, then you will get soft switching (can add a little C to the Drain-Source to help here too), now the peak rectifying effect on the sec is much reduced and the min DC load or zener clamp will be low in power, you will get your desired VDC out (less volt drop in the fets, Tx, and o/p diodes) but no current limit - add a fuse to the input.
Hope this helps...
 
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