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Full Bridge DC-DC Converter Operating Frequency

TheDannia

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Hi,

I'm trying to understand full bridge working principles and I wonder, if I choose operating frequency as 10kHz,

transformer working frequency will be 10kHz or 5khz?

Also output filter will be designed according to 10kHz or 20kHz because of frequency doubling caused by rectification?

Thank you.
 

KlausST

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Hi,

As with your previous post: draw a sketch including the signal/frequency informations.

We don't know where you input/expect which frequency.
Where do you "rectify"?

Klaus
 

TheDannia

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sorry I forgot to mention, it's a full bridge dc-dc converter with full wave bridge rectification:Blank circuit diagram.png

When I drive the switches at 10kHz, will transformer's operating frequency also be 10kHz or it will decreased to half because of turning into AC?

and

rectification doubles the frequency, will output filter be designed according to doubled frequency caused by rectifier or the switching frequency?

Thank you.
 

FvM

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For usual switching patterns (simple push-pull, phase shifted bridge), 10 kHz transistor switching results in 10 kHz transformer and 20 kHz output frequency.
 

KlausST

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Hi,

depending on your (not yet shown) driving scheme of the bridge you get different waveforms at the transformer.

But the transformer does not care if there is a rectifier at it's output or not.
Only after the rectifier you get somehow doubled frequency at L1 (again in details depending on your driving scheme)
R1 and C2 should see only minor residuals of AC (on a DC offset).

Klaus
 

TheDannia

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For usual switching patterns (simple push-pull, phase shifted bridge), 10 kHz transistor switching results in 10 kHz transformer and 20 kHz output frequency.
Hi,

depending on your (not yet shown) driving scheme of the bridge you get different waveforms at the transformer.

But the transformer does not care if there is a rectifier at it's output or not.
Only after the rectifier you get somehow doubled frequency at L1 (again in details depending on your driving scheme)
R1 and C2 should see only minor residuals of AC (on a DC offset).

Klaus
so the both L and C filter frequency parameter is 20kHz to be used in formulas right?
 

KlausST

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Hi,

again, again, we can not know what signals you have.. so it makes answering difficult, uncertain and full of doubts for us and for you.

In case you have a low duty cycle at the rectifier ouput you have lots of current ripple and you need a better filter
in case you have a 99% duty cycle ... your filter has almost nothing to do...

Klaus
 

TheDannia

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Hi,

again, again, we can not know what signals you have.. so it makes answering difficult, uncertain and full of doubts for us and for you.

In case you have a low duty cycle at the rectifier ouput you have lots of current ripple and you need a better filter
in case you have a 99% duty cycle ... your filter has almost nothing to do...

Klaus
duty cycle is between 0.6(Vin-max) to 0.9(Vin-min),
switching is like: M1 and M4 are conducting while M2 and M3 are not and visa versa, like inverter
and filter is going to be designed for lowest duty cycle of course,
but my question is not about these things I just wonder should I use the switching frequency or double switching frequency when I calculate filter values?
 

KlausST

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Hi,

I just wonder should I use the switching frequency or double switching frequency when I calculate filter values?
Why do you wonder?
From post#5:
Only after the rectifier you get somehow doubled frequency at L1
From post#4:
10 kHz transistor switching results in 10 kHz transformer and 20 kHz output frequency.
What is unclear?


Klaus

added:
but my question is not about these things
as a raw estimation:
* with half frequency you get about double the ripple voltage = 200%
* with a change in duty cycle from 90% to 60% you get a change in ripplevoltage of maybe 500% ...600% (1.5 times peak current x 4 times more OFF time).

--> In my eyes it´s more important to care about duty cycle than on frequency.

I recommend to use a simulation tool to test/verify your design.
 

d123

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Hi,

sprabw0c.pdf, 'Voltage Fed Full Bridge DC-DC and DC-AC Converter for High-Frequency Inverter Using C2000' might have interesting general overview information, maybe too basic for you.

If you're asking about the LC filter and how to calculate the component values, look for LC resonant frequency and SMPS LC filter design app notes and tutorials. Good luck.
 

c_mitra

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rectification doubles the frequency, will output filter be designed according to doubled frequency caused by rectifier
Yes, the ripple frequency will be doubled and the filter must be designed at double the frequency of the signal coming before the rectification stage.

The last output filter does not know or care about how the signal was produced; it can only see a DC + a ripple AC. The frequency of the ripple is twice that of the original AC before rectification.

When I drive the switches at 10kHz,
The meaning is not clear.

If you drive each pair with a 10kHz train of pulses (alternately to each pair) the effective frequency gets reduced by 2.

If you drive each input with two 10 kHz signal (out of phase) then the effective frequency will be same as the input (10kHz).

What exactly is the confusion?
 

Easy peasy

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The answer is 10kHz on switches = 10kHz on all mosfets and Tx primary - according to your firing control mentioned above
 

TheDannia

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Yes, the ripple frequency will be doubled and the filter must be designed at double the frequency of the signal coming before the rectification stage.

The last output filter does not know or care about how the signal was produced; it can only see a DC + a ripple AC. The frequency of the ripple is twice that of the original AC before rectification.



The meaning is not clear.

If you drive each pair with a 10kHz train of pulses (alternately to each pair) the effective frequency gets reduced by 2.

If you drive each input with two 10 kHz signal (out of phase) then the effective frequency will be same as the input (10kHz).

What exactly is the confusion?
driver is attached, each output is 10 kHz and complementary. I'm using same pwms for other half bridge with changing the positions.driver2.PNG
 

c_mitra

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driver is attached, each output is 10 kHz and complementary.
Based on the sketch you have given, I conclude that the operating frequency of the transformer will be the same as the driving frequency. The transformer shown is working at 10 kHz.
 

treez

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Filter is easy for full bridge because generally the output inductor means your output current into Cout is continuous.

You can even cheat if you want and do a bit of emperical swapping in the free LTspice simulator.

I attach a full bridge folder, full of full bridge ltspice sims for you to look at...if you give me your full bridge spec, i will do you a full bridge sim in the free LTspice now and send it to you here
 

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