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Half-bridge DC/DC converter?

Thread starter #1
Hello,
I have designed a half-bridge DC/DC converter, but I am not not sure if this works properly.
I asking cause the transformer is designed to give 380 V output (where the ratio is 7:14), but I only obtain around 277 V in the output of the converter.
 

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alec_t

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#2
Try L1=37u.
 

ronsimpson

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#4
(where the ratio is 7:14)
You have the inductance at 7:14. Which is the same as 1:2. (inductance)
turns=inductance squared
The turn ratio is 1:1.4 not 1:2 not 7:14
-------
If you really want a 1:2 turn ration then the inductance must be 1:4. (70uH : 280uH)
 

alec_t

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#5
Thread starter #7
Hmm, is it strange that the losses is higher for the half-bridge converter compared to the full-bridge?
I have read that in articles that the half-bridge should have higher efficiency and lower power losses in contrast to full-bridge.

I guess it has something to do with selection with the MOSFETs, i had IRFH5302 but this gave for example 1.82 W for M1.
and with R6020PNJ i got 8.48 W at M1. It should be a mosfet that is higher than 400 V I imagine?
 

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alec_t

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#8
I guess it has something to do with selection with the MOSFETs
Drain-source resistance plays a big part. You can test your theory by using the same FETS in both the half-bridge and full-bridge simulations.
 

ChrisP58

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#9
Hmm, is it strange that the losses is higher for the half-bridge converter compared to the full-bridge?
I have read that in articles that the half-bridge should have higher efficiency and lower power losses in contrast to full-bridge.
I'd be interested in reading those articles.

I've never really compared the relative efficiency of the two. Generally, you choose the half bridge for lower power because it a bit cheaper to build. But that's about the only advantage.

The disadvantages include:
  • You need two capacitors in series. And, since the capacitance of two equal caps in series is half what each would be alone, you'll need larger individual capacitors if you need a certain value of bulk filter capacitance.
  • The transformer primary winding in a half bridge gets only half the voltage that a full bridge winding would get. That means, to pass the same power, it needs to carry twice the current. So twice the copper cross sectional area. This leads to higher power loss in the transformer, both in IR losses, and magnetic losses due to eddy currents and proximity effects.
  • The two mosfets also need to switch twice the current. But they still need to be the same >400 volt part.
So, all else being equal, a well designed full bridge converter should be more efficient than a well designed half bridge converter.
 
Thread starter #10
I'd be interested in reading those articles.
Here you go: http://shodhganga.inflibnet.ac.in/bitstream/10603/134450/14/14_chapter 5.pdf

Otherwise that source that I found is wrong.. But where did you find your source ChrisP58? Could you share it please? :) I need it as source for my report...
In that case my simulation seems to be correct.

You need two capacitors in series. And, since the capacitance of two equal caps in series is half what each would be alone, you'll need larger individual capacitors if you need a certain value of bulk filter capacitance.
At the moment I have 47uF for each cap. But should it be larger than that you think?

--edit--
which one is the smallest one in terms of size? Half-bridge or full-bridge?
 
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ronsimpson

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#13
Please, do anyone knows, regarding post #11?

As you can see the full bridge puts full Vcc on the primary. The half bridge puts Vcc/2 on the primary. So the turn ratio must be different.
(Vcc at 1A)=(Vcc/2 at 2A) Same power.
From this can you answer your question?
 
Thread starter #14
As you can see the full bridge puts full Vcc on the primary. The half bridge puts Vcc/2 on the primary. So the turn ratio must be different.
(Vcc at 1A)=(Vcc/2 at 2A) Same power.
From this can you answer your question?
For my half-bridge, it have the ratio as we discussed before 7:14 (70u and 280u), where I obtain 380 V at the output.
For my full-bridge, it have the ratio 7:7 (70u and 70u), where I obtained 380 V at the output.

So if understand you correctly, the load should be the same for both cases cause I have the same output voltage.
 

ronsimpson

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#15
Maybe I don't know which point you are talking about.
The load resistance at the output is the resistor you are using for the load. (same)
I thought you might be talking about the load as seen by the MOSFETs. (not the same)
 
Thread starter #16
The load resistance at the output is the resistor you are using for the load. (same)
I thought you might be talking about the load as seen by the MOSFETs. (not the same)
I actually meant the load resistance at the output :) so then load is the same for the half and full if I understand this correctly.
 

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