Dc-Dc COnverter

Discussion in 'The Projects Forum' started by mermerzac, May 8, 2012.

Feb 9, 2012
30
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Hi all!! I'm doing isolated dc-dc converter now. Basically it will be a dc-ac inverter at the first part thn connected to a step down transformer for isolation purpose. The output from the transformer secondary circuit will be connected to a ac-dc rectifier and LC filter. Finally, I will get a dc output.

So how am I gonna to calculate the efficiency of this converter?

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2. strantor AAC Fanatic!

Oct 3, 2010
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That will depend on lots of factors, switching frequency, parasitic properties of the components you choose, how well you wind the transformer, etc.

I would say first step is to simulate it as accurately as possible. Really though you might just want to build the thing and see.

Feb 9, 2012
30
1
Oh well i've built the converter on the board. Just wanna understand where should i measure for Pout for calculating efficiency (as I know efficiency= Pout/ Pin) ? I'm confused whether I should measure the Pout on transformer primary or measure Pout after the filter??

4. strantor AAC Fanatic!

Oct 3, 2010
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I would measure it after the filter, or wherever your load connects to the circuit. treat it as a black box; how much power goes into the black box, and how much comes out?

Feb 9, 2012
30
1
However, due to my circuit has a step down transformer in between the inverter and rectifier, which means the turn ratio of secondary to primary will be smaller than 1. So I'll obtain output voltage which is stepped down.
The output value obtained after filter will be reduced according to the turn ratio i set to the step down transformer.

If I calculate the efficiency based on the Pout after the filter and Pin at the starting point of inverter, the efficiency seems to be very low.

So i might think my way of measuring is not so correct maybe?

6. strantor AAC Fanatic!

Oct 3, 2010
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Ok, imagine you have a perfect DC/DC converter. Let's say you're stepping down 100V to 10V. Your transformer turns ratio is 10:1, so your voltage will go down by a factor of 10 and your amp capacity will go up by a factor of 10 (remember perfect world). You apply power to the input, with no load on the output. You measure the current draw; you get 0amps. Now, you connect a 1ohm resistor across the output. You check the current draw at the output, 10A. You check the current draw at the input, 1A.
so,
input power = 100V * 1A = 100W
output power = 10V * 10A = 100W
Efficiency = Pout/Pin = 100%.

But your converter is not perfect. So let's sat you put that 1ohm resistor across the output; you measure 10A @ 10V, so still 100W. Now you check your amps at the input, and you read 1.5A. 100V * 1.5A = 150W. efficiency = Pout/Pin = 100w/150W = 66% efficiency.

So what is the efficiency that you calculate? if it is severely low, can I assume that your transformer is a standard hex head bolt with a bunch of insulated 12AWG building wire wrapped around it?

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