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[SOLVED] DC/DC converter input current

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sebas

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Hi,

Newbie's question, I was looking at the following DC/DC converter's datasheet: https://www.murata-ps.com/data/power/uwr15wa-series.pdf and I see that there is a parameter called Iin. I'm guessing that this is the maximum input current for the device, right? Why is this dependent on the load? It has a Iin parameter for full-load and one for no-load, why?

Thanks alot.
 

It is because of basic Ohm's Law and Power Equation P=V*I. If the device is 80-90% efficient then 'power in'*0.8='power out'. There is a minimum value because even under no-load conditions the device still consumes some power.
 
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    sebas

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templemark said:
It is because of basic Ohm's Law and Power Equation P=V*I. If the device is 80-90% efficient then 'power in'*0.8='power out'. There is a minimum value because even under no-load conditions the device still consumes some power.
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Ok, I understand the efficiency, I know what 80-90% efficiency means but what does this have to to with input current? What is the input current after all? I guessed it is the maximum current that being fed to the converter doesn't damage the device, am I wrong?
 

Say your 1.2V DC Regulator is running at Full Load @ 6 Amps.

So the Device outputs 1.2V * 6 = 7.2 Watts.

Your efficiency is say 0.8. So you need '7.2 Watts / 0.8' = 9 Watts to deliver 7.2 Watts. Now your nominal input voltage is 12V. So 9Watts/12=0.75 which would be your max current in.

Make sense?
 
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There will be a real IIN which depends on the efficiency at
use conditions, and the "constant power" relation. There
is also a maximum rated IIN beyond which you will damage
something (thermal or current stress in the high side switch
most likely). You'd have to determine which, by context and
how its conditions are spec'd.
 
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