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DC DC buck converter

Rajinder1268

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Hi all,
I am using AP63205 DCDC buck converter to generate 5V from a 12V input.
I measured 12V input and the output is 4.96V at the output under no load.
I have a 5R load ( wire wound resistor 5R 50W).
When I connect this why am I only seeing 450mA drawn from the output (I am measuring this with a meter in series). The output drops to 4.90V under load.

Should I not get I = V/R = 4.9V / 5R = 0.98A.

I have measured the resistance of the 5R and this is 5.2R. So with even this slight discrepancy I should be getting close to 900mA not half that value?

I have attached the data sheet. My circuit is as on page 9 of the datasheet (typical application at 5V).

Note I have tried using a different meter to measure the current but it is still the same.

Any suggestions?
--- Updated ---

Hi all,
Please disregard this post. I have sorted it out. Sorry.
 

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  • AP63200-AP63201-AP63203-AP63205.pdf
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Last edited:
I have a 5R load ( wire wound resistor 5R 50W).
When I connect this why am I only seeing 450mA drawn from the output (I am measuring this with a meter in series). The output drops to 4.90V under load.

I guess you're using your meter on the 1 or 10 or 20A current range?
It has its own resistance built into the measuring circuit. Probably in the vicinity of 1 ohm. It's included in the loop and causes an automatic drop in measured Amperes.

An alternate method of measuring current is to measure voltage across a known resistor whose value is a fraction of an ohm. You can assemble it from a few 1 ohm resistors in parallel.

Your load dissipates 4.5W (5V x .9A). That's enough to make it hot after several seconds.

Did you attach various test loads? Did you get expected voltage measurements?
 

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