1
\$\begingroup\$

I want to use JWK2524S05. Its input voltage range is 9V-36V and its nominal voltage range is 24V. I have an input voltage source of 12V.

Will it work if I supply 12V input? Will the converter will work as efficiently as when supplied with 24V input?

\$\endgroup\$
2
  • 2
    \$\begingroup\$ I’d expect some variance of the efficiency over the input voltage range, but only a few % at most. You can confirm this by measuring the input and output voltage/current and do the math. Y Watts going in, X Watts going out. The difference is the loss. \$\endgroup\$
    – Kartman
    Aug 5, 2021 at 10:49
  • \$\begingroup\$ They only list one efficiency for 9V to 36V range for each output. And nominal efficiency for all is typically 90%. \$\endgroup\$ Aug 5, 2021 at 18:20

1 Answer 1

2
\$\begingroup\$

OK, two questions to answer...

Question 1: Will it work? Answer: Yes! The data sheet explicitly allows an input range of 9 to 36V. You will be applying either 12V or, presumably, 24V. You should be good to go.

Question 2: Will it work at least as efficiently at 12V vs 24V? Answer: In general, the answer is yes. In fact it should operate slightly more efficiently at 12V vs 24V.

A higher input voltage yields a lower Duty Cycle, D ~= Vout/Vin. A lower Duty Cycle generally results in lower overall efficiency all else being equal. I would agree with @kartman that the difference in efficiency should be small and in the range of 1%-4%. This efficiency degradation with lower Duty Cycle can be readily seen in the data sheet efficiency graphs for a variety of SMPS devices from TI or Analog devices.

Here is a snippet from the Analog Devices (formerly Linear Technology) LT8461. Notice that the best efficiency is achieved with a lower input voltage.

LT8641 datasheet Efficiency graph for LT8641

I make no attempt to explain why the efficiency is lower for lower Duty Cycle, rather, only to state that it exists. It has to do with the ratio of switching losses to conductive losses in the pass transistor. For a detailed explanation of this effect, consult the internet :-).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.