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I recently acquired a Lantronix UDS1100 telnet-> serial converter. It came without a wall wart / power supply.

I have an extra power supply from an old modem that I think might work, but I want to be certain.

The Lantronix's stated specs are 9-30VDC, 9-24VAC, 1.5 watt max. It doesn't mention current.

My potential replacement lists output of +12V, 2.2A.

How does one determine if this is okay or not?


marked as duplicate by JYelton, Matt Young, Chetan Bhargava, Nick Alexeev Apr 22 '14 at 16:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ @JYelton thanks for your comment, I was searching for the link to include that in my answer. \$\endgroup\$ – Vladimir Cravero Apr 22 '14 at 15:38

You can easily calculate it starting from the power. Since \$P=V\cdot I\$, then \$I=\frac{P}{V}\$. In your case \$P_{max}=5W\$, while \$V=12V\$, then \$I_{max}=\frac{P_{max}}{V}=\frac{1.5W}{12V} = 0.125A = 125mA\$. Since your PSU can provide up to 2.2A you are safe. Keep in mind that your replacement is way too big, if you have space or possibly heat constraints you might want to buy a smaller PSU.

If you want to better understand how to choose a power supply have a read at this question.


1.5 Watt Max gives you current.

P = IV

1.5 W = I * 12V

I = 1.5 / 12 A = 125 mA

As long as your power supply is rated equal to or more than the current needed, you will be safe. However, that doesn't mean it will run.

Your 2.2A power supply will give enough current, IF the load is pulling enough for the power supply to run. Some supplies have a minimum current required to operate. You might have to hang a power resistor in parallel with the output to get it to run. You are generally better to have a supply with a closer match in power, for efficiency of the circuit.

You shouldn't damage anything by trying out your supply. But try a 12V car light in parallel if it doesn't work at first. This will load up the supply and tell you if it is a minimum load issue.

  • 1
    \$\begingroup\$ So this power supply would supply more amperage than the device requires, right? If I understood correctly from another question on SE, this situation would be okay, because the danger is when amperage is too low not too high. \$\endgroup\$ – Kirkman14 Apr 22 '14 at 14:47

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