I'm feeling like an idiot here. I've found this great little IC that will do the job I want but I can't seem to figure out what the max voltage is that can be supplied for sensing.

The IC runs on 5Vdc and the version I want to use is the 30A size.

I'm trying to sense when a 100W RF amplifier is working. The amplifier runs on 13.8Vdc and has a standby current of about 0.5A but when it's working it jumps to 21-23A.

Being fairly new to the whole thing I was hoping someone could point out where on the data sheet it says I'm good-to-go or DON'T DO IT MAN you're going to be making magic blue smoke.

Looking at the datasheet makes me think there is no friggin way a tiny IC can handle this much juice running through it without it melting. Am I missing something? It looks like they say it needs to be wired in series with the amplifier. ACS714-DataSheet

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    \$\begingroup\$ I've worked with the Allegro ACS71[2-4] ICs. Yes, all the current does go through the IC. \$\endgroup\$ Aug 7, 2013 at 23:18

2 Answers 2


Yes, you have it correct, the device should be wired in series with the amplifier. The pins 1,2 and 3,4 are isolated from the hall sensor circuitry. They are simply a conduction path with a nominal resistance of 1.2mΩ, so even at 30A the maximum dissipation is only 30^2 * 0.0012 = 1.08W

Hall sensing is an isolated technique, this IC just has the current p[ath fixed rather than needing to position it over a current carrying trace as with other hall ICs

This page on Allegros site explains:

Hall IC


66 mV/amp x 30 amp < 2 Volts on the output. Sound to me like you're good to go.

1.2mOhm resistance on the input gives \$ P = I^2 R \$ = a very small number for 30 amps and 1.2mOhm


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