# Current carrying capabilities of surface mount ICs

While putting together a PCB project, I have come across the ACS724 current sensor, for measuring AC/DC current in high ranges up to 50 amps. This chip uses hall effect for measurement and is a 'pass-through' type of sensor: No shunt resistor or any other external circuit is used for sensing.

Now here is where it gets amazing: This chip comes in 8-pin SOIC package! For those not familiar, this is very very small. So I started looking for information about the SOIC package standards, but couldn't find anything relating to electrical capabilities, only dimensions. Using the chip involves thermal management but still, I have a hard time believing that a single pin on an SOIC package can handle 25 amps?!?

• Does anyone know where to find electrical data on surface mount packages?
• Does anyone have advice on using such a chip in a circuit, without causing some havoc?
• I agree. But just a thought: the pin is just metal, heatsunk to the PCB, and not connected to anything else inside the IC unlike other ICs, so maybe it's okay? Apr 7 '20 at 22:35
• Allegro is a respectable maker and I do trus them but it just seems 'outworldly' to put 25a in such a small pin Apr 7 '20 at 22:39
• I don't know how they do it but if I had to do it, uninformed as I am, I would keep the pins the same (no choice after all) and require them to be well-heat sunk to the PCB but have the conductor considerably wider inside the IC. You get even crazier current ratings for MOSFETs (though I never expect to be able to run at those in practice, unlike the current sensor). Apr 7 '20 at 22:39
• I've played around in industrial VFD's and the switching semi's weigh up to pounds with their ceramic packages and are rated at 50 amps with 8mm screws for terminals; the whole allegro sensor fits on top of that screw with space to spare... That just clashes with my experience. Apr 7 '20 at 22:51
• There's a reason I use the CQ-3202 and not the ACS724. For the record, it seems to heat up to like 60-80C at 20A and it has massive terminals. Though...that heat could be coming from accidentally undersizing the trace leading to it. It's hard for me to be sure. Apr 7 '20 at 22:55

Assume the pins conduct about as well as AWG 28. That's a resistance of 0.106 ohms per meter.

The pins of an SOIC8 are approximately 2mm long.

You split the current between two pins, so you halve the resistance.

That'd be something like $$\ \frac{0.002m \times 0.213 ohms/m}{2}= 0.000213 ohms\$$

At 50A, that's a voltage drop of 0.01065V.

Total power dissipated in one pair of pins is then about 0.53 watts.

The 50A sounds scary, but the total power lost through the pins is fairly small. You still need to get rid of the heat, though.

That's why datasheet shows that the evaluation board has those great honking copper areas attached to those pins:

The values above aren't necessarily exact. They are intended to show that even though the current seems outrageously high, the total loss as heat will be much lower than a first glance suggests.

• The pin cross section only works out to be 0.04mm^2. 24 AWG is 0.2mm^2 Apr 7 '20 at 22:59
• The pin cross section is about 0.4mmX0.4mm=0.16mm^2. That's about AWG 25. 0.106 ohms per meter.
– JRE
Apr 7 '20 at 23:05
• My bad. It's 0.39 x 0.19mm^2 which is 0.0741mm^2. Not sure where you are getting the second 0.4 from. Two of them gets you in the ballpark of 24AWG I guess. Apr 7 '20 at 23:07
• My brains are mush. Too late at night here for this. I'll try it again.
– JRE
Apr 7 '20 at 23:10
• Hmmpf. I mixed Ll and T.
– JRE
Apr 7 '20 at 23:11