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Here is the schematic, except this uses the L293d enter image description here

I wanted to better understand how does the current split (if it does) between two of these IC chips wired as diagrammed with 2 electromagnets on each IC (total of 4)? If my external power supply is 3.0V ~0.85A.

Also, how can I adequately measure current between an electromagnet that's hooked up to an h-bridge driver? I notice the current at each of the motor pins (3, 6, 14, 10) differ when I do.

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I wanted to better understand how does the current split (if it does) between two of these IC chips wired as diagrammed with 2 electromagnets on each IC (total of 4)?

Providing the power supply can handle the current and both circuits are correctly wired (parallel) to the power supply, the two circuits can be though of as independent i.e. whatever current is taken by one chip does not affect the current (or operation) of the other chip and it's solenoids/motor loads.

If my external power supply is 3.0V ~0.85A.

Now you are in trouble because the logic supply voltage is 4.5V minimum. However, if you are talking about the power supply to the H bridges then you are barely going to get anything out of these devices at all. I refer you to this question and answer that explains why this device (and the L293D) are really crappy on low voltage H-bridge supplies. H-bridge supply minimum voltage is also specified as 4.5V for recommended operation: -

enter image description here

Also, how can I adequately measure current between an electromagnet that's hooked up to an h-bridge driver?

You use an ammeter in series with the solenoid.

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  • \$\begingroup\$ I'd like to have a 0.5A (minimum) current for each electromagnet and I see that the pins (3 and 6, for instance) may have two different current values. Another question that may help me understand is if they're wired in parallel and one chip doesn't affect another chip's current, then if the power supply is 0.8A, what will the current be for each chip? Then, the current that goes through the electromagnet, where is the source of that based on the IC? How is it being split? \$\endgroup\$ – user2899444 Dec 31 '15 at 17:37
  • \$\begingroup\$ Wait, in terms of measuring current you meant something like one probe to pin 3, another to solenoid leg... I don't know why that didn't register to me. Anyway, correct me if I am wrong, but the sn754410 chip's max current that it can handle is 1A? \$\endgroup\$ – user2899444 Dec 31 '15 at 18:10
  • \$\begingroup\$ If your total load takes 0.5A x 4 electromagnets then the power supply you envisage using will be no good. \$\endgroup\$ – Andy aka Dec 31 '15 at 18:24
  • \$\begingroup\$ My total load will be a total of 2A, so I need a power supply that can provide 2A, I think I understand that in theory. What I don't understand is isn't the chip's max current value 1A? That power supply would burn the chip out, wouldn't it? That happened to the L293d that was originally being used, I think. Also, I still want to know what I had asked in the first comment/question response regarding current. I don't understand what's going on with the current and the current of the electromagnets regarding the sn754410. \$\endgroup\$ – user2899444 Dec 31 '15 at 19:39
  • \$\begingroup\$ Just because a chip can supply 1A, it only will do if the voltage it can put across the load AND the load resistance dictate that 1A will flow. As it happens the 754410 is pretty s*it at low voltage driving so I would urge you to understand and recognize this in the link in my answer. You might just decide that the 754410 is not really good enough on low voltage supplies i.e. use mosfet bridges ala texas instruments. \$\endgroup\$ – Andy aka Dec 31 '15 at 20:08

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