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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.

user2899444
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1 Answers1

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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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Andy aka
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  • 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? – user2899444 Dec 31 '15 at 17:37
  • 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? – user2899444 Dec 31 '15 at 18:10
  • If your total load takes 0.5A x 4 electromagnets then the power supply you envisage using will be no good. – Andy aka Dec 31 '15 at 18:24
  • 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. – user2899444 Dec 31 '15 at 19:39
  • 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. – Andy aka Dec 31 '15 at 20:08
  • Maybe I misread the data sheet. Then at what current value would the chip start to malfunction? Also: "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?" – user2899444 Dec 31 '15 at 21:07
  • If you have loads that are more than 0.8A then hell will break lose if the supply cannot handle this and I'm not even going to guess what the voltage will dip to or speculate what might happen. The current is sourced from the power rails via transistors attached to the power rails. I think you need to take my advice and choose a better driver chip and a better power supply. – Andy aka Dec 31 '15 at 21:32
  • I've been playing with a variable DC power supply. I was concerned that 1A would burn out my chip since it seemed to happen with the L293D, which is why the values were 3.0V and 0.85A. – user2899444 Dec 31 '15 at 22:04
  • Both chips are really crap at doing what they say they do at low voltage supplies. They are the same silicon but just re packaged. Choose a better chip. – Andy aka Dec 31 '15 at 22:47
  • Okay, I'll review that thread again and ask questions. – user2899444 Dec 31 '15 at 23:22
  • Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/33729/discussion-between-user2899444-and-andy-aka). – user2899444 Jan 02 '16 at 01:42
  • @user2899444 Haven't got time. – Andy aka Jan 02 '16 at 11:49
  • Okay, so then I'll post that question here: To clarify/understand, based on that topic you had linked to regarding the chip not being able to work reliably at low voltages, does this hold true with power supply voltages above 6V? – user2899444 Jan 02 '16 at 12:29
  • Not really. It isn't well suited to any applications that require more than about 100mA load current. – Andy aka Jan 02 '16 at 13:39
  • Hey Andy, really sorry about that. Totally unintentional. – user2899444 Jan 08 '16 at 23:51