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I'm making my first circuit for a line following robot. I already made the prototype, soldered it together, and now I'm at the assembly stage.

Here is the circuit diagram I'm using: a line following circuit diagram

What should happen is that the LEDs on one side of the circuit (and the corresponding motor) switch on depending on which side of the circuit is receiving more light. The input from the photoresistors is compared by an LM393 comparator chip which switches on the appropriate side of the circuit. The two sides of the comparator chip have their inputs reversed so unless the light level is completely balanced both sets of LEDs should never be on at the same time.

During assembly one of the motors came disconnected so I had to resolder it (my initial soldering was not very good).

Afterwards, my dad connected the battery to test the circuit, but he put it in backwards. As soon as I noticed this I switched the power off.

Prior to the motor breaking, the circuit seemed to be working fine. However, now both sets of LEDs light up and both motors switch on. This is the case even when I completely cover one set of photoresistors. I tested the photoresistors and their resistance does change when I cover them.

I think the problem must be with the IC but I would have assumed that if reversing the battery had broken it then it would just have stopped conducting. Is it possible that it would just give the wrong output instead? I already checked that the IC is correctly wired, and in any case, it was working before the motor disconnected and my dad put the battery in backwards.

Has my IC been fried by the reversed battery or is there another possible problem here?

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    \$\begingroup\$ I like calling an op amp (or is it a comparator?) BRAINS :) \$\endgroup\$ – Gregory Kornblum Aug 18 '18 at 16:20
  • \$\begingroup\$ "I tested the photoresistors and their resistance does change when I cover them." That is an astonishing statement. Photoresistors are not polarized, so I would not expect them to be damaged by reverse polarity. Did you measure their resistance out of circuit, with no power applied? \$\endgroup\$ – vicatcu Aug 18 '18 at 18:00
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Has my IC been fried by the reversed battery

Most likely, yes, because as you said:

it was working before the motor disconnected and my dad put the battery in backwards

In general, ICs are very sensitive to reverse polarity of their power connections. Replace the LM393 and re-test.

If you didn't already use an IC socket to hold the IC, then remove the existing IC, solder an IC socket in its place, then insert the replacement IC (the correct way round) in that socket.


If there is an ongoing possibility of someone reversing the battery connections (depending on your battery type and how much force is applied to the battery connector!) then you could consider adding a suitable diode in series with the power input. Then, if the battery is connected again with reverse polarity, no significant current would flow and no components would be damaged.

Using a Schottky diode would have a slightly lower forward voltage drop than a typical silicon rectifier diode. Your choice depends on what is available to you and what voltage drop across the diode your design can accept (I expect that this is not critical on your design).

The chosen diode also needs to be rated to cope with whatever maximum current (IF(AV)) your design can draw from the battery (plus some margin) e.g. with the motors stalled. This means that, for example, a 1N4148 won't be suitable as its IF(AV) rating can be around only 150mA (depending on the diode manufacturer) and your motors might draw more than that, especially when stalled (you should check).

The chosen diode must also have a suitable voltage rating (VDC). However for a 9V power supply, it would be difficult to find a power diode rated less than that, so this is unlikely to be a limiting parameter.

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    \$\begingroup\$ +1 for diagnosis and fix. A diode in the type 1N4001 to 1N4007 series would be the obvious choice. There are many others and this is only needed for reverse polarity protection. \$\endgroup\$ – KalleMP Aug 18 '18 at 18:56

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