Good day all.

I think I might be just about to ask the impossible.

I have a DC motor controller, and I think something has gone wrong with the current sensing side of the circuit.

I don't know what the current sensing ic is, as it looks like it has not had any details etched into it, or it has been sanded off.

Most of the parts on the board are all smd.

I have tried to follow the current sensing traces on the pcb, and have come up with this schematic.

As said before I think there is something wrong, as pin7 without any load is showing 4.98-4.99volts.

So my questions are, does anyone have any idea what chip this might be, and what sort of voltage should I be seeing at the output with and without any load.

Thanks for your time.

Best Regards.

 

I'm not seeing all the components I'd expect, but I wonder if this is just a dual op amp. They're only using one of the amps, so the other one has its inputs grounded (pins 2 & 3) and its output unconnected. That makes sense, but then the + input of the active amp seems to have a fixed input at about 1% of the power supply, and there's no feedback resistor, though there is an input resistor with value 2K. That doesn't make sense! But it would be reasonable to have a simple gain circuit giving an output proportional to the voltage across the shunt. Are you sure the diagram is right?

Or maybe it's something else completely.

 

I guess John P could be right.

Could this just be an overload protection? The Opamp is then used as a comparator (or it is a comparator). No current, output 7 High, if the current is high enough to give a voltage drop over the shunt higher than the voltage at pin5 then pin 7 goes Low.

 

Thanks for your replies.

I think I have the schematic, but my eyes are not as good as they used to be.

Here is the PCB it's self.

I have updated the schematic a little bit more.

Any more thoughts?

Thanks for your time.

Best Regards.

 

I'm afraid that the photo isn't helpful at all, and the new diagram only proves that it's not possible to draw the circuit by looking at the unit. What you're showing there is that there are two MOSFETs that can never be turned on, because their gates are grounded. It just has to be wrong.

 

R3 and R4 off 5 volts gives about 45 mV at pin 5 of IC1. If IC1 is truly a dual op amp (with one side disabled) then the output on pin 7 should be normally high until the current gets "too large." Here "high" is close to 5V and "too large" means over about .45mV over the resistance of shunt S4. IC1-7 should go low for an over current which you can simulate by bringing IC1 pin 6 high.

If your schematic is correct you can just tap a wire with +5V on it to IC1-6 and see IC1-7 go near ground. If you have a 1K or so resistor on hand it would not hurt to put that in series with the 5V when and if you try that.

The junction of R14, R24 and R27 probably goes back to the PIC (or an intermediate driver) so the controller can turn on the load.

 

ErnieM: You was quite right, when I applied 5v via a 1k resistor to pin6 on the IC, pin7 of the IC did indeed switch from high (4.98v) to low (0.02v).

 

MP: so you have one part of the circuit working and can be ruled out: the OC sense seems to be working, and the unit is not in OC protection (or the IC1 output would be low).

How do the gates of the FETs look? Are they really close to zero, a little above, or sometimes seeming to pulse? All but the first mean they may be pulsed on at times.

Where does the motor control input come from? A pot or what?