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electronics noob here. Just looking for some advice on replacing a part for a ebike motor controller. I think I may have shorted one of the schottky diodes (part) by running current through the circuit without the piece of metal the mosfet & schottky barrier rectifiers contact.

Anyway, I tested a diode (sorry if this terminology is incorrect) by touching the red lede of multimeter to the middle pin and black lede to outer pin. The voltage steadily climbed to 1.9 and then read O.L. I flipped the ledes and it read .2 V, I guess some voltage leakage. I also tested the schottky diodes with power running through the circuit and did not get voltage drop on my multimeter.

Would replacing these schottky diodes fix the problem? My bike motor doesn't run anymore. First, it wasn't too bad, the motor would turn off in the middle of the ride. But finally it just quit. I attached a schematic for the motor controller I bought off ebay. I am hoping I didnt damage the motor through my folly. Your help is much appreciated!

enter image description here

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  • \$\begingroup\$ Lede and lead are two different words. Anyway, it's not clear what you're asking. What do you mean by "middle pin" on a diode? Diodes should only have one pin, no? \$\endgroup\$ – Hearth Apr 6 at 22:56
  • \$\begingroup\$ oh i think you're right. the part I'm looking at is a schottky barrier rectifier. I think T1 in the schematic. It''s actually made of two schottky diodes it appears. I was testing the output (middle pin) to the input of either diode. \$\endgroup\$ – st4rgut Apr 6 at 23:03
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    \$\begingroup\$ A "schottky barrier rectifier" is just another way of saying a schottky diode. Those things in the schematic are just regular diodes, they just have two of them in the same package, with one pin common between them. Now that you've clarified what you meant, you had your leads backwards at first; to measure the forward voltage you need to put the red lead on the anode--either one of the outer two pins--and the black one on the cathode. \$\endgroup\$ – Hearth Apr 6 at 23:06
  • \$\begingroup\$ Thanks, the forward voltage (.2 v) seems acceptable. When I flipped the leads, does the fact that I got a reading on the multimeter before it displayed O.L. suggest it is faulty? The multimeter displayed up to 1.9 v before O.L. \$\endgroup\$ – st4rgut Apr 7 at 0:10
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    \$\begingroup\$ On the contrary, that's exactly what diodes are supposed to do: prevent any current flowing in the wrong direction. However, 0.2V forward voltage sounds very low, even for a schottky diode. \$\endgroup\$ – Hearth Apr 7 at 0:13
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You haven't "shorted" the diode since the voltage is not zero on your multimeter, but you may have cooked it. The "piece of metal" might be a heatsink - without it the diode will overheat.

Your multimeter measurements sound suspicious, but there is a lot of variability in measurement method and multimeters. What you should get is:

  • Pin 1 - 2: 0.3V
  • Pin 2 - 1: O.L. within a second
  • Pin 1 - 3: O.L. within a second
  • Pin 3 - 1: O.L. within a second
  • Pin 2 - 3: O.L. within a second
  • Pin 3 - 2: 0.3V

where pins 1, 2 and 3 could be any of them - you'll have to test all six combinations, work it out from the way it is placed in the circuit, or try to identify the part.

The tests you did with power running are unlikely to give you anything useful. In fact, they're likely to cause damage - the diode mode of a multimeter actually injects voltage so will change the behaviour of the circuit under test. Not only could this activate the diode at the wrong time, it can affect other parts of the circuit that are connected too. If you used voltage mode, then that's different, but you need to be explicit.

The slow rise to O.L. and the low 0.2V the other way is suspicious. It's probably cooked, and if so it may result in the symptoms you describe. It's probably worth replacing it, but given the important role these devices play there may have been collateral damage, so I wouldn't be too confident about that being the fix.

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