I'm a materials reliability scientist, not an electrical/electronics engineer, so I could use some help. I am running an experiment where I pass between 0.5 and 5A through 50 DOT's and measure a voltage drop across the devices w.r.t. time, up to 500 hours.

At some time, a device will fail, causing an open circuit. If this happens during a weekend, I am not near the equipment to short around the device that failed, so that the other DOT's resume current testing. There are two issues with this:

  1. Lost test time when the whole circuit is open, and
  2. Possible partial recovery of the remaining devices. The voltage across the failed device is 42V, but I can adjust the current source to a lower maximum voltage.

I think adding a small circuit that will short out the failed device, so that the remaining devices continue to be exposed to the test current, might be possible. Including a simple mechanical switch and an indicator LED would be nice. Because I would need 50 of these circuits, they would need to be small and inexpensive.

I have looked at latching relay circuits on this forum, and they are all connected to ground. I don't think this is possible with what I want to do.

Here is a sketch of my test:

diagram of test configuration

Any help would be appreciated.

Thanks for the helpful replies. Here is another sketch, is something like this do-able?

relay resistor indicator

  • \$\begingroup\$ I think you feel that if you detect the maximum voltage across a device, then it is bad and should be shorted. But if two devices fail at the same time, you won't be able to make that determination. You may want to consider adding a large-valued parallel resistor to each device under test (series string, in effect.) This way there is always a galvanic connection and the measurements will be more robust against unusual events like this. \$\endgroup\$ – jonk Jan 27 '20 at 23:01
  • \$\begingroup\$ Thanks for the comment. However I think it is highly doubt that two DOT will fail in less than a few minutes of each other. If one DOT shorts at time a, and the relay shorts the failed DOT, then if a 2nd DOT fails at time a+b, its relay should short that failed DOT also. How big a parallel resistor are you suggesting? \$\endgroup\$ – Karen Jan 27 '20 at 23:22
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    \$\begingroup\$ I like Transistor's thoughts (almost what I was considering, by the way.) But another approach is to use a very fine fuse wire that will melt immediately if the voltage reaches anywhere near 42 V and then release a weight previously held against gravity by that fine wire which closes the relay bypassing the failed device. I often like to think of non-electrical solutions if they aren't difficult to conceive and can be reliable. Such wire is cheaply replaced when you come back to reset the process, again. \$\endgroup\$ – jonk Jan 28 '20 at 7:26
  • \$\begingroup\$ @jonk - I like the fuse-able wire solution you suggest. Thanks! \$\endgroup\$ – Karen Jan 28 '20 at 15:15
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    \$\begingroup\$ Thanks for the kind words. It may have been the "materials reliability scientist" part of what you wrote that sent my mind in that direction. Most of my life I've worked on creating new products at the leading edge, such as the first rewritable CD or creating new sensors for the Space Shuttle, that has put me continually working beside good physicists. I've learned to see the world more as they do from those beneficial relationships. It was more them speaking, than me. Best wishes on this! \$\endgroup\$ – jonk Jan 28 '20 at 15:36


simulate this circuit – Schematic created using CircuitLab

Figure 1. A possible solution.

How it works:

  • With 50 dots on a 42 V supply there is < 1 V across each DOT. This is not enough to light the LEDs in the opto-isolators.
  • When a DOT fails 42 V will be available for the opto-isolator and its series resistor. 3.9 kΩ has been chosen to limit the current to about 10 mA.
  • The opto-isolators' transistors can be wired to the monitoring controller which, when a failure is detected, will switch on the associate relay coil, energising the bypass contact.
  • The circuit should be able to respond in tens of milliseconds so given that you are testing for hundreds of hours it's unlikely that multiple simultaneous failures would occur.
  • \$\begingroup\$ Thank you very much!!!! \$\endgroup\$ – Karen Jan 27 '20 at 23:38
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    \$\begingroup\$ The "3k9" value for the resistor wasn't a typo, it's a common way to write the values and has the advantage that it is clearer on a third generation photocopy of a schematic its meaning is clear whereas in '3.9k' the decimal point may fade into oblivion and be read as 39k. I removed all ambiguity by using 39 k&Omega; which renders as 3.9 kΩ. Note that SI units get a space between the number and the units (in the same way that all numbers do so we don't write '5elephants'). \$\endgroup\$ – Transistor Jan 28 '20 at 16:47
  • \$\begingroup\$ Sorry, I did not know that. As I said I'm not an electrical engineer. Thanks again for your help, and hopefully you will help me again with some other project. \$\endgroup\$ – Karen Jan 28 '20 at 17:19

Your issue would be figuring out WHICH DOT failed. As you have them all in series, none will have any current. A latching relay like you're talking about is certainly possible, but you would need to short out one device at a time until things started working again.

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    \$\begingroup\$ turn on all relays, then turn off one at a time \$\endgroup\$ – jsotola Jan 27 '20 at 22:49
  • \$\begingroup\$ Thanks for your answer. In order to know which DOT failed, an indicator such as a small LED might help. \$\endgroup\$ – Karen Jan 27 '20 at 23:06

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