In this ULN2803 based relay driver schematic, why won't the back-EMF from the relays hurt the relay status LEDs?
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4\$\begingroup\$ No - the ULN2803 has internal 'flyback' diodes. \$\endgroup\$– brhansMar 14, 2017 at 20:13
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\$\begingroup\$ Even without that, it's likely that the resistor would still provide sufficient protection for the LEDs. They have some internal capacitance. \$\endgroup\$– David SchwartzMar 14, 2017 at 22:56
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2 Answers
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At face value it certainly appears that the LED's would be damaged by back EMF as generated by the relays on turn-off. However if you take a look at the datasheet for the ULN2803 you see it is designed specifically for driving inductive loads such as relays, and has integrated clamp diodes between each output and common. The common point is connected back to Vcc of the circuit, thus clamping any relay back EMF.
The rate of change of voltage (as the magnetic field collapses and a back EMF voltage is generated) will not be fast enough to overcome the reverse recovery time of the clamp diodes and other latency due to track impedance etc. If you had to put a scope directly across the LED, I suspect you would see a relatively fast rise in reverse voltage to -0.6V, with only a small overshoot to perhaps a volt or 2. Nowhere near the 7V breakdown voltage of the LED. The only caveat here would be for a very badly designed board (which is not the case here) where there are very long traces back from the relay to the clamp diodes in the IC. In that case the inductance of these tracks would impede the reverse recovery diodes ability to sufficiently clamp the voltage in time.
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\$\begingroup\$ Yes, but won't the back-EMF voltage be applied across the 1k resistor and LED too thereby causing the LED to be reverse biased? (I am probably not fully understanding how clamp diodes work in a parallel circuit.) \$\endgroup\$– VANMar 14, 2017 at 20:45
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4\$\begingroup\$ The LED will see a reverse voltage : but that voltage won't exceed the forward voltage of the protection diode (probably 0.7V) while the LED can tolerate about 5V reverse. \$\endgroup\$– user16324Mar 14, 2017 at 20:55
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\$\begingroup\$ Can I understand the same thing by saying the clamping diodes would be forward biased way sooner than the potential difference could reach anywhere near the LED's breakdown voltage and since the LEDs are in parallel to the clamping diodes, the LEDs won't see any more reverse voltage than the forward voltage of the clamping diodes. \$\endgroup\$– VANMar 14, 2017 at 22:01
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1\$\begingroup\$ Getting really technical now, but the short answer is yes, the clamp diodes will always clamp far sooner than any voltage approaches the breakdown voltage of the LEDs (LED's are usually good for about -7V reverse or blocking voltage). \$\endgroup\$ Mar 15, 2017 at 3:24
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\$\begingroup\$ Ha, this is stackexchange. Getting technical is welcome. Thank you for the clearer answer. \$\endgroup\$– VANMar 15, 2017 at 7:56
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In the schematic you provided you can see that pin 10 [COM pin] is connected to VCC. If you have a look at the internal schematic, you will see that every pin has it's own freewheeling diodes (flyback diodes) to protect the transistors from the back-EMF
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\$\begingroup\$ I can understand how the clamp diodes protect the Darlington pairs and the switching circuit before that, but my question is about the status indicator LEDs. \$\endgroup\$– VANMar 14, 2017 at 20:52
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4\$\begingroup\$ The leds are in parallel to the transistors so they bennefit from the same protection diodes. the back emf can't be larger than the forward voltage of the protection diode (0.7v) and a reverse voltage of 0.7V won't damage the leds. I hope this helps \$\endgroup\$– BruceMar 14, 2017 at 21:26
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\$\begingroup\$ Yes, I went and looked at how dissimilar diodes behave in parallel and now it's clear how the clamping diode will protect everything else in parallel to it. \$\endgroup\$– VANMar 15, 2017 at 7:59