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In any circuit where a transistor or FET is driving a DC relay, its common to see a reverse blocking diode added to suppress the inductive spikes, caused by fast current transitions (particularly the "OFF" transition). Over the years Ive seen such diodes placed across the relay coil, or alternately across Transistor. I don't think I've ever seen both in the same circuit, except in the cases of MOSFETs where a diode is often built in. In my own projects I've used both configurations, and both seem to work equally well for protection. I'm aware that putting the diode across the relay coil will slow down the dropout speed, which could be a problem in some designs. But from a pure spike protection perspective, is one diode placement more optimal than the other?

Almost feel foolish asking this, but I've seen both cases so many times, I have to wonder if there is some confusion or disagreement about it.

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Putting a regular diode across the transistor (reverse biased) will do nothing of value in a low-side switch of an inductive load. The intrinsic body diode in a MOSFET is thus of no use in preventing breakdown.

You can put a Zener diode across the transistor, of higher breakdown voltage than the supply, which may speed the relay drop-out (to the extent it allows the voltage to go higher than the supply), at the expense of being harder on the transistor during switching. You can also put a zener in series with a diode across the relay. Or a resistor in series with a diode across the relay.

Whether across the relay or the switch is better is mostly a matter of how the currents and voltages change when the switch opens and closes, for EMI caused by the coil (usually not that much compared to the contacts).

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  • \$\begingroup\$ Ah that makes sense. If the transistor is on the low side, a Diode across it won't protect it at all. What if the transistor is on the high side, and again assuming an ordinary diode. Would you say the diode in one position or the other assists in transient protection better? \$\endgroup\$
    – Randy
    Mar 29 '18 at 23:27
  • \$\begingroup\$ Exact same deal, just the polarities reversed. A regular diode across a high-side switch will not help. \$\endgroup\$ Mar 30 '18 at 0:56
  • \$\begingroup\$ But a diode across a high side switch will route any inductive spike back to the supply, which presumably has a capacitor that can better squelch the spike then the transistor could. \$\endgroup\$
    – Randy
    Mar 30 '18 at 14:28
  • \$\begingroup\$ No. The voltage at the switch will go below ground as the switch opens and a diode across the switch will remain reverse biased. It will go negative wrt ground until it either rings with the distributed capacitance of the coil or it breaks down the transistor. \$\endgroup\$ Mar 30 '18 at 14:40
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    \$\begingroup\$ @Randy: Any N-channel MOSFET will have a junction between the N-doped channel and a P-doped substrate. Within an integrated circuit, it's common for the substrate to be connected to the lowest potential in the circuit so that diode will never conduct, but doing that will make it necessary to have the source, gate, and drain all located on the same surface of the chip. Discrete MOSFETs often connect the substrate to the source, allowing the drain connection to be on the opposite surface from the source connection, thus doubling the surface area available for the connection. \$\endgroup\$
    – supercat
    Aug 9 at 22:03

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