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I have a project that uses several high power DC solenoids (50 VDC) that will have ~10 feet of wire between them and the driver board. This is for pinball machines. These solenoids typically come with flyback diodes pre-installed on their terminals which makes sense and is very common with any switching inductive load.

However, when I look at schematics for the old driver boards from their heyday in the 80's and 90's I see flyback diodes there too.

Is there a reason you would want flyback diodes at both ends of the wire run?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ It's nice to have solenoids with built-in flyback diodes. A lot of solenoids don't have built-in diodes, though. The diode on the driver board allows to use solenoids without built-in diodes. Additional flexibility in the selection of solenoids. \$\endgroup\$ – Nick Alexeev Jul 28 '18 at 0:20
  • \$\begingroup\$ see also electronics.stackexchange.com/questions/370952/… \$\endgroup\$ – Jasen Jul 28 '18 at 2:45
  • \$\begingroup\$ I've never heard of a 1N1004 diode. \$\endgroup\$ – Andy aka Jul 28 '18 at 9:40
  • \$\begingroup\$ The dual diodes may reduce the Radiated EMI. \$\endgroup\$ – analogsystemsrf Jul 29 '18 at 4:44
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To draw the circuit with some extra stray components

schematic

simulate this circuit – Schematic created using CircuitLab

Only D1 is necessary.

D2 can only control the kickback from L1. The kickback from L2 is not controlled by D2. If L2 is large enough, if there's enough wire or it runs in a loop rather than as a pair twisted with its return, then there could be enough energy stored in L2 to concern Q1.

Solenoids are often sold with D2 incorporated, to simplify the circuit. It doesn't hurt to have both D1 and D2, as it affords some measure of redundancy.

Some FETs have an 'unclamped avalanche' rating, for use as solenoid drivers without D1. They will safely take a specified kickback energy without failing. This is rarely the case for BJTs.

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  • \$\begingroup\$ Thanks for the info. With pinball machines, often four or more solenoids are wired up together sharing the same 50V line. The wires are all bundled together in a harness, but not twisted. So that could make sense on why they're on the driver board. \$\endgroup\$ – Geomancer Jul 28 '18 at 12:03
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Not really. The diode continues the coil current in the same direction so ideally it is across the transistor. But more important is to have the supply decoupled near the transistor and use a paired wire to avoid open loops.

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If the wires are short you don't need the diode at the driver end.

Having a diode from the transistor drain to Vdd is effective at protecting the transistor with or without the diode at the solenoid and regardless of wire length.

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D2 is not required.
The current in the wires does not change direction (which could make for a noise problem) and as Tony points out, the protection is there for the driving device.

The only time it makes sense to put the protection at the remote (inductor) end of the wires is where you need the field to collapse very quickly.

For example you might put D2 in series with a Zener diode (no D1) to maximize the losses when Q1 turns off.

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