Clock with a simple electric remontoire, consisting of a solenoid, with an arm, that kicks back the main spring, rewinding it.

Solenoid coil data:

  1. operating voltage: 12V-15V
  2. measured resistance: 3ohm
  3. theoretical operating current: 4-5A

(unable to measure current since the coil is energised just for a fraction of a second. Weirdly enough... a lot of current for a small clock).

  1. What would be a good flyback diode to use for this application?
  2. What are the rules/formulas helpful for chosing a flyback diode for other characteristics of solenoids (relays).


simulate this circuit – Schematic created using CircuitLab

enter image description here


  • 1
    \$\begingroup\$ Your circuit is not very clear. It either never closes the contacts, or, if somehow the contacts are closed, it will never open them again (until the battery dies, of course). What do you intend it to do? \$\endgroup\$ Feb 5, 2019 at 20:52
  • \$\begingroup\$ Added a drawing of the clock. The solenoid (4) is grounded to the chassis through the lever (5), completing the circuit for a brief time. The lever is kicked back, the solenoid is disconnected. When the spring unwinds the contacts touch again, the solenoid is grounded, it pulls the lever, kicks the spring, the solenoid is disconnected and the cycle repeats every 2-3 minutes. The solenoid only remains energised if the contacts stick for one reason or another, otherwise the solenoid automaticalliy disconnects itself by kicking the lever and rewinding the spring. \$\endgroup\$
    – Chip T
    Feb 5, 2019 at 21:09

1 Answer 1


The peak voltage across the diode is equal to the supply voltage.

The peak current through the diode is equal to the solenoid current.

The duty cycle of the current is very low, so the average current through the diode is negligible.


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