I am building a simple motor control application, driving a series wound motor that runs on rectified 230VAC. Originally, it was controlled by a pendant switch connected to a 7 pin connector. The idea is to modify the setup inside the motor as little as possible and build a remote controller that connects to that same connector.

The connector serves the rectified 230VAC signal and I plan to use this as the source for the remote controller. This rectified signal must come back into the motor housing to drive the motor, and I will be controlling the connection using power relays.

I am planning to use a RF remote switch to control the relay coils, and I have two versions available: a 220VAC version, which does not work with the rectified signal, and a 24VDC version. I order to use the latter one, I need to reduce the rectified signal to 24VDC. What would be the easiest way to do this?

I have been looking into several topologies and found one that may be suitable:


For the circuit I am considering, the input must be the output of the bridge rectifier, which is what is available through the cable connected to the 7 pin connector, so there is no possibility of using a transformer before the bridge. This rectified signal must also drive the universal motor, but this will be done through another circuit.

In this case, this circuit must not drive a large current, only enough to drive the Load. This load is the RF remote switch and the relay coils connected to it.

I am aware that the R should be a power resistor that must stand several Watts of power dissipation. Considering a Zener voltage of 24V and a current of 5mA, with an input of 310VDC, the resistor value must be 59.2k and has a power dissipation of 1.5W. This is without taking into account the current consumption of the load, which at first glance should not be very high. I think a dissipated power of several Watts is an acceptable value considering that the series wound electric motor that it drives has a 1.5kW rating (EDIT: connected to the output of the bridge rectifier through relays, not driven by this circuit).


A more elaborated schematic of what I meant is the following (with the motor part heavily simplified):

Topology with motor

I have been running some simulations and it seems that this topology is unsuitable. The LY2 24V relay I was planning to use has a coil resistance of 650 Ohm, resulting in a current of 37 mA when switched on. This produces a large voltage drop on the R1 resistor (and thus a large power dissipation). This causes the Zener to work below Vz. If the value of the resistor is reduced, the Zener voltage increases but also the power dissipation in the resistor.

The schematic is publicly available here.

As suggested in the comments, using a switched-mode power supply (SMPS) might be the way to go. I am still not sure if it will work correctly with a rectified voltage as the input.


It seems that this enclosed AC-DC converter is suitable for the application.

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    \$\begingroup\$ No way. You can't use this circuit for 1,5kW @ 24VDC. You need a large transformer. \$\endgroup\$ Commented Apr 7, 2021 at 6:49
  • \$\begingroup\$ You can't use a 59k resistor in series with a 1.5kW load at 230V, since 230V/59000Ohms = 0.003A. The maximum current trough a 59k resistor at 230V is 3mA. \$\endgroup\$
    – PMF
    Commented Apr 7, 2021 at 7:23
  • \$\begingroup\$ I gather that the motor is line voltage and you only want to drive the relays at 24v. Is the 24V relay rated to switch 310VDC? You should add schematics to show how you want to connect the components. \$\endgroup\$
    – K H
    Commented Apr 7, 2021 at 7:26
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    \$\begingroup\$ most 230VAC input DC supplies will work fine off rectified 230V (all the switoch-mode ones) so just use one of them. \$\endgroup\$ Commented Apr 7, 2021 at 7:58
  • \$\begingroup\$ The 1.5kW motor runs at the the rectified voltage, but it will be connected to the output of the bridge rectifier directly (through the relays, not shown in the schematics). This circuit is only meant to drive the relay coils using the RF remote switch @ 24V. In this circuit, the load is the switch and the relays. Sorry if it wasn't clear the first time. Will add schematics of the whole setup later. \$\endgroup\$ Commented Apr 7, 2021 at 8:02

1 Answer 1


A few thoughts:

  • There's no capacitor in there. Remember that the AC voltage drops to zero twice every cycle of the mains.
  • A "capacitive dropper", using a capacitor to limit the current, would generate a lot less heat than a big resistor. It's a very common circuit design, often used for driving LED lamps.
  • The circuit may generate an apparently safe 24V, but it's directly connected to the mains. Assume everything is live and insulate accordingly.
  • \$\begingroup\$ I guess it is a good idea to include a capacitor to stabilize the DC output, where should it be placed? Before or after the zener? A capacitive dropper uses a X2 rated capacitor before the bridge to limit the current, but in this case I am not able to modify that part of the circuit (the bridge is also being used to drive the motor). Could it be placed after the bridge? I will be careful and correctly insulate everything once this is done. \$\endgroup\$ Commented Apr 7, 2021 at 12:46
  • \$\begingroup\$ @GuillermoOliver The capacitor would be in parallel with the Zener. Before or after is down to how you choose to draw the circuit. \$\endgroup\$
    – Simon B
    Commented Apr 7, 2021 at 13:33
  • \$\begingroup\$ @GuillermoOliver A capacitive dropper would only work if the capacitor is before the rectifier, as it needs AC to work. So it may not work for you. \$\endgroup\$
    – Simon B
    Commented Apr 7, 2021 at 13:34
  • \$\begingroup\$ My bad, I meant before the resistor or after the Zener. I guess it would not be a good idea to put it before the resistor because it should have a much higher voltage rating than in parallel with the zener and there is no real need to stabilize the output of the bridge. \$\endgroup\$ Commented Apr 7, 2021 at 14:07

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