I'm looking in how to activate a current chopper in a motor application, when there is no power available from a power supply Vin. This should prevent the motor from turning too fast in power off mode.

Now there is a Vin DC voltage (+48VDC, current limited to 15A) and the intermediate circuit voltage Vbus. The intermediate circuit voltage can be increase by turning the motor, when there is no Vin voltage. On Vbus there is a current chopper Q1 with R1, that shorts Vbus in normal operation, when the Vbus > Vin + threshold and this is done by a uC, that applies a signal to the chopper. Now I designed the following concept:

enter image description here

The concept should allow a disconnected Vin to trigger Q1 and so short the Vbus, but I see some problems here, why this will not work:

  • At startup when Vin ramps up, the voltage will be shorted until it reaches the threshold to trigger the chopper off. It might work, since Vin is current limited with hiccup mode, but I'm unsure about this
  • The most critical part I think is the linear mode of Q1, when Vin is too low at startup. This might destroy Q1 and so somehow the braking should start only when there is enough Vbus voltage, in this case +12VDC.

So, has anyone an idea on how to improve the concept or maybe tackle the problem from a different angle?

Edit: Something else that came to my mind was a relay, but I would like to prevent electromechanical stuff for this:

Edit: I have one problem, there seems to exist no NC relay for 30A and more. So I need somehow another method.

enter image description here

Edit 2: Another solution that came to my mind is to use a MOSFET in diode mode and then trigger the gate connection with a NC relay. The thing is, there are huge losses in the FET, since a modern FET has a Vgth of around 1-2V and so for a current of 15A, the FET will probably has a 4V drop, which leads to 60W of losses. Since the high current through the resistor comes only with increasing voltage, this increasing voltage might be usable to fully turn on the FET.

If for example there are 5V voltage from the motor, there will flow 2.5A, a loss of around 10W in the FET. This 5V can then be used (threshold based to somehow turn the FET fully on. I'm still thinking of this logic to achieve this.

What I could do ist to directly source the MOSFET from the Power Input, so this might work without high losses, since only at very low voltage the losses are high, but the resistor stops too much current and so this should be fine. The relay can probably also be replaced by a PMOS

enter image description here

  • \$\begingroup\$ Why so complicated? Why not just a NC relay that connects a resistor across the motor? If you don't want a relay I suppose a power MOSFET could work too with a large gate-source storage cap with a quick and dirty zener regulator and diode to capture the charge when the motor starts generating. \$\endgroup\$
    – DKNguyen
    Jan 24, 2022 at 15:29
  • \$\begingroup\$ Yes, I already tought about that, see my edit. The thing is, R1 is only 2 ohm and it will short the input voltage at startup too much (power supply delivers maximum of 15A). It might work, If I use additional a buck converter and for example a 12VDC relay. Maybe there is still a better idea around. \$\endgroup\$ Jan 24, 2022 at 15:33
  • \$\begingroup\$ Th other catch I see is that at 2 Ohms, R1 is going to be make the generator output voltage so low that nothing can be powered off it. I don't think that would be enough to power the MOSFET idea I mentioned above, \$\endgroup\$
    – DKNguyen
    Jan 24, 2022 at 19:48
  • \$\begingroup\$ Yes, that is true, so I think the relay is the way to go \$\endgroup\$ Jan 25, 2022 at 6:46
  • \$\begingroup\$ Maybe an NTC would help with the short on power up. \$\endgroup\$
    – DKNguyen
    Jan 25, 2022 at 14:25

1 Answer 1


DC contactors with NC contacts are not easy to find. More commons are auxiliaries contactors, with 4 NC contacts, for AC current. Some manufacturers have available coils for 48 Vdc, that you can use directly. But you have to derate current of contacts to use them in DC according datasheet, and if not enough current with one contact, use parallel contacts connected to individual resistors, just to split current and not to be overloaded. You can keep the resistor to the MOSFET or use the new ones with diodes. See example for 2 poles.

enter image description here

Regarding the startup, a delay timer (KT) should switch DC power to motor controller after this contactor has been actívated, to avoid the initial short of braking resistors.

EDIT: Added diagram with logic. An auxiliary relay can activate MOSFET Q1 when power power shutdown (t4), if gate control can support this voltage. The braking relay KB should keep activated this time with a capacitor bank, so when the NC contact close (t3), low voltage will remain in Vbus. If a inrush current limiter is necessary in the aux. power supply because cap. bank, timer on delay t1 > t2. With this stuff, the brake relay will not switch with Vbus voltage neither in start or shutdown.


simulate this circuit – Schematic created using CircuitLab

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.