Momentary over-excitation voltage circuit

Question

I want to control an electromagnetic brake (fail-safe brake) used along with a servo motor. A microcontroller will control the trigger signal for the electromagnet coil.

Electromagnet requirements:

Overexcitation voltage - 24 V, 1 A for 200 ms.

After 200 ms the voltage steps down to 8 V with a current consumption of 0.3 A. Please refer the waveform images below.

The brake coil cannot sustain high current for a long time so we provide a momentary high voltage to excite the coil and then reduce the voltage to hold the electromagnet at its state.

The solution that popped up in my mind is to use a timer circuit which will provide the required 24 V for 200 ms and then switch to 8 V using a relay. But as the trigger signal generated is a continuous signal (not a trigger pulse) I don't know if this solution will work well for the required functionality.

Please suggest any other optimized approach.

Answer 1

You could use 2 relays. One a single pole relay, the other a double pole relay.

On the single pole relay, use an output from the micro controller to pull in the coil. Then wire the 12v source through the NO contacts on the single pole relay as well as a NC set of contacts on the double pole relay then on to the brake.

On the other pole of the double pole relay wire the 8v source through the NO contacts to the brake.

In your code put the output wired to the single pole relay to turn on when the you want the brake to come on. Now put in a 200ms delay timer in the output for the double pole relay with will come on after the 200ms timer is complete.

This will mechanically interlock the 2 signals so they won’t ever come on a the same time while making it as quick as the relay you choose.

Answer 2

"Mickey mouse" and only sort of right solution.

On switch over Vout is Vin + a charged cap, decreasing to Vin - a diode drop.
Vin could be higher with a series resistor, if desired. Cap size is based on the desire for about 5 J or energy.

Use: 0.1F supercap with a diode across the capacitor, non conducting when charging.

With switches as shown the capacitor charges.
When the switches are toggled the capacitor bottom switches to Vin and capacitor +ve switches to Vout.
Capacitor discharges into inductor.
When Vcap falls to one diode drop the inductor is supplied via the diode.

A higher Vin than 8V and a series resistor could be used.

^{simulate this circuit – Schematic created using CircuitLab}

NB !!! - Capacitor is 100 mF = 100 millifarad = 100,000 microfarad, NOT 100 microfarad.

Answer 3

A solution could be using a 555 timer.

• Use a 555 in monostable mode generate the 200 ms pulse for the 24V supply.
  Benefit of a 555 is that it can make relative large time pulses and you can easily change the timing.
• The brake signal has to be converted into a break pulse in order to have the 555 timer reset again (after 200 ms). This edge detection is mainly done by C3.
• Use two ORing diodes to smoothly switch between the 24V and 8V supply.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 4

How critical is the timing and the amplitude? My experience with these types of devices is that they need higher voltage (and hence higher current) just for the initial magnetic pull-in. Once the the brake is engaged, drop the voltage to minimize heating.

Similar to Russel's approach, I'd start with a large electrolytic capacitor. Since the sustain voltage is only 8 volts, it would take 11-12 diodes to drop 24V down to 8V. Use either a dedicated voltage regulator (7808) or a 5W power resistor.

Main power in is 24v. Add a 8 volt regulator to the 24V power to provide the 8V sustain voltage. Use a DPDT switch to connect both the 8 volt source and charged capacitor to the brake. Once to the capacitor has discharged, the 8 volts will sustain the brake