1
\$\begingroup\$

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

Electromagnet Requirements:

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

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

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

enter image description here

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

\$\endgroup\$
  • \$\begingroup\$ Mickey mouse and only sort of right:. 0.1F supercap. | Diode across cap non conducting when charghing. | Cap negative and positive each to a SPDT switch (relay or whatever). | Usually cap -ve switch goes to ground. |Cap +ve switch to Vin. Cap charges. || GO: Cap bottom switch to Vin. Cap +ve switc to Vout. | QED. sort of. \$\endgroup\$ – Russell McMahon Jul 17 '19 at 4:56
0
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\begingroup\$ Welcome to SE.EE! Please draw a schematic as it is more clear than describing it. There is an embedded Cicuitlab schematic editor in the toolbar. You don't need to have an account for it. \$\endgroup\$ – Huisman Jul 17 '19 at 9:23
0
\$\begingroup\$

"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.

schematic

simulate this circuit – Schematic created using CircuitLab

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

\$\endgroup\$
  • \$\begingroup\$ @Huisman (1) Many reasons. Some of them good :-). AG47 has been on site hours after this answer was given. It is common for people to ask questions but not then engage in a manner that encourages more effort. How long do you suggest should be spent? (2) Feel free to provide a superior solution. (3) V_cap cannot change instantaneously. I_l cannot change instantaneously. You can expect a reasonably well behaved turn on. (4) I have noted that a series R would allow Vin to be more than 8V. That has issues too. Use of a temporarily triggered boost converter is liable to be quite a good solution. \$\endgroup\$ – Russell McMahon Jul 17 '19 at 10:29
  • \$\begingroup\$ How long do you suggest should be spent? You're right: it should be proportional to the time OP has spent. I withdraw my critism. Feel free to provide a superior solution I really like your voltage doubler solution. Moreover, I'm an average EE, I would never claim superiority. \$\endgroup\$ – Huisman Jul 17 '19 at 11:55
  • \$\begingroup\$ Got the concept but I didnt get how will I get 24V momentary peak. The voltage doubler will simply take it upto 16V. Although a series resistor can be used to get higher voltage but wont that reduce the current required by the coil to actuate. \$\endgroup\$ – AG47 Jul 17 '19 at 13:00
0
\$\begingroup\$

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.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ Yes I followed the same approach to use 555 in Monostable multivibrator mode earlier. In fact my next step was to convert the brake signal into a short pulse. Good to know I was on track😁. \$\endgroup\$ – AG47 Jul 17 '19 at 12:52
0
\$\begingroup\$

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 Simple Brake Circuit

\$\endgroup\$
  • \$\begingroup\$ Exactly. The drop in voltage is meant to reduce heat while maintaining its state. Rather than so many diodes, zenor diode can be used instead to regulate voltage to 8V. \$\endgroup\$ – AG47 Jul 17 '19 at 12:46
  • \$\begingroup\$ Amplitude is more critical as the brakes won't actuate if it doesnt get sufficient current. I think it should handle some 50-100 ms time extra. \$\endgroup\$ – AG47 Jul 17 '19 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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