4
\$\begingroup\$

I've got a circuit with two DPDT relays controlling a motor - the diagram below is borrowed from a related but different question. RLY1 turns everything on and off, RLY2 switches the direction of the motor.

I've found that after a brief amount of use the relays are damaged. Both relays have the same problem - on one set of contacts common/NO/NC are all connected, even when the relay coil is unpowered. (The other set of contacts is as expected - ie. common/NC are connected, NO is connected to neither of the other two - so only "half" of each relay is currently damaged).

Questions: what is likely to have caused NO/NC terminals to be connected, and how should I have avoided that problem?

Gory Details (possibly not needed):

  • motor runs on 30V
  • PSU that came with the motor is rated 30V/2A
  • relays are rated for 2A/30VDC (model Meishuo MCB-S-205-C-M)
  • relay is controlled from an ATMega MCU
  • relays both have an IN4001 protection diode across the relay coil - not shown in diagram
  • I did manage to "fix" one relay by tapping it on the table. Unknown if that will stay fixed or if permanent damage is done. The other relay doesn't respond to this treatment.

    enter image description here

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Can't provide complete answer right now, but it sounds like your relay arced and welded shut. \$\endgroup\$
    – efox29
    Apr 8, 2016 at 18:03
  • 1
    \$\begingroup\$ When switching motor direction, do you first stop the motor by releasing RLY1? \$\endgroup\$ Apr 8, 2016 at 18:12
  • \$\begingroup\$ Measure the stall current of the motor.. \$\endgroup\$
    – user16324
    Apr 8, 2016 at 19:15
  • \$\begingroup\$ In addition to the answers given, notice also that the coils of the relays are also inductive loads and should have flyback diodes or other suppression across those coils to protect whatever is being used as the "Enable" and "Dir" switches. \$\endgroup\$
    – Tut
    Jul 19, 2016 at 18:49

5 Answers 5

8
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Snubbers on reversible motor.

Provided your relay contacts are break-before-make you shouldn't be causing a direct short. The problem is likely to be the inductive kick causing arcing of the contacts when they open.

One solution is to put diodes on the motor to shunt the current to the PSU. This may look a little strange but is really just a rearrangement of the standard H-bridge protection as shown in Figure 2.

schematic

simulate this circuit

Figure 2. The same circuit in H-bridge configuration.

schematic

simulate this circuit

Figure 3. Redrawn to use relay configuration on OP's schematic.

\$\endgroup\$
2
  • \$\begingroup\$ +1 for the comparison to Figure 2. (and the answer in general, of course!) \$\endgroup\$
    – bitsmack
    Apr 8, 2016 at 18:56
  • \$\begingroup\$ Thanks! The document that WhatRoughBeast links to mentions snubbing, but having an explicit example is useful \$\endgroup\$
    – Anthony
    Apr 11, 2016 at 17:23
2
\$\begingroup\$

Yes, you do have a problem. Start here. Your big problem is that inductive loads, like motors, want desperately to keep their current at whatever level it currently is, and the result is that opening a contact will invariably draw an arc, which will damage the contact surfaces of the relay. Then, when you close the switch or relay, if the motor is stopped, the motor will draw what is called stall current, which is much larger than running current. This results in welding the contacts if they have not burned out when arcing. From the linked article,

The result is a large in-rush current at “turn-on” and arcing at “turn-off.” When you are switching a motor load, typical industry practice is to derate to 20 percent of the resistive rating.

So, for a 2 amp motor, you should replace your relays with units rated at 10 amps minimum.

As to how the NO/NC contacts got shorted, frankly I haven't a clue, but if you take a hacksaw to your relays (carefully!) you should see some strange sights around the contacts.

\$\endgroup\$
4
  • \$\begingroup\$ Thanks, that document looks useful! I'm not sure if it is a 2A motor, all I know is what the PSU is (the motor is built in to a device, I can't get the specs). Seems that over-specifying just in case seems worthwhile to avoid problems though. \$\endgroup\$
    – Anthony
    Apr 11, 2016 at 17:10
  • \$\begingroup\$ To add: that document specifically says that manufacturers rate for resistive loads, so when using inductive loads you need to adjust the rating down to 40% of its resistive load rating. \$\endgroup\$
    – Anthony
    Apr 11, 2016 at 17:20
  • \$\begingroup\$ @Anthony - Nope. The document says, "When you are switching a motor load, typical industry practice is to derate to 20 percent of the resistive rating." \$\endgroup\$ Apr 11, 2016 at 17:28
  • \$\begingroup\$ Indeed, I only read as far as inductive loads, my fault! Thanks for correcting \$\endgroup\$
    – Anthony
    Apr 16, 2016 at 18:27
0
\$\begingroup\$

Although the PSU is rated 2A and the relays are rated 2A when the motor starts there is a very briff moment where the motor draws much more than 2A. Also, when the motor stops it might switch from being a motor to being a generator generating a current that can be higher than 2A. The PSU might be capable of dealing with the higher 'reverse current' but the relays might not.

The solution, change your relays to other relays that can do, at leasy, 6A.

\$\endgroup\$
0
\$\begingroup\$

Been there, done that.

Our solution was appropriate for the era we were in (late 80's, early 90's). We used N-channel Current-Sense MOSFETs (IRCZ44) for the bottom half of the bridge and SPST-NO relays (American Zettler AZ-2150A) for the top half of the bridge. A small CPLD handled the logic. Clamp diodes were installed across the top relay contacts (K to Batt +).

Startup was simple: close appropriate relay, delay, turn on proper FET. Turn-off was a little more complex: turn off FET, wait, turn off relay, wait, turn on both FETs for braking if needed.

MOSFET current was continuously monitored and appropriate action taken when needed. Mostly PWM-type current limit, with shutdown timer if the over-current persisted for too long a time.

This was used to drive a large linear actuator that changed the pitch of fan blades on large industrial earth-moving machinery. The pitch of the fan was constantly changing to keep the engine temperature constant as the engine load varied. We are talking about many fan-blade movements during every minute of run time.

The technique was extremely reliable.

\$\endgroup\$
0
\$\begingroup\$

While a redesign using FET transistors would make a more reliable solution you could also implement a simple solution by replacing RLY1 with a N-Channel MOSFET, and a 2K resistor to the gate. The output that drives the relay coil can just drive the 2K resistor that feeds the gate. The resistor slows the switching of the transistor enough to prevent generating RF noise and protects the MCU if the FET transistor fails with a short between the gate and the drain.

If isolation is needed between the control circuit and the motor circuit you can use an isolated FET Driver with a high input voltage 3-terminal regulator to supply the power to the FET driver. FET drivers will specify their ability to drive an ampere or more but this type of circuit will only uses 20-30 mA so it can easily be supplied by a 3-terminal regulator without any need for a heatsink. A 47 OHM resistor between the driver and the gate will work in this case to keep the RF noise down.

The components could easily fit on a small PC board that would fit in place of the relay. All that is needed is to make sure the direction relay is not activated or deactivated with power enabled and upgrade the relay to handle a minimum of 5 amps, 10 would be better to fix the other relay problem. The FET should be a logic-level drive FET if driven directly by the MCU. The FET should also be rated for at least 75 volts and handle more than 20 Amperes. FETs can be obtained for a very reasonable cost that meet these requirements and will give you a very long life because the transistor will not be challenged even by surge currents caused by the motor.

Adding a capacitor across the motor as close as possible to the motor will help keep electrical noise down which is a good thing for all the components in the circuit. I would go with a ceramic or foil capacitor 1 uF 150 Volts.

\$\endgroup\$

Your Answer

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

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