simulate this circuit – Schematic created using CircuitLab

Added requested schematic. Current circuit is on the left. Proposed additional circuit is on the right. I've never seen two solenoids put in parallel like this and am not sure what problems, if any, would result.

I may have found a solution to the problem below. Albright SW200 main solenoid/contractor works up to 72 V and 400 A of continuous current. It is a SPNO instead of the current SPDT, but leaving the solenoid output floating (so power is removed) should work to stop motor. Worse comes to worse, I'll add a second solenoid to put battery - across the motor (as opposed to leaving motor + floating) if I need it to stop the motor.

I still am interested in why or why not you can put two solenoid outputs in parallel if they have the same voltage levels. When solenoid is closed, it is basically a wire, so I don't see any reason not to, but I also have never seen/heard of it being done. I would still be interested in an answer even though I may have solved the problem with just a bigger solenoid.


I have a current limited DC motor.

Current circuit, starting at the battery + supply is to a normally open 586 series solenoid input. The normally closed solenoid input is the battery -. The solenoid outputs are wire or-ed together to the + motor wire. The - motor wire returns to the - of the battery. When the control to the solenoid goes on, the motor runs. When the control to the solenoid goes off, the motor has the battery ground (-) across it and stops. The solenoid is rated for 200 A continuous and the batteries supply 180 A continuous. The motors stall current is 390 A.

The problem is that I have some start up torque issues that are current limiting the motor. For a variety of reasons I can't increase the voltage, but I could put another set of batteries to supply more current (extra amp hours wouldn't hurt either). If this works, the 360 A of current are still below the motors stall current of 390 A.

What I am thinking of doing is putting a second 586 series solenoid in parallel with the first and run a second battery bank through it. The outputs of both solenoids would be wire or-ed together and go to the + wire of the motor. The - wire of the motor would wire or-ed and go to the two battery banks. Rest of the circuit would stay the same (?) I've never seen this done and am not sure that there is a problem waiting to release smoke from some component in the proposed system.

Question 1) Is there some reason not to run two solenoids in parallel to increase the current?

Question 2) Should I use the same control for both solenoids (the control goes to no other circuitry) or generate a second control signal in parallel for the second solenoid?

Question 3) What should I connect to the normally open input on the two solenoids, the wire or-ed - for the two banks of batteries, leave the normally open input floating or something else? I like being able to stop the motor quickly by putting the - battery across it.

Question 4) Is there a simpler way to solve the problem? I couldn't find a larger continuous current solenoid at the voltage I am using (my first thought).


  • 2
    \$\begingroup\$ Paragraphs? Schematic? \$\endgroup\$ Commented Dec 11, 2015 at 1:39
  • 1
    \$\begingroup\$ "I have some start up torque issues that are current limiting the motor." - what exactly is is 'current limiting' the motor, and what value is it limiting the current to? \$\endgroup\$ Commented Dec 11, 2015 at 2:31
  • \$\begingroup\$ The system is a robotic spinner. The motor is pulling 180 A now and you can hear it struggle to spin up to speed. Once at speed it works fine. I believe that with higher current it should spin up faster. \$\endgroup\$
    – John H
    Commented Dec 11, 2015 at 2:41
  • \$\begingroup\$ You didn't answer Bruce's question. What is limiting the current? If you have it directly connected to the supply (and not through a speed controller) then the only way to get more current is to raise the voltage. \$\endgroup\$
    – Transistor
    Commented Dec 11, 2015 at 8:24
  • \$\begingroup\$ Transistor - Maybe I don't understand how a motor works. As motor torque requirements increase, current draw by motor increases until stall. The torque required of the motor is high. My belief is that by enabling more current to the motor, I'll get more torque out of it. Total proposed current would still be below stall current. I could test your theory by adjusting battery to provide 90 A of current and see if situation is same (i.e. not current limited) or worse (i.e. current limited). \$\endgroup\$
    – John H
    Commented Dec 11, 2015 at 15:10

2 Answers 2


Your question is very difficult to follow without a schematic. If this answer does not answer the questions then please use the schematic editor to add a schematic to your question.

  1. Yes. The most wear and tear on contacts is caused by arcing at contact closure and contact opening. Parallel relays or contacts will never open or close at exactly the same time. The first to make and the last to break take all the pain.
  2. There is no problem paralleling relay coils - just the contacts.
  3. I don't understand this question without a schematic but I suspect that you are asking can you use the relay contacts to short out the motor for rapid braking. If you do this you are effectively short-circuiting a generator and the current spike could be relatively huge. You should probably switch in a high powered braking resistor to limit the braking current to a safe value.
  4. Schematic required.

Design a current feedback amplifier to reject back-emf and use a lead compensator to overcome the limitations of the intrinsic electrical time constant of your solenoid.

Place in series with your solenoid, a current sense resistor, and use a differential amp to read the voltage across it. Feed that current monitoring voltage back to an amp that can serve to provide a difference between the desired current and what you measure from the shunt. Feed this difference to a high gain lead compensator. Chose the zero of the lead to cancel the electrical pole of the solenoid (R/L) and chose the lead amplifier pole to replace with a faster pole.

The high gain feedback will serve to reject the effects of back emf by boosting the initial voltage when you apply a step to the control loop. 'Replacing' the solenoid pole with a faster 'virtual' pole also increases this boost effect. It reduces the effective time constant.

  • \$\begingroup\$ It sounds interesting, but I'm not sure I understand the whole thing. Any chance of a schematic? \$\endgroup\$
    – John H
    Commented Dec 11, 2015 at 20:24

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