I am having trouble getting enough current to drive a bi-directional solenoid with two leads. I am using a half of a L293 (Quadruple Half-H Driver) to drive the solenoid back and forth. The solenoid has to overcome some stiction so it barely flinches if there isn't enough current. I have gotten it to work, albeit slowly, with two standard 9V batteries in series.

I didn't think the two 9V batteried were pumping out enough current, so I switched over to a 11.1V, 5000mAh - 35C LiPo battery. This battery should be overkill but it's still not driving the solenoid as fast as it needs to.

In the diagram below, VCC2 is the connected to the positive terminal of the battery. Pins 2 and 7 (driver inputs) are connected to a LM556 dual timer circuit (shown below) which creates a timed pulse and only allows one of these pins to be high at a time. The switching circuit works well.

Questions: 1. Is there some way I can further boost the current going to the solenoid?

  1. Do I need a flyback diode on the solenoid? If so, how do I do that with a bi-directional solenoid which only has two terminals?

  2. Is there a simpler circuit that will get the job done?

It's also worthy noting that since I'm using a breadboard and can't properly heat sink the L293, have to unplug after using it a few times. Two times now, it has gotten hot and the circuit oscillates after one of the buttons is pressed. After it cools down, everything works again, so I don't think I let the magic blue smoke out of anything ... yet.

Circuit Diagram 556 Timer Circuit

  • \$\begingroup\$ Do you own an oscilloscope or multimeter? You could use one of those to measure the voltage at the solenoid (and potentially various other points) to help you see what's happening. \$\endgroup\$ – JohnSpeeks Apr 21 '16 at 2:26

Your issue is voltage under load.

Your two 9V batteries were at a nominal 18V (prob a bit more). When loaded, they probably dropped to 15V or so, depending on the load.

Your lipo is 11.1V, probably a bit more.

  1. Put another lipo on series and you should hold your voltage up at load. You might want to experiment a bit using a DC power supply to determine your actual voltage and current requirements.
  2. Yes, you need a flyback diode. Inductive kickback can be pretty severe.
  3. As far as your circuit, section 8.2 in the datasheet shows you how you should probably wire your diodes.

Your heat issue may be from too much current. Your part can only handle 600mA or 1A in the best of conditions (depending on the variant). You should find a way to measure the current. Testing current draw with a DC motor is similar.

  • \$\begingroup\$ No, the issue is the that the L293D is unsuited to this task. That is why it is getting so hot. The voltage of the supply is already many times what is needed, if a suitable switching device were used. \$\endgroup\$ – Chris Stratton Apr 21 '16 at 2:45
  • \$\begingroup\$ I don't know the solenoid model, but that is definitely a possibility. \$\endgroup\$ – slightlynybbled Apr 21 '16 at 2:54
  • \$\begingroup\$ Note that the L293D part has the protection diodes built in, (see spec sheet). The similar L293N part (and other non"D' types) do not have the built in diodes. Also the non-"D" type parts are the higher current versions, (L293D types are 600ma, while non-D is 1A). So be sure about which chip you have. One additional option: For a higher current part try using an L298. This part has a 2A driver (or 4A if outputs are paralleled). See this app note: positron-libre.com/archives/moteurpasapas/… \$\endgroup\$ – Nedd Apr 21 '16 at 5:16

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