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I built the circuit described in a previous question/answer which I will repost here:

556 timer and solenoid circuit

It's been running quite well, actuating the solenoid at regular intervals as designed. The TIP122 transistor is barely warm, the solenoid is within normal temperature for its duty cycle, but the 556 timer IC is too hot to touch. It's been operating continuously for 24+ hours without any failures, but I am concerned because I think that this IC should not be warm at all.

Is there something I overlooked in this schematic that should be done to prevent excess heat in the IC? I haven't (yet) measured current usage by the IC alone, but the total circuit draw is 125 mA on standby and 2A when the solenoid is actuated.

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    \$\begingroup\$ You are running your 666 timer from 12 V. Have you checked the datasheet whether that is OK, and if so, what current you can expect it to draw? Maybe everything is working as it should? All in all, this circuit seems like a long way to go to avoid using a PIC 10F200. \$\endgroup\$
    – Olin Lathrop
    Feb 7, 2013 at 19:51
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    \$\begingroup\$ @Olin Yes, the evil timer is supposed to operate on 5 to 15V. We are using a B&K Precision regulated bench power supply set to 12V. I'm more familiar with AVR than PIC, I thought about simply using an ATTiny, but I have all these timers laying around... I thought I'd use one here. :) \$\endgroup\$
    – JYelton
    Feb 7, 2013 at 19:52
  • \$\begingroup\$ Can the 556 drive that transistor with a large enough current? Even if it's within the maximum output current, it might still be a high enough current to heat up the 556, it's not really designed to dissipate any heat. Maybe consider using a mosfet, or a secondary transistor to drive the first one, or a darlington which is essentially the same thing \$\endgroup\$
    – c10yas
    May 3, 2019 at 6:44

2 Answers 2

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According to the datasheet for the NE556, the no load supply current should be a maximum of 30mA at 15V supply. Since you are reading 125mA and the unit is getting hot suggests something is wrong somewhere.

I can't see any obvious issue with your schematic at a glance that would cause this. So a few things to check:

  • Check that there are no shorts/low resistances anywhere (particularly from each output to ground)
  • Check polarised capacitors are the right way round.
  • Check resistor values are correct (in particular R4)
  • Try swapping the IC in case it's faulty
  • Make sure only the solenoid is pulling the other 1.75A when activated (e.g. put multimeter in series with it only)
  • Disconnect solenoid, test, then disconnect first timer from second, test, etc, until the current drops to a reasonable value.

If you can't find anything, trying one of the other circuits suggested might be worth a go.

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    \$\begingroup\$ R4 was indeed the fix. I used a 1K arbitrarily, and never went back to calculate a more appropriate value. Using a 10K significantly reduced the current and the heat. Thanks. \$\endgroup\$
    – JYelton
    Feb 8, 2013 at 0:30
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Maybe that 1KΩ resistor was actually just 100 Ω, thus accounting for the 125mA current,- but this would have only been the case if the 556 output was high combined with the solenoid circuit being opencircuit somewhere, thus not to actuate. In my most recent 555 timer project I added an emitter follower onto the output of a 555 timer to drive a bigger transistor (2N3055) to turn on and off a 12 volt pump motor. My 555 timer got hot after I accidentally shorted it's output to ground for just a few seconds and I had to replace it. So be careful not to short the output.

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  • \$\begingroup\$ Ve is ground, and Vbe is one diode drop. With the voltage rail at 12 V, when the timer output cycles high, you've got over 11V dropped through R4. The solenoid has no effect on that. \$\endgroup\$
    – Ben Voigt
    May 3, 2019 at 3:38

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