I have a Motorola MC14011CP Quad-dual input NAND IC in which Vdd = 12v. I'm trying to use it as a NOT gate by tying the inputs of one of the gates together. When the inputs are not both true, the output is changing to ~0.5v instead of Vdd +- 0.5v as listed on the datasheet. Could it be something wrong with the chip or have I perhaps made an error in either my component selection or wiring?

Schematic: The Vdd on my IC is tied to +12v and the Vss to 0v. The load is an electromagnet.


simulate this circuit – Schematic created using CircuitLab

  • 2
    \$\begingroup\$ You've broken the device. The MC14011 is rated for VSS at 0V/ground and VDD at +5 to +15V, +18V absolute maximum. This is not what you want to use to drive an electromagnet. \$\endgroup\$ Jul 9, 2016 at 23:48
  • \$\begingroup\$ When you have a single supply, you have +12V & 0V, not +12V & -12V (which would give you 24V in total). If you really did connect a 4000 series IC to 24V you'd have seen the magic smoke leak out. And as others have said, a 4000 series IC is NOT suitable for driving a solenoid. \$\endgroup\$
    – brhans
    Jul 10, 2016 at 1:50
  • \$\begingroup\$ So it seems I made a pretty obvious error in my description of the circuit, the voltage source has a potential difference of 12v. My positive terminal is at +12v and my negative at 0v. Regardless, I seem to have the wrong chip for the job. Are there any that you would recommend or should I alter more than just the chip in this circuit? \$\endgroup\$ Jul 10, 2016 at 13:54

1 Answer 1


Your problem is that your load resistance (an electromagnet) is far too small. The 4000 series should not be asked to put out more than a few mA at 12 volts. Try disconnecting the load and measuring the voltage. Without knowing how much current your magnet requires, I can suggest the following circuit:


simulate this circuit – Schematic created using CircuitLab

Q1 is just about any n-type power MOSFET, although you'll need to be sure that it can handle the current of the magnet. Likewise, almost any diode will do as long as its current rating is greater than the magnet current.

  • \$\begingroup\$ To clarify, "... your load resistance (an electromagnet) is far too small ...". The load is too great for the NAND gate on its own. \$\endgroup\$
    – Transistor
    Jul 9, 2016 at 21:46
  • \$\begingroup\$ The magnet is rated to operate at 12v DC and 150 mA. Disconnecting the load give me an output from the NAND around 5.6v \$\endgroup\$ Jul 9, 2016 at 22:14
  • \$\begingroup\$ Also, I take it that the inductor in your diagram is the electromagnet? \$\endgroup\$ Jul 9, 2016 at 22:19
  • \$\begingroup\$ @TheEngineEar - Yup. And I suspect that trying to drive the magnet has damaged the gate. Try one of the others in the package. \$\endgroup\$ Jul 10, 2016 at 2:41
  • \$\begingroup\$ Will do, I have just one question regarding the diagram you posted which is: Why the 100 ohm resistor? \$\endgroup\$ Jul 10, 2016 at 15:49

Your Answer

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

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