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I'm doing a project with an SN74HC00N NAND gate and when I wired it up the chip got really hot really quickly.

I immediately unplugged everything and tested just with 5V and GND plugged in and none of the other pins connected to anything. it's doing the same thing -- gets really hot really quickly.

Am I doing something wrong or is the IC just bad?

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  • \$\begingroup\$ Electrostatic damage is another possibility. \$\endgroup\$
    – user16324
    Aug 4, 2015 at 9:58

4 Answers 4

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The chip is probably bad now. What most likely caused it to become bad was not having all of the input pins connected to either a Logic HI or Logic LO level in conjunction with a beefy power supply.

If your power supply had been current-limited to a few mA, no damage would likely have occurred.

If you have any more of those chips available, try using 10k resistors from each input pin to Ground. The output pins should all be HI (Vdd) and the chip should be running very cool - the quiescent current consumption is very low micro Amps.

If you confirm this to be the case, it means that you most likely have everything wired correctly. Now substitute the first chip that you were using (the one that got very hot). If it now runs cool and the output pins are reading the correct voltages, there is a good chance that the chip is still good.

To confirm that the chip is in fact good, use 1k resistors to pull each pair of input pins to Vdd. When both pins for one gate are HI, the output should go LO.

There is a good reason for using the resistor values that I mention. 10k is a reasonable pull-up resistor (0.5 mA); 1k is a stiff pull-down (5 mA). If you get either resistor connected to an output instead of an input, no damage will occur.

Additionally, the two resistors (10k pull-down, 1k pull-up) forms a voltage divider that puts the pin at about 4.5 Vdc. This is considered to be a logic HI whether the chip has TTL-compatible inputs or CMOS (symmetrical threshold) inputs.

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  • \$\begingroup\$ interesting. is this because I let the inputs float? why does that cause damage to the chip? \$\endgroup\$
    – user358829
    Aug 4, 2015 at 3:11
  • \$\begingroup\$ @user358829 because 74hc is a CMOS device, with a very large impedance and it picks up all kind of electric fields from the air.en.wikipedia.org/wiki/Latch-up \$\endgroup\$
    – ilkhd
    Aug 4, 2015 at 19:19
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Okay, I'll add yet another opinion. There is a common theme here, though, barring completely wrong connections (and even then..) your chip is toast and thus getting toasty.

I don't think it happened from floating inputs or shorting outputs.Floating inputs may cause excessive power consumption, but not usually enough to damage the chip. Shorting outputs doesn't normally damage the chip either, at least relatively brief shorts at room temperature. Fairchild says in AN375:

When bench testing 54HC/74HC devices, for example, you can short one output for several minutes without harm.

Rather, it probably happened from violating the input voltage limits or supply voltage limits. For example, applying a voltage to an input (sans a series current limiting resistor) with Vdd = 0 can damage the chip. Some signal generators can source enough current to cause damage (a couple hundred mA). A test probe or screwdriver or free wire slipping can cause similar issues, depending on what it touches. As Brian says, ESD damage could also cause this.

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I dont believe this was caused by letting the inputs float.

What is more common for experiences to go bad is when you accidentally join two outputs. If one is giving 0 and the other 1, kaboom, unlimited current flowing from 0V to 5V, iirc, the full short-circuit current of the source will develop on the IC, wich cannot deal with so much current and will heat and melt internally.

Now the IC is ruined. You will have to get another one.

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Are you sure you got the polarity correct? Pin 14 is for the positive 5V and pin 7 for the negative.

Pin 14 is on the opposite side of the chip from the pin 1 marker (little dot or mark).

I would be surprised if a 74HC00 would get hot just because you left the inputs open.

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  • \$\begingroup\$ I once killed an AVR microcontroller by letting the inputs float. \$\endgroup\$
    – ilkhd
    Aug 4, 2015 at 19:14

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