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I have the above IC and I want to check them before using them. I don't have access to a lab currently and I only have access to the usual stuff (LED, resistors) and a simple ammeter/voltmeter. What can I do to test them? Obviously(?) the below circuit doesn't work.

enter image description here

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3 Answers 3

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7400 series ICs take a bit of getting used to, as they are not as straightforward as, for instance, 74HC00 types.

Here are circuits to use with, for instance, the 7404 and 7400.

schematic

You'll note that the input switches provide a LOW input when closed, and the output LEDs are ON when the output is LOW. If this gives you too much trouble mentally, use unused inverters to switch the input and output polarities. But keep the switch and LED connections as they are.

TTL inputs are best driven by pulling them down or letting them be pulled up with a (obvious name, here) pullup resistor. usually in the range of 1k to 10k. And TTL outputs are much better at sinking current rather than sourcing it, so don't try to drive an LED output to ground.

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If your parts are really 7404, 7408 and 7432 (with no letters other than possibly L or LS in the middle, like 74LS04) they are bipolar TTL logic, and require a 5 volt power supply for correct operation.

The inputs of bipolar TTL parts will usually appear as logic High (1) when not connected. You must connect the inputs directly to ground to have them seen as a Low (0).

If the parts have a C in the middle (74C04 or 74HC04), they are CMOS, and the inputs are very high impedance, and will be in an unknown state when not connected - they must be connected to the positive supply for a High, and to ground for a Low.

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  • \$\begingroup\$ Hd74hc04p, hd74hc32p, sn74hc08n \$\endgroup\$
    – studious
    Commented Apr 11, 2016 at 17:33
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    \$\begingroup\$ It is important to give the correct part numbers! Those parts are CMOS, so my last paragraph applies. The 74HC parts can use a power supply between 2 volts and 6 volts. \$\endgroup\$ Commented Apr 11, 2016 at 17:38
  • \$\begingroup\$ Then shouldn't the diagram I drew above work? \$\endgroup\$
    – studious
    Commented Apr 11, 2016 at 17:41
  • \$\begingroup\$ You do need a current-lilmiting resistor (470 Ohms or so) in series with the LED, but with pin 1 tied to Vcc, the output of the inverter on pin 2 will be Low, so the LED won't light. Also, ALL inputs on a CMOS chip (even if you aren't using the associated gate) MUST be connected to either Vcc or Ground (or to a logic output). \$\endgroup\$ Commented Apr 11, 2016 at 17:51
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Seems like you got an arduino. You can proceed like this for 7404. Get the pinout correct. Here is the pinout diagram:

7404 IC

Wire up pin 7 to GND and pin 14 to VCC of arduino. You are halfway through.

Connect an LED to pin 1 and pin 2 each (along with current limiting resistors. 470 ohms will work fine). Now take a wire and connect pin1 to VCC. LED on pin1 should light up and LED on pin2 should remain off. Now remove the wire and connect pin1 to GND this time. LED on pin1 should be off and LED on pin2 should light up. This verifies a working gate.

You need to repeat it six times to check all gates.

Similarly, you can think about all other ICs. It's just logic High and Low. Write the truth table, get the pinout and follow the same pattern.

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  • \$\begingroup\$ Where should I connect the LED on pin 2 (of 7404) ? \$\endgroup\$
    – studious
    Commented Apr 11, 2016 at 18:18
  • \$\begingroup\$ Positive of LED to pin 2. Negative to the GND via a resistor. i.e. Negative to resistor, resistor to GND. \$\endgroup\$ Commented Apr 11, 2016 at 18:25

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