0
\$\begingroup\$

I am absolute beginner and I purchased SN74HC08 Quad 2 input AND gate . Hoping it will work as expected I connected it to the breadboard and used the 1A and 1B inputs connected to a push button and the 1Y output connected to a LED with a 220ohm resistor to the ground. The Vcc and GND are connected properly using Arduino 5V and GND pins. But when I power the circuit the LED is always high. it doesn't seem to care about the inputs. When I read the datasheet and some articles, they say that all unused pins must be either Vcc or GND. I did as per the instructions but still the gate wont work. I finally decided to post this question on SO.

If you see the attached image the Vcc on the IC is not connected and neither are the 1A and 1B inputs. all other pins are connected to Vcc and GND pin to ground. but the LED is still on.

Please help me I confused as to why this doesn't work. Did I damage the IC??

enter image description here

\$\endgroup\$
3
  • 2
    \$\begingroup\$ You have so many wiring errors, but it might not be damaged yet. a) Vdd pin 16 (open ) b) Outputs shorted to Vdd (dd=drain Vcc= collector old school TTL) c) no switches wired up to input with bias resistors (1k~1M) to 0V and jumper to 0V removed. With NO normally open switches High=1=5V and LED to 5V driven on low side. you now have negative Logic NAND gate = OR gate by De Morgan's rule. So then either Input ON turns ON LED as logic OR \$\endgroup\$
    – D.A.S.
    Commented Oct 14, 2019 at 18:32
  • \$\begingroup\$ Thank you for pointing out all the errors. I will fix all of them and post my response. \$\endgroup\$
    – Nilesh
    Commented Oct 14, 2019 at 18:41
  • 1
    \$\begingroup\$ Thank you again. I fixed the errors you mentioned and got it working. \$\endgroup\$
    – Nilesh
    Commented Oct 15, 2019 at 3:19

1 Answer 1

Reset to default
1
\$\begingroup\$

CMOS inputs have protection diodes to ground and supply pins. In essence you are powering the chip via the inputs that are connected to 5V.

Also the inputs must be either GND or 5V, it must not be disconnected. A simple pushbutton is normally disconnected and you connect by pushing, so it is not enough. At least you need a resistor for each pushbutton to set the voltage while button does not connect.

Worst thing is you have connected some of the outputs to supply voltages. Unused outputs must be left unconnected. The chip might drive too much current and break. So disconnect them immediately.

\$\endgroup\$
6
  • 2
    \$\begingroup\$ And "outputs" do not need any special treatment (do not connect unused outputs.) \$\endgroup\$
    – rdtsc
    Commented Oct 14, 2019 at 18:20
  • \$\begingroup\$ Indeed! Will edit. \$\endgroup\$
    – Justme
    Commented Oct 14, 2019 at 18:23
  • \$\begingroup\$ I will try this today. That is definitely stupid of me to connect outputs to Vcc. i will take care of that. \$\endgroup\$
    – Nilesh
    Commented Oct 14, 2019 at 18:30
  • 1
    \$\begingroup\$ @Nilesh, don't be hard on yourself. You're interested in learning and you're working at it - it's great to see. As a next step, try editing your question and adding a schematic of your circuit. The schematic editor here is easy to use and it is valuable learning. Sketch it on paper first if it helps and you'll find plenty of examples for it on the Internet. \$\endgroup\$
    – TonyM
    Commented Oct 14, 2019 at 20:51
  • \$\begingroup\$ Thank you @TonyM for the motivation. I will use of the schematics in my next post. For this question, I was just testing the AND gate and could not get it to work. \$\endgroup\$
    – Nilesh
    Commented Oct 14, 2019 at 21:06

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.