I do not understand how the AND gate circuit shown functions (with high measured as ~6V and low as ~0V at the voltmeter PR1). Primarily I think I'm confused about the interaction between both power supplies and how that affects the behavior of the circuit around the diodes.

With both switches open (as in diagram), the top most supply (labeled as V1) drops its 6V across resistor R1. This is also the case if one of the switches but not both are closed. Can someone explain this behavior?

AND gate

  • \$\begingroup\$ Your last sentence in your second paragraph is answered in your second paragraph. \$\endgroup\$ – Harry Svensson Feb 16 '18 at 8:45
  • \$\begingroup\$ the switches are "closed" in both positions ... you have to differentiate the two states by using different language \$\endgroup\$ – jsotola Feb 16 '18 at 8:45
  • \$\begingroup\$ @jsotola sorry, i think i've clarified now \$\endgroup\$ – 5uperdan Feb 16 '18 at 8:51
  • \$\begingroup\$ there is no interaction between the two power supplies because they are both at the same voltage ... you could remove V2 and run everything from V1 \$\endgroup\$ – jsotola Feb 16 '18 at 8:56

As long as one of your switches is open (means here to be connected to ground), the connected diode will pull PR1 to ground+0.7V Diode forward voltage. Only if both switches are closed you'll have the 6V on PR1.

  • 1
    \$\begingroup\$ oh wow, it really is simple, i wasn't really paying attention to the detail that switches aren't "open" or "closed". One route connects to the left side of the circuit and the other route connects to ground. Thanks \$\endgroup\$ – 5uperdan Feb 16 '18 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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