0
\$\begingroup\$


(source: electronics-tutorials.ws)

I recently constructed a NOR gate using the following schematic but when one of the transistors is set to HIGH an led I connected to the output of the gate glows brighter, but the other one works just fine. What's causing this?

\$\endgroup\$
9
  • \$\begingroup\$ First, why do you think this circuit implements a NOR gate? It looks like AND to me. Second, how was the LED connected? From the output to +6V or from the output to ground? \$\endgroup\$
    – The Photon
    Commented Nov 17, 2012 at 1:03
  • \$\begingroup\$ Sorry, wrong schematic, fixed it and the LED was connected at OUT to the ground of an arduino which supplied the power for the gate \$\endgroup\$ Commented Nov 17, 2012 at 1:16
  • \$\begingroup\$ The question is not clear to me. You connected an LED to the OUT terminal of the gate, and when one of the inputs (let's say A) is HIGH (so from the truth table, OUT should be LOW), the LED glows brighter, and the other one works just fine? What is the 'other one', and what would 'fine' behaviour be for you? \$\endgroup\$ Commented Nov 17, 2012 at 1:59
  • 1
    \$\begingroup\$ @Samuel, looks like NOR to me. The two transistors' c-e paths are in parallel, not series --- though it might be hard to see if you're looking at it on your phone or something. \$\endgroup\$
    – The Photon
    Commented Nov 17, 2012 at 5:47
  • 1
    \$\begingroup\$ @kaptainkernel, nothing in the schematic explains the behavior you're describing. Can you post a photo of the physical circuit? \$\endgroup\$
    – The Photon
    Commented Nov 17, 2012 at 5:48

1 Answer 1

1
\$\begingroup\$

It looks like you have a NOR gate now, albeit non-standard RTL input levels. TTL is two diode drops and HCMOS is Vcc/2 threshold. The saturated gain of a transistor is only 10 to 20. So choose R values for a gain of 10 for reliable current switching.

CMOS NOR Gate

This is a simplified CMOS NOR gate with fancy triangular inputs running at 100 & 110 Hz to give a 10 Hz beat frequency on the output. Purpose is to show the linear input threshold. This uses n and p Mosfets to give low impedance for both output states, although the high side usually has a resistor to current limit it.

\$\endgroup\$

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.