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I'm afraid I have a rudimentary stupid question but here it goes...I have a circuit which produces a signal through a series of NOR gates. That signal, I input in a 2n2222 BJT to drive an LED. When I do so, the voltage of the gate drops to below 1V indicating a current leak. When I disconnect the BJT the voltage of the gate is 5V. I'm afraid there's something I'm missing about these guys, any help?

Here's the part of the circuit that's giving me problems: enter image description here

PS: In the same manner, if I connect an IRLZ44N MOSFET, I have no problem. So what's the obvious thing I'm missing here? haha

Thank you!

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    \$\begingroup\$ The base of a BJT looks like a diode to the driving circuit. It is expected that the voltage would be low. If you put a resistor in series with the base, the NOR output should remain in its normal high state while still allowing enough current to turn on the LED. What is the output voltage of the 4001B NOR gate? Is it 5V also, or ?? \$\endgroup\$ – mkeith Aug 4 '18 at 5:15
  • \$\begingroup\$ You hit it right on target. That's what happens when I try to play with wires in the wee hours of the night haha Indeed, the output of the NOR gate is 5V. Once I connected a 47K resistor in series with the base everything worked normally. Thank you for the fast reply! \$\endgroup\$ – Stratos Aug 4 '18 at 5:42
  • \$\begingroup\$ I mean, what is the output voltage SUPPOSED to be? Or, another way to ask the same question, what is the VCC of the NOR gate? \$\endgroup\$ – mkeith Aug 4 '18 at 5:45
  • \$\begingroup\$ It is 5V. So the output is supposed to be 5V on high which is when the 2N2222 is energized. Excuse my cloudy mind as I'm both tired and dangerously knowledgable but isn't the series resistor with the base going to be a "current limiting resistor"? It may well be I'm missing the basic theory here but the way I understand it, in layman's terms, is that the BJT amplifies current so I need a limiting resistor so I don't kill the 4001B. Am I in the ballpark? I know it worked, but I want to make sure I understand why :) \$\endgroup\$ – Stratos Aug 4 '18 at 5:51
  • \$\begingroup\$ Yeah. As I said, the base looks like a diode. Diode current goes up exponentially with voltage. It will blow up before it ever gets to 5V. In the case of the 4000 series NOR gate, the base is almost like a short-circuit load. For this type of circuit, assume the base will want to be at 0.7V when the transistor is on. And choose your base resistor so that base current is about 10% of the current going through the load you are switching. In other words, if the resistor was completely on, how much current would flow through the collector? Then pick your base resistor to get 10% of that (approx). \$\endgroup\$ – mkeith Aug 4 '18 at 5:56
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The base of a BJT looks like a diode to the driving circuit. It is expected that the voltage would be low. If you put a resistor in series with the base, the NOR output should remain in its normal high state while still allowing enough current to turn on the LED.

In this type of arrangement, where the NPN transistor is being used as a switch, you want the base current in the transistor to be around 10 percent of the collector current. In your case, the collector current would probably be around 13mA when the transistor is on, depending on the type of LED. So you can target 1.3mA for the base to make sure the transistor is on.

So that means (5V - 0.7V) / Rb = 1.3mA. So, Rb = 4.3V / 1.3mA = 3.3k. Just pick the closest value that you have to 3.3k.

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  • \$\begingroup\$ Thanks man, i appreciate the effort and speedy response! It appears that my understanding ended up in the correct ballpark. Thanks again! \$\endgroup\$ – Stratos Aug 4 '18 at 5:57

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