Electronics > Beginners
Voltage drop accross resistor?
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hhoka:
I'm having trouble with this circuit right here: https://imgur.com/a/AH4JsxF It's supposed to be an OR gate but it doesn't seem to be working correctly. The input on B reads 2V on my multimeter but only 0.6V past the resistor. This is not enough to trigger the transistor, so the voltage needs to be higher, but I can't figure out why it's dropping so much. I apologize if this is an extremely basic/obvious/stupid question, but I'm only a beginner in electronics. Thanks!
piguy101:
What are you measuring your voltage with respect to? The ground shown it the circuit? Do you have the 4.7 kΩ resistor in the circuit?
rstofer:
That's a horrible circuit!  At low voltages...

Start at the emitter and assume the transistor is saturated such that Vce = 0.2V so the emitter is 2.8V.  The base needs to get to 0.6V above the emitter to saturate the transistor which is 3.4V or 0.4V above the 3V rail.  Not going to happen.

Just 2 items to consider:  What is the voltage drop across the transistor, best case, and what is the base emitter voltage.

Another way to look at it:  If the emitter voltage is going to be 3.0 - 0.2V or 2.8V, the base has to be 3.4V.

The way this gate is usually used is with a much higher Vcc than the assumed logic levels.  If Vcc was 12V then it isn't necessary to get the transistors into saturation to have a logic level of, say, 10V.  When you deal with low voltages, the base emitter voltage and the collector emitter voltage become a significant percentage of Vcc
hhoka:
I just raised the supply to 5V and it appears to work now, no voltage drop.  :-//
rstofer:
What you probably have, given a 5V Vcc and a 5V input is an output of around 4V.  Which goes to the next logic gate in the chain and results in an output of 3V and so on.

I don't know if you are using LTspice but I modeled just one of the transistors (the other doesn't change the results) and attached the .asc file.

A better configuration might be to use two transistors in parallel sharing a common resistor at the collector with the emitters grounded.  Turning either input on causes the shared output to go low.  This will be a NOR gate - the output goes low when either input goes high.  Then you run this output of these transistors to the base of another transistor (through a resistor, of course) to invert the output and you have an OR gate.  The point of all this is to keep the logic voltages at some nominal Vcc and Gnd.  It does take an extra transistor for the inversion.

As an exercise, you have what you need.  You just needed enough voltage such that Vce and Vbe didn't prevent the circuit from working.
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