Voltage drop accross resistor? - Page 1
Forum section: electronics / beginners

author: hhoka
author_rank: Newbie
timestamp: January 30, 2019, 03:38:45 pm
content: I'm having trouble with this circuit right here:https://imgur.com/a/AH4JsxFIt'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!

author: piguy101
author_rank: Regular Contributor
timestamp: January 30, 2019, 04:02:28 pm
content: 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?

author: rstofer
author_rank: Super Contributor
timestamp: January 30, 2019, 04:11:04 pm
content: 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

author: hhoka
author_rank: Newbie
timestamp: January 30, 2019, 04:30:54 pm
content: I just raised the supply to 5V and it appears to work now, no voltage drop.

author: rstofer
author_rank: Super Contributor
timestamp: January 30, 2019, 06:45:51 pm
content: 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 Vceand Vbedidn't prevent the circuit from working.

author: rdl
author_rank: Super Contributor
timestamp: January 30, 2019, 07:51:59 pm
content: The same example circuit was used in a Forrest Mimms book, except with 6 volts and 10k resistors.

author: rstofer
author_rank: Super Contributor
timestamp: January 30, 2019, 09:21:55 pm
content: As a textbook exercise, the circuit works and uses few parts. It is worth prototyping and taking measurements. It is not worth using in a practical circuit with a few levels of logic.The circuit is easy to understand and worthwhile as an exercise in "what happened to my voltage?". That's a worthwhile analysis for any transistor circuit.Another 2 transistor solution might drive the logic signals through diodes (1N914) with their cathodes connected to the base transistor through a resistor. Any signal passing through the diode would turn the transistor on and the collector would pull toward 0V. Obviously, the emitter is grounded.Again, this forms a NOR gate and another transistor is needed for inverting the output. The advantage is that regardless of the number of inputs (one diode each), the circuit only requires two transistors.It is also possible to make OR and AND gates without transistors using just diodes and resistors.http://people.seas.harvard.edu/~jones/es154/lectures/lecture_7/pdfs/215ln03.pdfThe Burroughs B5000 mainframe used a lot of diode logic:http://www.decadecounter.com/vta/articleview.php?item=1283

author: LukeW
author_rank: Frequent Contributor
timestamp: February 01, 2019, 05:15:12 am
content: 20k at the base seems a bit high.Start with a basic single-BJT switch and try making a high-side switch and a low-side switch circuit.In basic terms, for good results, use a NPN BJT in the low-side switch and a PNP BJT in the high-side switch. Then combine what you have learned to make the logic gate.Let’s assume 0 is 0V and 1 is 5V (or whatever)If you make this circuit with two PNP high-side switches you will turn the transistor on when that input is low - so what Boolean function will you get?