Sadly, as a computer science major, I never took EE201 so I'm a bit clueless on how to figure this out.
Take a look at the following:
If the pot is exactly in the middle of its rotation (assuming a linear pot), then I know that at point A my voltage is 2.5 volts. But what is it at point B? I can argue myself in two different directions. Either it's at the same potential or there is a voltage drop across the resistor.
The more I think about it and chew on it the more I think that maybe it depends on what's after point B? Like, if it's the input to a CMOS opamp (5-10 Mohm?) then the signal has quite a ways to go to get to ground. And if it's driving the base of a BJT (10k ohm?) then maybe point B would be at 1.25v?
(And forgive me for not building the circuit and measuring but I'm using up all of my breadboards in a massive Don Quixotian synthesizer project. )
Please help me calibrate my intuition
Moderators: adafruit_support_bill, adafruit
Please be positive and constructive with your questions and comments.
- westfw
- Posts: 2008
- Joined: Fri Apr 27, 2007 1:01 pm
Re: Please help me calibrate my intuition
exactly correct. It'll depend on the current through R6. (Va - 10k*I) If you're going into something like a CMOS op-amp, there will be very little current and B will be at the same voltage as A...The more I think about it and chew on it the more I think that maybe it depends on what's after point B?
- zener
- Posts: 4567
- Joined: Sat Feb 21, 2009 2:38 am
Re: Please help me calibrate my intuition
Yes. Although, in your second example the voltage at point B would be .77 volts. The reason is that you are also drawing your output current through the top half of the pot, so the midpoint of the pot is no longer 2.5V. To solve this you can either breadboard it, simulate it or do simultaneous equations (nodal analysis.)
- stkmtd
- Posts: 13
- Joined: Wed Mar 24, 2010 12:05 pm
Re: Please help me calibrate my intuition
You have to specify what the voltage at that point is relative to. If we're relative to ground, then you drop 2.5V across half the pot and have 2.5V at A (relative to GND).
Because you have no current path from B, I = 0.
V = IR
V = (0A)(10k)
V = 0V
I think that's the most appropriate answer. If B were connected to ground, you'd drop 2.5V across the 10k resistor, and B would be 0V (same node as GND).
I'm just beginning to study nodal analysis. Is it even possible to do nodal analysis on something that doesn't form a loop/have a current path?
Because you have no current path from B, I = 0.
V = IR
V = (0A)(10k)
V = 0V
I think that's the most appropriate answer. If B were connected to ground, you'd drop 2.5V across the 10k resistor, and B would be 0V (same node as GND).
I'm just beginning to study nodal analysis. Is it even possible to do nodal analysis on something that doesn't form a loop/have a current path?
- zener
- Posts: 4567
- Joined: Sat Feb 21, 2009 2:38 am
Re: Please help me calibrate my intuition
No. The 10K would be in parallel to the bottom half of the pot. B would be zero but you would not have 2.5 volts across the 10K. You would have about 1.1V across the 10K resistor in that case.stkmtd wrote:If B were connected to ground, you'd drop 2.5V across the 10k resistor, and B would be 0V (same node as GND).
There is no path? Through the power supply.stkmtd wrote:Is it even possible to do nodal analysis on something that doesn't form a loop/have a current path?
- stkmtd
- Posts: 13
- Joined: Wed Mar 24, 2010 12:05 pm
Re: Please help me calibrate my intuition
Happily corrected. You are right.Zener wrote:No. The 10K would be in parallel to the bottom half of the pot. B would be zero but you would not have 2.5 volts across the 10K. You would have about 1.1V across the 10K resistor in that case.stkmtd wrote:If B were connected to ground, you'd drop 2.5V across the 10k resistor, and B would be 0V (same node as GND).
(for the bottom)
(R_T)^-1 = (R_1)^-1 + (R_2)^-1
(R_T)^-1 = (1/25000) + (1/10000)
(R_T)^-1 = 0.00014
R_T = 7k1
7k1 * (5V/32k) = 1.1V
I meant using point B. If you went from 5V to B there's nowhere to go from there. You could look at the loop from 5V to GND, but how would you involve point B in that long enough to isolate it in a KVL equation?Zener wrote:There is no path? Through the power supply.stkmtd wrote:Is it even possible to do nodal analysis on something that doesn't form a loop/have a current path?
- zener
- Posts: 4567
- Joined: Sat Feb 21, 2009 2:38 am
Re: Please help me calibrate my intuition
If B is floating then it is the same potential as A.
-
- Posts: 58
- Joined: Sun Dec 06, 2009 11:12 pm
Re: Please help me calibrate my intuition
There is some voltage drop across R6. Remember the ever-important Ohm's law, V = IR.
The voltage drop across R6 = IR, where R = 10 k and I = whatever current you're drawing through there.
Voltage B = Voltage A - voltage drop.
With nothing else connected up to R6, the current through R6 is zero, and therefore the voltage drop is zero, and therefore voltage B equals voltage A.
The voltage drop across R6 = IR, where R = 10 k and I = whatever current you're drawing through there.
Voltage B = Voltage A - voltage drop.
With nothing else connected up to R6, the current through R6 is zero, and therefore the voltage drop is zero, and therefore voltage B equals voltage A.
- stkmtd
- Posts: 13
- Joined: Wed Mar 24, 2010 12:05 pm
Re: Please help me calibrate my intuition
The math makes sense... but I still want to breadboard this one now. My intuition is thoroughly violated (it's not usual to see that voltages on 2 sides of a resistor are equal, at least from what I've experienced so far)minerva wrote:There is some voltage drop across R6. Remember the ever-important Ohm's law, V = IR.
The voltage drop across R6 = IR, where R = 10 k and I = whatever current you're drawing through there.
Voltage B = Voltage A - voltage drop.
With nothing else connected up to R6, the current through R6 is zero, and therefore the voltage drop is zero, and therefore voltage B equals voltage A.
- cstratton
- Posts: 294
- Joined: Wed Sep 29, 2010 3:52 pm
Re: Please help me calibrate my intuition
Just remember that the input impedance of whatever you measure it with will effectively be a load resistance to ground. If that's a megaohm and the test resistors are in the 10's of K you won't notice it. 100k test resistors and you'll start to see an error.stkmtd wrote:(it's not usual to see that voltages on 2 sides of a resistor are equal, at least from what I've experienced so far)
I suppose you could measure across the series resistor, in which case you shouldn't see any voltage drop since there wouldn't be a load.
-
- Posts: 127
- Joined: Wed Sep 23, 2009 5:48 pm
Re: Please help me calibrate my intuition
Take a resistor out of your parts box, put your meter in voltage mode, and measure the voltage across the resistor, without connecting the resistor to anything except the probes. If your meter doesn't read 0.0V, your meter is broken.(it's not usual to see that voltages on 2 sides of a resistor are equal, at least from what I've experienced so far)
Now, connect one side of the resistor to a voltage source, leave the other side unconnected, and measure the voltage across the resistor. Once again, it better be 0.0V across the resistor.
Ohm's law says the voltage across the resistor will be proportional to the current flowing through the resistor. If there's zero current, the voltage across the resistor will be zero. And if one side of the resistor is unconnected, Kirchoff's current law says no current can flow through the resistor. Therefore, the voltage on both sides must be equal.
As others have pointed out, as soon as you connect node B to something that can source or sink current, then things get more complex. There will be a voltage across the resistor proportional to the current, and furthermore, the potentiometer is no longer a simple voltage divider, because the current in the top half isn't the same as the current in the bottom half. Kirchoff's current law says that the current going through the resistor creates an imbalance between the currents in the top and bottom half of the potentiometer.
- adafruit_support_mike
- Posts: 67454
- Joined: Thu Feb 11, 2010 2:51 pm
Re: Please help me calibrate my intuition
Bit late to the party, but nobody else seems to have mentioned this: Check the internet for 'Thevenin equivalents'.. that's the intuition tool you want.
Basically you can reduce the entire circuit to a voltage source in series with a single resistor.
In this case, the first step is to find the voltage at point A.. you gave an example of 2.5v, right in the middle of the pot's range. In that case, your Thevenin equivalent voltage is 2.5v.
The next step is to find the total resistance looking into the pot from R6. To do that, you short the voltage source to ground, thus putting the two halves of the pot in parallel with each other. Two 25k resistors in parallel have a combined resistance of 12.5k, so what R6 sees at its left end is electrically equivalent to a 2.5v battery followed by a 12.5k resistor.
To characterize the right end of R6, you simply put R6 in series with the Thevenin equivalent for the pot. A 12.5k resistor in series with a 10k resistor gives 22.5k total resistance, so your whole circuit is electrically equivalent to a 2.5v battery followed by a 22.5k resistor.
So what's the voltage at point B? It depends entirely on what comes next. If you just short point B to ground, you get the short-circuit current for the Thevenin equivalent circuit (2.5v / 22.5kR == 111uA), also known as the Norton equivalent current (A Thevenin equivalent is always a voltage source in series with a resistor. A Norton equivalent is always a current source in parallel with a resistor. If you know one, it's trivial to find the other). In other words, the most current your circuit can possibly supply is about 100 microamps.
(BTW - shorting the 10k will cause voltage at point A to drop as well.. to 1.11v (10k * 111uA). You can check that by going back to the original circuit and putting the 10k in parallel with the 25k of resistance at the bottom of the pot. 25k in parallel with 10k is roughly 7k, so you have a voltage divider with 25k on top and about 7k on the bottom. 25+7=32, so the voltage across the bottom of the divider will be 7/32 of 5v. That's 7*5/32 or 35/32v, which is about 1.1. That's close enough to confirm the accuracy of the Thevenin equivalent circuit)
If you connect another 10k resistor between point B and ground, you get a voltage divider. The voltage at point B will be roughly 1/3 of 2.5v (2.5v * 10k / (10k + 22.5k)) or about .77v. If you connect a 50k resistor, the voltage at point B will be roughly 2/3 of 2.5v (2.5v * 50k / (50k + 22.5k)) or about 1.72v.
If you change the pot so the voltage at point A is 4v, its Thevenin equivalent resistance changes too. You get a 10k resistor in parallel with a 40k, which equals 8k. Putting R6 in series with that gives you an equivalent circuit made of a 4v voltage source in series with a 18k resistor. The Norton equivalent current would be 222uA.
Cranking the pot all the way up so the voltage at point A is 5v demonstrates why you put the two halves of the pot in parallel. If the left end of R6 connects directly to the voltage source, the resistance across the pot is irrelevant. The two components are in parallel.
At any rate, Thevenin equivalents give you a good (and precise) intuition for the way a resistive circuit will behave.
Basically you can reduce the entire circuit to a voltage source in series with a single resistor.
In this case, the first step is to find the voltage at point A.. you gave an example of 2.5v, right in the middle of the pot's range. In that case, your Thevenin equivalent voltage is 2.5v.
The next step is to find the total resistance looking into the pot from R6. To do that, you short the voltage source to ground, thus putting the two halves of the pot in parallel with each other. Two 25k resistors in parallel have a combined resistance of 12.5k, so what R6 sees at its left end is electrically equivalent to a 2.5v battery followed by a 12.5k resistor.
To characterize the right end of R6, you simply put R6 in series with the Thevenin equivalent for the pot. A 12.5k resistor in series with a 10k resistor gives 22.5k total resistance, so your whole circuit is electrically equivalent to a 2.5v battery followed by a 22.5k resistor.
So what's the voltage at point B? It depends entirely on what comes next. If you just short point B to ground, you get the short-circuit current for the Thevenin equivalent circuit (2.5v / 22.5kR == 111uA), also known as the Norton equivalent current (A Thevenin equivalent is always a voltage source in series with a resistor. A Norton equivalent is always a current source in parallel with a resistor. If you know one, it's trivial to find the other). In other words, the most current your circuit can possibly supply is about 100 microamps.
(BTW - shorting the 10k will cause voltage at point A to drop as well.. to 1.11v (10k * 111uA). You can check that by going back to the original circuit and putting the 10k in parallel with the 25k of resistance at the bottom of the pot. 25k in parallel with 10k is roughly 7k, so you have a voltage divider with 25k on top and about 7k on the bottom. 25+7=32, so the voltage across the bottom of the divider will be 7/32 of 5v. That's 7*5/32 or 35/32v, which is about 1.1. That's close enough to confirm the accuracy of the Thevenin equivalent circuit)
If you connect another 10k resistor between point B and ground, you get a voltage divider. The voltage at point B will be roughly 1/3 of 2.5v (2.5v * 10k / (10k + 22.5k)) or about .77v. If you connect a 50k resistor, the voltage at point B will be roughly 2/3 of 2.5v (2.5v * 50k / (50k + 22.5k)) or about 1.72v.
If you change the pot so the voltage at point A is 4v, its Thevenin equivalent resistance changes too. You get a 10k resistor in parallel with a 40k, which equals 8k. Putting R6 in series with that gives you an equivalent circuit made of a 4v voltage source in series with a 18k resistor. The Norton equivalent current would be 222uA.
Cranking the pot all the way up so the voltage at point A is 5v demonstrates why you put the two halves of the pot in parallel. If the left end of R6 connects directly to the voltage source, the resistance across the pot is irrelevant. The two components are in parallel.
At any rate, Thevenin equivalents give you a good (and precise) intuition for the way a resistive circuit will behave.
-
- Posts: 46
- Joined: Tue Dec 22, 2009 11:25 pm
Re: Please help me calibrate my intuition
Wow! Lots of really great info here. I have a ton of questions but I think I know where to go look for the answers now. I looked up Thevenin equivalent circuits, fearing big complicated math but it's not too bad at all. And my circuits are simple so that helps.
Thanks! If I have any more questions I'll come back and ask.
Thanks! If I have any more questions I'll come back and ask.
Please be positive and constructive with your questions and comments.