Basic Potentimeter question?

Why is it that 10kohm pot can take a voltage from the high of the input voltage down to zero.

But if I put a 68k ohm inline from a 3.1v source it only drops it to 2.9V?

I am guessing that a pot is really working like a voltage divider, not a single in-line resistor.


So that leads me to my next question.

I have a game where the player plugs connects 4 circuits. Normally you just hardwire a predefined path but I want to randomize the solution.

So I was originally planning on putting various size resistors on all 8 pins.

But my concern is if all 8 pins have a voltage divider, will I get unpredictable results when two voltage dividers are put in series.

. . . . . . . . . .

Originally I wanted to do this (conceptually)

voltage
reduced
by
Loc 1 .4v
Loc 2 .8v
Loc 3 1.2v
Loc 4 1.6v

loc 5 .02v
loc 6 .04v
loc 7 .06v
loc 8 .08v

so then I could determine which connection was made from any of the top 4 to any of the bottom 4
with an analogread().

ex:

Loc1 plugged into Loc5
3.3v source - (.4v drop + .02 drop) = 3.3 - .42 = 2.88v

or

Loc3 plugged into Loc7
3.3v source - (1.2v drop + .06v drop ) = 3.3v - 1.26v = 2.04v


if I do need to go the voltage divider route, any suggestions on some starting points to achieve my objective. I am aware of various web divider calculators, but there are a lot of possible scenarios. Any help to whittle down resistor values would be helpful.

Am I going about this correctly?

Capture.JPG

Capture.JPG

The pot is still a resistor with an added wiper (third pin) that can travel from the top of the resistance to the bottom. That would give you the input voltage at one end and depending upon how the potentiometer is wired in the circuit (say to ground) 0V on the other end. Ohms law still applies to the resistance to ground, that being the current across the resistance.

[post]But if I put a 68k ohm inline from a 3.1v source it only drops it to 2.9V?[/quote]

Again, ohms law still applies. If that resistor is from source to ground, it drops the entire 3.1V across the resistor. That means you have 3.1V on one side, and 0V on the other, provided the source can provide the 46uA without sagging. I'm not sure where you got 2.9V from.

As for the latter part of your question, I have no idea what you are looking for. I'm just not following what you are asking.

The results of your experiment with the 68k resistor and the 3.1 volt source will be dependent on the internal resistance of the source and the load you put across it, including the internal resistance of your meter.

If your objective with the game is to determine between which contacts the leads are connected, there may be a better way. Connect each contact to an Arduino pin. In a loop, define one pin as output and set it high. The remaining pins are defined as inputs and you test each one. If one of the input pins is high, you know it is connected to the pin currently defined as output. Now define the next pin as output and so on. You’ll need pull down resistors on the pins.

What you describe sounds like an analog keypad. But, I could be wrong.

There's also an example of this technique in the Arduino projects book.

Let me explain it better

Capture.JPG

Requirements:

8 inputs.

4 jumper wires.

4 correct complete circuits

4 pins.

Correct solution (4 jumpered pairs) are dynamically created with random() function.

User can press a button to see how many correct circuits they have so far.

(there can be 200+ possibilities)


So in the multi button video (resistors in series), that is only testing one circuit at a time. and is a many-to-one test.

This is a many-to-many game.

So to narrow down the complexity, 4 jacks will be feed 3.v with various resistors
and
4 jacks (with various resistors) will go back to arduino (esp32) pins.

The problem is when I put a resistor on each jack and complete a circuit, I am not getting a drop in voltage unless I do a voltage divider.

In the image, I have completed a circuit from 3.3v > 4.2k resistor > pin32 and the analogread(pin32) shows a value of 4095 (8 bit resolution).


My objective was to have an array of 16 values that represented the 16 combinations of analogread() values. Actually it was going to be a range allowing for fluctuations. This way I could do a random() function to put various patch solutions after each attempt and help the user by showing how many they currently have correct (think Mastermind, the board game).

There are 200+ combinations to the user since they don't know which 4 are 3.3v and which 4 are coming back into the esp32 pin.

6v6gt:
The results of your experiment with the 68k resistor and the 3.1 volt source will be dependent on the internal resistance of the source and the load you put across it, including the internal resistance of your meter.

If your objective with the game is to determine between which contacts the leads are connected, there may be a better way. Connect each contact to an Arduino pin. In a loop, define one pin as output and set it high. The remaining pins are defined as inputs and you test each one. If one of the input pins is high, you know it is connected to the pin currently defined as output. Now define the next pin as output and so on. You’ll need pull down resistors on the pins.

actually, this is a good idea but uses 8 pins. I tweaked a little and got 6 pins. I can leave 2 jacks unwired as red hearings to the player. they will never know. I will just ask them to complete three circuits rather than 4. .

Thanks. Simple and elegant. No resistors, bla bla bla.

If you want the full eight pins back for the game, you could employ an I2C port expander.

dougp:
If you want the full eight pins back for the game, you could employ an I2C port expander.

another good idea. I have i2c already with the lcd 20x4.


Ok one more question.

I will provide 4 different colored patch jumpers. I would like which color patch wire is in each port.

One way would be to throw a different level resistor in the middle.

Is there another method, perhaps with pwm? the constraint is there will be only 3.3v and no ground. anyway to change a pwm signal / duty cycle with some small inline device without ground?

DivinerGregg:
Is there another method, perhaps with pwm? the constraint is there will be only 3.3v and no ground. anyway to change a pwm signal / duty cycle with some small inline device without ground?

Sorry, can't help with that.

DivinerGregg:
Why is it that 10kohm pot can take a voltage from the high of the input voltage down to zero.

But if I put a 68k ohm inline from a 3.1v source it only drops it to 2.9V?

I am guessing that a pot is really working like a voltage divider, not a single in-line resistor.

I think the issue here is - you haven't properly described what you mean by "put 68k ohm inline from a 3.1v source it only drops it to 2.9V".

Eg. 'what' exactly drops to 2.9V? That is, what voltage difference are you measuring?

A circuit diagram and more info is needed here.

If you measure 3.3volt through a 68k resistor with an old analogue meter (50kohm/volt, 10volt scale),
then you would measure almost exactly 2.90volt.
You would get 3.278volt if you measure that with a DMM with a 10Megohm impedance.
Leo..

Wawa:
If you measure 3.3volt through a 68k resistor with an old analogue meter (50kohm/volt, 10volt scale),
then you would measure almost exactly 2.90volt.
You would get 3.278volt if you measure that with a DMM with a 10Megohm impedance.
Leo…

Thank you. So I am reading the data correctly with my el-cheapo multimeter.


And this goes back to my original question. Why can a 10k pot take the voltage to zero but a simple in-line 68k resistor only take the voltage down to 2.9v?

My assumption was a pot was nothing more than a variable resistor. But obviously that it is not the case. The pot functions more like a voltage divider.

I still don’t understand how a pot is fundamentally different than a resistor. A 10k pot gives a different output voltage than a 10k resistor.

Tinman has explained it so now I need to understand it. My understanding from Tinman is a pot has a ground that enables it to take the voltage to zero. In my case, the current flows back into the ESP32 pin and there is no ground so the voltage won’t drop anywhere near as much. I’m sure there is more to it than that (olm’s law), but if I had to explain it to someone who doesn’t know much about electronics, that is how I would explain it.

tinman13kup:
The pot is still a resistor with an added wiper (third pin) that can travel from the top of the resistance to the bottom. That would give you the input voltage at one end and depending upon how the potentiometer is wired in the circuit (say to ground) 0V on the other end. Ohms law still applies to the resistance to ground, that being the current across the resistance.

A potentiometer is a variable voltage divider by definition - the total resistance is constant, the
proportion in each part of the divider is continuously variable.

The original potentiometers were long pieces of wire forming a Wheatstone bridge: Wheatstone Bridge

A variable resistor is a resistor of variable value, almost invariably implemented as a pot. with
one terminal unused, since there is no advantage to selling a two-terminal version.

In the original bridges the idea was to null out the current to zero when the bridge is balanced,
so that the voltage ratio matches the wire-length ratio, thus measuring the potential (voltage).

This idea is still common today, volume controls rely on the amplifer input taking 'negligible'
current from the wiper, ie not disturbing the ratio appreciably.

The reason a pot gives you more voltages than an inline resistor is that its connected to more
than one potential in the circuit in the first place.

DivinerGregg:
And this goes back to my original question. Why can a 10k pot take the voltage to zero but a simple in-line 68k resistor only take the voltage down to 2.9v?

My assumption was a pot was nothing more than a variable resistor. But obviously that it is not the case. The pot functions more like a voltage divider.

I still don't understand how a pot is fundamentally different than a resistor. A 10k pot gives a different output voltage than a 10k resistor.

You're probably not going to understand what you're talking about until you show a diagram that illustrates the measurement that you're doing.

This following sentence sounds incoherent..... "Why can a 10k pot take the voltage to zero but a simple in-line 68k resistor only take the voltage down to 2.9v?"

Same with this one......... "A 10k pot gives a different output voltage than a 10k resistor."

It's probably a case of incorrect idea/understanding about potentiometers and resistors (while you are believing that you have the right idea about them). Just sketch a diagram of your circuit and show what you're measuring, and we'll point you in the right direction.

It's all about voltage dividers. Google it.

68k and the 500k internal resistance of your analogue meter make a 2-resistor voltage divider.
Voltage across the (500k) analogue meter is a 500/(68+500) ratio of the supply voltage.
500/568 * 3.3volt = 2.9volt.

When you remove that voltmeter (no divider), voltage on both sides of the 68k will be 3.3volt.
Or when you use a digital voltmeter with 10000k internal resistance, that voltage behind the 68k resistor will be closer to 3.3volt.

What is the impedance (and max input voltage) of your ESP-32.
If you don't know, you will have a hard time calculating/measuring dividers with an analogue meter.
Leo..

I think I got it now. Here is my schematic.

The schematic is voltage divider. The different resistors for each colored wire patch/jumper function just like a pot.

I use 2k pulldowns and added a different resistor for each jumper wire. I can now do all the requirements.

  1. I can see which specific contacts are made by cycling through defining pinmode to outputut and setting to high, then reading the other gpios with an analog read.
  2. 2k pulldowns mean I read 0 when nothing is connected (not floating)
  3. Adding a resistor to the patch / jumper gives good separation of values so I know what exact color of wire there is is.

This stage of the game is now 2 levels.
First the player tries different patch combinations. They press a button and I let them know how many patches are correct.

The second level is for the player to swap colored patch wires around which I can do with the different resistor values.

So problem solved.


Where I struggled is I don't know the current of the esp32 and based on omhs law I would need it.

I know how to calculate the drop for voltage dividers. I have done that before.

Tinman has explained it so now I need to understand it. My understanding from Tinman is a pot has a ground that enables it to take the voltage to zero. In my case, the current flows back into the ESP32 pin and there is no ground so the voltage won't drop anywhere near as much. I'm sure there is more to it than that (olm's law), but if I had to explain it to someone who doesn't know much about electronics, that is how I would explain it.

You are not following what I was saying. A pot does NOT have a ground. It has 2 leads for the resistor and 1 for the wiper. The resistor is just that. If it is a 10K pot, it will be 10K across those two leads, just as the 68K resistor you have is 68K across the 2 leads.

With a pot, you also get a wiper. Just like it sounds, it wipes across the surface of that resistive material. That turns it into a divider on the wiper. From the wiper pin to either of the two leads, the resistance will change as you turn or screw the pot (depending upon what type you have). The resistance can go from 0 ohms to 10Kohms. If you then take the meter lead off and attachit to the other pot lead, you would read the balance of the original 10K resistor. For eg; you adjust it so one way it reads 3Kohm from wiper to lead 1, then moving the meter lead to pot lead 3 would read the remaining 7Kohm. What you just did was read both sides of the divider.

If you take the resistor and the pot and connect both up with one end of the resistor to 5V, and the other to ground, BOTH will read the same voltage across them, being 5V. The difference at this point would be the current. Ohm's law says the 68K resistor will have 74uA through it and the 10K will have 500uA.

Now for the "wiper". Because it is variable, and there is no load attached, it will simply allow you to read a voltage somewhere between the 5V and 0V, depending upon where it is. Since there is nothing attached to it at this point, the current through the 10K resistor remains the same at 500uA.

Following this yet?

tinman13kup:
You are not following what I was saying. A pot does NOT have a ground. It has 2 leads for the resistor and 1 for the wiper. The resistor is just that. If it is a 10K pot, it will be 10K across those two leads, just as the 68K resistor you have is 68K across the 2 leads.

With a pot, you also get a wiper. Just like it sounds, it wipes across the surface of that resistive material. That turns it into a divider on the wiper. From the wiper pin to either of the two leads, the resistance will change as you turn or screw the pot (depending upon what type you have). The resistance can go from 0 ohms to 10Kohms. If you then take the meter lead off and attachit to the other pot lead, you would read the balance of the original 10K resistor. For eg; you adjust it so one way it reads 3Kohm from wiper to lead 1, then moving the meter lead to pot lead 3 would read the remaining 7Kohm. What you just did was read both sides of the divider.

If you take the resistor and the pot and connect both up with one end of the resistor to 5V, and the other to ground, BOTH will read the same voltage across them, being 5V. The difference at this point would be the current. Ohm's law says the 68K resistor will have 74uA through it and the 10K will have 500uA.

Now for the "wiper". Because it is variable, and there is no load attached, it will simply allow you to read a voltage somewhere between the 5V and 0V, depending upon where it is. Since there is nothing attached to it at this point, the current through the 10K resistor remains the same at 500uA.

Following this yet?

Watched a few youtubes. I get it. Thanks.

Just returning to the game that you are creating, you have shown 2 different type:
Type 1 in the OP where it appears that patch wires can go between any 2 sockets. Type 2 in later pictures seems to imply that the patch wires can go only from one group of 4 sockets into the other group of 4 sockets (which is an easier problem to solve).

Another way of implementing this without using special resistor values, and still identifying the wires used would be to have sockets like two pin audio sockets
In the plugs you'd build in a DS2401 serial number device which requires only 2 connections. That way, you would not even need to connect pairs of plugs with wires. You'd simply colour code them. 2 reds, 2 blues etc.

Update. The schematic posted worked like a champ. The only issue was I ran out of adc capable pins on the esp 32 (fyi this chip is the real deal and will only get better with more library support). Since the esp 32 has 7 touch capacitance pins that also double as adc inputs, I had to kill one requirement. So I have only 4 adc pins. That means I couldn't do the requirement of having all cables plugged in for a test.

Thanks to all who the posted advice. It is appreciated.