# voltage signal systems

Discussion in 'Homework Help' started by jimjdanger, Jun 30, 2015.

1. ### jimjdanger Thread Starter New Member

Jun 14, 2015
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How does the voltage go round this circuit?. I thought that the electrons came from the negative side of the battery coming round to the positive. How can I read the indicator when the electrons haven't gone through the potentiometer for me to get a reading?.
Any help would be appreciated.
Thanks, Jim.

2. ### WBahn Moderator

Mar 31, 2012
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You can look at this a couple of different ways. First, you are measuring voltage and not current and there doesn't have to be any current flow in order to have a voltage difference. Consider a tank of water on the roof of your house with a hose coming down to a valve on the ground. Even with the valve shut off (no water flow) there is still pressure in the hose just before the valve. The pressure is roughly analogous to voltage in that it is a measure of the potential to cause current to flow.

Another way to look at it is that your voltmeter has a very high resistance, such as 10MΩ, while the potentiometer might have a resistance of, say, 10kΩ. So almost all of the current goes through the potentiometer. You can view the potentiometer as being two resistors in series (not exactly in series since there is the wiper, but if the current flowing in the wiper is very small compared to the currents in the other two terminals, we can ignore this). So say your battery is 12V and the wiper is 3/4 of the way to the top (about what is shown in your diagram), then their will be 9V across the lower part of the pot and 3V across the top part. The voltmeter is connected to read the voltage across the lower part, so it will read 9V.

In approximate, but very close, terms you will have a current of 1.2 mA flowing through the potentiometer and 1.2 μA flowing in the meter.

3. ### MikeML AAC Fanatic!

Oct 2, 2009
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Forget electrons. Current flows from the positive pole of the battery (the long bar) into the top of the potentiometer. It divides at the wiper position; some flows out of the bottom of the pot, the rest flows out of the wiper to the + end of the meter. Current out of the bottom of the meter and the bottom of the pot add together and flow into the negative pole of the battery (the short bar). Current out of the positive pole must equal current into the negative pole.

btw- this is not the way most fluid level measuring systems (e.g. car fuel gauges) work. They do not use a three wire potentiometer; instead they have only one wire to a rheostat, the bottom end of which is grounded to the car body.

4. ### ian field Distinguished Member

Oct 27, 2012
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AFAICR: a rheostat is a 2 terminal device, so there is no bottom end to connect to the metal work, potentiometer is a shortened/(and/or fancy) way of saying potential divider, that is a 3 terminal device.

The last car tank sender I saw was a basic 2 terminal rheostat, but I couldn't state with any certainty that they were all that way - and it was a long time since I last took a car to bits, so the technology probably moved on since then.

5. ### WBahn Moderator

Mar 31, 2012
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You would connect one end to ground and the other end to the monitoring circuit and the monitoring circuit merely responds to the current flowing through the rheostat to ground.

6. ### ian field Distinguished Member

Oct 27, 2012
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But there is no "other end" to connect to ground - at least there wasn't on the one I took to bits to see how it worked.

In the days when I spent my time taking cars to bits, the instrument cluster voltage regulator was just a bi-metal strip with some turns of resistance wire wrapped around it (basically a simmerstat). The meter movement in the fuel gauge was also a bi-metal strip wrapped with heater wire - none of it was precisely calibrated to the extent of needing a potentiometer in the sender unit.

The last instrument cluster I opened had flexiprint and a house coded 3-terminal regulator - but I think some of the gauges were still bi-metal strip type.

Another peculiarity is some automotive time clocks were still clockwork till not all that long ago - a limit switch detected when the spring became unwound and started a motor to wind it up again.

7. ### WBahn Moderator

Mar 31, 2012
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Could you post a sketch of the circuit you are talking about? It sounds like you are saying that the sensing element had just one electrical contact, which I don't think is what you are trying to say. The alternative is that you are saying that both terminals went back to the instrument cluster, which seems quite possible but I would think that they would only do that if necessary.

The big advantage of these is that they weren't a constant drain on the battery. At least in some of the older models the clock winding motor was on the ACC circuit so if the key was out the clock would not get wound. Of course that often meant the clock stopped working if the car sat undriven for more than a few days. Since the clock would wind as soon as the key was turned few people ever noticed that the clock had actually stopped, so they just assumed that the clocks in cars couldn't keep time worth a damn. I know I never bothered to set mine and completely ignored it as a complete waste of time (no pun intended).

8. ### ian field Distinguished Member

Oct 27, 2012
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782

Somehow I never got around to shoving all the tools & grease aside and sketching out circuit diagrams.

Its the sort of simple thing I'd remember easily enough anyway - the bi-metal strip regulator (basically a simmerstat) supplied about 9V (average) rail for various instruments - the fuel gauge was yet another bi-metal strip in series with the sender rheostat.

That's all there was to it - eventually the simmerstat regulator was superseded by a 3-terminal regulator, AFAIK: the bi-metal gauges carried on for some time longer - I've no idea what they use now.

9. ### jimjdanger Thread Starter New Member

Jun 14, 2015
4
0

MikeML- I thought electrons went from negative side of the battery?. It said at the
of the 1st textbook that all circuits in the books used Electron flow noatation.

Also the circuit is in chapter 9 of textbook 1 on this site in voltage system signals if this picture that i've posted doesn't work.

How can the change of potentiometer resistance be shown in the indicator when it is on the wrong side of the circuit?.

Forgive my ignorance, i've only been doing this for a month.

10. ### WBahn Moderator

Mar 31, 2012
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Well, if you can't describe the circuit well enough to put together even a rough sketch there's no point trying to discuss it because what you have in mind and what I am picturing from your description are probably going to be two different things.

11. ### MikeML AAC Fanatic!

Oct 2, 2009
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I have the fuel bladder out of one of my airplanes as we speak. I took out the fuel sender just yesterday. It has one wire going to the gauge and a ground strap to airframe. Same in the fuel tank on my Jeep, which I had apart about a year ago...

Oct 2, 2009
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13. ### WBahn Moderator

Mar 31, 2012
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Which textbook? The E-book here?

Electrons are, by definition, negatively charged. So moving so many electrons from point A to point B actually moves a negative amount of charge from point A to point B, which means that a positive amount of charge has moved from point B to point A. In all but very rare cases, whether you have a positive charge carrier moving from B to A or a negative charge carrier moving from A to B is completely irrelevant since what is of interest isn't the flow of charge carries but rather the flow of charge.

Electrical current is measured in amperes which, by definition, is the flow of one coulomb of (positive) charge per second through whatever area is of interest (such as a wire). Conventional Current is merely current flow according to this definition, namely the flow of charge and not charge carriers.

People that use "electron current" try to have it both ways and end up being internally inconsistent -- the E-book here is no exception. They want to talk about 1A of current flowing from the negative terminal of the battery to the positive terminal but yet when asked how much charge has flowed into the positive terminal over a span of 10 seconds and they will tell you -10 coulombs (because they acknowledge that electrons carry a negative charge and aren't claiming that it should be positive). So how do you have a +1 coulomb of charge per second entering the positive terminal of battery for 10 seconds and yet end up with a negative amount of charge entering it? To make things work out they end up having to apply magical mystery minus signs here and there.

You can do the work using either convention, but the electron current convention often results in you applying mental minus signs where needed which increases the risk of making mistakes as circuits get more complex.

Learn and use Conventional Current.

What's on the wrong side of the circuit.

Start with a two-resistor voltage divider and get comfortable with that. Then replace the two resistors with a potentiometer and realize that all you have done is take a device that effectively has two resistors in series with a third wire attached to the junction between them. The key to understanding it is understanding that two-resistor voltage divider.

14. ### ian field Distinguished Member

Oct 27, 2012
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Three 2-terminal components in series - how hard can it be!!!!!

15. ### WBahn Moderator

Mar 31, 2012
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Then how hard can it be to sketch a diagram showing why it is not possible to ground one side of the sensor in the same way that most other circuits on a car are grounded?

16. ### ian field Distinguished Member

Oct 27, 2012
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It is possible to ground one end of a 2-terminal rheostat - and sometimes that's the way its done. Alternatively the rheostat could be in the supply feed between the regulator and the gauge.

A potentiometer would *HAVE* to have a ground connection, and I've never seen it done with a potentiometer.