# Potentiometer schematic

Discussion in 'General Electronics Chat' started by uzair, Jan 14, 2008.

1. ### uzair Thread Starter Active Member

Dec 26, 2007
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I am pretty confused.Is there any fault in the above potentiometer figure, if you consider conventional current flow?

2. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
The confusing part is that there is no ammeter nor voltmeter in the circuit.

At the top, "E" is your battery or voltage source.

At the bottom is your potentiometer.

The lower the value of the resistance of the potentometer, the more current will flow in the circuit, which could be measured using an ammeter inserted into the wire on either side of the current loop.

Since in this schematic, the potentiometer is also RL (load resistor), all voltage "drop" will be across the resistor.

3. ### kubeek Expert

Sep 20, 2005
5,563
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Really interesting picture. Is that your work?
The only fault in that circuit is that it does absolutely nothing, except for variable draining of the battery.

4. ### uzair Thread Starter Active Member

Dec 26, 2007
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Consider that we have a voltmeter placed at the bottom line of the circuit, then what do you say about it?

5. ### beenthere Retired Moderator

Apr 20, 2004
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You will read the battery voltage. The pot is configured as a variable resistor, so all it will do is control current in the circuit.

6. ### kubeek Expert

Sep 20, 2005
5,563
1,083
Do you mean INSTEAD of the bottom line in the circuit?
Than it is a simple divider.

7. ### uzair Thread Starter Active Member

Dec 26, 2007
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No, not instead of bottom line, but in the bottom line.

8. ### Ron H AAC Fanatic!

Apr 14, 2005
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Then the battery current will be constant, and the voltmeter will read from zero to the value of E as you move the wiper from full left to full right.

9. ### Audioguru Expert

Dec 20, 2007
11,202
1,329
The circuit is a smoke producer when the resistance of the pot is turned down. It has nothing to limit the max amount of current when the resistance is turned down. The pot might even light up for a moment.

10. ### uzair Thread Starter Active Member

Dec 26, 2007
110
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If we consider the conventional current, i think there is a fault in this schematic.

11. ### uzair Thread Starter Active Member

Dec 26, 2007
110
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You mean this circuit is not a potentiometer?

12. ### Ron H AAC Fanatic!

Apr 14, 2005
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With the voltmeter in the bottom line, it is a potentiometer, not a rheostat, because voltmeters ideally draw no current.

13. ### uzair Thread Starter Active Member

Dec 26, 2007
110
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Hey everybody!!
I am trying to find out that do this circuit performs the function of potentiometer, when we consider the conventional current only??!!!

14. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
If you wish to be able to vary the output voltage, the wiper arm must not be connected to either side of the voltage source, and your load needs to be outside of the current loop. I'll upload a schematic in a few minutes.

Please look at the attached file, Potentiometer.png, by clicking on the link below. I've added an ammeter and a voltmeter to the schematic, and isolated the wiper arm so that the voltage potential can be measured.

Note that the current in the circuit is 1 mA, or 1 milliampere, or 0.001 Amperes. This is in keeping with Ohm's Law, as the voltage source is 10 volts, and the resistance in the loop is 10,000 Ohms:
I = E / R (Current = Voltage / Resistance)

Note that the potentiometer in the schematic is set to 40% of it's possible value of 10K Ohms, and the voltmeter reads 4.0 volts. In this particular circuit, were the potentiometer set to any value between 0% and 100% of it's rating, the voltage read from the wiper to the minus side of the Vsource would reflect that change. The voltmeter DOES use a very, very small amount of current, but it's so low (as modern DVM's are) that it's current draw does not reflect in the ammeter reading.

The GROUND symbol (triangle made of dashed lines) was added because that is a requirement of SPICE analysis programs in order to function; a ground reference point is a must.

Does that make it more clear?

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Dec 26, 2007
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16. ### uzair Thread Starter Active Member

Dec 26, 2007
110
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And basiclly my problem is the attachment of wiper of p. meter with the negative terminal in the circuit.Because my limited knowledge suggests that it should be joined at opposite terminal(+ ve).

17. ### Ron H AAC Fanatic!

Apr 14, 2005
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What are you wanting to do with a potentiometer? If you connect the wiper to either battery terminal, the pot will be destroyed when you turn the wiper all the way to the other end. Furthermore, when the wiper is connected to one end of the resistance element, the circuit is properly called a rheostat. If you leave the wiper disconnected, it forms a voltage divider between the two ends of the resistive element, and is a true potentiometer.

18. ### uzair Thread Starter Active Member

Dec 26, 2007
110
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In the link i have given above you can see the potentiometer, i am just asking that there is one node only in that schematic, if we simply put that single node on the upper edge(just as it was on the lower edge) then what would happen? If you see my schematic you will notice only one change between my schematic and the schematic of the e-book i.e. which i have discussed above.
I think by doing this, we destroy true functioning of p. meter as potential divider.

19. ### Ron H AAC Fanatic!

Apr 14, 2005
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681
I don't know how we can explain it any better. Several people have given you really good explanations. I suggest you reread all of them.
You are correct, though. When you connect the wiper to either end, you no longer have a potential divider. You just have a variable resistor, otherwise known as a rheostat. Have you read the Wikipedia entry on this subject?

20. ### uzair Thread Starter Active Member

Dec 26, 2007
110
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Ok I will do it thanks