# Potentiometric Voltmeters/ Null Detectors - HELP!

Discussion in 'General Electronics Chat' started by QCesarJr, May 1, 2009.

1. ### QCesarJr Thread Starter New Member

May 1, 2009
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I've been trying to understand this section, but I am thoroughly stuck.

How does Null Detection work? Here's what I've gathered:

- The Voltage entering R2 is 24V; exiting is 12V, equalling a 12V drop.

- What should the Adjustable Voltage Source be set to?

- If its at 24, theres no difference in Voltage along the bottom path, but wouldn't 24V be dropped across the Voltmeter, allowing for 0V to enter point labeled 2 and 12V to be entereing point labeled 1?

- On the other hand, how is the Voltmeter not forming a parallel resitance with R2? Is it because the A.V.S. is indeed set to 24V, effectively stopping any current flow from heading from the A.V.S to R2? If this is the case, it just goes back to the previous point's question.

Ugh. I hate being stuck; I just can't move on if I am. I'll go learn some Ruby Programming while someone (hopefully) clears this up.

Thanks everyone for your help in advance, and sorry to post such a n00b question.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Suppose (for the moment) that you break the circuit at the point where the arrow meets the terminal (node) 1 - junction of R1 & R2.

What would the voltage across R2 be? It would be 12V since R1 & R2 form an equal value voltage divider of the 24V source.

For a null reading to occur when you reconnect the circuit, the AVS would have to equal 12V so that no current can flow between nodes 1 & 2 via the "null detector".

Because you have this null condition it doesn't matter that the Voltmeter + "null detector" sits in parallel with R2. No current flows - therefore no loading effect occurs.

What you would possibly notice is the effect of the Voltmeter loading as you approach the null condition when adjusting the AVS. It's worth thinking about what that process would be like if the Voltmeter resistance was 10kΩ vs 10MΩ for instance. Not that's its likely you would use/find a 10kΩ Voltmeter these days!

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Hmmm - silly question ....

In reality - If the AVS is an ideal voltage source there would be no effect on the null process since the AVS itself shunts the Voltmeter.

4. ### QCesarJr Thread Starter New Member

May 1, 2009
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I EDITED THIS POST AND MADE SOME CORRECTIONS TO MY QUESTIONS. I MISWROTE THE FIRST TIME AROUND

Thanks for the reply. I mostly comprehend the concept. I'm still a little lost, however:

With the A.V.S. set to 12V, what happens with that difference in Voltage between the two voltage sources?

Concerning the Voltmeter creating a parallel resistance with R2: as I understand, a Voltmeter includes a built in resistance (whether from its wires or by multipliers). Wouldn't this resistance form a Parallel Resistance with R2?

Okay, now, lets say that indeed there is no loading from the voltmeter (which I assume is true, I just don't understand, as explained above ). As you pointed out, Node 1 does indeed have a 12V Potential due the original 24V being cut in half by R2. However, on the voltmeter's side, 12V are passing through only 1 resistor (the voltmeter's built in resistance) which would effectively give all of the path connected to Node 2 a potential of 0V. Obviously, when the voltages at the Null Detector are compared, Node 1 will be bringing in a Voltage of 12V while Node 2 will bring in 0V. Apparently, this is somehow NOT the case.

I know my reasoning has to be wrong somehow, I just don't see it!

(Thanks for responding so fast, by the way.)

Last edited: May 1, 2009
5. ### eblc1388 AAC Fanatic!

Nov 28, 2008
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This only occurs if the voltage at node 1 is exactly the voltage at node 2.

If this condition is not met, i.e. A.V.S is set to a voltage either higher or lower than 12V, then there is certainly a loading, via current through the NULL detector.

The method works like this. The Null detector is not connected to node 1. The operator "guess" about the voltage at node 1 and setup the A.V.S. to this voltage. He then briefly connects the NULL detector to node 1 and observes if there is any movement/reading on the NULL detector. If there is, then he opens the connection again.

He then either increase/decrease the A.V.S. voltage and repeats the same procedure until he reaches the point that there is nothing happens when he connects or disconnect the NULL detector.

At this moment, the A.V.S. will be set to the SAME voltage as node 1 and there is NO loading on the circuit as there is no current in the NULL detector.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I understand what you are saying. You think Node 2 is at zero volts when the AVS is set to 12V.

This is not the case.

The voltage at Node 2 at is always equal to the AVS setting. Connecting the meter or any circuit to a voltage source does not collapse the source to zero volts. The only thing which will collapse a source to zero volts is a short circuit.

Imagine you connect a 1.5V battery to a small light globe. The globe lights. You take a voltmeter and measure the voltage across the battery. A fresh battery will read give a reading of 1.5V. The same voltage appears across the globe as well. The battery (source) voltage equals the globe (load) voltage.

7. ### QCesarJr Thread Starter New Member

May 1, 2009
5
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Sorry everyone, but even with all your explanations, I still see the same thing in my head. (The image is also attached, in case its not large enough).

I do understand that connecting a meter across a circuit will not bring the source voltage down to zero, but it WILL drop the voltage. And, in a case where a meter is the onnly resistance, it will drop the entire voltage acroos that one resistor, fulfilling KVL, right?

This is what I am imagining. Obviously, in my drawing Null Detection doesn't work, so its wrong. So, can you guys (and girls) please show me where my errors are on my schematic, please?

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8. ### eblc1388 AAC Fanatic!

Nov 28, 2008
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Ok. I'll try.

Voltage is not acting like current. Its a measurement of potential difference across any two points.

To ease calculation, a Common or reference point is usually assigned first. Everything is then measured w.r.t. this reference or common. This reference can be any point on the circuit but usually the negative terminal of the power source is selected, as shown in the image.

After this common is assigned, then every other voltage figure now got real meanings. It means its potential w.r.t. the common.

If one choose another point as common, one can. Every calculation and current value will ends up be the same, only now the voltage figures are not because a difference common point is used.

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9. ### QCesarJr Thread Starter New Member

May 1, 2009
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Thanks very much for those last two images. They helped me understand. I completely see how it works now. I also had to ask my dad to explain to me how the thing worked, but the images completely helped. There's a couple of things I would like to point out:

This model, using Conventional Flow methods, makes sense %100. 12V are dropped across R1. Then, with the AVS set to 12V as well, the Null Detector finds no difference in voltage, showeing that the 2 volatges are equal. 12V are dropped across R2 and 12V are dropped across the Voltmeter. The bottom path is common, and BAM! Nothing but magic. Well, okay, not magic, but science.

Sometihng my dad had to explain to me was this: Current is not actually passing THROUGH the Null Detector, but just being DETECTED or MEASURED. This is how the Voltmeter is not actually in Parallel with R2. Is that the proper way to explain it? Or does anyone have better wording for it?

So, this first image is great. Thanks.

However, when we come to the image using actual Electron Current Flow, (as is used in the E-Book) we have problems.

In this one, we have -24V moving across. However, in this scenario, it seems that the Original Voltage overtakes the AVS voltage, making that whole path Common to -24V. In this case, it also seems that R2 is indeed in parallel with the Voltmeter, since there is nothing separating them. This casues 2 problems. 12V will not be dropped across R2 because of the loading caused by the Voltmeter. And, even IF the Voltmeter caused no loading: R2 would drop 12V, allowing for the Null detector to register -12V. However, the -24V across the Voltmeter have to be dropped as well. However, ALL -24V would be dropped here, causing the resulting voltage to be 0V. The Null detector would register a difference in Voltage: -12V entering Node 1, and 0V entering Node 2. Obviously, this seems like it doesn't work. However, its still the same setup. Can anyone explain why it doesn't work when viewing it one way, but it does when viewed from another way?

10. ### eblc1388 AAC Fanatic!

Nov 28, 2008
1,543
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You are nearly there, just a tiny little step remains.

The circuits causes you confusion because they are connected together by the null detector. Each of them has voltage source and there are interactions between each other.

I'll give you another two images where I had broken off one connection of the null detector. Now, even after I have done that, in the first image, all the voltages will still be the same and not affected by my action. So all the "blue" voltage readings are all correct. Try to understand this circuit first.

Then in image below I have increased the AVS voltage a bit more to 12.4V. Please note how the voltages changes and the voltage on the R1/R2 node remains unaffected by this increase in AVS voltage.

If I then re-connect the null detector, oh dear, then there will be current through the null detector(one end of it is -12V, the other end is -11.6V) and the null detector "will load" the R1/R2 node and changes its true voltage in doing so. This is precisely the condition that we don't want to happen.

So what should we do? We break the null detector connection again, change the AVS voltage up or down, then do the test again to see if there is any reading in the null detector. If by luck we have set the AVS to exactly -12V, like in the image above, then when we reconnect the null detector, nothing happens. We have found out the true voltage at node 1.

A null detector will load the circuit if there is current passing through it, in contrast to what your dad had told you. Sorry dad!.

I'll show you one additional image where I have chosen the common to be on the side of the null detector. If I made this choice of common to the first image of this post, then node1 voltage =0V and the voltage on the 24V battery is +12V and -12V respectively. In this last image, after I increase the AVS to 12.4V, the node1 voltage "will change" along with the changing of the AVS voltage even if the null detector is disconnected. You can see that the node1 voltage has changed from 0V(AVS=12V) into -0.4V(when AVS=12.4V) . All I'm telling you is voltage is always a relative measurement.

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11. ### QCesarJr Thread Starter New Member

May 1, 2009
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First, let me say that you are GREAT for explaining this to me. I know how it is explaining simple things to someone when you already know so much. I know you have a lot patience, so I thank you.

I understood all of your circuits and all of the concepts. I see that voltage is relative too!

Now, I just ask if you would explain 2 details that are bugging me about the circuit.

1. When the original voltage source (-24V, in this case) meets the Adjustable voltage source (-12.4V), what happens to the -12.4V? It seems they Original Voltage Source (OVS, from now on) overtakes the AVS. That is why I get confused as to the Voltmeter loading the bottom Resistor. But I am guessing that eventhough the OVS overtakes the AVS, its still the AVS supplying all Voltage to the Voltmeter? Could you explain this a little more please?

2. In this (same) circuit, I used your altered AVS setting of +/-12.4V so that I could understand the loading you say will indeed take place when there is a voltage difference measured by the Null Detector. In this circuit, the Null Detector will obviously measure a difference of +/-0.4V (depending on how the Null Detector is actually connected). You stated this will indeed cause loading. Will you please explain HOW it actually causes loading. I tried figuring out, but my brain just did a short circuit. (Good joke?... No? Okay )

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12. ### eblc1388 AAC Fanatic!

Nov 28, 2008
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A voltage source, by definition, maintains its voltage regardless of any external connection. If there is a short circuit across it, then current will be infinite, by definition.

The 12.4V AVS output will remains at 12.4V because it is a voltage source, as well as the 24V too. So the voltage of arrow "A" is fixed at -24V. Arrow "B" is fixed at 12.4V so Point "C" or arrow "C" is also fixed at -11.6V. The voltmeter will indicate 12.4V as it is connected across this AVS voltage source directly. This is true whether the null detector is connected or not.

I don't like the word "overtake" but prefer to use "interact" instead. What happens in a real circuit now depends on its configuration, values of the voltage sources and all other resistance values.

When the connection of the null detector is made, and there is a voltage difference between its two ends, thing get a bit messy and some mathematics will be required to deal with this situation. It is not until 1845 that Kirchhoff had devised his famous Kirchhoff's circuit laws to tackle particular these type of situations where there are interactions.

One of the laws is easy to understand. Sum of all current going into a node is equal to the sum of current going out out it. I have indicated this by different color arrows to show you what happens. Each color represent a certain current.

Now immediately we have a problem. You will notice that the current going through the R1 resistor(cyan arrow) is not the same as that of the R2(white arrow) because of the involvement of a yellow arrow. This damages the balance of the original circuit and we can no longer know about easily the voltage at node "D" without some additional calculation.

Because the connection of the null detector causes the yellow arrow to become part of the circuit, we say that the null detector "load" the circuit and change the voltage of the node we want to measure in the first place.

The only time where the above statement is not true is when the yellow arrow is zero, i.e. the current=0. When it is zero, the node D voltage will not change and no loading is taken place.

How can we make the yellow arrow zero? By changing the AVS such that its output is exactly equal to the node D voltage. Then Ohm's Law tell us current(yellow arrow) = (Volt_difference) /(Resistance_of_Null_detector)= {-12V -(-12v)}/some_R = 0.

If one can't make the AVS to match the node D voltage, then we'll need to apply the above two Kirchhoff's Laws to get the answer.

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