# Simple effective resistance of a circuit at low voltages

1. Apr 30, 2017

### zapnthund50

I passed small voltages (0-5V) through a known resistance, measured the current each time, and got different "effective resistances" for each case. However, when graphed, the effective resistances appear to approach the known resistance of the circuit as a limit at higher voltages.

What is the reason for this? Nobody I ask can give me a satisfactory response. I've been told resistance never changes .... is it due to the inner workings of the ammeter I'm measuring it with?

Thanks!

2. Apr 30, 2017

### baldbrain

It's not entirely true that resistance can't vary. Resistance of a wire depends on the temperature of the wire as well as it's temperature coefficient of resistance (which indeed depends on the material of the wire).
Rt=R0(1 + αt)

where Rt is the resistance at t°C, R0 is the resistance at 0°C,
t is the temperature of the wire when the current of known potential difference is applied
α is the temperature coefficient of resistance depending on the material of the wire.

So, as the potential difference varies, the temperature of the wire changes slightly and deviates slightly from the experimental value

3. Apr 30, 2017

### jim hardy

You've asked us to evaluate data that you haven't shown..

"I've got a secret" is a game, not a request for help.

4. Apr 30, 2017

### zapnthund50

Sorry didn't mean to be secretive, lol. Here's a pic of the data and the graph I constructed from it (I just uploaded it here as a jog)

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• ###### input resistance.jpg
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5. Apr 30, 2017

### zapnthund50

Hmm, so the resistance depends linearly on temperature...but this doesn't match my results, I'm still confused

6. Apr 30, 2017

### zapnthund50

Wait a second I might have stumbled on part of my answer. P=I2R, so if temperature itself is dependent on the current (as I think it should be, since the temperature increase in a resistor arises from the resistance to current flow), then this means that the temperature is proportional to power, but itself approaches an upper temperature limit. But this would only work for a nonlinear resistor, so I'm still in the dark lol. Thoughts?

Last edited: Apr 30, 2017
7. Apr 30, 2017

### sophiecentaur

Where are you measuring the voltage across the resistor? PD should always be measured independently across the terminals and 'inside' the connections to the power supply. You have to eliminate contact resistance and other such nasties.
What sort of temperatures are you experiencing? A tungsten filament can increase in resistance by a factor a ten, when it is at full power. What metal is the wire made of?
In fact, it would improve your chances of a good answer if you gave a full description of your experiment (circuit diagram and even a photo of the layout). You can't assume we know exactly what you have ben doing and the details can be very important.

8. Apr 30, 2017

### zapnthund50

The circuit was built on this breadboard by Digilent Inc (http://store.digilentinc.com/electr...-one-usb-oscilloscope-multimeter-workstation/). It was simply a few resistors in series, the voltage was measured from one resistor leg to the other, across the whole resistor chain. Since I was only working on 0-5V with only 0.02-0.18 mA current, the temperature rise must be so tiny its insignificant. I know the explanation must be something fundamental that I'm just missing.

Again, I'll post the pic of my results, which has the resistance graph on it.

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• ###### input resistance.jpg
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9. Apr 30, 2017

### milesyoung

Here's a plot of your data with a polynomial degree 1 fit:

Doesn't look that bad. Fit gives an effective resistance of 30.85 kohm.

I'd take a guess that your measurements are a bit inaccurate down at those microamp levels. Have a look at the datasheet for your equipment for its accuracy at whatever range setting you're using.

Alternatively, lower the values of your resistors.

10. Apr 30, 2017

### zapnthund50

Yes, that is the plot I got as well ... and I do understand that we get resistance from the slope. I guess what I'm failing to grasp is why the Input Resistance varies, while the real resistance remains constant. My professor could not really explain it to me, he studied mechanical, not electrical lol. Quite simply, here is my confusion. Take the first data point, 0.1 volts, 0.02 milliamps. Then according to V=IR, the resistance should be 5000 ohms. But obviously the resistance was 31 kohms, so ... what is the concept I'm failing to comprehend?

11. Apr 30, 2017

### milesyoung

The equipment you're using to measure this isn't perfect. Typically, the lower the value of something you're measuring, the more inaccurate your equipment gets. That's only 20 microamps!

Try redoing your experiment with lower values of resistors. Aim for an effective resistance of a couple hundred ohm or something like it. Make sure you don't go too low and fry your resistors, although this is a rite of passage that you'll inevitably go through at some point.

12. Apr 30, 2017

### zapnthund50

Okay thanks, I will retry the experiment with lower resistors and higher current. The thing is though, I think what I'm describing is an attribute of Linear Resistors. Ohms law is just a description of linear resistors, right? V=IR, this can be looked at in the basic linear form y=mx+b, where b=0 and m=R. Then y=Rx. We both graphed voltage on the x-axis, and current on the y-axis, so the equation of our line is actually x = y*(1/R), or I=V*(1/R). So no matter what value resistor I measure, I should always get the linear VI relationship, right? Because that's why we call them linear resistors.

Is it a property of linear resistors to have a very small resistance at low voltages, but at high enough voltages, they approach their rated resistance value? That's the question that's bugging me. I've used the fitted equation that we both got from the data to extrapolate the current out to about 100 volts, and I've attached the resulting graph. As you can see, resistance starts at zero, but quickly climbs to the rated value. Thoughts on this?

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13. Apr 30, 2017

### milesyoung

No, there are significant nonlinear effects that depend on temperature and (high) voltage, but none you should be worried about here. You're probably just seeing the effects of measuring a very low level of current with a piece of equipment that's not designed for measuring that range accurately. If you had such a piece of equipment, I'd wager you'd get the results you initially expected.

There doesn't actually seem to be a datasheet with metering accuracies listed available for the kit you're using, which is a bit disappointing. I guess they expect you to go through the schematics and components yourself, or maybe I'm just not looking in the right place.

Edit: I see there's a voltmeter function in the kit. How are you measuring current?

14. Apr 30, 2017

### zapnthund50

With a fluke (on ammeter setting) in series with the resistors. I'll build a circuit with about 500mA of current, that should be more than enough to minimize any errors that the instrument gives. Will keep you posted.

15. Apr 30, 2017

### milesyoung

That's a lot of current. Make sure you calculate the power you'll be dissipating in each resistor before you turn anything on (they have a maximum power rating).

What Fluke model would that be and what range are you using it on when measuring the 20 microamps?

16. Apr 30, 2017

### zapnthund50

Yeah good point, I'll recalculate it for close to the max power that the resistors can handle. Don't want to fry anything. I'll have to get back to you on the Fluke model, its a good one though, part of the campus equipment.

17. May 1, 2017

### sophiecentaur

You should be aware that the breadboard system was never intended to be a 'measurement system'. It's a way of testing circuit designs in which the range of values of circuit components are assumed to swamp any of the incidental (parasitic) components. You can't tell what sort of contacts that your components / wires are making. (We still need to see what your actual connections and layout consisted of.) Half of your measuring system is hidden underneath the plastic cover. This is the sort of experiment that would probably best be carried out with wires and good-ol' crocodile clips. I have had classes of A level students, 'verifying' that Ohm's Law applies to out-of-the-drawer resistors (using that crude equipment). They were getting 'good' straight line graphs over a range of 10:1 supply volts. So it's quite possible to do a lot better than your results suggest. Doing it with visible connections would give you a chance to verity that there are no unexpected voltage drops anywhere, due to bad contacts. Fault finding can be a very satisfying exercise and can really help with gaining a deeper level of understanding.

18. May 1, 2017

### rbelli1

Were you using the mA or uA setting on the meter?

This Fluke 87 meter http://support.fluke.com/find-sales/Download/Asset/2161164_6116_ENG_B_W.PDF which is fairly common as far as flukes go can't measure many of the values you are trying to measure on the mA setting. The first one is for example will have an accuracy of +-0.2% +-0.04mA. This is more than double the value you measured. The last digit has nearly no meaning except for comparative measurements.

BoB

19. May 1, 2017

### zapnthund50

The fluke has an automatic current range selector, there is no way to select milliamps or microamps... This was a class experiment that was checked by staff. I don't know the exact model of the fluke, but I will let you know as soon as I have access to it again.

20. May 1, 2017

### jim hardy

That's a pet peeve of mine. Beginners should start out with simple analog instruments.

This will measure 20 microamps with ease. With no ambiguity as to what range it's on.

old jim

21. May 1, 2017

### rbelli1

It can be set to manual ranging if you need that.

BoB

22. May 1, 2017

### zapnthund50

I have such an analog multimeter, present from my dad! They're great, but I'm spoiled by the instant readings digital meters spit out.

23. May 2, 2017

### jim hardy

I use them when i want precision
but their least significant digit must not be taken seriously.

24. May 2, 2017

### sophiecentaur

I wonder what that consisted of. What level are you working at and are the staff EE specialists? This stuff is harder than people acknowledge.
You should challenge the staff about this. 'How can they expect you to measure such low values of current with that meter?"
On the other hand, they may be wanting you to spot the problem (which you have) and to identify which of your measurements should be used to find the resistance value. (Choice of range of current values)
But I still have a problem with that definite curve you are getting (even when you ignore the lowest value). Inaccuracy at the low current range would hardly be responsible for such a systematic trend.
Did you choose the value of the resistor that you measure or was it randomly picked from a box full?
You have my sympathy. I appreciate that, in the class situation, there is seldom time to do these things 'properly' and you have to take your measurements and then pack away. Much of the time can be involved in just putting the equipment together (often for the first time) and tidying up before you have done any processing of your data. It's always a good idea to get as many measurements in as possible - in this case, using more than just one resistor - and to try graph plotting as you go. I wouldn't mind betting that you would not have had to come to PF if you had been issued with a 500Ω resistor.

25. May 2, 2017

### jim hardy

When experimenting
i like to plot my raw data.

Observe that your plot is volts / amps with a very small denominator.
With such a small denominator a teeny error there exaggerates error in the result. Arithmetic just behaves that way.

Let's use hindsight, which is always 20/20
Looks like your resistor turned out to be somewhere around 27k ohms, a standard 5% value.

With 0.1 volts applied to 27kΩ you ought to get 3.7 microamps .
You wrote down 20 microamps. Probably the 2 is least significant digit on your voltmeter so shouldn't be believed. Consider writing instead "Too small to measure".

With 1 volt applied you ought to get about 37 microamps .
You wrote down 50 microamps. Least significant digit on ANY meter is ±1 count, so on your plot instead of a dot, how about drawing a vertical bar from 40 to 60 microamps and labelling it "limit of resolution". Observe that 37 rounds to 40, which is how your meter would report 37 were it a dead-accurate meter.

With 2 volts applied you ought to get 74.1 microamps and you wrote down 80.
Again your meter's resolution is ± 10 microamps so a vertical bar from 70 to 90 would represent what you can measure.
Observe as your measured variable becomes larger compared to your meter's resolution your error band becomes a smaller fraction of what you're measuring.

That's why i like analog meters better than digital for beginners. That the needle is so near zero makes it obvious to the eye you're at the limit of measurability.

We had a saying in my shop: " Twice as many digits makes your errors twice as accurate.".

Form the habit of drawing error bars around any measured data point.
Try it with your experiment - plot your measured currents on a linear scale and draw a best straight line through all the error bars,
see if its slope comes any closer.to the marking on your resistor.
27K is likely marked red-purple-orange, silver stripe = ±10%, gold = ±5%

Beware of small denominators. And overly smart voltmeters.

old jim