Hi,

I picked up a couple 0.1Ω 10w resistors to track current usage with an Arduino on a battery powered project but I don't have a precision ohmmeter, how can I extrapolate the resistance value? I only have a cheap multimeter and Arduino at my disposal and already I see a huge error ratio with the multimeter.

Thanks

 

This is not a trivial measurement to make. Contacts and cable resistance alone will result in inaccurate measurements.

 

Hi,

I picked up a couple 0.1Ω 10w resistors to track current usage with an Arduino on a battery powered project but I don't have a precision ohmmeter, how can I extrapolate the resistance value? I only have a cheap multimeter and Arduino at my disposal and already I see a huge error ratio with the multimeter.

Thanks

Extrapolate is the wrong word for this situation and has nothing to do with what you are trying to accomplish. If you could measure data in another range and infer the resistance, that would be extrapolation. What you mean is how can I accurately measure the resistance of these resistors with the limited equipment that I have. The answer is you probably can't. The laboratory approach would be to obtain a precision voltage source and measure the current. You have neither a precision voltage source, nor a way to accurately measure small currents. I think the best you can do is put a bunch of them in series to get an average. What is the alleged tolerance on these guys ±10% -- hopefully.

Just for completeness, an example of extrapolation would be taking a set of collector characteristics for a transistor and using the slopes to infer the Early Voltage.

 

The usual way to measure small resistances is to put a known current through the resistance (such as 1A) and measure the voltage drop directly across the resistor (should be 100mV for a 0.1Ω resistor).
That's called a 4-wire measurement, which takes the wire resistances out of the measurement.

The current can be measured with the multimeter first with the power supply adjusted in the constant-current mode, so that the current doesn't change when you remove the multimeter to measure the voltage.

 

The usual way to measure small resistances is to put a known current through the resistance (such as 1A) and measure the voltage drop directly across the resistor (should be 100mV for a 0.1Ω resistor).
That's called a 4-wire measurement, which takes the wire resistances out of the measurement.

The current can be measured with the multimeter first with the power supply adjusted in the constant-current mode, so that the current doesn't change when you remove the multimeter to measure the voltage.

Doing this with a 4 1/2 digit Fluke is different than doing it with a $5.99 Centek from Harbor Freight.

 

Doing this with a 4 1/2 digit Fluke is different than doing it with a $5.99 Centek from Harbor Freight.

That's obvious.
The Fluke may even have the 4-wire measurement built in.
But I have found those Harbor Freight meters to be fairly accurate when measuring voltage and current and will likely be sufficient for the TS's purposes.

 

That's obvious.
But I have found those Harbor Freight meters to be fairly accurate when measuring voltage and current and will likely be sufficient for the TS's purposes.

Obvious only to the members of the cognoscenti.
I have also found them useful, but it remains to be seen if they are useful enough for the TS's purposes. He didn't mention anything about having a constant current supply with is also a requirement for this method.

 

how can I extrapolate the resistance value? I only have a cheap multimeter and Arduino at my disposal and already I see a huge error ratio with the multimeter.

Extrapolate is calculating an estimate of a value outside a range of data values. Interpolate is calculating an estimate of a value within a range of data values.

 

My meter's leads are 0.3 ohms when measured by themselves. Then if I measure a 0,1 ohm resistor it will show (0.1 + 0.3=) 0.4 ohms.

 

The usual way to measure small resistances is to put a known current through the resistance (such as 1A) and measure the voltage drop directly across the resistor (should be 100mV for a 0.1Ω resistor).
That's called a 4-wire measurement, which takes the wire resistances out of the measurement.

The current can be measured with the multimeter first with the power supply adjusted in the constant-current mode, so that the current doesn't change when you remove the multimeter to measure the voltage.

I tried the 4-wire measurement prior, I was hoping to utilize the Arduino as it is the heart of the whole operation

I'll just have them measured next time I'm at the nerd shop

and thanks to everyone correcting my English my French is much worse

 

You need to do 4-terminal or kelvin type measurements which means you need to measure the current through it and the voltage across it. Values as low as this might actually be a 4-terminal resistor.

 

Obvious only to the members of the cognoscenti.
I have also found them useful, but it remains to be seen if they are useful enough for the TS's purposes. He didn't mention anything about having a constant current supply with is also a requirement for this method.

No I don't have a power supply yet, I've been working with 5v usb and battery power.

Realistically speaking is resistor in series with a battery a current source? Ignoring changes in voltage over time due to discharge I could use the multimeter to find resistance, current and voltage then use some resistor network and ohms law to find my resistor value. It was this thinking that inspired this thread.. how did the pioneers find these values using primitive methods.. after all it was their work their forged todays benchmarks.

 

to track current usage with an Arduino on a battery powered project

How much current? The resistor might get hot enough to change it's own resistance.

 

No it's not. The battery and resistor will put out different currents with the meter in the circuit and the meter out of the circuit, because you are changing the total resistance across the voltage source. the whole idea of a constant current source is to put out a constant current regardless of the actual load. Of course Ohms law will require a certain voltage across the load is that is the case and that is precisely what you want beeause the high impedance voltmeter will have only a small effect on the parallel combination of 20K or more in parallel with 0.1Ω

 

You could add an accurate resistor in series with the 0.1Ω resistor to give about 1A current with the supply or battery you have.
You then measure the voltage across the large resistance to get the current and the 0.1Ω resistor to get its resistance.

 

I push the zero ohm button on my trusty old Micronta.

 

As pointed out, you can do this using the 4-wire measurement method.

Here is a similar technique along the same idea using a battery, resistor and your DMM.
Firstly, you need a known low resistance. For example, suppose you have a 1Ω ±10% resistor.
You create a circuit with your battery in series with the 1Ω resistor R1 and your 0.1Ω DUT R2 (device under test).



With your DMM in voltage range, measure the voltage V1 across R1 and V2 across R2.
By Ohm's Law analysis:

I = V1 / R1 = V2 / R2

R2 = R1 x V2 / V1

independent of the voltage of the battery.

 

Make sure the resistor can handle the power, in watts, that will be delivered by the battery. I'd be a bit reluctant to take 1 Ω resistor for a test drive without knowing something about the batteries capabilities.

 

Yes, I intentionally left out the gory details for the uninitiated to discover.

For, example, don't expected this to work with any conventional battery.
A 9V battery, for example, cannot deliver 9A!

Ideally, you want R1 and R2 to be about the same order of magnitude.
You can insert an additional resistor R3 in series to limit the current. It is not necessary to know the value of this resistor.

 

Yes, I intentionally left out the gory details for the uninitiated to discover.

For, example, don't expected this to work with any conventional battery.
A 9V battery, for example, cannot deliver 9A!

Ideally, you want R1 and R2 to be about the same order of magnitude.

In particular -- don't try this with a 1Ω 1/2 watt resistor. A standard D-cell has more than enough energy to give the 1/2 watt resistor a smoking headache