Author Topic: Can we measure voltage and resistances, or are we really only measuring current?  (Read 2743 times)

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Offline rjardinaTopic starter

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I do understand we can use a multimeter to measure voltage and resistance.

Try to keep this simply stupid for me. If there is some math involve that is fine. But more curious about the idea or concept.

1) If I measure a resistor, the meter is outputting a current at a known voltage, and measuring the current?

2) If measuring voltage with a meter. If I have a known resistance (10 mega ohm impedance), and can measure the current, I we can extract voltage from that?

is this correct?

When I look inside analog meters, All I really see are voltage dividers, and therefore limits the current going through the needle's coil for different ranges.

Do digital meters work under the same principal?

LCR meters are putting out a known voltage at a known frequency (time) and calculates what the current does over time? To get L and C (and other characteristics.)

Same thing with a scope (X1 or X10 probes). Measuring a very small amount of current and graphing it? We are not really measuring voltage are we?

Am I trying to under simplify this? I realized it more complex then that
 

Online bdunham7

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An old-school analog meter would typically have a meter movement that measured current.  50µA full scale was a common spec for a high-quality sensitive meter.  In those it was true that resistive dividers were used to convert voltage into a current and something slightly more complex to convert resistance into a current, although not linearly which was why you would see the separate non-linear resistance markings on the scale.  So yes, if you have a Simpson 260, your ultimate measurement--the movement of the needle--was due to and proportional to current.

With more modern instruments it is more complex and this is not really true anymore.  The 10M input impedance of a meter may be entirely arbitrary and you could substitute a 5M or a 20M resistor and there wouldn't be any change.  Or, there may be voltage dividers, but the ultimate measuring circuit--the ADC--is still actually measuring voltage, not current.  And for high-impedance meters which are common in better bench meters, there may not be a direct correlation at all between input voltage and current.  It is true that if you look at the actual operation of a dual or multi-slope ADC, it may look like it is measuring current, but this is the result of an internal process that is typically completely buffered from the input voltage.  So no, I don't think it would be technically accurate or helpful to say that modern DMMs or oscilloscopes are actually measuring current.  In addition, in many cases, the real input impedance is different enough from a resistor that a specific voltage may result in different currents depending on things like frequency, temperature or specific devices, but the voltage reading will still be correct in each case.
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline TimFox

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Back when Simpson 260s were popular, if you wanted a more precise voltage measurement, you compared the unknown voltage to that of a “standard cell”, using a precision “potentiometer”, which was a resistive voltage divider.  It might be a two-step comparison, where a stronger battery was first compared with the standard cell, and then the unknown was compared with that battery.  To effect the comparisons, the voltage divider was adjusted to get a null voltage difference between one voltage and the divider output, although the null was often done with a sensitive galvanometer (low-current meter).  In principle, that is the same as the automated process with the ADC.
 

Offline David Hess

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1) If I measure a resistor, the meter is outputting a current at a known voltage, and measuring the current?

2) If measuring voltage with a meter. If I have a known resistance (10 mega ohm impedance), and can measure the current, I we can extract voltage from that?

For modern digital meters, resistance measurement is done by supplying a current and measuring the voltage, and the voltage is measured with charge and time.  The input resistance is arbitrary and can be made whatever is convenient like 10 megohms or effectively infinite.

As bdunham7 points out, analog meters measure current.
« Last Edit: June 14, 2022, 09:12:20 am by David Hess »
 

Offline Doctorandus_P

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Most current measurements are actually voltage measurements over a shunt resistor with a known value.
Autoranging DMM's usually have an Amps, a mAmps, and a uAmps range, and all it does is change the shunt resistor for the right measurement range.

Also, if you buy separate ADC's, they almost always have a voltage input, and so do the ADC's in microcontrollers.
 

Offline David Hess

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Also, if you buy separate ADC's, they almost always have a voltage input, and so do the ADC's in microcontrollers.

They have a voltage input, but internally they are measuring charge, which is why those converters are called "charge redistribution analog-to-digital converters".  They are successive approximation converters with capacitors replacing the resistors and current sources.  The sampling capacitor is charged to the input voltage resulting in a charge proportional to voltage, and then charge redistribution is used to measure the charge.

For multimeters, some type of integrating converter is used for higher resolution and accuracy, which includes slope integrating converters but also now delta-sigma converters.  In both cases a combination of charge and time are used to make the measurement.  For instance a voltage or current is applied to an operational integrator for a defined time, creating a charge proportional to the voltage or current.  Then the time is measured to de-integrate that charge.  Most of the earliest ADCs in multimeters also rely on measuring charge in one way or another.
 

Offline strawberry

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1) to measure a resistor, the meter is outputting a known current at a unknown voltage, and measuring the voltage
2) meter reading 10V across 10Mohm can extrapolate as 1uA current. or if known source voltage is 100V can extrapolate Rx = 90Mohm
 

Offline BeBuLamar

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An old analog meter would measure current in measuring voltage, resistance or current.
A typical modern digital meter would measure voltage in measuring voltage, resistance or current.
 

Offline Doctorandus_P

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  The sampling capacitor is charged to the input voltage resulting in a charge proportional to voltage, and then charge redistribution is used to measure the charge.

They do indeed charge the sample capacitor to the input voltage, but in essence it is still a sampling capacitor, which means very high DC input impedance, but short current peaks during sampling, especially when combined with a MUX and sampling different input voltages.

In the end you have to always take all the peculiarieties of your ADC into consideration when designing a frontend.
With these charge redistribution things for example, you can put a 10Meg series resistor in front of it if you combine it with a capacitor to make a low-pass filter. The capacitor should then be several orders of magnitude bigger then the sampling capacitor, so it coompletely swamps it, but as the sampling capacitor is typically in the pF range this is easy to do.

A fun experiment with such an ADC (whether in a uC or something external such as ADS1115)

1. Put a 1uF film capacitor between an input and GND.
2. Measure it continuously. (You can charge and discharge it by putting one finger on either GND or 5V and the other on the input)
3. Connect another channel to GND, and a third to +5V.
4. Reprogram it to alternately sample GND and the capacitor (This will discharge the capacitor)
5. Reprogram it to alternately sample +5V and the capacitor (This will charge the capacitor).
 

Offline eugene

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I just want to remind of some basic physical facts. The only quantities that any meters of this kind have to work with are voltage and current. They have to accomplish every goal by measuring those alone. They can provide a source of voltage (battery or power supply) but supplying current can only be done indirectly by first supplying voltage which then results in current.

Measuring resistance directly is impossible. The only way that any instrument (DMM or otherwise) can do it is to supply a voltage, measure both the voltage and the current, then do the math for you.
90% of quoted statistics are fictional
 

Offline TimFox

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I just want to remind of some basic physical facts. The only quantities that any meters of this kind have to work with are voltage and current. They have to accomplish every goal by measuring those alone. They can provide a source of voltage (battery or power supply) but supplying current can only be done indirectly by first supplying voltage which then results in current.

Measuring resistance directly is impossible. The only way that any instrument (DMM or otherwise) can do it is to supply a voltage, measure both the voltage and the current, then do the math for you.

Resistance can be measured using a Wheatstone bridge, which compares the resistor being measured to a standard resistor and a pair of resistors whose ratio is adjusted.
The bridge needs a voltage excitation, and a null voltage or current sensor, but the measured resistance is independent of the voltage.
Some digital meters use basically the same technique, comparing the unknown resistor to a standard resistor by voltage ratio measurement, independent of the voltage.
 

Offline Mechatrommer

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no current, no measurement... current is base quantity, the rest are derived...
Nature: Evolution and the Illusion of Randomness (Stephen L. Talbott): Its now indisputable that... organisms “expertise” contextualizes its genome, and its nonsense to say that these powers are under the control of the genome being contextualized - Barbara McClintock
 

Online bdunham7

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Measuring resistance directly is impossible. The only way that any instrument (DMM or otherwise) can do it is to supply a voltage, measure both the voltage and the current, then do the math for you.

I'm not sure I would say that method is indirect because V/I is the very definition of resistance.  The concept and its quantity doesn't really exist in a more direct form.
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline WattsThat

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Resistance can be measured using a Wheatstone bridge, which compares the resistor being measured to a standard resistor and a pair of resistors whose ratio is adjusted.
The bridge needs a voltage excitation, and a null voltage or current sensor, but the measured resistance is independent of the voltage.
Some digital meters use basically the same technique, comparing the unknown resistor to a standard resistor by voltage ratio measurement, independent of the voltage.

Back in 70’s, Julie Reasearch Labs build a resistance measuring meter named the “Digibridge”. It was an automated KVD. Each decade had ultra low TEMF reed relays and resistors in hermetic cans. It took several seconds for a conversion. IIRC it was 8-1/2 digits and everything internally was four wire.

It sure beat setting up a bridge and transfer standards to do everyday measurements on 0.005% tolerance resistors. Only problem was failing relays as the meters aged, they didn’t hold up in production environments.
 

Offline eugene

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I just want to remind of some basic physical facts. The only quantities that any meters of this kind have to work with are voltage and current. They have to accomplish every goal by measuring those alone. They can provide a source of voltage (battery or power supply) but supplying current can only be done indirectly by first supplying voltage which then results in current.

Measuring resistance directly is impossible. The only way that any instrument (DMM or otherwise) can do it is to supply a voltage, measure both the voltage and the current, then do the math for you.

Resistance can be measured using a Wheatstone bridge, which compares the resistor being measured to a standard resistor and a pair of resistors whose ratio is adjusted.
The bridge needs a voltage excitation, and a null voltage or current sensor, but the measured resistance is independent of the voltage.
Some digital meters use basically the same technique, comparing the unknown resistor to a standard resistor by voltage ratio measurement, independent of the voltage.

True, the Wheatstone bridge is a brilliant device that compares an unknown resistance to known. But, I would argue that in the end, we still apply a voltage and observe the current.
90% of quoted statistics are fictional
 

Offline TimFox

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I just want to remind of some basic physical facts. The only quantities that any meters of this kind have to work with are voltage and current. They have to accomplish every goal by measuring those alone. They can provide a source of voltage (battery or power supply) but supplying current can only be done indirectly by first supplying voltage which then results in current.

Measuring resistance directly is impossible. The only way that any instrument (DMM or otherwise) can do it is to supply a voltage, measure both the voltage and the current, then do the math for you.

Resistance can be measured using a Wheatstone bridge, which compares the resistor being measured to a standard resistor and a pair of resistors whose ratio is adjusted.
The bridge needs a voltage excitation, and a null voltage or current sensor, but the measured resistance is independent of the voltage.
Some digital meters use basically the same technique, comparing the unknown resistor to a standard resistor by voltage ratio measurement, independent of the voltage.

True, the Wheatstone bridge is a brilliant device that compares an unknown resistance to known. But, I would argue that in the end, we still apply a voltage and observe the current.

I disagree.  You do not observe the current, you adjust bridge balance to obtain a null voltage and the resistance value reported by the device is independent of the applied voltage.
Until modern digital meters, this was the most accurate and frequent way to measure a resistance.  Simple ohmmeters (such as in the Simpson 260) did apply a voltage and measure the current.
Similarly, some impedance measurements also use a bridge, which requires two adjustments (resistive and reactive legs) to obtain a null voltage, and the results depend on a standard resistance and reactance and are independent of the magnitude of the applied voltage (but maybe not the frequency).
Also note that the current standard definition of the ohm is based on a quantum-mechanical measurement.  Before that, the legal ohm was defined in terms of a set of wire-wound resistors maintained at NIST. 
See  https://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=17477  for details on NIST calibrations.
« Last Edit: June 14, 2022, 04:51:03 pm by TimFox »
 

Offline eugene

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I disagree.  You do not observe the current, you adjust bridge balance to obtain a null voltage

Yes, you are correct.

In case it gets lost in this discussion, the point of my first post in this thread was that all we have to work with is an applied voltage and measurements of resultant voltages or currents. The Wheatstone bridge cleverly balances two voltages so that the not only is the applied voltage unimportant (as long as it's not zero) but moreover, the meter we use to measure the voltage across the bridge doesn't require any calibration so long there are no offsets i.e. zero voltage can be identified as zero. Of course it does require calibrated reference resistances, but that doesn't change the story.

There are other ways to be clever when designing instruments. We might even design a constant current source. But in every case, it starts by applying a voltage and observing resultant voltage(s) and/or current(s).
90% of quoted statistics are fictional
 

Offline TimFox

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Going back to the original question about measuring voltage, there is another method that involves neither resistors nor current.
The term "electrostatic voltmeter" refers to a voltage sensor that indicates the electrostatic force between electrodes when a voltage is applied between them.
The usual use case is measuring high voltages.
(It can be considered analogous to a d'Arsonval ammeter for measuring current that indicates the force between the moving coil and a static magnetic field.)
See  https://www.electricaldeck.com/2021/05/types-of-electrostatic-voltmeters-quadrant-attracted-disc-type.html
(Note that another usage of the term "electrostatic voltmeter" refers to a different instrument for measuring surface charge distribution.)
 

Offline Kleinstein

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Old style analog meter based on a coil primarily measured current.
With DMMs it is usually the voltage that is converted to the digital result. They usually have a voltage reference and not a current reference. When the temperature changes the input current at the same voltage can change quite a bit, but the voltage reading can still stay valid. With many cheaper meters there is a more or less fixed resistor at the input and thus an input current about proportional to the input voltage. With high impedance meters the input impedance can be more capacitive and with a variable voltage (e.g. low frequency AC) the current and voltage are no longer in phase. The reading in voltage moder still follows the voltage and not the current.

With an old anlog meter there can be some inductance and the reading is more like following the current and not the voltage - though the response speed was usually not fast enough to really see the difference.
 

Offline Berni

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You almost always have both voltage and current. Not just one of them

You need a superconductor to have current while having 0 volts at the same time.
You need a infinite resistance insulator to have voltage while having exactly 0 current.

For this reason pretty much every device has a certain input impedance. Sometimes this impedance is very low, like in an ampmeter where you have lots of amps with very little voltage (less the better). For voltmeters you want high input impedance so that you can poke any voltage and have very little current flow (less the better). To make things consistent voltmeters and scopes have standardized to a certain input impedance (10Meg or 1Meg) since having too high of a input impedance can be a bad thing (shows a voltage with the inputs disconnected). With oscilloscopes you also have to consider input capacitance, in the MHz ranges this causes many times more current draw than the resistive part.

There are bench DMMs that have the option of turning on high impedance mode. This disconnects the 1MOhm load and feeds the internal voltage buffer directly, resulting in high GOhm range impedance, getting you really close to being an open circuit. This makes 1V push only ~100pA into the meter. At this point even the insulation on cables can be an issue. Heck even the cable being dirty can make it leak more current. Once you go to sensitive enugh equipment (electrometers) you can even notice that some colors of cable insulation are more conductive than others.
 

Offline David Hess

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1) to measure a resistor, the meter is outputting a known current at a unknown voltage, and measuring the voltage
2) meter reading 10V across 10Mohm can extrapolate as 1uA current. or if known source voltage is 100V can extrapolate Rx = 90Mohm

Multimeters use an "ohms converter" circuit which is simply a precision current source applied to the output when in voltage measurement mode.  The output current is scaled by the range and the basic voltage measurement, typically 0.2000 or 2.000 volts for a 3-1/2 digit instrument, is used to make the measurement.  Since the input decade divider is used to scale the current, it is unavailable as a 1 or 10 megohm load to the input, so the input resistance is infinity and does not interfere with the resistance measurement.

I am not sure about the exactly configuration for autoranging multimeters which use a singled ended decade divider, but they do something similar.

The sampling capacitor is charged to the input voltage resulting in a charge proportional to voltage, and then charge redistribution is used to measure the charge.

They do indeed charge the sample capacitor to the input voltage, but in essence it is still a sampling capacitor, which means very high DC input impedance, but short current peaks during sampling, especially when combined with a MUX and sampling different input voltages.

I was only discussing charge redistribution ADCs which are commonly used in microcontrollers and known as successful approximation converters, although this also covers delta-sigma converters which have a similar input structure.  There are now some buffered delta-sigma converters which have low charge injection and work with moderate (10k) impedance sources.

Quote
In the end you have to always take all the peculiarieties of your ADC into consideration when designing a frontend.

With these charge redistribution things for example, you can put a 10Meg series resistor in front of it if you combine it with a capacitor to make a low-pass filter. The capacitor should then be several orders of magnitude bigger then the sampling capacitor, so it coompletely swamps it, but as the sampling capacitor is typically in the pF range this is easy to do.

With some early exceptions, multimeters buffer their input but charge injection from the input multiplexer still exists.  The input multiplexer is required to implement the automatic zero loop for the input buffer to remove offset and common mode errors.  Some early multimeters have a separate precision buffer outside of the automatic zero loop however this places considerable demand on the performance of the buffer.

The early exceptions are interesting because their discrete analog-to-digital converter is high impedance without a buffer or multiplexing so they do not suffer from any charge injection on their input which can be useful in certain applications.  How do you make a multimeter with no multiplexing, no precision buffering, and still achieve better than 100 microvolt precision?  They did it by using a non-inverting operational integrator, which serves as the integrator for the analog-to-digital converter, while the non-inverting input of the operational integrator is the high impedance input.  The only special part required is a JFET differential pair.

Measuring resistance directly is impossible. The only way that any instrument (DMM or otherwise) can do it is to supply a voltage, measure both the voltage and the current, then do the math for you.

I have measured resistance before by measuring the Johnson noise from the resistor.  I did this testing low noise DC amplifiers, and no applied voltage or current was required.  Accuracy was comparable to a good handheld multimeter.
 

Offline TimFox

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Short form answers:
To measure either voltage or current, one has several choices:
1.  Measure a mechanical force due to the voltage or current directly (as in a current balance, q.v., or a d'Arsonval meter;  or with an electrostatic voltmeter, as I discussed above).
2.  Compare the unknown voltage or current to a known voltage or current (as in a potentiometer for voltage, or DCCT current transducer for current) by measuring the ratio.  This is the basis for most digital voltmeters, as well.
3.  Convert the unknown to the other type, as in passing the voltage through a "multiplier" resistor in series with a current meter, or measuring the voltage across a "shunt" resistor for current.
For resistance, one also has choices:
4.  Apply a known voltage to the unknown resistance in series with a current meter.  As in a Simpson 260.
5.  Apply a known current to the resistance and measure the voltage across it.
6.  Using a Wheatstone bridge, measure the ratio between the unknown resistance and a standard resistance. 
Digital resistance meters usually use #5 or a variation of #6, where the unknown and standard resistors are in series and the voltmeter measures the ratio between them.
 

Offline julian1

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In a recent Amp Hour interview, Chris was talking with the guest about an adc based on a voltage-controlled ring-oscillator. But the design was not explained very well. Voltage is converted to the frequency domain, which is ok. But what are the electrical parameters/phenomena involved?

I think in simplest form, it is still a current/charge measuring device - principally the charge on the fet gates. But I think VGSon and RDSon will be involved too.




https://upload.wikimedia.org/wikipedia/commons/7/75/Ring_osc_5.png
« Last Edit: June 15, 2022, 11:32:29 pm by julian1 »
 

Offline TimFox

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I see an output, presumably the frequency proportional to voltage, but where is the input?
That just looks like a fixed-frequency oscillator.
VFCs are useful circuits for measuring voltage, and generally involve charging a capacitor with a current proportional to the input voltage, where the voltage across the capacitor moves between two defined levels to make a cyclic oscillation.
« Last Edit: June 16, 2022, 02:33:37 am by TimFox »
 

Offline David Hess

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In a recent Amp Hour interview, Chris was talking with the guest about an adc based on a voltage-controlled ring-oscillator. But the design was not explained very well. Voltage is converted to the frequency domain, which is ok. But what are the electrical parameters/phenomena involved?

Conversion to frequency circuits work with different inputs.  Voltage can be directly measured if it is applied to a comparator to make a trigger point.  Current can be measured by integration.  Charge can be measured by pumping.  These different methods lead to different transfer functions like linear, 1/x, power, etc.
 


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