This cite originates from https://www.electrical4u.com/digital-voltmeters-working-principle-of-digital-voltmeter/

is on how digital voltmeter works:

  1. Unknown voltage signal is fed to the pulse generator which generates a pulse whose width is proportional to the input signal.

  2. Output of pulse generator is fed to one leg of the AND gate.

  3. The input signal to the other leg of the AND gate is a train of pulses.

  4. Output of AND gate is positive triggered train of duration same as the width of the pulse generated by the pulse generator.

  5. This positive triggered train is fed to the inverter which converts it into a negative triggered train.

  6. Output of the inverter is fed to a counter which counts the number of triggers in the duration which is proportional to the input signal i.e. voltage under measurement. Thus, counter can be calibrated to indicate voltage in volts directly.

Reference: https://www.electrical4u.com/digital-voltmeters-working-principle-of-digital-voltmeter/

It makes some sense, but it is far from being precise.

I understand that finally we get our measure by ADC the signal that we know is proportional to the voltage. Question is how is this proportional signal produced and what relations are used. What gets measured really, is this the current measured by the number of charges per time, and how it is measured?

  • \$\begingroup\$ Well it is a digital voltmeter, so the pulse should be proportional to the voltage of the input signal. I'm not going to lie, when I read your post I also wondered for a second "input signal could be.... amplitude of the voltage? Or current? Or... something else?... Well it's obviously the voltage". \$\endgroup\$ – Harry Svensson Jul 23 '17 at 14:39
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    \$\begingroup\$ That article may not be the best: "Now-a-days digital voltmeters are also replaced by digital millimeters due to its multitasking feature i.e. it can be used for measuring current, voltage and resistance." \$\endgroup\$ – Transistor Jul 23 '17 at 14:42
  • \$\begingroup\$ @HarrySvensson You have a flow of charge in the wire. How do you know what is the voltage of this flow? All you can sense is number of charges per time, correct (by counting them somehow, how?)? \$\endgroup\$ – 4pie0 Jul 23 '17 at 14:48
  • \$\begingroup\$ Flow of charge is current. If that is running through some resistance then Ohm's law determines the voltage produced. "How do you know what is the voltage of this flow?" Flow doesn't have voltage. It's the resistance to flow that does. \$\endgroup\$ – Transistor Jul 23 '17 at 15:02
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    \$\begingroup\$ Possibly because you have quite a bit of detail on a poorly written article. Your opening line makes a statement that reads as though they all work that way (which they don't). Then at the end you appear confused between voltage and current. I'll update my answer later in light of your edit. \$\endgroup\$ – Transistor Jul 23 '17 at 16:50

Many (probably most) current digital voltmeters are dual-slope integration. To translate to the hand-waving description of your article the point is to generate a pulse of width proportional to the ratio of the unknown voltage to a reference voltage. If you assume the reference is fixed, then it reduces to being proportional to the unknown voltage.

Here is the idealized waveform at the integrator output for a common series of chips:

enter image description here

The integrator is an amplifier pair and passives like this one (somewhat simplified):


simulate this circuit – Schematic created using CircuitLab

The input is switched between the unknown voltage and the reference. Timing is controlled by a clock. The clock frequency is non-critical, however it must remain stable during the integration and de-integration phases. In other words, slow variations (say from temperature) are fine, but short-term stability must be better than the desired stability of the reading.

The clock first is used to time the integration phase (integrator connected to the unknown voltage), which ideally starts from 0V, and then is used to measure the de-integration phase when the integrator has the reference connected to the input. The latter terminates when the integrator output returns to zero.

So the resulting measurement count is ideally 1000 * Vin/Vref, provided the integrator does not saturate, and assuming the integration time is 1000 clock cycles. Note that the count is not limited to 1000.

If the integration time (which is a fixed 1000 cycles) is chosen to be an integral number of cycles of a known periodic disturbance that effect can be minimized. For example, a time of 100ms will reject 50Hz, 60Hz, 400Hz etc. Practical chips typically include an auto-zero phase which eliminates the large offsets typical of CMOS op-amps, and there is a bit more logic needed to handle bipolar inputs and overrange, and fully differential inputs, but this is basically 'it'.

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  • \$\begingroup\$ What does the integrator count? Does it count the amount of charge flowing in the signal? Is there anything else you can count in the signal - it all boils down to electrons flowing. What is physically measured and how? \$\endgroup\$ – 4pie0 Jul 23 '17 at 15:39
  • \$\begingroup\$ The integrator integrates the input voltage with respect to time. Eg. If R1 is 100K and C is 1uF and you apply -1V after 100ms the output voltage will be 1.00V higher than at the start of the time. It's strictly an analog circuit, it does not count anything. \$\endgroup\$ – Spehro Pefhany Jul 23 '17 at 16:06
  • \$\begingroup\$ Do you mean usage of "charging of capacitor equation" and RC=tau? Is capacitor used there in this case? \$\endgroup\$ – 4pie0 Jul 23 '17 at 16:36
  • \$\begingroup\$ @4pie0 I supplied the schematic of an integrator above. If the capacitor starts off at 0V, the voltage at the output at time t is the - 1/RC * integral of input voltage wrt time. Since reference and unknown are both integrated with the same capacitor and resistor the capacitor and resistor values drop out from the ratio, as does the clock frequency. The capacitor current is input voltage v(t)/R1 since the op-amp input is a virtual ground. \$\endgroup\$ – Spehro Pefhany Jul 23 '17 at 18:00
  • \$\begingroup\$ This is a typical arrangement for a multimeter. Cheap voltmeters sometimes use a successive approximation converter in a low end microcontroller. The principle is quite different- using switched capacitors. \$\endgroup\$ – Spehro Pefhany Jul 23 '17 at 18:04

Suppose you have 1MHz crystal oscillator.

We will charge a capacitor from 0 volts to some voltage, using an R+C+opAmp integrator. We don't know the voltage, but we allowed the charging to occur for exactly one second. Exactly.

Now, using a precision voltage reference inside our DVM, we charge the cap back to zero volts, exactly, while TIMING the delay.

Then we perform some easy math: Vin = V_reference * Delay / 1 second.

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The linked article doesn't give an adequate explanation of the operation of the scheme they are describing.

enter image description here

Figure 1. Typical voltmeter block diagram. Source: Radio Electronics.

The scheme of Figure 1 is much more typical of digital voltmeters.

  • A sample and hold reads the applied voltage and holds it (usually on a capacitor).
  • The Successive Approximation Register (SAR) outputs a 50% signal on its digital output.
  • The DAC converts this to an analog value which is fed back to the comparitor. The comparitor signals to the SAR whether the input is higher or lower than 50%.
  • The SAR now switches to 75% or 25% depending on the previous result.
  • This process continues successively homing in on the input value.

An alternative method is to generate a "staircase" output from the DAC by feeding it with an increasing binary count. When the DAC output exceeds the input signal the count value is latched and converted to a digital reading.

enter image description here

Figure 2. Pulse counting. Source: Blogspot.

Your referenced article may be describing a system similar to Figure 2. A ramp signal is generated. (This will be a staircase as described in the previous paragraph.) Pulses are counted until the staircase crosses the input value. The pulse count is then converted into a meter reading.

I understand that finally we get our measure by ADC the signal that we know is proportional to the voltage.

That's fine and is explained above.

Question is how is this proportional signal produced and what relations are used.

That is also explained above. We generate a binary pattern, feed that to the ADC and compare the ADC output with the signal to be measured. Then adjust the ADC and try again.

What gets measured really, is this the current measured by the number of charges per time, and how it is measured?

Charges per unit time is current. The definition of the ampere is one coulomb per second. \$ 1\;A = 1\;C/s \$.

Your voltmeter measures potential difference which is not the same. Current is the flow but voltage is the pressure causing the flow. For resistors the two are related by Ohm's law, \$ V = IR \$. If a current is flowing between two points but there is no resistance to flow then there will be no voltage drop measurable between those two points.

Voltage, electric potential difference, electric pressure or electric tension (formally denoted ∆V or ∆U, but more often simply as V or U, for instance in the context of Ohm's or Kirchhoff's circuit laws) is the difference in electric potential energy between two points per unit electric charge. The voltage between two points is equal to the work done per unit of charge against a static electric field to move the test charge between two points. This is measured in units of volts (a joule per coulomb). Source: Wikipedia, Voltage.

One of the beauties of digital voltmeters is that their input impedance is very high - typically 1 or 10 MΩ. They steal tiny amounts of current from the circuit they are measuring.

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  • \$\begingroup\$ This is not what I am asking about. I do understand what is current, voltage and resistance. What I was missing is that you use capacitor for the measurement of voltage of a signal. How do you determine end of count (zero crossing) in terms of hardware? \$\endgroup\$ – 4pie0 Jul 23 '17 at 17:41
  • \$\begingroup\$ Most voltage measurement circuits do not use a capacitor so I'm not sure where you got this idea. End of count is determined by a comparitor. \$\endgroup\$ – Transistor Jul 23 '17 at 17:43
  • \$\begingroup\$ "A sample and hold reads the applied voltage and holds it (usually on a capacitor)." \$\endgroup\$ – 4pie0 Jul 23 '17 at 17:49
  • \$\begingroup\$ That just keeps the voltage steady during the reading so that it's not changing while the SAR tries to home in on it. For a slowly changing voltage (relative to the sampling time) it wouldn't be required. \$\endgroup\$ – Transistor Jul 23 '17 at 18:04
  • \$\begingroup\$ Thank you. What i was missing is voltage comparator. This is answer to my question. The comparator does the raw job of sensing (well... comparing) voltage and this is the foundation of how it works. Thanks for good pointers. \$\endgroup\$ – 4pie0 Jul 23 '17 at 22:07

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