Author Topic: How a multimeter measures PWM voltage?  (Read 23567 times)

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Offline karkoonTopic starter

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How a multimeter measures PWM voltage?
« on: May 30, 2018, 05:47:54 pm »
Hi there,

I am a hobby electronics guy. I am unable to understand how a multimeter measures pwm voltage (average voltage).

I am trying to understand it by finding analogy with an oscilloscope. As per my understanding an oscilloscope takes multiple samples every second and then plots them on the the screen. Using various mathematical operations it can determine duty cycle, frequency and voltage levels etc. 

I generated a 5v pk-pk square wave with 50% duty cycle. I fed it to an oscilloscope and a multimeter. Oscilloscope showed all the measurements, pk-pk, duty cycle and frequency and average voltage too. Multimeter on the other hand just showed me 2.4-2.5v and nothing else. I was wondering how it works? How multimeter averages this? Is there a sample interval which is constant across all the multimeters? e.g. 100ms gate time?

When a pwm wave is applied to a resistor, to calculate voltage drop, do you consider the pk-pk voltage or average voltage as input voltage? Can you please point me in the right direction? I am little confused as wave always have 5v and 0v levels but the multimeter reads an entirely different reading. Earlier I thought when you apply pwm to a LED it just flickers and thats why it appears as dimmed or brighter. Seems thats not the case.

Thanks.
 
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Offline GEOelectronics

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Re: How a multimeter measures PWM voltage?
« Reply #1 on: May 30, 2018, 06:28:19 pm »
What is the make and model of the DVM/DMM?

Geo>K0FF
 

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Re: How a multimeter measures PWM voltage?
« Reply #2 on: May 30, 2018, 06:43:46 pm »
Yes, a sample time, usually 6Hz, the LCM of common world mains frequencies -- so you don't see mains ripple/hum.

Read up on multi-slope ADCs.  These are the most common technique used.

It is an averaging filter function, so that the oscilloscope will also read 2.50V (give or take) average.  The scope just has a harder time doing it, because it only samples discrete points in time and runs a bunch of math on it.  The multi-slope ADC gets it for free in the analog domain. :)

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Online IanB

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Re: How a multimeter measures PWM voltage?
« Reply #3 on: May 30, 2018, 07:08:44 pm »
Yes, a sample time, usually 6Hz, the LCM of common world mains frequencies -- so you don't see mains ripple/hum.

6 Hz or 600 Hz?

Quote
Read up on multi-slope ADCs.  These are the most common technique used.

It is an averaging filter function, so that the oscilloscope will also read 2.50V (give or take) average.  The scope just has a harder time doing it, because it only samples discrete points in time and runs a bunch of math on it.  The multi-slope ADC gets it for free in the analog domain. :)

Surely this discussion is not complete without consideration of averaging vs. true RMS multimeters? I think the result with a square wave will be quite different in either case.
 

Online IanB

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Re: How a multimeter measures PWM voltage?
« Reply #4 on: May 30, 2018, 07:15:46 pm »
When a pwm wave is applied to a resistor, to calculate voltage drop, do you consider the pk-pk voltage or average voltage as input voltage? Can you please point me in the right direction? I am little confused as wave always have 5v and 0v levels but the multimeter reads an entirely different reading. Earlier I thought when you apply pwm to a LED it just flickers and thats why it appears as dimmed or brighter. Seems thats not the case.

If a PWM signal (or any waveform) is applied to a resistor then the instantaneous current and voltage drop will correspond to the instantaneous magnitude of the applied voltage. For example, if a 0 to 5 V PWM signal is applied to a 10 Ω resistor, then the current will be a PWM waveform varying between 0 A and 0.5 A.

If the PWM has a duty cycle of 50% then the power in the resistor will be 50% of the power it would see with a continuous 5 V applied. (This is because the maximum power is there for half the time and absent the other half of the time. You average them to get 50% power.)
 

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Re: How a multimeter measures PWM voltage?
« Reply #5 on: May 30, 2018, 08:16:43 pm »
Yes, a sample time, usually 6Hz, the LCM of common world mains frequencies -- so you don't see mains ripple/hum.

6 Hz or 600 Hz?

GCM*. Same thing, just... opposite. I always mix them up... ::)

Quote
Surely this discussion is not complete without consideration of averaging vs. true RMS multimeters? I think the result with a square wave will be quite different in either case.

I think he's referring to DC mode only.

He would indeed get a different result on AC range for average vs. peak vs. quasi-peak vs. RMS readings; and different still on the oscilloscope, which does not filter DC from its measurement! :)

Tim
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Offline bhishmar

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Re: How a multimeter measures PWM voltage?
« Reply #6 on: September 07, 2018, 04:07:15 pm »
I am a hobby electronics guy. I am unable to understand how a multimeter measures pwm voltage (average voltage).
................
This is an old post, & IanB has answered it somewhat.
But let me expand on his answer.


I am also an electronics hobbyist and engineer, & pondered this same issue some years earlier, when working with various DMMs & scopes.  I got a better handle on this after I worked with True RMS meters along with older conventional Averaging DMMs.  But got refined when I started working with PSPICE Tools & circuit simulations, to better my understanding.

SCOPE:
In a scope you will see the actual (instantaneous) waveform value in real-time always.  So there is not much to say any more about it.

DMMs:
What you see in a DMM, is a single representational value, which is computed  depending on following factors.
I am answering for a general case first, & then narrowing to the specific case asked by OP.

1.  Shape of waveform (Square/ Sine/ Triangular/ Irregular).
2.  If it is a Square waveform what is the duty cycle %
3.  If the waveform is bipolar & symmetrical across the zero-line.  (say -2.5V to +2.5V, for 5Vpp)
4.  Else how much DC offset is there on the case-3 waveform.
5.  Is the DMM configured, in DC mode or AC Mode?
6.  If DMM in AC Mode, then is it an "Averaging" AC-Mode or "True-RMS"?

The special case OP's question is as below, for earlier 6 factors.
1. Shape: Square Wave
2. Duty Cycle : 50%
3. Symmetry across zero-line:  NO. Waveform is Unipolar [ 0V to +5V].
4. DC Offset : +2.5V DC offset, on a bipolar +/- 2.5V Square wave (Vpp = 5V)
5. DMM DC/AC Mode : Not specified. (Context indicate AC mode)
6. True-RMS/ Averaging : Not specified.

# Item-4 is simply another way of seeing item-3. (Not a diff parameter)

Answer for the above special case & the varying cases specified in 5,6 items:-

A. If DMM is  in DC Mode.
The DMM in DC-Mode reads the mathematical average of the input wave form, over the timeline,  which is +2.5V.
Most current DMM's have a display update of 3 readings per sec.  At very low  frequency probably flicker effect or inconsistent reading may occur, since the wave form value may change during the dual-slope conversion time of the DMM- ADC.

B. If the DMM is in AC mode  (True-RMS AC-Mode).
The DMM reads  +2.5V, since the RMS value of the 5Vpp Unipolar Square wave = 2.5V @ 50% duty cycle.

C. If the DMM is in AC mode  (Averaging AC Mode).
The DMM reads about +2.776V = (2.5 * 1.1107).  1.1107 being the form factor of a Sine wave.
This is because the Averagng-AC-Mode  DMM is factory calibrated to show  1.1107 times the average value of the full-wave precision-rectified input waveform, so that without complex TrueRMS circuitry inside, it will show approximate RMS value of any Sine wave input waveform.

But here, since you are inputting a square wave (& the DMM circuit is unaware of the wave shape), it tries to apply the same scalefactor, to come up with earlier reading.

When a pwm wave is applied to a resistor, to calculate voltage drop, do you consider the pk-pk voltage or average voltage as input voltage?
................
Typically the internal dual-slope ADC always use the average value of the input wave form across multiple  wave form cycles. The conversion time of the dual-slope ADC is in the range of 100 to 300 msec only. That is why it is only able to provide 3 readings per sec.  So the averaging occurs over several cycles (periods) of the input waveform. There may be special circuitry to correct boundary conditions (not whole number of conversion  cycles), for precise readings.
This is for DC Mode.

In the case of Averaging -AC mode,  the scenario is not much different, since the wave form presented to the dual-slope ADC is the precision-full-wave-rectified signal.  Rest is same except for built in form-factor correction (1.1107).

In the case of the True-RMS the circuit is totaly different, the input wave form is presented to the RMS circuitry first, before the ADC.
« Last Edit: September 07, 2018, 04:40:04 pm by bhishmar »
 

Offline vk6zgo

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Re: How a multimeter measures PWM voltage?
« Reply #7 on: September 08, 2018, 06:08:35 am »
Hi there,

I am a hobby electronics guy. I am unable to understand how a multimeter measures pwm voltage (average voltage).

I am trying to understand it by finding analogy with an oscilloscope. As per my understanding an oscilloscope takes multiple samples every second and then plots them on the the screen. Using various mathematical operations it can determine duty cycle, frequency and voltage levels etc. 

I generated a 5v pk-pk square wave with 50% duty cycle. I fed it to an oscilloscope and a multimeter. Oscilloscope showed all the measurements, pk-pk, duty cycle and frequency and average voltage too. Multimeter on the other hand just showed me 2.4-2.5v and nothing else. I was wondering how it works? How multimeter averages this? Is there a sample interval which is constant across all the multimeters? e.g. 100ms gate time?

When a pwm wave is applied to a resistor, to calculate voltage drop, do you consider the pk-pk voltage or average voltage as input voltage? Can you please point me in the right direction? I am little confused as wave always have 5v and 0v levels but the multimeter reads an entirely different reading. Earlier I thought when you apply pwm to a LED it just flickers and thats why it appears as dimmed or brighter. Seems thats not the case.

Thanks.

A "square wave", by definition, has a 50% duty cycle.
Other duty cycles are "rectangular waves".

As your waveform spends half its time at 5.0volts,( Vpk) , the other half at zero volts, the average voltage =Vpk/2 or 2.5 volts.

RMS voltage means "the Root of the Mean Of the Squares"
If you take n samples of one cycle of your waveform, you will have
n/2 samples at  5.0 volts & n/2 samples at 0 volts.

Squaring all the samples we have n/2 squares of value 25 & 
n/2 squares of value 0.
As the "mean" is the same as "average" the mean of the squares will be 12.5.
If we now take the square root of 12.5,  we get a figure of 3.5355

A true RMS meter will read  around 3.54 volts.

If you change your duty cycle & follow the same process, you will find that both the average & RMS voltages will change.

This is why Pulse Width Modulation (PWM) works.
 

Offline bhishmar

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Re: How a multimeter measures PWM voltage?
« Reply #8 on: September 09, 2018, 05:58:39 pm »
A true RMS meter will read  around 3.54 volts.

Sorry, this is not correct! 
Almost all True-RMS Meters including the high end flukes will read 2.50V, as I pointed out in my earlier post, in point "B".
I am talking about the  Unipolar SquareWave varying from 0V to 5V (i.e Peak-to-Peak value  = 5V).

But all the other things what you are saying, including  the MATHS of the TrueRMS is very correct.
i.e the actual  RMS value of the above waveform (as per the theoritical mathematical definition of RMS) is indeed = 3.5355V, as you say.  No dispute there.

I am only saying that such a waveform when applied to a True-RMS proclaiming DMM will give a reading = 2.5V only.

Why this discrepancy between Theory & practice ?
The Answer is simple.  All DMMS including high-end ones, will connect an internal series coupling capacitor, when they are configured to read AC voltage  (both Averaging & True-RMS) ,  to block any DC content in the input signal.  @@  (See below)

So essentially what the DVM reads is not a Unipolar (0V - 5V) square wave, which you are feeding externally.   Instead the internal TRMS circuit  of the DMM  will be seeing only a (-2.5V to +2.5V) bipolar symmetrical squarewave after the series-coupling capacitor, blocks out the DC-content= 2.5V.

Now if you apply your RMS math on this bipolar (-2.5V to +2.5V signal).  You will see that it is also 2.5V.

So ultimately there is no difference between Theory & Practice. ;D

So now the question is how much is true in the "True-RMS"?
Ok, You can see it as:
it is only the  True-RMS of the AC-content alone (after blocking DC-content)

@@: See the : Fluke 287/289 UserManual : Page-77, Detailed specs/ True-RMS
Quote
True-rms:
AC mV, AC V, AC µA, AC mA, and AC A specifications are ac-coupled, true rms, and are specified from 2 % of range to 100 % of range,
except 10 A range is specified from 10 % to 100 % of range.
« Last Edit: September 09, 2018, 06:42:15 pm by bhishmar »
 
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Offline agaelema

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Re: How a multimeter measures PWM voltage?
« Reply #9 on: September 09, 2018, 06:32:38 pm »
Just complementing,

As bhishmar said, a True-RMS DMM will show 2.5V in the AC scale because of the coupling capacitor, but, if the DMM has an AC+DC scale (sometimes named as DC-RMS), the result of the measure will be 3.5355V related to the sum of AC and DC value.

V = sqrt(AC^2 + DC^2)
 
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