Post Go back to editing

KCC's Quizzes AQQ285 about analog voltmeter expertise

1. Quote of the week: "If you think nobody cares if you're alive, try missing a few loan payments" - unknown

Source: Wikimedia

2. New quiz AQQ285 about the analog voltmeter expertise

We are back to our fundamental area: electronics...

Sources:

https://upload.wikimedia.org/wikipedia/commons/0/01/Galvanometer_scheme.svg

https://cdn2.picryl.com/photo/2015/12/24/spiegelskala-1-b45d8a-1024.JPG

We all remember from our first electronic lab, the use of analog multimeters to measure voltages, currents or resistances.

They are based on the galvanometer invented by d’Arsonval in which the interaction of a current (to be measured) with a permanent magnetic field moves a needle in front of graduated scales.

Due to mechanical inertia, the current measured is only sensitive to its average value.

Therefore measuring AC signals, like for example pure sinewaves, will indicate zero normally. Multimeters overcome this by rectifying the AC signal before to be measured.

The average value obtained is somehow fake since the original signal has been altered. However, such biased measure can reflect the RMS value of the signal.

This is why a second scale called “AC” mode, usually flagged in red, is calibrated to give the RMS value. However, such scale is only valid for pure sinewaves.

Questions:

1. That “AC” scale is slightly different vs the “DC” scale. The factor is around 1.11

How is it determined?

2. With the same voltmeter defined above, if you enter the following periodic signal, what are you going to read in the AC scale? 

Good luck and try to be among the first ones...

P.S. As usual, please spread the quizzes among your friends and colleagues! 

Please share your answer to view other submitted answers
Parents
  • 1. The difference comes from the fact that an analog voltmeter measures average value of AC voltage, supposing is full-wave rectified.

    And what we need is a RMS value.

    Thereby, for sinusoidal waveform, after integrating it for period of PI (fully rectified), Vavg= 2xVpeak/PI.

    The RMS value is Vpeak / SQRT(2).

    The ratio of Vrms/Vavg = (Vpeak / SQRT(2)) / (2 x Vpeak / PI) = PI / (2 x SQRT(2)) = 1.11072 (The correction value)

    Thereby, this calibration is valid for sinusoidal waveform only. For other forms, there is an error.

    2. For the square signal with period of 20 ms, there is no duty cycle defined.

    Thereby, for any duty cycle of positive pulse, D=Tpos / T, means Vavg = ((50 x D)  + (25 x (1-D) )) / T.

    The analog instrument will show Vavg x 1.11072 = (50 x D + 25 x (1-D) ) x 1.11072.

    True RMS would be SQRT( 50**2 x D + 25**2 x (1-D)).

    For instance, assuming D = 0.5 ( 50 % duty cycle) we got Vavg =  (50 x 10 ms + 25 x 10 ms) / 20 ms = 37.5 V.

    The analog voltmeter will display Vavg x 1.11072 = 41.652 V.

    The RMS value would be 39.5285 V, thereby the error would be Vrms / Vavg = 1.054, means the error of 5.4 %.

  • Hmm..The error would be (1- Vdisplay/Vrms) x 100% = 5.32%.

Reply Children
No Data