Measuring superimposed voltage

Discussion in 'Homework Help' started by Thevenin's Planet, May 15, 2011.

  1. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    Hello

    Why do I get different Voltages when the meter probes are reversed on the components?

    Component | A | B | C | D | Results | Direction
    VR1 + - 3.3rms Forward
    VR1 - + 23.94rms Forward
    Vs1 - + 0.6 rms Reversed
    Vs1 + - 19.8 rms Forward
    Vs2 - + 12.06 rms Forward
    Vs2 + - 12.0 rms Forward

    D.C.V. Meter Scale

    Componet | A | B | C | D | Results | Direction
    VR1 + - 9.2 DCV Forward
    VR1 - + 9,25 DCV Forward
    Vs1 - + 9.2 DCV Forward
    Vs1 + - 9.2 DCV Reversed
    Vs2 - + 0.00 0.00
    Vs2 + - 0.00 0.00
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    1. What is your meter's model number and manufacturer?

    2. What is the frequency of your AC component and is there any offset?

    3. Do you have access to an oscilloscope?

    4. What did you "expect" to see before you started your measurements?
     
  3. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    No oscillascope.Don't know frequency.I expect readings that is suppose to show that a.c. is above and below the off set or D.C. voltage. If a. c. is larger than D.C. then a.c. should be seem as an a.c. voltage. If a.c.is smaller than D.C. then there should be a deflection in one direction.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The meter probably has a half-wave rectifier to the meter movement on AC range. A good meter would have full-wave rectifier and the indications would make more sense.

    With half-wave rectification the DC bias from the DC source would have a decided influence. I guess the manufacturer assumes that any AC measurements are "pure" AC with no DC offset.

    If necessary I can probably give a reasoned argument with some numbers to clarify what is probably happening.
     
  5. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    Hello There
    The meter is FET analog multitester.Cat # 22-220A that have a Freq.response of 45 Hz to 1 KHz at +/- 1% up to 30 volts and a 1 KHz to
    10 KHz at +/-3% up to 30 volts.It is a Radio Shack and was purchased in the year of 1994.
    Don't forget that the a.c. transformer has a 12 volt a.c. secondary and the D.C. Battery is a 9 volt instead of the 1.5 volts that is on the schmatics addional a 2 K ohms resistor,if you are attempting to place the circuit on your simulator.
     
    Last edited: May 23, 2011
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I did see the other (duplicate?) post which noted that the DC bias was 9V rather than 1.5V

    It's not too hard to equate what you are observing with my point that the meter is probably behaving as a half wave rectified input case for the AC range.

    A half-wave rectified AC range basically scales up to an indicated RMS equivalent of the half wave rectified mean value.

    The scaling factor is ∏/√2=2.22

    This means a 9V DC source would register as 9x2.22=20V on the AC range.

    A half wave rectified 12V rms AC source offset by +9V DC has a mean value of 10.63V which when scaled up by a factor of 2.22 would be indicated as 23.6V.

    Reversing the meter leads in this case would effectively give a half-wave rectified output for 12V rms with an offset of -9V DC. In this case the mean value would be 1.7V and this would scale up to an indicated value of 3.73V.

    As you might observe these values are very close to what you are seeing with your meter for the particular measuring cases.
     
  7. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If you have had experience with meter circuits, you will perhaps know that simple meters usually have an internal rectifier included in the measuring path on AC ranges. The rectifier converts the AC input to a DC or mean value. The DC value is then usually scaled to provide an equivalent RMS indication on the scale. The AC input range waveform is usually assumed to be sinusoidal for the scaling to indicated RMS to be "correct".
     
  9. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    Hi
    Is this the concept your refering to,"The mean is pertaing to the root-mean-square,"Which is said to equal the square root of the average value of the squares of all the instantaneous values of current/voltage during one-half cycle." Also it was said that the voltage/current increases from zero to peak value and then back to zero,the average value must be somewhere between zero and the peak value.Is this where the scaling factor Pi/square root of 2 comes from?
    Another puzzle, there can be an infinite amount of instantaneous points on a rising voltage/current from 0 to 90 degrees.
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    A half-wave rectified sine wave has a mean [average] value of Vm/∏. The RMS value of the same sine wave is Vm/√2.

    So the ratio of RMS-to-Mean value of a sine wave is (Vm/√2)/(Vm/∏)=∏/√2

    I'm not sure what your other question relates to - "Another puzzle ..."
     
  11. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You seem to be referring to the process of integration. It is normally introduced as a summation of evenly spaced rectangular blocks under the curve of interest. As the step or space size is reduced the number of blocks increases in proportion as they become "thinner". The smaller the block width then (notionally) the more accurate the estimate of the total area becomes. In the limit, the block width is regarded as infinitely small and the number of blocks becomes infinite. At this stage the summation (integration) becomes continuous rather than discrete. The continuous case is therefore the true estimate of the area.
     
  13. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    Hi

    I was thinking did that had something to do with getting the mean (average ) value of the sine wave (or is it just the peak and not the peak-to-peak of a sine wave).
    You had a thread that was originally posted by tom 123 it was about halve wave rectified wave and D.C.off-set.You posted your response, a equation about harmonic spectrum for the halve -wave rectified part of the wave.I was guessing it is what you called intergation.It was posted 10-31-2010 @08:59 PM.General Electronics Chat,Work sheet help RMS bias calculation?
    Regarding the discussion about the mean value in the explaination "a halve wave rectified 12 rms a.c. source off- set by +9 DCV has a mean value of 10.63 DCV which when scaled up by a factor of 2.22 would be indicated as 23.6 V. Are you using the mean concept that is adding 12 +9 =21 and dividing 21/2 =10.5 ?
     
  14. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The derivation of the average value of a half-wave rectified 12V rms sine wave offset by 9V DC requires some more complex maths that simply adding 12V rms to 9V DC and so forth. I didn't post the math part. I simply gave the result.

    A more obvious question from your perspective, is to ask why with only the 9V DC battery connected to the AC range of the meter [with the leads in one orientation] you measured about 20V, and about 0.6V with the leads reversed.
     
  15. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    Since it is suggested at this moment,Please do explain, "why the meter registered 0.6 V in one direction and 20v with probes reversed?" This question is definitely within the range of original concern of the post.
     
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    My suggestion is that it would be beneficial for you to answer that question yourself.

    The various comments I have made thus far point to the answer.
     
  17. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    So... therefore,I conprehend that there is a point that is Frustration.Hmmmm....Should go over the mountain or through the mountain? I guess this discusstion should put to at rest for now. I not sure how much memory you guys process
    This has been a very interesting conversation that I have not participated in long time of discorvering vital infomation about the meter anology.Yea,.... this is GREAT.How the meter is adjusted to register root-mean- square.But, I insist that you give me the math about the mean square someday...until the mean time ,"I see ya later allegator!!
     
  18. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    Good Day

    It took some while but I believe that the Diode in the meter has something to do with the .06 volt on the a.c. scale.
    #1. Using the equations [Vdc^2+Vrms^2]^2 had a result of 15 volt not 23.6 volts rms, according to your calculation.
    #2.Using the Superimposed equation,[A1^2+1/2(Ao-A1)^2]^2, also had a RMS of 15volts and not 23.6 volts if this is one of the equation you used.

    The two drawings are what it's all about.They indicate that if DC is greater than a.c. peak value is a sine wave that would never reverse polarity.Does that define on the meter D.C voltage?
    The other drawing implies that a.c. is grater than DC which would have a portion of the wave in the negative polarity.Would that read as a.c. on the meter?
     
  19. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You seem to be on the right track but applying the wrong equations. This is clearly not a true RMS meter so you can't apply the square root of the sum of squares to calculate the expected RMS reading.

    I've posted the notes I made when thinking about the problem a couple of weeks back.

    Based on your comments concerning AC amplitude and DC offset I would suggest another experiment to confirm what is going on.

    Use a lower RMS AC source value or a higher DC offset value to ensure the DC offset exceeds the AC peak amplitude.

    For instance a 5V RMS AC source in series with a 9V DC source would ensure this condition was met. Connecting the meter across this composite AC+DC source would give a reading of about 20V with the meter leads in one orientation on the AC range. With the leads reversed I would expect about 0V.

    Concerning the diode effect - you mention a value of 0.06V in your last post but noted a value of 0.6V in your first post - which is correct?
     
  20. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    Hello

    The 0.6 volt is the correct reading for the Vs1 which is the 9 volt D.C. battery at a particular position of the prods.
    Yes, this integral calculation,which I am not proficient at it do enlighten my understanding much more then the previous threads.Now, concerning the Theta one,which seems as phase lagging which came about from (sine 9/17). I am not sure if you meant that the wave is ascending or it abruptly begins at 0,0 as collapsed part of wave or what??? Although it has a 32.0 degrees or as you have it -0.558 rads.

    #2 When the ac goes below zero volts the diode goes off for 3.7 rads or at 3.7 rads. Is this at 9-17=-8 a.c.v.?

    #3 The last of the f(x) intervals can't grasp how did 5.02 come about? That is 33.3+14.42+5.02+14.42.
     
Loading...