Difference in Voltage drops

Discussion in 'General Electronics Chat' started by paul_alan, Nov 27, 2011.

  1. paul_alan

    Thread Starter Member

    Nov 5, 2011
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    What voltage would be measured across each resistor if you setup a circuit on a breadboard using two 10MΩ resistors with a 20V p-p applied and a 200Hz signal from the Function Generator? I set it up with two 1MΩ resistors with a 5V p-p using the same setting on the Functon Generator and measured 3.2V across each resistor. Is there an easy way to make these calculations without always setting up acircuit?
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    The potential division formula would normally answer this question. If an input voltage Vin is applied to two resistors R1 and R2 in series, and the an output voltage Vout is taken to be that appearing across R2,

    then Vout = Vin*R2/(R1+R2) This link gives more details: http://en.wikipedia.org/wiki/Voltage_divider

    Your measurements do not seem to line up with this result however, as if there was really 3.2V across each resistor for 5V input, the total of the voltages across each resistor comes to more than the input.

    Perhaps your measurements were not quite right, or maybe the large resistor values allowed errors to occur, for instance due to stray capacitance.
     
  3. paul_alan

    Thread Starter Member

    Nov 5, 2011
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    I'm sorry, I put in some incorrect values. I set up a Sine Wave output of 10v p-p at a frequency of 20Hz with a DC offset of zero voltsusing two 1MΩ resistors. I calculated the voltage drops to be 5V a piece and measured a voltage drop of 3.2V across each resistor with my DMM.
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
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    In this case, it is very likely that the input impedance of your DMM had a loading effect on the resistor voltages.

    A potential divider made up as you describe would have an effective output impedance of 500kΩ. (Have you covered Thevenin's Theorem?)

    If your meter has 1MΩ input resistance, the expected output voltage would fall from the expected 5V

    to 5V*1MΩ/(0.5MΩ+1MΩ), or 3.333V. This is pretty close to your 3.2V

    This shows that even with a high-impedance instrument, loading effects become significant if the circuit impedance is high enough.
     
  5. BillB3857

    Senior Member

    Feb 28, 2009
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    Does your multimeter read peak, peak to peak, or RMS values? What is the input resistance of your multimeter. Do you know how to convert peak to peak value to RMS value? Do you understand the concept of test equipment loading on a circuit? Using such high value resistors for your experiment can lead to some very mysterious readings.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    If you have an analog voltmeter (From the sound of the readings, you do), they only have 10kΩ/Volt input resistance.

    A new DMM of quality will have 10MΩ/Volt of internal resistance.

    The only accurate "needle type" Meters are panel meters built for the task, or the type with a FET input, which put them in the 1MΩ/volt draw
     
  7. Yako

    New Member

    Nov 24, 2011
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    You cannot properly test caps and transistors with a DVM.

    HFE is not a proper test either.
     
  8. Adjuster

    Well-Known Member

    Dec 26, 2010
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    No-one would claim that the capacitance and hFE measurement features offered by some DMMs are foolproof, but they can be useful under some circumstances.

    Note however that the OP is trying to get to grips with basic measurements. Diverting into a discussion of other topics may be confusing.
     
  9. paul_alan

    Thread Starter Member

    Nov 5, 2011
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    I appreciate the respone to my question with a bunch of questions BillB3837. If all of you are senior members, this should come easy to you because, like I said before, I'm a new Biomed student and this is the beginning of the term...so I assume it's basic stuff. Using a DMM (which means Digital Multimeter), setting up a circuit with two (2) 10M Ohm resistors, an output of 200Hz from the function generator with a sinewave of 20V p-p on the graticule of the Oscilloscope. All I'm asking for is a simple way to calculate the equation. It doesn't matter about converting RMS, the internal resistance of the DMM or anything. It's just a simple experiment using basic student equipment.
     
  10. Yako

    New Member

    Nov 24, 2011
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    Do you want the truth?

    It is not even a test, all they provide you with is measure of farads and βeta / hFE.
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Your DMM will 'correctly' respond to the rms value of a pure sine wave.

    If the source really is 10V p-p then each resistor will have 5V p-p. That's 2.5V peak or 1.768V rms. If the resistors are 1MΩ and your DMM has an internal resistance of 10MΩ you would expect a reduction to 10/11 of the true voltage drop as your indication - that being a value of 1.607V rms. Double this value and you have 3.214V which is close to what you are reading. It's most likely you actually have 20Vp-p input rather than 10V p-p.

    When you say converting values (p-p, rms or whatever) and including the meter loading impedance doesn't matter you are completely wrong.

    EDIT: My error - the reduction factor isn't 10/11 rather it's 0.953 giving 1.685V rms with a 10Vpp source or 3.37V with a 20Vpp source.
     
    Last edited: Nov 27, 2011
  12. Yako

    New Member

    Nov 24, 2011
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    If it is a true RMS meter that is. Otherwise it displays the 'average' value.
     
  13. BillB3857

    Senior Member

    Feb 28, 2009
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    Sorry if I came on a little strong, Paul. I re-read the entire thread and didn't see where it said you were a new Biomed student. As far as trying to calculate what you should see with your DMM, it is important to be able to know that Peak to Peak voltage is different than what you meter will indicate. Most meters are calibrated to indicate RMS voltages for a sine wave input. Some, more sophisticated units, will indicate True RMS for non sinusoidal waveforms. To convert from Peak to Peak, multiply Peak voltage by 0.707 to get the RMS. Therefore, with a 10 V peak, the RMS would be 7.07. 50% of 7.07 would be more in the neighborhood of 3.535 volts. If your meter has an input of 10 Meg ohms, that, in parallel with one of the 1 Meg resistors of your circuit, would effectively reduce that leg to 9.09 megohms. If you calculate your expectations using a series circuit of 10 meg in series with 9.09, you will see the effect of test equipment loading vs the expectation of having two 10 meg resistors..
     
  14. Yako

    New Member

    Nov 24, 2011
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    Either way, this cannot be properly measured without a scope. OP is talking high frequencies.
     
  15. BillB3857

    Senior Member

    Feb 28, 2009
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    200 Hz is high frequency? Use of the scope was a very good way to verify the signal from the signal generator. Again, it important to understand the relationships between, Peak, Peak to Peak, Average, and RMS when dealing with AC.
     
  16. Yako

    New Member

    Nov 24, 2011
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    I must be going nutz. I thought I read 200KHz.
     
  17. BillB3857

    Senior Member

    Feb 28, 2009
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    200Hz is high frequency? He is talking in Low Audio freqs. Use of the scope to verify the signal amplitude from the signal generator is a very good move. I went to the very top of the page to the VOL-II - AC tab and then clicked on "Simple AC circuit calculations

    " . It has some interesting comments about having all measurements made in the same terms, ie., Peak, Peak to Peak, RMS, etc.

    The link above it goes into a lot of detailed explanation.
     
  18. Yako

    New Member

    Nov 24, 2011
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    Your typical AC-measuring DVM is meant for mains frequency mate.
     
  19. Adjuster

    Well-Known Member

    Dec 26, 2010
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    If you mean to set up two resistors in series, driven by a generator at one end, and then monitor the voltage at the mid-point with an oscilloscope, then the internal resistance of whatever is used to make that measurement does matter.

    I would be obliged if you could confirm that the set up is two 10MΩ resistors in series driven by 20Vp-p. If this is the case, then with nothing else connected to the mid-point of the two resistors, the voltage will be as given by the equation in my first post.

    That is Vout = Vin*R2/(R1+R2), so where R1= R2, Vout = 0.5*Vin, or 10Vp-p

    If however you connect an oscilloscope or other instrument to measure the mid-point voltage, the result will be reduced by loading caused by the instrument's input resistance. The output impedance of such a divider is equivalent to the two resistors considered as being in parallel, which in this case comes to 5MΩ.

    The loaded output can also be found using potential division, if we know the input impedance of the instrument concerned. Typically most oscilloscopes will present a 1MΩ input impedance at low frequencies, but some 10X voltage probes will increase this to 10MΩ. We can give estimates of the divider output for each of these two cases:

    For a 1MΩ instrument, the output is 1MΩ/(1MΩ+5MΩ) of the unloaded value, namely 1/6 as much or 1.666Vp-p.

    For a 10MΩ instrument, do the maths for yourself.
     
  20. Yako

    New Member

    Nov 24, 2011
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    ^ he's right. Knows what he is saying.
     
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