input impedance

Discussion in 'Homework Help' started by PG1995, May 26, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi :)

    Please have a see Example 1-5 on the given link:
    http://img854.imageshack.us/img854/8376/inputimpedance.jpg

    The example mentions input impedance. What is it in simple words? The book says: Because this input impedance is across the measured terminals , a small current flows flows through the multimeter....

    What is book saying? Please help me with it. Thank you very much for this.
     
  2. jegues

    Well-Known Member

    Sep 13, 2010
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    From wikipedia,

    "The input impedance, load impedance, or external impedance of a circuit or electronic device is the Thévenin equivalent impedance looking into its input."

    It is also commonly described as the opposition against the flow of a given signal current at the input of a circuit or load.
     
  3. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thanks, jegues.

    I think I understand it now. Say, voltmeter measures how much voltage is appearing across its terminals. In the Example 1-5 open circuit voltage is 9 volts and a Rth of 1.5 K. Measuring with a multimeter which has impedance of 200 K ohm will give a reading of 9 * ( 200 K / 201.5 K) or 8.933 volts (I've used voltage divider rule). Error is 0.067.

    In the scanned example in my first post the author used the phrase "measured terminals", what does the author mean by it?

    The author also says that moving coil has typical sensitivity of 20k ohm per volt. What does the part "20k ohm per volt" mean? That would mean for two volts it would be 40k ohm.

    The author also suggests to use a field-effect transistor. What does this mean? If it's complicated, then you can ignore it because perhaps I won't be able to understand it.

    Please help me with the above stuff. Thanks a lot. Things would have been really difficult without your help.
     
  4. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,395
    1,607
    Correct analysis and answer. (I would just prefer to state it as "Error is 0.067 volts.")

    By "measured terminals" he means the terminals you are measuring across, A & B. In other words, what the meter is across.

    Nope, it means the full scale voltage of the meter range setting: when on the 10V scale the voltage is 10V, for the 100V scale the voltage is 100V.

    A field effect transistor (or FET) does not take any current to drive it so it acts as a very high input impedance.

    While you'll soon be learning about FETs in general I would be surprised if you ever see a FET-VOM as they are now obsolete; the cheapest digital multimeter is probably more accurate and less costly to produce.

    Anytime :D
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    A sensitivity of 20kohm per volt means that the meter input impedance will be 20kohms multiplied by the full-scale voltage for the range selected. For a full-scale voltage of 10V the input impedance would be 20kohm/V * 10V = 200kohms.
     
  6. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thank you, ErnieM, Adjuster.

    As it has been said that a moving coil multimeter has typical sensitivity of 20k ohm per volt. Therefore, when the scale range is set 5V (which I think means that you would be able to detect voltage from 0 to 5 volts. I have not seen a moving coil multimeter in real life). For 5V range, the impedance would be 100k ohms. For higher range scale such as 10V, there would be more impedance.

    But why more impedance as the scale range increases?

    Say, voltmeter measures how much voltage is appearing across its terminals. In the Example 1-5 open circuit voltage is 9 volts and a Rth of 1.5 K. Measuring with a multimeter which has impedance of 200 K ohm will give a reading of 9 * ( 200 K / 201.5 K) or 8.933 volts (I've used voltage divider rule). Error is 0.067 volts.

    If we had used a multimeter which had impedance of 50 K ohm then it would have given a reading of 9 * ( 50 K / 51.5 K) or 8.737 volts (I've used voltage divider rule). Error would have been 0.262 volts.

    So, using more impedance with large scale range offsets the error.
     
    Last edited: May 26, 2011
  7. Adjuster

    Well-Known Member

    Dec 26, 2010
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    The old moving-coil meters were based on micro-ammeters. These typically required between 1mA full-scale deflection (quite insensitive) and maybe 20μA full-scale (relatively sensitive).

    These meters were made to read voltages by means of suitable series resistors, calculated to increase the meter resistance so that the full-scale deflection current would flow at whatever full-scale voltage was required. A 1mA meter movement would thus require the total resistance to be 1kΩ per volt of full-scale deflection, but a more sensitive 50μA meter would give 20kΩ per volt.
     
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Get a simpson 260 manual. You can trace the common and plus connections to see how an analog meter measures voltage. Change the scle and see the change in the series resistance.
     
  9. PatM

    Active Member

    Dec 31, 2010
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