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Calculating the output impedance of a function generator

  1. Jan 1, 2013 #1
    Hi everyone. I'm doing an experiment where I measure the capacitance of a capacitor. However I want to make sure that the output impedance of the function generator is low (to get accurate results). My circuit is attached. How do I calculate the impedance from the information provided?

    Help is much appreciated.

    Kindest
     

    Attached Files:

  2. jcsd
  3. Jan 1, 2013 #2

    AlephZero

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    You can't "calculate" the impedance of the function generator. The information will be in the manual, and probably on a data plate somewhere on the actual device.

    The output impedance should be negligible compared with your 100K resistor. A common value is 50 ohms.
     
  4. Jan 1, 2013 #3
    Hello physicsforme, welcome to Physics Forums.

    Since you are doing an experiment you can measure the output impedance of your signal generator, using the same method as you are doing with your resistor and capacitor forming a potential divider.

    You need a couple of resistors about 100Ω to do this.

    Measure the output across a single 100Ω resistor, put a second one in parallel and measure the output again.

    Can you see how to calculate the output impedance from these two measurements?
     
  5. Jan 2, 2013 #4

    NascentOxygen

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    In making your calculations, you should not overlook the input impedance of your oscilloscope.
     
  6. Jan 2, 2013 #5
    A very good point.
     
  7. Jan 2, 2013 #6
    On second thoughts, would calculating or measuring the impedance of the function generator or oscilloscope matter because I am measuring the voltage across the capacitor such that I measure the peak voltage and various voltages in time along the discharging part of the curve - then I would be calculating the ratio of Vo/V for various times and use the exponential formula to calculate the capacitance.
     
  8. Jan 2, 2013 #7
    The input impedance of the scope should not matter, unless it is an unusual scope ( there are some with 50 ohm input impedances) or the frequency is very high.

    The input impedance of most scopes is high - in the 1MΩ to 10MΩ in parallel with a few picofarads range.

    So even with you 100k resistor the shunting effect will not be very much and effectively non existent with my 100Ω resistors.

    I think the point was for you to consider this and establish that its effect is negligible.

    Given the above can you think of a situation where the effect is not negligible?
     
  9. Jan 2, 2013 #8
    Would a radio be an example, when you are tuning it to a high frequency?

    Also I am using a square wave form to display the results - is that because the wave form is going simulating an 'on and off' form even though it will be exponential.
    Also would background noise effect the measurements from the oscilloscope and function generator . Will there be any other uncertainties other than the uncertainty in the measures voltages and time due to precision limit of the oscilloscope?
     
  10. Jan 2, 2013 #9
    I am assuming you can do simple series and parallel calculations and also calculate the reactance of a capacitor at any given frequency { Xc = 1/2∏fC }.

    Let us take a worst case cheap scope with input impedance of 1M // 10pF.

    You can calculate the effect of the 10pF on your circuit at 10kHz, 100kHz, 1 MHz, 10mHz, 100Mhz and 1000MHz (not that your scope would have that range but here's wishing)

    This is your project , I am trying to promote your obvious interest.

    The capacitance will round the corners of a square wave as it contains components of much higher frequency than the fundamental to make the corners.
     
  11. Jan 2, 2013 #10
  12. Jan 2, 2013 #11

    NascentOxygen

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    Measurement of the exponential will give you the product C.R, so unless you know the effective R you cannot isolate the value of C. Whereas the circuit shows one resistor, there are in effect 3 resistances that you must consider.
     
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