Large Input Resistance

Discussion in 'General Electronics Chat' started by Digin8918, Nov 7, 2009.

  1. Digin8918

    Thread Starter New Member

    Sep 5, 2009
    In my electronics class we talk about amplifiers with BJT's and Emitter Follower circuits and such, and I always hear about large input resistance. What is the advantage to having a large input resistance/impedance? I suppose I would just like for anyone to clarify for me the usefulness of designing around different input or output resistances and impedances. Thanks for reading.
  2. Wendy


    Mar 24, 2008
    Think in terms of voltage or current measurments. If you are measuring a voltage that is feeding through a 100KΩ resistor as part of the circuit (maybe it is as close as you can get) then a meter with 1MΩ input impedance will throw the measurement off by around 90%. Most meters have 10MΩ impedance, so the reading will be 99% correct.

    Older equipment was a lot worse, VOMs were lucky to have 100KΩ impedance, and could be a lot worse.
  3. SgtWookie


    Jul 17, 2007
    In "the good old days", a quality multimeter like the trusty Simpson 260 had an impedance of 10,000 Ohms per volt. You had to calculate in the impedance of the meter, or your voltage readings could be significantly in error.

    Some of the cheap multimeters were only 1,000 or 2,000 Ohms per volt; it was tough to get an accurate reading using them.
  4. hobbyist

    Distinguished Member

    Aug 10, 2008
    The reason for the design parameters to focus on input and output impedances, is because the signal amplitude can be DEamplified rather than amplified as it transfers from one stage to the next.


    You have a signal generator with an output impedance of 800 ohms, this means that the generator has an internal resistance of 800 ohm (for sake of example ) assume it to be a resitor,
    so if you feed a signal of for example, 10mV. to a stage with a 800 ohm input impedance, than half of the signal voltage has been attenuated, where the amp will only see coming in 5mV. instead of the full 10mV,

    so lets say that this first stage amplifies it 10 times, so now we have 50mV coming out , instead of 100mV, so now as this signal is going to the next stage and if the output of this first stage has a 1K ohm output resistance, and the next stage has an input resistance of 1K ohms

    then again the signal coming into the second stage will be 1/2 as it was before, so this second stage will see only 25mV coming into it instead of the full 50mV.

    So if both stages together were designed to have a total voltage gain of 100,
    the 10mV. signal from the generator should be 100 times greater, or 1v.

    But because each stage attenuated the signal by 1/2, then the actual total gain is only 250mV, which is 1/4 of what it should be.

    That's why it is important to make the input impedances as high as possible with respect to the output impedance of the preceding stage that is feeding into it,

    And it is important to make the output impedances as small as possible with respect to the input impedance of the next stage it is feeding into.

    So as to keep as much as possible little attenuation of the signal as you can.
  5. lmartinez

    Active Member

    Mar 8, 2009
    It can simply be explained by saying: keep a mismatched resistance at the connection points of the stages of the amplifier. This shall include a mismatched at the source signal coming into the amplifier. The input resistance of each stage should be high compare to the output resistance of the driving stages.Why? it keeps the THEVENIN'S equivalent RCs small. It is a goood thing. WIDEBAND........ AND OTHER GOOD BENEFITS
  6. Digin8918

    Thread Starter New Member

    Sep 5, 2009
    Maybe I am forgetting very simple things, but could you elaborate on this? Why is it exactly that if the input-stage has the same impedance as the preceeding output stage that the 10mv will be cut in half?
  7. lmartinez

    Active Member

    Mar 8, 2009
    Think of a voltage source with a resistance of 800 ohms in series with it. Then connect the output of this circuit in series to another 800 ohm resistor, now measure the voltage across the last resistor added to the circuit. The measured valued should be equal to half of the input voltage to the circuit. It is governed by the voltage divider rule.

    Senior Member

    May 26, 2009
    Yes indeed.

    The voltage divider rule is: \frac{R2}{R1+R2}V_I_n=V_O_u_t.
  9. Duane P Wetick

    Active Member

    Apr 23, 2009
    The basic reason why you want to look into a large input resistance is to not load your circuit excessively and thereby change its operation.

    Cheers, DPW