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Hight input impedance DC buffer ?

Discussion in 'General Electronics Discussion' started by Externet, Dec 31, 2018.

  1. Externet

    Externet

    710
    145
    Aug 24, 2009
    Hi.
    I believe am loading my circuit too much and collapses. Am I seeing it wrong ?

    A 7555 IC as astable has 100KΩ as 10μF capacitor charge path and 100KΩ at its discharge path for ~ 0.2 Hz
    I want to use the oscillation at the capacitor, that goes from ~1VDC to ~2VDC (1/3 to 2/3Vcc) triangle waveform to drive a 5KΩ load.
    Using a transistor common emitter as amplifier, the triangle wave is overloaded and stops working.
    What is a convenient single 3V supply circuit that may work ? I guess with a >1MΩ input resistance.

    Like this ? How is Re set ?

    [​IMG]
     
    Last edited: Dec 31, 2018
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    9,378
    1,908
    Nov 17, 2011
    From Wikipedia: Rin ~ ß0*RE
    Knowing ß0 and the expected Rin allows calcuation of RE.
     
  3. BobK

    BobK

    7,643
    1,662
    Jan 5, 2010
    If you really want an exact copy of the input, a rail-to-rail opamp would do the trick.

    Bob
     
    davenn likes this.
  4. Externet

    Externet

    710
    145
    Aug 24, 2009
    The voltage follower works well. Re=100K yields 15MΩ loading to Vin with a 150beta 2N2222A. No more overloading of the triangle wave. Output is near desired range.
    What would be the effect of adding some resistance to the collector ? Would it produce less or more attenuation at Vout ?
    [​IMG]
     
  5. AnalogKid

    AnalogKid

    2,307
    649
    Jun 10, 2015
    As long as you are taking your output from the emitter, adding *reasonable* resistance (not 1 ohm, not 1 Meg) to the collector leg will not affect the output. It does shift some heat from the transistor to the resistor.

    Note that your circuit has a *very* asymmetrical output impedance. Pulling the output up is at a low impedance. Pulling the output low is the 100K resistor.

    It sounds like the -0.6 V shift in the output waveform is not a problem for your application. If it is, there is a relatively simple way around it.

    ak
     
  6. duke37

    duke37

    5,253
    726
    Jan 9, 2011
    The output impedance will only become asymmetrical when the transistor current has reduced to zero so will depend on signal amplitude and load.
     
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