Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Small signal diode modelling

  1. Apr 12, 2010 #1
    small signal diode modelling i understand what it is but when i get asked a question like:

    if vin(t)=2.5+cos(2*pi*100*t) for 0<t<30

    use the small signal model of the diode to predict the output voltage of the circuit i have added. assume the opamp is ideal R=2000 Is=10e-16

    so i know that the small signal equation is

    id=(ID/VT)*vd where VT=0.0252

    now since the opamp is ideal zero current is drawn in and since the configuration is in the inverting configuration then Vout=-vD

    now i know i have to somehow dervive and expression for Vout using the small signal diode model equation which is

    id(ac current)=(ID(dc current)/0.0252))*vd(ac voltage)

    i also know that ID=Is*e(vD/VT)
    =(10e-16)*e(vD/0.0252)

    subbing into the small signal diode model equation

    id=((10e-16)*e(vD/0.0252)/0.0252)*vd

    and now i get lost and stuck because i dont know what to do or if i am doing the right thing in order to derive and expression for vout using the small signal diode model

    please help!
     

    Attached Files:

  2. jcsd
  3. Apr 12, 2010 #2

    dlgoff

    User Avatar
    Science Advisor
    Gold Member

    For a very small signal, the diode behaves more like a resistor. How could you use this fact in the op-amp circuit?
     
  4. Apr 12, 2010 #3

    marcusl

    User Avatar
    Science Advisor
    Gold Member

    I'm not so sure. The input voltage is +/-2.5V, so the diode quickly becomes fully forward biased on each positive half cycle with usual 0.6 to 0.7V drop, assuming silicon. Output waveform is asymmetric.
    EDIT: Sorry, my mistake. I misread it as 2.5 * cos() instead of 2.5 + cos()
     
    Last edited: Apr 13, 2010
  5. Apr 12, 2010 #4
    I think we are dealing with an inverting amp.

    For DC analysis, one can calculate the diode bias current which is related to R,

    and then for AC analysis, replace the diode with a dynamic resistance. Then it looks easy.
     
  6. Apr 12, 2010 #5
    would it be correct to say iD=(VD0-Vout)/rd for the circuit where VD0 is the voltage across the diode
     
  7. Apr 12, 2010 #6
    Yes, but for AC analysis.

    For simplicity, start with doing DC analysis first. If

    v_in = 2.5V,

    can you find I_d?

    Hint: the negative input of the op-amp is a ground.
     
  8. Apr 13, 2010 #7
    well for DC we could use the diode equation and say iD=vin/R and since the diode equation is iD=Ise(vD/VT) we could rearrage it to

    Vout= -VT*ln(vin/R*Is) for DC

    and then you have stated replace the diode with the dynamic resistance are you referring to rd the resistance across the diode?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook