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Diode incremental model help

  1. Sep 21, 2014 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    r_d= V_T/I_D, V_T=25mV

    3. The attempt at a solution
    i alreayd find the operation of diodes, D1 off and D2 on (2nd circuit). the current through this diode i also calculate as 34mA, also r_d=0.7353 ohm. I am having trouble using incremental model, in my book it gives voltage divider like V(r_d/(r_d+R)) where V is ripple in voltage (+1 here) and R is I guess the eq. resistance. except I don't think I can voltage divider here, I try to change the V source to 3.7 and 5.7V (+1) and I get the diode curent change by 1mV each time.. but not sure how to prove this. Using node voltage I find V at node is -20V, the diode current is (4.7+1+20)/1k + (20)/1.93k this also give change in diode current +1mA for +1V of the 10V source, but is this the right way to do it?
    Last edited: Sep 21, 2014
  2. jcsd
  3. Sep 21, 2014 #2


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    Staff: Mentor

    Your operating point current and diode resistance look about right. You might consider letting the 10V source be represented by a variable and then finding the Thevenin equivalent for the network as "viewed" from the diode D2's point of view. So the Thevenin voltage will be an expression that depends upon ##V_s##.

    With the incremental model diode attached as its load you'll have a series circuit with ##V_{th}, R_{th}, r_d, ## and ##v_d##. Writing an expression for the diode current with respect to ##V_s## will be simple then.

    You might even consider finding ##\frac{dI}{dV_s}## for that expression. Being a linear circuit, it should be equivalent to ##\frac{\Delta I}{\Delta V_s}##.
  4. Sep 22, 2014 #3
    I'm not sure if this is right but I get Vth = 3.096V (or 23.096V?) and Rth = 1k||1.93k= 658.703. so with Vth and Rth in series with Rd, do I need to include the -20V or not? If I just calculate first change in diode voltage then current from ripple, +1V*(Rd/(Rd+Rth)) = 1.11mV, Id=Vd/Rd = 1.11mV/0.7353 =+1.51mA... something like that?
  5. Sep 22, 2014 #4


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    Staff: Mentor

    Ah, you left the 10V source as 10V and didn't make it a variable. The problem is, any variation in that 10V will be scaled by the resistor network in the same fashion that the 10V gets scaled when it gets transformed into its contribution to the Thevenin voltage. I think what you've done above is apply your +1V to the Thevenin voltage itself. That will result in a larger effect on the current than if it were the 10 V source varying by 1V. I was thinking to carry a variable for the 10V source through the Thevenin model in order to retain the network scaling factor. But no matter...

    Another approach is to take advantage of superposition to determine the effect on the current of just the noise source of the 10V. Consider your circuit re-drawn as follows:

    Now, ##V_s## represents the noise on the 10 V supply. If you suppress ALL sources except for ##V_s## then calculating your Thevenin model should be straightforward.
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