I have been given a circuit and asked to find the voltage across the diode (both the DC and AC parts).

enter image description here

I believe the right approach is the ignore the ac source initially and find the IDQ. I know there is 0.6 V across the diode so I found IDQ to be 14.4 V/1 kΩ = 0.0144 A by analyzing the resistor.

I know there is an equation for dynamic resistance: $$r_d = \frac{nV_T}{I_DQ}$$ I am given n = 1 and VT = 26 mV so I get rd = 1.86 Ω. I interpret this as: in this setup, because the ac signal is small and the diode will always be forward biased, we can effectively treat the diode as a resistor of value rd.

But if I look at the voltage across the diode at the Q point using the voltage divider: $$V_D = \frac{15r_d}{r_d + 1\mathrm{\ k\Omega}} = 0.028\mathrm{\ V}$$ But it should equal 0.6 V. Have I done something wrong in my calculations? Or am I interpreting rd incorrectly?

I know I haven't finished the given question yet, but I thought I would make this check and it would show me if I am on the right track or if I have already made a mistake.

  • \$\begingroup\$ The dynamic resistance you are using is relevant to when supplying with an AC source - it doesn't give you the full picture on DC. \$\endgroup\$ – Andy aka May 8 '14 at 7:14

Your misunderstanding is around how the small-signal AC characteristics combine with the large DC characteristics. If VD is the DC (ie: average) voltage across the diode, you've already assumed that to be 0.6V.

The value rd represents only the incremental resistance of the diode if you add or subtract a small amount of current (ie: add or subtract a small amount of voltage at the input, like 0.1 cos(wt) ). So you use rd to calculate the voltage resulting only from that small change in current.


vin = 0.1 cos(wt), and the resulting small signal effect across the diode is:

$$v_d = \frac{v_{in} r_d}{r_d + 1\mathrm{\ k\Omega}}$$

and Vdiode_total = VD + vd

  • 1
    \$\begingroup\$ I agree to the above comments. Knowing that for each non-linear device you have to discriminate between STATIC and DYNAMIC resistances it is clear that you are not allowed to add both parts. For a diode, the dynamic resistance is the inverse of the SLOPE of the Id=f(Vd) curve (at the Q point) and the static resistance is the inverse slope of the line between the the origin and the Q point. \$\endgroup\$ – LvW May 8 '14 at 7:39
  • \$\begingroup\$ That clears it up! I went ahead with my calculations and ended up with the total output voltage across the diode to be \$0.6+1.8*10^-4*cos(wt)\$ which seems consistent with your explanation. However, I have simulated this circuit on PSPICE and it gives me an output of \$0.7436+4*10^-4*cos(wt)\$. Why does the simulation show a different output to what I have derived (I'm certain the simulator is correctly set up)? Is it because of assumptions during the derivation? \$\endgroup\$ – Sam May 8 '14 at 7:57
  • \$\begingroup\$ Yes - of course. There is - at first - a very rough assumption of 0.6 V (which coud be also 0.65...0.7V) and - secondly - the formula for the dynamic part of the diode resistance does not yet consider the ohmic part of the Id=f(Vd) characteristic. As you probably know, this curve is NOT a pure exponential function. And all these "parasitics" are included in the simulation model. \$\endgroup\$ – LvW May 8 '14 at 8:05
  • \$\begingroup\$ I am curious how the simulation arrived at a V_D of 0.7436V at only 14mA. Regardless, that moves you to a steeper part of the diode's V vs I curve, so that small changes of current (here: input voltage) cause larger changes in v_d, as your simulation reported. \$\endgroup\$ – gwideman May 8 '14 at 8:19
  • \$\begingroup\$ @gwideman I also thought 0.7436V was too high as we have been taught that it is somewhere between 0.6 and 0.7 but I have triple checked all of the component values in the simulator... \$\endgroup\$ – Sam May 8 '14 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.