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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.

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  • \$\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
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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.

So:

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

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    \$\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

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