DC bias current and small-signal resistance

[JessicaHelena]

Homework Statement

Consider the diode D_1 shown in Fig 7-1.

The diode is a non-linear device with the following I-V characteristics:

$i_D = I_S \cdot (e^{\frac{v_D}{v_{TH}} - 1)$

where i_D is the current through the diode, I_S is the reverse bias saturation current, v_D is the voltage across the diode, and V_TH is the thermal voltage. Assume $I_S = 8*10^{-12}$ and $V_{TH} = 20mV$.

Although the diode is a non-linear device, it can be modeled as a resistor for small signals. Calculate the numerical value of the DC bias current I_D in mA such that its small-signal resistance is 5 Ohms.

Relevant Equations

iD = ID + id
(i.e., total variable iD = DC output ID + small signal id)

also,
$I_D + \delta i_D \approx$ (iD evaluated at v_D) + (derivative of iD evaluated at v_D)*$\delta v_D$
$I_D =$ (iD evaluated at v_D)
$\delta i_D =$ (derivative of iD evaluated at v_D) *$\delta v_D$

We are given that $i_D = 8\cdot 10^{-12} (e^{v_D/20m} - 1)$

Hence $i_D' = e^{50 v_D}/2500000000$ and $i_D'' = e^{50 v_D}/50000000$

Then I have that $\delta i_D \approx\frac{ e^{50 v_D}}{2500000000} \cdot \delta v_D = \delta v_D / 5$ Cancelling $\delta v_D$ from boh sides and solving for the v_D, I have that $v_D = 0.400602$.

Plugging this value into the original function for iD, I can get I_D, which gives me 0.004. However, this is not right. What am I doing wrong?

 

[berkeman]

I didn't go through your calculation in detail, but I'd be inclined to manipulate the diode equation to get Vd as a function of Id, and differentiate to get the impedance, then substitute in the point Id = 20mA...

$$Z(Vd) = \frac{dVd}{dId}$$

 

[gneill]

Or just differentiate the $I_D$ function w.r.t. $V_D$. Conductance is the reciprocal of resistance.

 

[gneill]

@JessicaHelena : is it possible that you're running into significant figures issues when entering your results? It looks like your $I_S$ value is only specified to one sig fig.

Edit: I ask because I've run the calculations and come up with results that match yours. So I have to wonder where the underlying issue lies.