4
\$\begingroup\$

Consider the following simple circuit: enter image description here

In the above circuit, D1 and D2 are not identical. The reverse leakage I of D1 is 0.1 pA and the reverse leakage I of D2 is 5 pA. The voltage V is 3 volts.

It is required to calculate I and V of both diodes.

As the diodes are series, so the current is equal for both of them. And if we assume that 3v is lower than the break-down voltages of the diodes, then we can conclude that I is - 0.1 pA (lowest reverse leakage).

But I have problem with the voltages. How can I calculate voltages of diodes in this circuit? If the diodes were completely identical, then the voltage of each diode was 3/2. But in this case they are different.

\$\endgroup\$
14
  • \$\begingroup\$ @KingDuken "The voltage V is 3 volts" \$\endgroup\$ – Joren Vaes Jul 27 '17 at 19:00
  • \$\begingroup\$ Is that really all the info you have on these diodes? If so, this is impossible to answer. But I think you're only giving us part of what you know about D1 and D2. \$\endgroup\$ – Marcus Müller Jul 27 '17 at 19:05
  • \$\begingroup\$ the question would be interesting for ideal diodes... \$\endgroup\$ – Eugene Sh. Jul 27 '17 at 19:08
  • \$\begingroup\$ @MarcusMüller That's all we have about the diodes! \$\endgroup\$ – Abraham Jul 27 '17 at 19:09
  • \$\begingroup\$ @EugeneSh. well, define "ideal". An ideal diode, imho, would have 0 reverse current up to breakdown voltage, and after, 0 resistance. \$\endgroup\$ – Marcus Müller Jul 27 '17 at 19:09
6
\$\begingroup\$

So, a common approximation for the current through a diode is (the Shockley diode equation) at voltage \$V\$ is:

$$ I = \left(e^{\frac{V}{nV_T}}-1\right) I_S$$

with \$V_T\$ being the thermal voltage (temperature-dependent), \$n\$ being a quality factor (device-dependent) and \$I_S\$ being the saturation voltage.

So; since we know that

$$I_{D1}=I_{D2}$$ we can infer that

$$ \left(e^{\frac{V_1}{n_1V_{T1}}}-1\right) I_{S1}=\left(e^{\frac{V_2}{n_2V_{T2}}}-1\right) I_{S2}$$

Now normally, these diodes will not be at the same temperature, they wouldn't have the same quality factor, and thus, this equation would be underdefined, but with \$V_{T1}=V_T=V_{T2}\$ and \$n_1=n_2=n\$, as you imply in your comment with:

I think we can suppose that the only difference between the diodes is the Is and everything else is identical!

we get, also noting that \$V_2 = V_0 -V_1\$:

$$\begin{align} \left(e^{\frac{V_1}{nV_{T}}}-1\right) I_{S1}&=\left(e^{\frac{V_0-V_1}{nV_{T}}}-1\right) I_{S2}\\ %\implies\\ %\frac{e^{\frac{V_1}{nV_{T}}}-1}{e^{\frac{V_0-V_1}{nV_{T}}}-1} &= \frac{I_{S2}}{I_{S1}} \end{align}$$

Now, this equation is hard to solve analytically, but both sides are really easy to plot (use something like 30 mV for \$nV_T\$). Just find the intersection of these two curves!

Analytically, we can move forward with: $$\begin{align} \left(e^{\frac{V_1}{nV_{T}}}-1\right) I_{S1}&=\left(e^{\frac{V_0-V_1}{nV_{T}}}-1\right) I_{S2}\\ &=\left(e^{\frac{V_0-V_1}{nV_{T}}}-1\right) 50I_{S1}\\ \implies\\ e^{\frac{V_1}{nV_{T}}}-1&=50e^{\frac{V_0-V_1}{nV_{T}}}-50\\ \implies\\ \ln \left(e^{\frac{V_1}{nV_{T}}}-1\right)&=\ln\left(e^{\frac{V_0-V_1}{nV_{T}}}-1\right)+\ln 50\\ \end{align}$$

\$\endgroup\$
3
  • \$\begingroup\$ Great! Thank you. But I was thought that the Shockley diode equation is for forward-biased diodes only, isn't it? \$\endgroup\$ – Abraham Jul 27 '17 at 19:28
  • 2
    \$\begingroup\$ It's not! You just have to take care, \$V_0 = -3\,\text V\$, and your \$V_1<0\$, \$V_2 <0\$, too! \$\endgroup\$ – Marcus Müller Jul 27 '17 at 19:30
  • 2
    \$\begingroup\$ +1 as I loved the analytical approach (of course.) I added an answer only because I felt it was useful to show a closed, relatively simple approximation that still is fairly precise. \$\endgroup\$ – jonk Jul 27 '17 at 20:25
5
\$\begingroup\$

You have to solve two equations simultaneously:

$$\begin{align*} I_{sat_1}\left(e^{\frac{V_{D_1}}{n V_T}}-1\right) &= I_{sat_2}\left(e^{\frac{V_{D_2}}{n V_T}}-1\right)\\\\ V_0&=V_{D_1} +V_{D_2}= -3\:\textrm{V} \end{align*}$$

This follows from the fact that the currents in both diodes must be equal and that the sum of their voltages must match with the supply voltage. Pretty obvious, really.

Solving these simultaneously is a bit tricky. You could attempt it, iteratively. Or you could attempt it with the Lambert-W (aka ProductLog) function. (Doable for a closed solution, but still takes some work.)

But we can use a symmetry argument to claim that -1 term can be ignored. This allows for a very simple solution:

$$\begin{align*} V_{D_1} &\approx -\frac{1}{2} \left[V_T\: \operatorname{ln}\left(\frac{I_{sat_2}}{I_{sat_1}}\right) + V_0\right]\\\\ V_{D_2} &= V_0 - V_{D_1} \end{align*}$$

Using \$V_T=26\:\textrm{mV}\$ and \$I_{sat_1}=0.1\:\textrm{pA}\$ and \$I_{sat_2}=5\:\textrm{pA}\$, this instantly gives the correct answer to quite a few places: \$V_{D_1}\approx -1.5508563\:\textrm{V}\$. This should check out okay.

You can "read" the above solution equation to say:

Start out with the assumption that the voltage is divided in half. Then apply a correction which will be one half of \$V_T\$ times the logarithm of the ratio. (The sign of the correction will, of course, depend upon which saturation currents are used in the numerator and denominator.)

It's my suspicion that this is the approach you were supposed to take because it focuses on what's important (the saturation current ratios) and avoids getting hung up on numerical solutions or overly mathy discussions which distract from rather than shed light on the subject.


A note about a huge assumption in all this is about the nominal temperature. It's enough to simply say "room temperature" and use a value for \$V_T\$ that is commonly used (somewhere from about \$25\:\textrm{mV}\$ to perhaps \$26\:\textrm{mV}\$ is often picked.) However, to imagine recomputing that for any temperature to get the right behavior from the above equations by just substituting in the new value for \$V_T\$ is wrong. It turns out that the saturation currents are a function of \$T^3\$ to \$T^4\$ and so they vary as well. In fact, they vary so much that they overwhelm the effect of \$V_T\$ enough to reverse the sign of the effect!

So it is probably fine to just assume that the saturation currents one is given are meant at "room temperature." But that's all. If the model is to apply over a wide temperature range, the variation of the saturation currents must also be incorporated into the model. And that is a whole other issue.

\$\endgroup\$
5
  • \$\begingroup\$ Yeah, that's very true – especially in reverse, \$I_S\$, so I've been told, is the main contributor to temperature-dependent behaviour; makes sense when you consider that square intrinsic carrier concentration is involved, iirc \$n_i^2 = N_s e^{-\frac{E_{gap}}{2k_B\,T}}\$, and the number of available states \$N_s \propto T^{\frac32}\$, so we get an inverse-exponential function of the inverse temperature multiplied with a power (>1) of \$T\$; that's bound to rise pretty strongly :) \$\endgroup\$ – Marcus Müller Jul 27 '17 at 20:57
  • \$\begingroup\$ and, to make matters worse: diode conductivity rises with temperature. Temperature rises with current. Current rises with conductivity. \$\endgroup\$ – Marcus Müller Jul 27 '17 at 21:02
  • \$\begingroup\$ @MarcusMüller (I could post a typical saturation current equation here, I suppose.) ... and to just continue expanding the discussion, another problem is also that some diodes will have a very pronounced increase in reverse biased currents due to leakage across the surface of the diode and because new carriers can be generated by collisions in the transition region of the PN junction. \$\endgroup\$ – jonk Jul 27 '17 at 21:06
  • \$\begingroup\$ ?! That sounds a bit surprising; wouldn't the impulses of these carriers have to be massively different to avoid recombination in a collision? \$\endgroup\$ – Marcus Müller Jul 27 '17 at 21:09
  • 1
    \$\begingroup\$ @MarcusMüller I think you are right when talking about modern diode devices. Note that I was careful to say "some diodes." I was remembering some very old literature regarding those older diode devices. I don't even think the surface leakage matters much for modern devices. You'd just stimulated some old memories, is all. (Though as a totally irrelevant side note I have had problems with the bulk impedance of epoxy packaging and was forced into wire-bonding dies to avoid it -- basically "dead bugging" it with dice.) \$\endgroup\$ – jonk Jul 27 '17 at 23:16

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.