Question: Design the circuit in this Figure to provide an output voltage of 2.4 V. Assume that the diodes available have 0.7-V drop at 1 mA.

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My approach: At first we assume that V0 is 0.7*3 = 2.1V .
Current across the circuit , I = (10-2.1)/R mA (R is in Kilo-Ohms)
V2 = voltage across diode when current is I2
V1 = voltage across diode when current is I1
V2 - V1 = 2.3*n*VT*log(I2/I1)
Here , V2 = 0.8V
V1 = 0.7V
I2 = 7.9/R mA
I1 = 1 mA
VT = 25 mV
n = 1
So , I get R = 139 Ohms
I am confused with my approach in selecting V2 & V1 . I have just made an average for 3 diodes while working with 2.4V and 2.1V which are 0.8V and 0.7V respectively.

  • 1
    \$\begingroup\$ If 1mA gives a forward drop of 0.7V, what current will give a forward drop of 0.8V. (hint - it will be higher than 1mA) Using this current value R = 7.6 / I \$\endgroup\$ Jul 20, 2017 at 12:46
  • \$\begingroup\$ Where does your 2.3 factor come from? And it looks almost as though you were trying to solve the first circuit before trying to figure out the answer (you use "10-2.1" in your "approach"), except that I can't really follow well the rest of what you doing, or why. So it's muddled. \$\endgroup\$
    – jonk
    Jul 20, 2017 at 14:31
  • \$\begingroup\$ V2- V1 = nVTln(I2/I1) = 2.3*n*VTlog(I/I1) \$\endgroup\$
    – Utshaw
    Jul 20, 2017 at 14:40
  • \$\begingroup\$ I've edited the question . please check that out @jonk \$\endgroup\$
    – Utshaw
    Jul 20, 2017 at 14:43
  • \$\begingroup\$ @Utshaw I just wrote an answer. And I think I understand something I did NOT beforehand. You are using a base-10 logarithm. I really didn't understand that before. \$\endgroup\$
    – jonk
    Jul 20, 2017 at 15:36

2 Answers 2


Your basic equation for each diode's voltage, in this problem where you are told that when \$I_D=1\:\textrm{mA}\$ that \$V_D=700\:\textrm{mV}\$, should look like:

$$V_D= 700\:\textrm{mV} + n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)$$

This is easily read as two terms: the first term being the assumed voltage when the current actually is \$1\:\textrm{mA}\$ and the second term being the difference from that assumed voltage when the current itself is \$I_D\$. Note that if \$I_D=1\:\textrm{mA}\$, the logarithm will be zero so there will be no adjustment.

Now you are asked to find the value of the resistor if you want to achieve a stacked diode voltage of \$V_o=2.4\:\textrm{V}\$. Clearly, this means that the voltage across each diode (same current in each, all diodes assumed to use the exact same model) must be \$\frac{1}{3}\$rd as much, or \$V_D=800\:\textrm{mV}\$.

So, you are trying to solve for \$I_D\$ in the case where:

$$\begin{align*} 800\:\textrm{mV}&= 700\:\textrm{mV} + n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)\\\\ 800\:\textrm{mV}-700\:\textrm{mV} &= n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)\\\\ 100\:\textrm{mV} &= n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)\\\\ \frac{100\:\textrm{mV}}{n\cdot V_T} &= \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)\\\\ e^\frac{100\:\textrm{mV}}{n\cdot V_T} &= \frac{I_D}{1\:\textrm{mA}}\\\\ I_D&=1\:\textrm{mA}\cdot e^\frac{100\:\textrm{mV}}{n\cdot V_T} \end{align*}$$

For diodes with an emission coefficient of \$n=1\$ and where you are specifying that \$V_T=25\:\textrm{mV}\$ (I use the slightly higher \$V_T=26\:\textrm{mV}\$ as a rule), then I get \$I_D\approx 54.6\:\textrm{mA}\$. From this, I would compute: \$R= \frac{10\:\textrm{V}-2.4\:\textrm{V}}{54.6\:\textrm{mA}}\approx 139.2\:\Omega\$.

As you can see, I didn't use the 2.3 factor you included. But I did arrive at a very similar place. So, this probably just means I don't understand where you got that factor and how you applied it in the rest of your work.

Please note that I would write the following:

$$\begin{align*} V_{D_1}&= n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_1}}{I_S}\right)\\\\ V_{D_2}&= n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_2}}{I_S}\right)\\\\ \therefore \Delta V_D=V_{D_2}-V_{D_1}&=n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_2}}{I_S}\right)-n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_1}}{I_S}\right)\\\\ \Delta V_D&=n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_2}}{I_{D_1}}\right) \end{align*}$$

I still don't see where that 2.3 came from or how you applied it. But we got to similar places.

EDIT: Never mind. I think you are using LOG10, right? I think I understand better, now!

  • \$\begingroup\$ Thank you . Yes , I am using LOG10 for which I used the 2.3 factor. \$\endgroup\$
    – Utshaw
    Jul 21, 2017 at 2:16

I was struggling same example with you and after read your post I have figured it out.

For first iteration you have to apply ohms law: 10V-2,4/R=I2

Then you will use the equation : V2=V1+0.1log(I2/I1)

It will gives you 0.8-0.7=0.1log(10-2.4/R*1)

From that equation you will find the R=0.76Kohms=760ohms which is the answer of the example.


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