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I have a question regarding the voltage drop across a diode when it is connected in series with others. For example, I will use the following exercise:

Each diode has Vd = 0.7 V, id = 1mA and n=1. I need to calculate the value of R so that the voltage on node V1 equals 3. I have the answer to this question but I do not understand how the voltage drop across each diode is calculated to find the current through them. Any help is appreciated. Thanks a lot.

schematic

simulate this circuit – Schematic created using CircuitLab

Answer:

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enter image description here

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    \$\begingroup\$ If each diode drops .7V, then shouldn't V1 = 2.8V? \$\endgroup\$ – Phil N DeBlanc Jun 19 '18 at 20:31
  • \$\begingroup\$ @PhilNDeBlanc how have you reached this result without knowing the value of the resistor? \$\endgroup\$ – Shivalnu Jun 19 '18 at 20:33
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    \$\begingroup\$ So what is the question? You have the target volatge drop on four diodes. Divide it by 4 and get the individual drop. From it you can calculate the current (using the exponential model). Now you have the current and voltage on the resistor, so you can apply Ohm's law. \$\endgroup\$ – Eugene Sh. Jun 19 '18 at 20:42
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    \$\begingroup\$ but I do not understand how the voltage drop across each diode is calculated That's the first line in the answer: 3 V / 4 = 0.75 V. Then use the diode formula to find the current needed for that 0.75 V. There's 10 - 3 = 7 V across the resistor. The current follows from the 0.75 V and diode formula. Then use Ohm's law to calculate the resistor. Each diode has Vd = 0.7 V, id = 1mA That should be read as: at Id = 1mA, Vd is 0.7 V. Obviously 4 x 0.7 V = 2.8 V so that does not make 3 V. A larger current Id must flow to make Vd = 0.75 V \$\endgroup\$ – Bimpelrekkie Jun 19 '18 at 20:42
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    \$\begingroup\$ @Peter. Do not get too caught up in the .7 Vdrop per silicon diode. At low current some may have a 0.65 Vdrop. Power rectifiers with many amps of current flowing may have a 1.0 Vdrop. High-voltage rectifiers of 10 KV rating may have a 100 volt drop due to 'stacking'. \$\endgroup\$ – Sparky256 Jun 19 '18 at 22:02
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I pretty much agree with your approach. Setting \$I_\text{CAL}=1\:\text{mA}\$ and \$V_\text{CAL}=700\:\text{mV}\$ and given \$\eta=1\$, the complete equation (using only Shockley diode equation) I get it, is:

$$R=\frac{10\:\text{V}-3\:\text{V}}{I_\text{CAL}\cdot\frac{e^{V_D/V_T}-1}{e^{V_\text{CAL}/V_T}-1}}$$

(Noting that \$V_D=750\:\text{mV}\$.)

Discounting the \$-1\$ terms, this simplifies a little:

$$\begin{align*}R&\approx\frac{7\:\text{V}}{I_\text{CAL}\cdot e^{{\left(V_D-V_\text{CAL}\right)}/V_T}}\\\\&\approx \frac{7\:\text{V}}{1\:\text{mA}\cdot e^{{\left(750\:\text{mV}-700\:\text{mV}\right)}/V_T}}\\\\&\approx 7\:\text{k}\Omega \:\cdot e^{{-50\:\text{mV}}/V_T}\end{align*}$$

Depending on the room temperature value for \$V_T\$ that you use (Spice defaults to about \$25.876\:\text{mV}\$), you will roughly get \$950\:\Omega\lt R\lt 1020\:\Omega\$.

I've rounded your value, which I take to be at the low end of this range. But I think your approach is solid.

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