# Voltage drop across a diode [closed]

For the sake of simplicity, the voltage drop across a diode, $V_d$ made of silicon is assumed to be $0.7\mathrm{V}$ after the input voltage exceeds this voltage; and $V_d$ is assumed to stay constant regardless how large the input voltage, $V_i$ is after the input voltage exceeds the threshold. In practical, there will be an increase in $V_d$, albeit a very small one, as compared to the large increase in $V_i$. There is an ideal diode equation: $$I=I_s\left(e^\frac{qV_i}{kT}-1\right)$$ which relates the current flowing through diode $I$ with the input voltage $V_i$. So a little increase in $V_i$ will result in a large increase in $I$, as $I$ is an exponential function of $V_i$.

I didn't express my question clearly in my previous attempt. Here is my question:

We consider a circuit consists of a diode is connected to a DC power supply. If we were to connect a voltmeter parallel to the diode, and plot the reading of voltmeter(voltage drop across the diode) against the value of the DC input, how will the graph look like? Is there any equation to describe the graph(like the ideal diode equation relating $I$ with $V_i$)?

Should the equation look something like this? $$V=IR_1+V_t\ln{\left(1+\frac{I}{I_s}\right)}$$ based on an article in Wikipedia under the title Shockley diode equation.

• You are kinda mixing two models. Keep also in mind that $V_d$ has no physical equivalent, i.e. it is chosen more or less arbitrarily. Starting from the true diode equation you can't possibly derive $V_d$. – Vladimir Cravero Apr 27 '16 at 14:48
• In this case $V_d = V_i$ – G36 Apr 27 '16 at 14:53
• I think the answer you are looking for is that, unlike resistors, capacitors and inductors, diodes are non-linear devices. You can't use one equation to describe how they work across commonly used voltage / current ranges. – st2000 Apr 27 '16 at 14:56
• @VladimirCravero Agree, I didn't realize I had been mixing two models. Is there any equation concerning the voltage drop across the diode? I have been looking up different semiconductor devices textbooks to no avail. – Dave Clifford Apr 27 '16 at 15:07
• My suggestion would be to measure the diode drop for a number of different currents and plot up your results. (say 100 nA to 1 mA) – George Herold Apr 27 '16 at 16:21

Let me explain a mathematical origin of the magical number 700 mV. This may help you to understand what's wrong with your question.

Let $V_T = \frac{kT}{q} \approx 26\,\text{mV}$ at room temperature. For $V_d > V_T$ we can rewrite the diode equation this way: $$I = I_s \left[\exp\left(\frac{V_d}{V_T}\right) - 1\right] \approx I_s \exp\left(\frac{V_d}{V_T}\right) = 1\,\text{A}\cdot\exp\left(\frac{V_d + V_T\log(I_s)}{V_T}\right)$$

A typical value of $I_s$ for a silicon p-n diode is around $1\cdot10^{-12}\,\text{A}$, so $\log(I_s) \approx -28$.

The numerator becomes positive if $V_d > -V_T\log(I_s) \approx 28\cdot26\,\text{mV} = 728\,\text{mV}$. Nothing abrupt happens at this point; just a border between negative and positive values of $x$ in $\exp(x)$, where $x = \frac{V_d + V_T\log(I_s)}{V_T}$.

Define $V_d = 728\,\text{mV}$ as "the threshold voltage", and the magical number is ready.

Diode is a "current driven" devices. And you should never connect a diode directly across the voltage source without current limiting device in series with the diode. Otherwise the diode will burn. Because the ideal diode act just like a "short circuit" when diode is in forward biased or like a voltage source with constant voltage (0.7V).

As for voltage across the diode versus current we have this $$Vd=\frac{kT}{q}*ln(\frac{Id}{Is})$$

And notice that the diode forward voltage have to be increased by only 60mV in order to increase the diode forward current by the factor of 10. And this is why in most application we assume Vd = 0.7V http://www.ittc.ku.edu/~jstiles/312/handouts/The%20Junction%20Diode%20Forward%20Bias%20Equation.pdf http://www.ittc.ku.edu/~jstiles/312/handouts/The%20Junction%20Diode%20Curve.pdf