What determines the forward voltage drop for a diode?

Question

I have always had the idea that the forward voltage drop in a semiconductor diode was related in a simple way to the bandgap energies in the semiconductor. However this is apparently not the case:

• germanium has a bandgap of 0.66 eV, but germanium diodes have a typical forward drop of 0.2 V

• silicon has a bandgap of 1.12 eV, but silicon diodes have a typical drop of 0.6 V

I'm aware of the Shockley equation describing the current in a diode as a function of diode voltage drop $V_D$ and temperature $T$, $$ I(V_D) =I_0 \left( e^{eV_D/kT} -1 \right) \approx I_0 e^{eV_D/kT} $$ where the scale current goes something like $$ I_0 = A \,T^3 e^{ -{E_\text{gap}}/{kT} } $$ and the constant $A$ depends on the geometry of the diode, the degree of doping, the charge mobility, and probably some other stuff, too.

I recognize that there's some arbitrariness to the approximation of a "turn-on voltage": the exponential grows so fast that if your choice for the current threshold differs from mine by a factor of a thousand, we'll only disagree about the turn-on voltage by about a couple tenths of a volt. However, I've had for years the impression that there's something fundamental about silicon that gives silicon diodes a forward drop of roughly 0.6V. Is that the case? Or is there some constellation of design choices that conspire to give the same drop both across most p-n diodes and across the p-n junctions of bipolar transistors?

I was motivated to ask this question by a similar question about forward voltage drops in LEDs. I was expecting to answer that question with some data from a student using LED turn-on voltage and the wavelength of light to measure the Planck constant. However those data are a lot more complicated than I expected: in fact, most of my LEDs apparently emit multiple wavelength components with comparable strength, and there doesn't seem to be much correlation the turn-on voltage between and the most prominent color in the LED spectrum. I don't seem to be able to say much more than "LEDs have turn-on voltages between two and three volts."

I have read a little bit about band-bending diagrams on Wikipedia, which suggest the potential barrier $\phi_B$ across an interface is different from the band gap, but I can't figure out why.

Answer 1

Lets get something out of the way first: The threshold, or turn-on voltage, is not really an intrinsic device property per se. It originates more from a desire by circuit designers to have a rule of thumb about how much a diode has to be forward biased to get it into conduction mode. As such, one takes the inherently non-linear current vs voltage response of the diode, pick an operating regime where enough current is flowing for your purposes, and linearly project back down to the voltage axis. You are now approximating a diode by being off (no conduction) up to the threshold, than a resistor (linear I vs V) at voltages above that. Given this, it is not obvious why or how the threshold should be related to semiconductor physics in a simple way.

First, a digression on Shockly-Read-Hall generation/recombination theory: Sze covers this in chapter 1, giving in equation 58 the recombination rate for a single defect level as (lets hope my Tex-fu is up to this):

$$U = \frac{\sigma_p \sigma_n v_{th} (pn-n^{2}_{i})N_t}{\sigma_n\left[n+n_i \exp(\frac{E_t - E_i}{kT})\right] + \sigma_p\left[p+n_i\exp(-\frac{E_t-E_i}{kT})\right]} $$

where $\sigma$ is the hole/electron capture cross section, $v_{th}$ is the carrier thermal velocity, $N_t$ is the trap density, $E_t$ is the trap energy level, $E_i$ is the intrinsic Fermi level, and $n_i$ is the intrinsic carrier density.

With that horror equation written down, one might begin to see how un-simple this problem is. Furthermore, that ugly equation assumes a single trap state. However, there may be more than one trap, and the trap concentrations may be a function of the Fermi level, temperature, etc., piling horror on horror.

Now, lets first look at two operating modes of the diode.

1. Reverse biased. Here, the junction is reversed biased, not so hard as to lead to avalanche breakdown. There is the built-in potential plus the applied potential across the depletion region. The diode is now under non-equilibrium conditions where $pn \ll n^{2}_{i}$ so there are no free carriers to get significant drift or diffusion contributions to the current in the junction itself. Instead, the reverse current comes from carrier generation in the depletion region, and those generated carriers are then swept out by the applied field. Throwing out the $p$ and $n$ terms, one is left (eq. 47 in chapter 2 of Sze) with the generation rate being proportional to the intrinsic carrier concentration $n_i$ and a bunch of remaining parameters that can look like a lifetime. So far so good. So, the reverse current can now be represented as the generation rate ($n_i/\tau$) and the width of the depletion region. But wait, the width of the depletion region depends on the doping profiles and the applied voltage, so it can vary as $(V_{bi} + V)^{1/n}$ where $n$ can be between 2 and 3. Furthermore, there may be diffusion components in the neutral regions, leading to (equation 50 in chapter 2 of Sze): $J_R = q\sqrt{\frac{D_p}{\tau_p}}\frac{n^{2}_{i}}{N_D} + \frac{qn_i W}{\tau}$. Which term dominates depends on the intrinsic carrier concentration, diffusion coefficient, and generation lifetime.

2. Forward bias. Here, the major SRH term is the capture processes as holes and electrons are driven together in the junction to recombine, leading to the observed current flow. We now have in the junction that $pn = n^{2}_{i}\exp[\frac{q(\phi_p - \phi_n)}{kT}]$ where $\phi_p$ and $\phi_n$ are the quasi-Fermi levels for holes and electrons. Plugging that in to the SRH equation above yields an unholy mess I won't attempt to replicate here, but is equation 51 in chapter 2 of Sze. Under some simplifying assumptions, one gets that $U = \frac{1}{2} \sigma v_{th} N_t n_i \exp(\frac{qV}{2kT})$. But, the diffusion current portion is still proportional to $\exp(\frac{qV}{kT})$ so the overall forward current is proportional to $\exp(\frac{qV}{nkT})$ where n can be between 1 and 2 depending on how important diffusion vs recombination is in a particular semiconductor material and device geometry.

So, the threshold voltage is somehow related to how the reverse biased case above transitions into the forward bias case. One would expect it to vary depending on the material and the junction design, as well as the 'normal' operating condition that the designer is calibrating their rule-of-thumb to.

So, stick with 0.7V for a silicon diode, unless of course it doesn't work out for your particular diode and circuit...

Answer 2

The contact potential (aka built-in voltage) of a p-n junction can be related to the material band-gap by combining two standard expressions:

First, the intrinsic carrier density $n_i$ of a pure semiconductor is given by the usual exponential formula $$ n_i = N_0 e^{-(E_g/q)/(2V_t)} $$

Here $N_0$ is a temperature-dependent factor, $E_g/q$ is the band-gap voltage (0.67V for Ge, 1.12V for Si), and $V_t$ is the thermal voltage $kT/q$, with $k=1.38\times10^{-23}$ Boltzmann's constant in Joules/Kelvin, $q=1.602\times10^{-19}$ the electron charge in Coulombs, and $T$ the temperature in Kelvin. At $T=300\,\mathrm K$, $V_t=25.8$ mV.

Besides depending on temperature, $N_0$ also depends on the material. Backing out values at $T=300\,\mathrm K$:

• Germanium ($n_i=2.50\times10^{13}/\mathrm{cm}^3$): $N_0=1.07\times10^{19}/\mathrm{cm}^3$
• Silicon ($n_i=1.50\times10^{10}/\mathrm{cm}^3$): $N_0=3.86\times10^{19}/\mathrm{cm}^3$

So it's not much of a constant. Carrying on regardless...

Second, the contact potential $\phi$ of a p-n junction is determined by balancing the opposing current flows of drift (from the electric field that produces $\phi$) and diffusion (from charge carrier density gradients). The result is: $$ \phi = V_t \ln \left( \frac{N_a N_d}{n_i^2} \right)= V_t \left( \ln{\frac{N_a}{n_i}} + \ln{\frac{N_d}{n_i}} \right) $$

Substituting for the intrinsic carrier density $n_i$ from our first result, we find: $$ \phi = \frac{E_g}{q} - V_t \left( \ln{\frac{N_0}{N_a}} + \ln{\frac{N_0}{N_d}} \right) $$

For typical doping levels $N_a=3.29\times10^{19}/\mathrm{cm}^3$ , $N_d=1.60\times10^{15}/\mathrm{cm}^3$, the contact potential is about 0.3V smaller than the band gap voltage.

Answer 3

The answers above all overcomplicate the problem. The reason there is a turn on voltage is because $Vt$ is $1/40 V$, so if there is a $25mV$ change in diode voltage the diode conducts more than twice as much.

This implies many things,

first that the plot of the diode "exponential" response is effectively like a brick wall.

Second the diode, when on, is not (cannot be) current limiting. The resistor is.

Third the Is of the diode sets the turn on voltage irregardless of what Is depends on (In fact Is can be a constant). This can be seen if you plot the solution of a diode in series with a resistor: You can move that resistor value around all you want (that's the intersection of the R I-V curve with the y-axis), or even the value of the voltage source (the intersection with the x-axis), but the turn on voltage will stay roughly the same. The current on the other hand varies widely.

However, change the value of the diode's Is, and the solution's voltage will change drastically. The current, however, will stay roughly the same.

Therefore, if the diode is in the on state, the resistor limits the current and the diode has a set voltage drop!

Finally, Ge diodes have different values of Vd because their Is is different.

Answer 4

By inverting the Shockley equation of the diode, you can easily obtain the forward voltage drop as a function of the current:

$$V_D(I) \approx \frac{kT}{e} \,ln \frac{I}{I_0}.$$

The "turn-on voltage" idea is just a rough simplification of this equation to make the calculations easier in diode modelling, but it doesn't have any physical meaning. It is possible because the logarithm present in the formula makes the changes in $V_D$ with $I$ very small. But still, to have a better description of the diode, the value of the turn-on voltage would require to be changed according to the current range of the particular calculation or simulation in which it is used.

However, the voltage forward drop $V_D$ does have some relation with the bandgap. Because the saturation current $I_0$ depends also on the bandgap:

$$I_0 = A \, e^{ -{E_\text{gap}}/{kT} },$$

semiconductors with higher bandgaps tend to have higher forward drops. But, of course, many other material parameters (and also temperature) contribute to shape the saturation current.