# Voltage across diode, Shockley equation

Voltage across diode, Schottky Shockley equation

So I know how a diode works and everything. However when I stumbled upon this equation I just happened to have this silly doubt. Vd it says (in the equation) is the voltage across the diode. I know that every diode has a small potential difference and only when the applied voltage is greater than that voltage across the diode, electrons are able to cross the barrier.

Now my question is : is the applied voltage same as the voltage across the diode? If not how do we calculate voltage across the diode is a voltage of 5V is applied. You can consider a simple circuit with just one diode and a battery.

• It’s the externally applied voltage. – Andy aka Feb 8 '18 at 14:34
• I'm sorry, I didn't get you. Could you elaborate? – user406653 Feb 8 '18 at 14:36
• Yes , as it says $V_D$ is the diode voltage, which in datasheets is Vf@If rated forward (min-typ-max) which is affected by Rs mfg tolerances for bulk resistance that depends on size of chip-terminations and chip etc... so when Vd rises , current rises exponentially up some rated current where Vf@If comes from Vth * n exponent which depends on material as in LEDs. Since V * I=P produces heat, temperature rise imposes a limit. Not shown in equation is bulk series resistance – Tony Stewart EE75 Feb 8 '18 at 14:42
• So are Vd(diode voltage) and applied voltage (V) different from one another? – user406653 Feb 8 '18 at 14:54
• Read again after "Where" – Tony Stewart EE75 Feb 8 '18 at 14:59

There are various diode models of varying fidelty to reality and I think you are mixing two of them.

The model you describe as

I know that every diode has a small potential difference and only when the applied voltage is greater than that voltage across the diode, electrons are able to cross the barrier.

differs from the model given by the Shockley equation. So don't be surprised if you get incompatible results if you mix them.

The Shockley equation is more accurate because in reality a diode is not either on or off but it conducts already little bit below the threshold voltage and also above it current increases more and more.

See following i-vs-v graphs of diode models that get more and more realistic. Note that although they may give you different results none of them is totally "wrong" or "right" just more or less accurate.

is the applied voltage same as the voltage across the diode?

Applied voltage is always the voltage across the device you apply it to. So this question doesn't make sense.

I guess you mean "Is the voltage across the diode the same as $V_D$?"

• For the simple models (top row) if the diode is on the voltage across the diode is always $V_D$ (constant; independent of current). There is nothing to calculate.
• Only for the more accurate models (bottom row) you can calculate the voltage for a given current and it is not the same as $V_D$ (depending on the current).

Now my question is : is the applied voltage same as the voltage across the diode?

In the equation, the term $V_D$ is the voltage that's across the diode's terminals. Note that you can either

• use a voltage source to apply a constant voltage of $V_D$ Volts across the diode's terminals and then measure the current $I_D$ that's flowing through the diode, or
• use a current source to inject a constant current of $I_D$ Amps through the diode and then measure the voltage $V_D$ across the diode's terminals.

For what it's worth, the commonly-used form of the Shockley diode equation (as shown in your question) does not make evident that

• the value of the reverse saturation current term $I_S$ depends upon the diode's junction temperature T—i.e., $I_S$'s value changes as the diode's junction temperature T changes, and
• $I_S$ plays a critical role in determining how the diode's current changes $\partial I_D$ as the diode's junction temperature changes $\partial T$—i.e., the rate $\partial I_D / \partial T$ is heavily dependent upon $\partial I_S / \partial T$.

For the sake of argument, let's assume $I_S$'s value is constant (more-or-less), and the values for $q$, $V_D$, $\eta$, and $k$ are all constants. In fact, for simplicity, let's define a new constant $x$ whose value is $x=(q\,V_D/\eta\,k)$, so that

$$I_D = I_S(e^{x/T}-1)\;\;\;\;\;\;\;\;(1)$$

For the conditions mentioned above, Eqn. (1) predicts that as the diode's junction temperature T increases, the current through the diode $I_D$ tends to zero:

$$\lim_{T \to \infty}I_S(e^{x/T}-1)\\ \Rightarrow I_S \cdot (e^{x/\infty}-1)\\ \Rightarrow I_S \cdot (e^{0}-1)\\ \Rightarrow I_S \cdot (1-1)\\ \Rightarrow I_S \cdot (0)\\ \Rightarrow 0$$

Of course in a real diode, if $q$, $V_D$, $\eta$, and $k$ are more-or-less constant, we know (from making measurements with test equipment) that the diode's current $I_D$ increases as its junction temperature T increases; this contradicts the result shown above which says that $I_D$ should decrease toward zero as the junction temperature T increases.

Therefore, we can conclude that the $I_S$ term a) cannot be constant-valued, and b) must be the term in the Shockley diode equation that causes the diode current $I_D$ to increase as the junction temperature T increases. In other words, the term $(e^{(q\,V_D/\eta\,kT)}-1)$ tries to decrease $I_D$ as T rises, and the term $I_S$ tries to increase $I_D$ as T rises, and for a given change in junction temperature $\partial T$ the following relation must hold if $I_D$ is to increase for increasing T:

$$\left | \frac{\partial I_S}{\partial T} \right | > \left | \frac{\partial}{\partial T} \left (e^{(q\,V_D/\eta\,k\,T)}-1 \right ) \right |$$

i.e., for a given change in junction temperature $\partial T$, the $I_S$ term has greater influence on the change in diode current $\partial I_D$ compared to the $(e^{(q\,V_D/\eta\,k\,T)}-1)$ term.

For what it's worth, Eqn. (2) is a commonly used formula (model) for calculating the reverse saturation current term $I_S$ as a function of junction temperature T:

$$I_S = I_K \cdot e^{(-q\,E_g/\eta\,k\,T)}\;\;\;\;\;\;\;\;(2)$$

And so an improved model for the diode current $I_D$ would be Eqn. (3):

$$I_D = (I_K \cdot e^{(-q\,E_g/\eta\,k\,T)}) \cdot (e^{(q\,V_D/\eta\,k\,T)}-1)\;\;\;\;\;\;\;(3)$$

For more information here's a useful reference that provides descriptions for the terms $I_K$ and $E_g$ in Eqn. (2).