iv Characteristic of a Diode

Q1: How can I understand the iv characteristic of a diode?
A1: You could use the Shockley diode equation to relate $i_D$ to $v_D$:
$i_D=I_S(e^\frac{v_D}{nv_T}-1)$

• $I_S$ should be familiar to you. For a rough idea, $I_S$ can range from $10^{-9}$ to $10^{18}$A
• n is a constant ideality factor that ranges from 1 to 2. We'll largely assume the ideal case where n=1.
• $v_T$ is a constant thermal voltage. At room temperature, $v_T\approx 26mV$

Let's confirm that the Shockley diode equation is consistent with what you already know:

• When the diode is forward-biased, the current through the diode increases dramatically.

$v_D$ is sufficiently large $\rightarrow$ The exponential dominates the expression inside the parentheses 

$i_D=I_Se^\frac{v_D}{nv_T}$

• When the diode is reverse-biased, a small current flows in reverse.

$v_D$ is sufficiently small $\rightarrow$ The -1 dominates the expression inside the parentheses 

$i_D=-I_S$

But note that the Shockley diode equation does not account for a diode's breakdown region.

Q2: Why should I be concerned that the Shockley diode equation describes a nonlinear relationship between voltage across and current through a resistor?
A2: Because the presence of a nonlinear element in the circuit means that you can't solve the circuit analytically. You could use numerical or graphical methods of solution. But the method we'll look at is the piecewise linear model (PWL) of the diode:

1. Replace the exponential that takes off past the knee voltage $v_{D0}$ by a vertical line.

• For silicon, $v_{D0}$ has an approximate range of 0.6-0.7V

1. Replace $I_S$ with 0 (assumes the reverse saturation voltage is negligible).

We can define 2 states: 

1. On: $v_D=v_{D0}$
2. Off: $v_D<v_{D0}$

Again, like the Shockley diode equation, the PWL model doesn't account for the diode's breakdown region. 

But unlike the Shockley diode equation, the PWL model gives us a straightforward algorithm by which diode circuits may be solved:

Basic Method of Solving Diode Circuits

1. Write down the circuit equations and simplify as much as possible
2. Assume the diode is on or off to get a diode equation. Use the diode equation to solve the circuit and find $i_D$ and $v_D$
3. If $i_D$ or $v_D$ satisfy the inequality associated with the state, then you have solved the diode circuit. Otherwise, repeat Step 2 assuming the diode is in the other state.

Q3: How is the basic method of solving diode circuits limited?

A3: The basic method of solving a diode circuit only gives a solution that assumes that the diode is in a particular state. To give a more general solution, you could solve the diode circuit parametrically (i.e. treating the diode as a parameter).

Q4: In the PWL model, a vertical line replaces the exponential at knee voltage. Why not replace the exponential with a non-vertical line that has a positive slope?

A4: There are two reasons for why diodes are not usually modeled that way:

1. The model introduces another parameter (slope), whose value is somewhat arbitrary
2. A positively-sloped line is not that much of a better fit for the exponential compared to a vertical line

Q5: So the PWL model seems pretty handy. But can I account for the diode's breakdown region?

A5: You can use a Zener piecewise linear model of the diode. The Zener PWL is the same as the PWL except that:

• At the breakdown voltage or Zener voltage $v_Z$, the exponential rise in negative current is replaced by a vertical line.

Hence, we may define 3 states (the first 2 of which are familiar):

1. On: $v_D=v_{D0}$
2. Off: $-v_Z<v_D<v_{D0}$
3. Zener: $v_D=-v_Z$

Again, if you want to know why a sloped line isn't used in place of a vertical line at $-v_Z$ refer back to the explanation in A4.

Usually, hitting the Zener voltage is catastrophic. But there are diodes that are specially made to operate in this breakdown region: the Zener diodes. Zener diodes are useful to serve as reference voltages.