Diodes

Due to the fact that there exist few freely movable charge carriers in the depletion region around the p-n junction, the conductivity is very poor. However, if certain voltage is applied to the two ends of the material, the conductivity may change, depending one the polarity of the applied voltage:

• Reverse bias (negative to p-type, positive to n-type)

  The negative voltage applied to the p-type will repel electrons in n-type and attract holes in p-type so that both carriers are moving away from the p-n junction. As the depletion region becomes thicker than before, there is no current through the p-n junction and the conductivity is zero.

• Forward bias (positive to p-type, negative to n-type)

  The positive voltage applied to the p-type will attract electrons in n-type and repel holes in p-type so that both carriers are moving towards the p-n junction. As the depletion region becomes thinner, the conductivity is improved and there is current through the p-n junction. The conductivity increases as the voltage becomes higher.

The voltage-current behavior of a p-n junction is described by

$$I_D=I_0 ( e^{V_D/\eta V_T}-1 ), \;\;\;\;\mbox{or}\;\;\;\; V_... ...V_T\;ln (\frac{I_D}{I_0}+1)=\eta V_T\;ln (\frac{I_D+I_0}{I_0})$$

where

• $I_0$ is the reverse saturation current, a tiny current that flows in the reverse direction when $V_D \ll 0$, due to the minority carriers. As this current is limited by the minority carriers available, it is called saturation current. $I_0$ is about $10^{-10} \sim 10^{-12}$ A for Si and $10^{-4}$ A for Ge.
• $V_T=kT/e$ is the thermal voltage, where $k=1.38\times 10^{-23}$ Joules/Kelvin is Boltzmann's constant, $e=1.602\times 10^{-19}$ coulomb is the charge of an electron, and $T$ is the temperature in degree K. For room temperature $T=300K$, $V_T=kT/e=26\; mV$.
• $\eta$ is the ideality factor which varies between 1 and 2, depending on the fabrication process and semiconductor material. In many cases $\eta$ can be assumed to be approximately equal to 1.

In particular, when $V_D=0$, $I_D=0$, when $V_D \ll 0$, $I_D=-I_0$, when $V_D\gg 0$, $I_D=I_0 e^{V_D/V_T}$.

The resistance of an electrical device is defined as $r=\Delta V/\Delta I$. For a diode, as $V_D(I_D)$ is not a linear function, the resistance $R_0=dV_D/dI_D$ is not a constant, but a function of $I_D$:

$$R_0=\frac{d}{dI_D}V_D=\frac{d}{dI_D} [\eta V_T\;ln (\frac{I... ...0}=\eta \; \frac{V_T}{I_D+I_0} \approx \eta \; \frac{V_T}{I_D}$$

The last approximation is due to the fact that $I_D \gg I_0$, i.e., $I_D+I_0\approx I_D$. We assume $V_T=26\;mV$, $\eta=1$, then if $I_D=1\;mA$, $R_0=26\;\Omega$, but if $I_D=2\;mA$, $R_0=13\;\Omega$. In other words, the resistance $R_0$ of a diode is not a constant, but a function of the current $I_D$, i.e., a diode is not a linear element.

Models of diodes:

• Ideal model: if $V_D<0$, then $I_D=0$, else $I_D=V/R$
• Ideal model with a voltage threshold $V_0=0.7V$: if $V_D<V_0$, then $I_D=0$, else $I_D=(V-V_0)/R$
• The model above in series with a resistance $R_0=20\Omega$: if $V_D<V_0$, then $I_D=0$, else $I_D=(V-V_0)/(R+R_0)$
• The model above in parallel with a current source that simulates the reverse saturation current.

$I_0$	1 mA	10 mA	100 mA
$V_D$ for Si ($I_0=10^{-10}$, $\eta=1.4$)	0.58 V	0.67 V	0.75 V
$V_D$ for Ge ($I_0=10^{-4}$, $\eta=1.0$)	0.06 V	0.12 V	0.18 V

In general, when the forward voltage applied to a diode exceeds 0.6 to 0.7V for silicon (or 0.1 to 0.2 V for germanium) material, the diode is assumed to be conducting with very little resistance.

Example 1: In the half-wave rectifier circuit shown below, $R=1000\Omega$, $V=3V$, and $D$ is a silicon diode. Find the current $I_D$ through and voltage $V_D$ across $D$.

• Method 0: The simplest model is to assume the diode is an ideal rectifier with infinite resistance when it is reverse biased but zero resistance when it is forward biased. As the diode is forward biased, the current is $I_D=V/R=3mA$.

• Method 1: Since the diode is forward biased, we can assume the voltage across the diode is $0.7V$ and the current can be determined by Ohm's law to be $I_D=(V-0.7)/R=2.3/1000=2.3mA$.

• Method 2: The current $I_D$ and voltage $V_D$ have to satisfy two equations simultaneously:
  
  $$\left\{ \begin{array}{l} I_D=I_0(e^{V_D/V_T}-1) \\ V=V_D+RI_D \end{array} \right.$$
  
  
  The first equation relates the current $I_D$ through and voltage $V_D$ across a diode, while the second is obtained by KVL. Substituting the first equation into the second, we get $V=V_D+RI_0(e^{V_D/V_T}-1)$, which can be solved numerically for $V_D$, then $I_D$ can be found.

• Method 3: The two simultaneous equations above can also be solved graphically. The first equation $I_D=I_0(e^{V_D/V_T}-1)$ is the characteristic curve of the diode, while the second equation $V=V_D+I_DR$ is the load line, a straight line passing through $(V_D=0, I_D=V/R)$ and $(I_D=V, I_D=0)$. The intersection point of the two curves is approximately at $(V_D=0.75V, I_D=2.4mA)$.

Example 2: Design a converter (adaptor) that converts AC power supply of 115V and 60 Hz to a DC voltage source of 14 V. When the load is $R_L=10\;K\Omega$, the variation (ripple) of the output DC voltage must be 5% or less.

• The peak of the secondary output is $14V$ with RMS value $14/\sqrt{2}=10V$, the ratio of the transformer should be 115:10.
• When the load is $R_L=10\;K\Omega$, the load current is $I=V/R_L=14/10 =1.4\;mA$.
• During the period between two peaks $T=1/f=1/60=16.7\;ms$, the charge on the capacitor is reduced by $\Delta Q=IT=1.4\;mA\times 16.7\;ms =23.4\;\mu C$.
• The voltage across the capacitor is therefore dropped by $\Delta V=\Delta Q/C < 14\times 5/\%=0.7V$.
• Solve above equation for $C$, we get $C=\Delta Q/\Delta V =23.4\;\mu C/0.7V=33.4\;\mu F$.

This is an approximation based on the assumption that the load current is constant, as the voltage drop is small. Otherwise the exponential decay of the voltage across capacitor should be used, and the current is:

$$i(t)=\frac{V}{R_L} e^{-t/\tau}$$