Diodes

A diode is formed by a PN-junction with the p side called anode and the n side called cathode. Due to the fact that there exist few freely movable charge carriers in the depletion region around the PN-junction, the conductivity is very poor. However, when external voltage is applied to the two ends of the material, the conductivity may change, depending one the polarity of the applied voltage.

• Forward bias (positive to P-type, negative to N-type)

  The positive voltage applied to the P-type will pull electrons in N-type and repel holes in P-type so that both carriers are moving towards the PN-junction. As the depletion region becomes thinner, the conductivity increases due to the drift current through the PN-junction from the P side to the N side, formed by the majority charge carriers (both electrons and holes) driven by the applied voltage. The conductivity increases as the applied voltage becomes higher.

• 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 PN-junction. As the depletion region becomes thicker than before, there is no current through the PN-junction from the P side to the N side. However, there exists a very small current $I_0$, called the reverse saturation current, due to the minority carriers. The carrier velocity increases as the applied voltage becomes higher. However, as the voltage further increases, the velocity will reach a maximum level called saturation velocity.

The non-linear voltage-current relationship of a PN-junction is described by

$$I_D=I_0 \left( e^{V_D/\eta V_T}-1 \right)\approx I_0\,e^{V_D/\eta V_T}, \;\;\;\; \;\;\;\; V_D=\eta V_T\;ln \left(\frac{I_D}{I_0}+1\right)\approx\eta V_T\;ln (I_D/I_0)$$ (2)

where

• $I_0$ is the reverse saturation current, a tiny current that flows in the reverse direction when $V_D < 0$, due to the minority carriers. As this current is limited by the minority carriers available, when all of them are contributing to this current, higher voltage $V_D$ does not result in greater current, i.e., the current is saturated. $I_0$ 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 the Boltzmann constant, $e=1.602\times 10^{-19}$ coulomb is the charge of an electron, and $T$ is the temperature in degree K. At room temperature $T=300 K$ ( $27^\circ C$), $V_T=kT/e=26\; mV$.
• $\eta$ 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 $V_D < 0$, $I_D\approx -I_0$;
• when $V_D=0$, $I_D=0$;
• when $V_D>0$, $I_D\approx I_0 e^{V_D/V_T}$.

The voltage $V_D$ across the diode is a function of the current $I_D$ through the diode. In the range of 5 mA to 20 mA, $V_D$ is about 0.7 V:

$$\begin{array}{c\vert\vert c\vert c\vert c\vert c\vert c} \hli... ... & 0.06 V & 0.10 & 0.12 V & 0.14 & 0.18 V \\ \hline \end{array}$$ (3)

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 $R_D=dV_D/dI_D$ can be found as

$$R_D=\frac{d\,V_D}{d\,I_D}=\frac{d}{d\,I_D} \left[\eta V_T\;ln \le... ...D+I_0}\frac{1}{I_0}=\eta \; \frac{V_T}{I_D+I_0} \approx \eta \; \frac{V_T}{I_D}$$ (4)

The approximation is due to the fact that $I_D \gg I_0$, i.e., $I_D+I_0\approx I_D$. We assume $V_T=26\;mV$, $\eta=1.4$, the resistance of the diode is not a constant, but a function of the current $I_D$, i.e., a diode is not a linear element:

$$\begin{array}{c\vert\vert c\vert c\vert c\vert c\vert c\vert ... ...mega & 36\,\Omega & 18\,\Omega & 7\,\Omega\\ \hline \end{array}$$ (5)

Models of diodes:

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

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 low resistance.

Example: 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 $I_D=V/R=3\;mA$.

• 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 $I_D=(V-0.7)/R=2.3/1000=2.3\;mA$.

• Method 2: The diode is modeled as a series combination of $V_0=0.7\;V$, $R_D=20\Omega$, and an ideal diode. $I_D=(3-0.7)/(1000+20)=2.255\;mA$, $V_D=0.7+I_DR_D=0.745\;V$.

• Method 3: The current $I_D$ and voltage $V_D$ have to satisfy two equations simultaneously:

  $$\left\{ \begin{array}{l} I_D=I_0(e^{V_D/V_T}-1) \\ V_D=V-RI_D \end{array} \right.$$ (6)

  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 $V_D=V-RI_0(e^{V_D/V_T}-1)$. Substituting this into the first equation we can then find $I_D$.

• Method 4: 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 $(V_D=0.75\;V, I_D=2.4\;mA)$.

Diodes are typically used as rectifiers which convert an AC voltage/current in to a DC one, such as shown in the following example.

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 desired DC voltage is 14 V, but we also need to compensate for the voltage drop of 0.7 V due to the forward biased diode, so the peak of the secondary output is $14.7\;V$ with RMS value $14.7/\sqrt{2}=10.4V$, the ratio of the transformer should be $115/10.4=11$.
• When the load is $R_L=10\;K\Omega$, the load current is approximately $I\approx V/R_L=14/10=1.4\;mA$.
• During the period between two peaks $T=1/f=1/60=16.7\;ms$, the charge on the capacitor is reduced by $\Delta Q\approx IT=1.4\;mA\times 16.7\;ms=23.4\;\mu C$.
• 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:

$$i(t)=\frac{V}{R_L} e^{-t/\tau}$$ (7)

More diode rectification circuits are shown below: