Speed increases collide with valence electrons bonds are broken.

Avalanche Breakdown

• If you doped this p-n junctions with much larger concentrations, then what happens?

• Then you have depletion layer is very thin.

• Depletion layer is thin, ~ $\mu m$

• Electric field is high in Reverse bias condition.

• Valence electrons going from valence band to conduction band.

• This is known as zener breakdown.

• The p-n junctions which use this zener breakdown for the application, they are known as zener diodes.

• Zener diodes can be used as a voltage regulator.

• The current increases almost without increasing the voltage.

• Different amount of currents for the same voltage across it.

• Power supply is giving DC.

• Shunt resistance: $R_S$

• Put a zener diode in reverse bias condition.

• The zener diode: 5V for the breakdown

• 2 watts as the power rating.

• $V_A = 10$ volts

• Let us first do an analysis where no load is connected.

• $V_s = 10 \text{volts} - 5 \text{volts} = 5 \text{volts}$

• $I_s = \frac{5 \text{volts}}{R}$

• $P=2W$ (Zener Diodes)

• $V = 5 \text{volts}$ (Zener Diodes)

• $P = VI$

• Minimum Resistance $R_s$, so that the Current in the diode does not exceed $400 ~mA ?$

• What should be the minimum resistance?

• Voltage across $R_s = 5V$

• Current $=0.4 A$

• $R_s = \frac{V}{I} = \frac{5}{0.4}\Omega = 12.5\Omega$

Current Through $1k \Omega$

$V = 5 \text{Volts}$

$R = 1 k\Omega$

$I_L = 5 mA$

$\underline{\text{Current in} {~R_s}}$

$I_s =\frac{5\text{volts}}{12.5\Omega} = 400mA$.

• So, now you can calculate how much is the current through the diode.

$I_D = 400mA - 5mA = 395 mA$.

• So, when the load was not there the current through the zener was 400 mA when the load is there the current is 395 mA.

• As you have upper limit on the current in zener, you also have a lower limit on the current in zener.

• In reverse bias breakdown condition you need a certain current.

$\text{Current} {~I_L} = \frac{5V}{1k\Omega} = 5mA$.

$R_s= 8-5 = 3 \text{Volts}$

$I_s=\frac{V}{R_s}=\frac{3V}{12.5\Omega} = 0.24A = 240 mA$

$I_D= 235 mA$.

Source voltage should be greater than the Zener Breakdown.

Current

${~I_L} = \frac{5V}{1k\Omega} = 5mA$

Voltage across, $R_s= 8-5 = 3 \text{Volts}$

$I_s=\frac{V}{R_s}=\frac{3V}{12.5\Omega} = 0.24A = 240 mA$

$I_D= 235 mA$

Source voltage should be greater than the Zener Breakdown.

• $E=\frac{hc}{\lambda}$

• hc = 1240 eV.nm

• $\lambda = 400 ~nm$

• $E=\frac{1240~eV.nm}{400 ~nm} = 3.1 eV$

• $\lambda = 620 ~nm$

• $E = \frac{1240 eV.nm}{620 nm} = 2 {~eV}$

Visible Region

• $\lambda\quad 400 - 700 ~nm$

• $E \quad 1.8 ~eV - 3.2 ~eV$

• $\lambda = 620 ~nm,$

• $E = \frac{1240 ~eV.nm}{620 ~nm} = 2 ~eV$

Current increases with more charge carriers created by the light falling on the junction increase in current is proportional to the intensity of light.

Recombination of electron - hole pair

$E_g$ should be in visible range

$E_g \sim$

$1.8 \mathrm{eV}-3.2 \mathrm{eV}$

{Ga As},{Ga As P},{Ga Ni},{Ga I},{Sic}

• 2014 Nobel Prize for Blue light emitting Diode.

$h\nu > E_g$

$e-h \hspace{2mm}$ Pair generation

• Pairs which are generated are swpet away by the Electric field.

• The voltage which is generated is called as a open circuit voltage

0.6 volts.

• Maximum Circuit ~ mA

• Short Circuit Current

Current given by the Cell

$p - n$ Junction

$V - I$ Current Through The Junction