Field and Potential in P-n Junction
Special Purpose P-N Junction
Avalanche Breakdown
Speed increases collide with valence electrons bonds are broken.
If you doped this p-n junctions with much larger concentrations, then what happens?
Then you have depletion layer is very thin.
Higher Concentration
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.
I-V Curve Showing Avalanche and Zener Breakdown
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.
Voltage Regulator
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 $=0.4 A$
$R_s = \frac{V}{I} = \frac{5}{0.4}\Omega = 12.5\Omega$
Problem
Current Through $1k \Omega$
$V = 5 \text{Volts}$
$ R = 1 k\Omega$
$ I_L = 5 mA$
$\underline{\text{Current in} {~R_s}}$
$I_s =\frac{5\text{volts}}{12.5\Omega} = 400mA$.
$I_D = 400mA - 5mA = 395 mA$.
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, $R_s= 8-5 = 3 \text{Volts}$
$I_D= 235 mA$
$E=\frac{hc}{\lambda}$
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} $
Photodiode
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.
Light Emitting Diode: LED
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}
Solar Cell
$h\nu > E_g$
$e-h \hspace{2mm}$ 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
Thank you