EEEE3089 Seminar Lecture examples
lecture 5 examples
clear variables
L = [350 400 650 750]*1e-9;
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H = 6.62e-34;
e = 1.6e-19;
Eev = E./e;
val = table(1e9*L.’,F.’,E.’,Eev.’);
val.Properties.VariableNames= {‘wavelength nm’,’Frequency Hz’,’Energy J’,’Energy eV’};
temp=([1e9*L.’,F.’,E.’,Eev.’].*eye(4));
Light and Energy
Complete the missing values from in the below table
val2 = table(temp(1,:).’,temp(2,:).’,temp(3,:).’,temp(4,:).’);
val2.Properties.VariableNames= {‘wavelength nm’,’Frequency Hz’,’Energy J’,’Energy eV’}
val2 = 4×4 table wavelength nm
Frequency Hz
350 0 0 0 0 7.5000e+14 0 0 0 0 3.0554e-19 0 0 0 0 1.6550
Recall some useful equations and constants
• H = 6.62e-34;
• e = 1.6e-19;
• C = 3e8;
val = 4×4 table wavelength nm
photon flux
Frequency Hz
8.5714e+14 7.5000e+14 4.6154e+14 4.0000e+14
5.6743e-19 4.9650e-19 3.0554e-19 2.6480e-19
3.5464 3.1031 1.9096 1.6550
What is the photon flux for the following light sources? Photon flux is the number of photons per second emiited (or recieved) by a device.
power = [0.1e-3 25e-3 2];
L = [480 532 1064]*1e-9;
Q = table(1e9*L.’,power.’);
Q.Properties.VariableNames= {‘wavelength nm’,’power W’}
Q = 3×2 table wavelength nm
0.0001 0.0250 2
480.0000 532 1064
we need to go from wavelength to frequency to get energy of a single photon. Then divide the power of the source by the energy of the photon to get the number of photons per second needed to produce this amount of power.
PF = 3×1 table
Photon Flux /s
PF = table((power./E).’);
PF.Properties.VariableNames= {‘Photon Flux /s’}
2.4169e+14 6.6969e+16 1.0715e+19
Bandwidth and Photodiodes
Calcualte the bandwidth of a photodiode with the following parameters 0.4mm square device, with , a saturation velocity (V_s) of 10×10^5 m/s and a depletion region width (w) of 1 microns. The load (R_L) it is connected to has an impedance of 50 Ohms.
plug in the numbers, remember to calculate the area od the device from the size given, it’s sqaure so 0.4mm*0.4mm is the area.
%recall the equation for the bandwidth is
% $$ BW = \frac{1}{2 \pi (\tau_R + \tau_RC)}$$
Vsat = 10e5; %m/s
E0 = 8.85e-12;
A = 0.4e-3*0.4e-3; %SQ
Eps = E0 * Er;
C = Eps * A/w;
TRC = C*RL;
TR = w/Vsat;
BW = 1/(2*pi*(TR+TRC))
BW = 1.9190e+08
BW_MHz = BW*1e-6
BW_MHz = 191.9009
we see TR is ~ 800 times smaller than TRC and so TRC dominates the response of the photodiode.
ans = 828.3600
What couild we change about this diode to improve the bandwidth? Both in terms of the use of this diode and in terms of changing it’s structure?
suggestions?
What is the responsivity at 532nm is the quantum efficieny of the photodiode is 65%
QE = 0.6500
We also have the short form, as long as we have QE as 0-1 scale and wavelength in microns
What is the Quantum Efficieny at 800nm if the responsivity at this wavelength is 0.5 A/W?
R = 0.5000
QE = 0.7750
This gives the QE to be 0.775 or 77.5% which is typical for a silicon photodiode in the near infrared. This means for each 100 photons recieved 77.5 electrons are produced.
Noise examples
Calculate NEP for 2 devices below,which noise source is dominant in each case? You may assume that the BW = 1 Hz, and they are operating at room temperature.
shunt = [500 1]*1e6;
Id = [1e-9 1e-8];
Res = [0.5 0.65];
Devices = table(1e9*Id.’,Res.’,1e-6*shunt.’);
Devices.Properties.VariableNames= {‘Dark Current (Id) nA’,’Responsivity (R) A/W’,’Shunt Resista
Devices = 2×3 table
H = 6.62e-34;
e = 1.6e-19;
QE = 65/100 % convert to number from percent.
R2 = QE*e/(H*C/L);
R2_2 = QE*L*1e6/1.24;
L = 0.800; %in microns
QE = R*1.24/L % use the short version and it’s very quick to work out
Dark Current (Id) nA
Responsivity (R) A/W
Shunt Resistance (Rd) Mega-Ohms
sn = 1×2 10-13 ×
0.1789 0.5657
jn = sqrt(4*k*T*BW./shunt)
jn = 1×2 10-12 ×
0.0056 0.1259
tn = sqrt(sn.*sn+jn.*jn)
tn = 1×2 10-12 ×
0.0188 0.1380
NEP = tn./Res
0.5000 0.6500
• T = 298; 25 deg C is room temperature • e = 1.6e-19; charge on electron
• k=1.33e-23; Boltzmans constant
T = 298; %25 deg C is room temperature
e = 1.6e-19; %charge on electron
k=1.33e-23; %boltzmans constant
sn = sqrt(2*e*BW.*Id)
NEP = 1×2 10-12 ×
0.0375 0.2124
Device 1 has lower NEP.
Shot noise dominates for device 1 ~10x more weight than thermal noise due to sum of squares. Device 2 the thermal is 2x higher but both contribute, though shot noise contribution is low.
3.1768 0.4493
If the devices are used in difference applications where the bandwidth and the temperature is as given below. How does this change the NEP of the devices? Device 1 BW is 100 MHz @52 degrees operating temperature Device 1 BW is 100 kHz @52 degrees operating temperature.
e = 1.6000e-19
k=1.33e-23
k = 1.3300e-23
sn = 1×2 10-9 ×
0.1789 0.0179
jn = sqrt(4*k.*T.*BW./shunt)
jn = 1×2 10-10 ×
0.5880 0.4158
tn = sqrt(sn.*sn+jn.*jn)
tn = 1×2 10-9 ×
0.1883 0.0453
NEP = tn./Res
BW=[100e6 100e3];
T = [325 325];
e = 1.6e-19
shunt = [500 1]*1e6;
Id = [1e-9 1e-8];
Res = [0.5 0.65];
sn = sqrt(2*e.*BW.*Id)
0.3766 0.0696
3.0420 0.4302
Device 2 has a lower NEP now, the increased bandwidth operation of device 1 makes it have more noise.
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