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[Moved]: dark and illuminate current of a photo diode in a

时间:03-29 整理:3721RD 点击:
how can see the dark current vs illuminate current in silvaco atlas. please help

Code:
region num=5 material= InP y.min=2.3
region num=6 material=air  y.min=-0.3 y.max=-0.1


electrode name=anode material= aluminum  x.min=16 x.max=19 y.min=-0.3 y.max=0.1
electrode name=cathode material= aluminum y.min=5.7 y.max=5.75

doping uniform concentration=1e11  n.type y.min=0 y.max=0.95
doping gaussian concentration=1e18 characteristic=0.02 p.type y.min=0 y.max=0.2 x.min=0 x.max=15.5
doping gaussian concentration=1e18 characteristic=0.15 p.type y.min=0 y.max=0.1 x.min=15.5 x.max=20
doping uniform concentration=1e11 n.type region=5
#doping uniform concentration=1e18 n.type y.min=5.7 y.max=5.9

material AUGN=9.0E-31 MATERIAL=InP
material AUGP=9.0E-31 MATERIAL=InP
material AUGN=7.0E-29 MATERIAL=InGaAs
material AUGP=7.0E-29 MATERIAL=InGaAs

material TAUN=6.0E-12 MATERIAL=InP
material TAUP=6.0E-12 MATERIAL=InP
material TAUN=8.0E-8 MATERIAL=InGaAs
material TAUP=8.0E-8 MATERIAL=InGaAs

material BN=3.45E6 MATERIAL=InP
material BP=2.785E6 MATERIAL=InP
material BN=1.95E6 MATERIAL=InGaAs
material BP=2.27E6 MATERIAL=InGaAs

IMPACT AN1=1.0E7 MATERIAL=InP
IMPACT AP1=9.36E6 MATERIAL=InP
IMPACT AN1=5.15E7 MATERIAL=InGaAs
IMPACT AP1=9.69E7 MATERIAL=InGaAs

INTERFACE OPTICAL region=6 AR.INDEX=2.05 AR.THICK=0.0634 

beam num=1 x.origin=1.0 y.origin=2 angle=10.0 GAUSSIAN MEAN=0.5 SIGMA=0.05 RAYS=400

save outf=t1_0.str
tonyplot t1_0.str -set apd.set
#quit

#solve init
#method gummel newton

method  newton autonr trap

solve b1=0 outf=s1
solve b1=1 outf=s2

load infile=s1
log outfile=diodeex01.log
Solve name=anode vanode=2 vstep=1 vfinal=55

load infile=s2
log outfile=diodeex02.log
Solve  name=anode vanode=2 vstep=1 vfinal=55
tonyplot diodeex01.log diodeex02.log
quit

Do you measure its output as current, or as voltage? To measure output as current, you need a loop of some kind for the current to flow around. It's likely that photodiode output is easier to measure as a voltage, instead of as a current.
Photodiode schematics commonly show one terminal going to a high-impedance input (such as an op amp). A voltage must be present in order for you to measure current. Your photodiode may need a certain bias voltage or bias current, to bring it to a 'good operating point' in certain lighting situations.

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