A diode junction acts as a photodetector: Light creates electron-hole pairs, and

Question

A diode junction acts as a photodetector: Light creates electron-hole pairs, and therefore a current through the external circuit. When the amount of the input current through the photodiode is 2 mA, the output voltage of the op amp is -10 volt voltage. Determine the feedback resistor R_F. According to the diode characteristics, the diode current I_0 can be described as follows: Where is the reverse saturation current for the diode. You can determine I_as by measuring the diode current I_d when V_D rightarrow -infinity, or Experimentally, this can be realized by connecting a circuit with the diode's cathode connected to +15 V and the anode connected to a 10M resistor that is then connected to ground. Determine the reverse saturation current I_gs when the voltage drop is 125 mV across the 10M resistor measured by the DMM whose internal resistance is 10 M Ohm. Given the following circuit and I_BS determined above, find the output voltage for an input dc voltage of 1 V. Assume that k_n T/q is approximately 25 mV at room temperature.

Explanation / Answer

Q 4a)

using the virtual ground concept at -ve node:

voltage at negative node =0

voltage across Rf=10 volts

hence Rf=voltage across Rf/current through Rf=10/0.002=5000 ohms

Q4.b)

the internal resistance of DMM is connected in parallel across the resistance of 10 M

using current division, if diode current is I, then current through DMM=I*10M/(10M+10M)=I/2

hence (I/2)*10*10^6=0.125 volts

==>I=25 nA

hence Irs=Id=25 nA

Q4.c)
using virtual ground concept, voltage at -ve node=0 volts

then current in the ciruit=(Vin-0)/1000

=1 mA

using the diode formula:

1 mA=25 nA*(e^(Vd/(k*T/q))-1)

==>10^(-3)=25*10^(-9)*(e^(Vd/0.025)-1)

==>e^(40*Vd)-1=40000

==>e^(40*Vd)=40001

==>40*Vd=10.59666

==>Vd=0.264916 volts

hence output voltage=voltage at negative node-voltage difference across the diode=0-0.264916 =-0.264916 volts

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