# Photoresistor characteristics

In the figure below are being given the I-V characteristics for a photo-resistor before ($Phi_0 =0$) and after ($Phi_0 >0$) illumination. If one connects in series with this photo-resistance a voltage source of $U =10 V$ and a load resistance of $R_L=2.5*10^3 Omega$ find using only graphical methods:

a) The voltage drop on the photo-resistor at dark and when illuminated.

b) The intensity of the current that flows in this series circuit at dark and when after illumination.

c) The variation of voltage drop on the load resistor, when illuminated.

a) If $I$ is the current through the circuit and $U_L =I*R_L$, respectively $U_{p h}$ are the voltage drops on the load and photo-resistance, then one can write

$U =R_L*I +U_{p h}$  so that $I = -frac{1}{R_L}*U_{p h} +frac {U}{R_L}$

In the above relation $U$ and $R_L$ are constants and $I$ and $U_{p h}$ are variables, depending on the intensity of the incident light on photo-resistance. The above equation is the equation of a line having a slope $m=-1/R_L$ and the x intersection of $x0 =U/R_L$.

This is called the load line of the photo-resistance (the black line above having the intersection points with I-V characteristics at $A_1$ ($U_{p h} =8V$; $I_0 =0.8 mA$) and $A_2$ ($U_{p h} =3 V$; $I_0=2.8 mA$). Now we have the answers to the questions:

a)

The voltage drop on the photo-resistance at dark is $U_{ph0} =8 V$ and when illuminated $U{p h} =3 V$

b)

At dark the intensity of the current in the circuit is $I_0 =0.8 mA$ and when illuminated is $I =2.8 mA$

c)

The voltage drop on the load resistor at dark is $U_{L0} =2V$ and when illuminated is $U_L =7 V$.

The variation of the voltage on the load is $Delta(U_L) =7-2 =5 V$