Solutions for HW2
Divija Gogineni
September 24, 2019
(2.14)
(a)
Suppose
ID1=Is(e
VD1
VT−1)
ID2=Is(e
VD2
VT−1)
By KCL, ID1=ID2=I;
Then (e
VD1
VT−1) = (e
VD2
VT−1) => VD1=VD2=VD
2
Therefore,
I=Is(e(VD/2)/VT−1)
Therefore, a series combination can be viewed as a single two-terminal
device with exponential characteristics.
(b)
Suppose Vi=initial VD,Need 10x.
Vf=final VDincrease in I
=> 10 = IS(eVf/VT−1)
Is(eVi/VT−1) ≈e(Vf−Vi)/VT
Therefore, ∆V =Vf−Vi=VTln(10)= (26mV )ln(10)= 60mV
(2.24)
Given IS= 3 ∗10−16A,find VD1
By KCL, Ix=VD1
R1+ID1=VT
R1ln(ID1
Is) + ID1
Since Ix, VT, R1and Isare known, this can be solved directly with
programs or by graphing calculators. Students can also use itera-
tions.
The final answer is: Ix= 1mA, VD1= 0.717V ,
Ix= 2mA, VD1= 0.755V
Ix= 4mA, VD1= 0.780V
Note: As Ixincreases, ID1increases, while (VD1/R1)stays relatively
the same because of the exponential characteristic.
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