Digital Electronics Solutions for Homework #1
Question #1:
Solution:
ts =pln(1+If/Ir)
If is the current through R when the diode is forward biased = (Vf-Vd)/R
If = (2-0.65)V/1k = 1.35mA
Ir = instantaneous reverse current through R (in the opposite indirection).
Ir = (0.65 - -20)V/1k = 20.65mA
so…
s = 1s*ln(1+1.35mA/20.65mA) = 63.33 ns
Question #2:
Solution:
Ts = P*ln(1+If/Ir) = 167ns*ln(1+10mA/20mA) = 67.7 ns
Once the diode begins to discharge its excess carriers, the current is:
iD(t) = IR e-t/r where
r = RC = 500*5pF = 2.5 ns
We need to find Tr such that:
0.1*Ir = Ir e(-Tr/r)
so Tr = -r *ln(.1)
tr = -2.5ns*ln(0.1) = 5.76 ns
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Question #3:
Solution:
For two diodes in series, the current through each must be equal. Therefore:
Is1 eV1/Vt = Is2 eV2/Vt
We know that V1 + V2 = 1V so V2 = 1 – V1
Substituting this in, we get:
Is1 eV1/Vt = Is2 e1-V1/Vt
(eV1/Vt)/(e(1-V1)/Vt) = Is2/Is1 = 100
e(V1 – 1 –V1) /Vt = 100
(2*V1 – 1) /Vt = ln(100)
V1 = (VT*ln(100) + 1)/2 = 0.559V and V2 = 0.441V
I = Is1 eV1/Vt = Is2 eV2/Vt = 51.38A
Question #4:
Solution:
Part A:
VD = VT ln(NAND/ni2)
ni = 1.5x1010
VT = 0.026V or 0.025V (see page 139)
VD = 0.026V * ln(1015*1017/(1.5x1010)2 = 0.697V
OR
VD = 0.025V * ln(1015*1017/(1.5x1010)2 = 0.671V
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Cjo = sA/W where W = ((2/q) * ((NA + ND)/(NA*ND))*(Vo – VD))1/2
at zero bias, VD = 0
Therefore:
W = ((2*(1.8)*(8.854x10-14F/cm)/(1.6x10-19C) * ((1015 + 1017)cm-3/(10151017cm-6))*(0.697V))1/2
W = 95.9 x 10-6 cm
Units: (F/cm)*(1/C) (cm-3/cm-6)*V
F = C/V
A = (40x40)m2
Cjo =(11.8*8.854x10-14 F/cm *16x10-6cm2)/(95.9x10-6cm) = 174.3x10-15 F = 174fF
Cj(av) = 2*Cjo/(1+k)1/2 where k = |Vr|/Vo
here k = |-10|/0.697 = 14.3 so
Cj(av) = 2*(174 fF)/(1+14.3)1/2 = 88.8fF
Part B:
VD = -10V recall W = ((2/q) * ((NA + ND)/(NA*ND))*(Vo – VD))1/2
W = ((2*(1.8)*(8.854x10-14F/cm)/(1.6x10-19C) * ((1015 + 1017)cm-3/(10151017cm-6))*(0.697-(-10))V)1/2
W = 375.6x10-6cm
Cj =(11.8*8.854x10-14 F/cm *16x10-6cm2)/(375.6x10-6cm) = 44.5x10-15 F = 44.5 fF
at VD = 0.5V
W = ((2*(1.8)*(8.854x10-14F/cm)/(1.6x10-19C) * ((1015 + 1017)cm-3/(10151017cm-6))*(0.697-(0.5))V)1/2
W = 50.9x10-6cm
Cj =(11.8*8.854x10-14 F/cm *16x10-6cm2)/(50.9x10-6cm) = 327.9x10-15 F =326.9fF
Question #5: Switching Diode:
Solution:
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a. IDf = (5-.7)V/2k = 2.15 mA
b. IDr = (-10-.7)V/4k = -2.675 mA
c. and d. Graph of ID & Graph of VD:
ID in mA
2.15mA
t
Time Constant
is RCT(av)
2.675mA
Ts
0.7V
t
t1
TR
e. Ts = *ln(1+IDf/|IDr|) = 10ns * ln(1+2.15/2.675) = 5.9 ns
f. TR = 5*rct(av)
RCt(av) = R*Ct(av)
Ct(av) = 2*Ct(0)/(1+k)
where k = |VR|/Vd = |-10|/.7 = 14.29
Ct(av) = 2.2pF/(15.29)1/2 = 1.023pF
RCt(av) = 1.023pf*2k = 4ns
TR = 20 ns
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g. Q= CV
V = -10V
C = Ct(-10) = Ct(0)/((Vd - -10)/Vd)1/2 = 0.5pF
Q = 0.5pF*-10V = -5 pC
Question #6: Switching Diode
Solution:
A. Plots of Vi and VD:
Vi
t
VD
t
Td
Ts
Tr
B. Cj(.7V) = 4pF
Cj(OV) = 2pF
Cj(-10V) = 2pF/(1+10V/.7V)1/2 = 0.51pF
C. Ir(av) = (Ir(initial) + Ir(final))/2
Ir(initial) = (-10V -.7V)/1k = -10.7mA
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Ir(final) = (-10V +10V)/1k = 0mA
Ir(av) = (-10.7mA + 0)/2 = -5.35mA
D. Q1 = 4pF * 0.7V = 2.8pC
Q2 = 0.51pF * -10V = -5.1pC
Q = –5.1pC – 2.8pC = -7.9pC
TR = t = Q/Ir(av) = -7.9pC/-5.35mA = 1.48 ns
E. As already determined Ir(init) = -10.7A and If = (10-.7)V/1k = 9.3mA
F. Ts = *ln(1+If/Ir) = 10ns*(1+9.3/10.7) = 6.25ns
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