Digital Electronics Solutions for Homework #2
Question #1:
IC(SAT) = 4.8mA
IB(EOS) = 48A
IBf = (5-0.7)V/5k = 860 A
IBr = (0-0.7)V/5k = -147A
 I Br  I Bf 
 (140  860) A 
Ts   o * ln 
  10ns * ln 
  16.7ns
I

I
(

140

48
)

A
 Br


B ( EOS ) 

TR 
[0.8 * Q A  C jC ( av) * VBC ]
[ I Br  I B ( recombo) ]
Where:

[0.384 pC  0.4 pC ]
 4.78ns
[140A  24A]
QA = BF*IB(eos) = 10ns *48A = 480A
VBC = 4V (10% to 90%)
CjC(av) = 0.1pF
IB(recombo) = QF/BF = QA/(2*BF) = 24A
TF 
[0.8 * Q A  C jC ( av) * VBC ]
[ I Br  I B ( recombo) ]
Part B:
Digital Electronics Solutions #2 page 1

[0.384 pC  0.4 pC ]
 0.937ns
[860 A  24A]
Question #2:
a. IC = Vcc/Rc = 10V/1k = 10mA
IB(EOS) = IC/ = 10mA/100 = 100A
b. IBf = (Vcc – VBE(sat))/RB = (10-0.8)V/22k = 420A
c. IBr = (VCE – VBE(sat))/RB = -0.8V/22k = -36A
d. QA = BF*IB(EOS) = 100A *20ns = 2pC
e. Qs = s*(IBf – IB(EOS)) = 20ns * 320A = 6.4pC
 I Br  I Bf 
 (36  420) A 
T


*
ln

20
ns
*
ln


o
 (36  100) A   24.196ns
f. s
 I Br  I B ( EOS ) 


g. TR 
[0.8 * Q A  C jC ( av) * VBC ]
[ I Br  I B ( recombo) ]

[0.8 * 2 pC  0.1 pF * 8V ]
 66.7ns
[36A]
Where VBC = .1010 to .9010 = 8V and I B(recombo) is ignored
h. IBr = 0 in this case so
 I Br  I Bf 
 (0  420) A 
Ts   o * ln 
  20ns * ln 
  28.70ns
I

I
(
0

100
)

A



B ( EOS ) 
 Br
Question #3:
When Vi = -%V, the BJT is OFF. Therefore, no current is flowing. This means
VB = -5V, VC = 5V and VE = 0V. From the equation sheet we know that
Digital Electronics Solutions #2 page 2
1/ 2
 V  V2 

  d
C j (V2 )  Vd  V1 
C j (V1 )
where V1 and V2 are the voltages across the voltage in
question (either base-emitter or base-collector. Don't make the mistake of
assuming either V1 or V2 is the potential at some terminal with respect to
ground!
Since we already know CjE(0) and CjC(0), we can set V2 = 0V and the equation
 Vd  V2 


C
(
V
)

C
(
0
)
*
j
simplifies to: j 1
V

V
1 
 d
1/ 2
Now all we have to calculate are the
values of V(CE) and V1(C). We can easily observe that VBE = -5V and that
VBC = -10V. Since these are the voltages across the B-E and B-C junctions,
they are also V1(E) and V1(C).
We can now solve for CjE(V1E) and CjC(V1C): Remember, for the VBE junction, we
were given Vd = 0.9V.
 0.9  0 
C jE (5V )  1.5 pF * 

 0.9  5 
1/ 2
 0.586 pF
For the VBC junction, we were given Vd = 0.7V so we calculate:
 0.7  0 
C jC (10V )  0.7 pF * 

 0.7  10 
1/ 2
 0.179 pF
We know that Q = CV (V is still the voltage across the capacitor, not the
voltage from a terminal with respect to ground!) So:
QBE = CjE(-5V) * -5V = .586pF * 5V= -2.93pC
QC = CjC(-10V)*-10V = .179pF * 10V = -1.79pC
Digital Electronics Solutions #2 page 3
Question #4:
The problem is essentially the same as the previous one, except for one major
difference: in this case, the transistor is ON and we can assume that the BJT
is saturated! (IB = (5-0.7)V/10k = 0.43mA and IB(EOS) = IC(SAT)/
IB(EOS) = (5-0.2)V/1k/ = 0.048mA. IB > IB(EOS) so, yes, we are in saturation.
See the equation sheet for original equations.
VBE(sat) is roughly 0.7V and VCE(sat) is roughly 0.2V (given). From here, we can
perform the same calculations as we did in Question #3:
 0.9  0 
C jE (0.7V )  1.5 pF * 

 0.9  0.7 
1/ 2
 3.18 pF
V1C = VBE – VCE = 0.7V –0.2V = 0.5V
 0.7  0 
C jC (0.5V )  0.7 pF * 

 0.7  0.5 
1/ 2
 1.3 pF
Therefore: QBE = 3.18pF * .7V = 2.22pC and QCE = 1.3pF * 0.5V = 0.650pC
Question #5:
If the BJT is saturated, the current is:
Ic = (Vcc – VCE(sat))/Rc = (10-0.2)V/1k = 9.8mA
How do we know where in our range of  to look? Let's think about it by first
calculating IB(EOS) = IC(sat)/. For  = 30, IB(EOS) = 9.8mA/30 = 0.327mA. For  =
100, IB(EOS) = 9.8mA/100 = 98A.
We want the MAXIMUM overdrive factor to be 10 which beats 10*I B(EOS) = IB.
Digital Electronics Solutions #2 page 4
Consider the case where IB(EOS) = 98A ( = 100). IB1 = 9.8A which is too small
to drive the transistor with  = 30 into saturation, let alone into overdrive. It
stands to reason, therefore, that we should use the IB(EOS) corresponding to  =
30 as our IB(EOS) value.
Another way to consider this is to examine the I B necessary to overdrive the
BJT with  = 30. IB1 = IB(EOS)=30 *10 = 3.27mA. This current is large enough to
saturate any BJT within the 30 <  < 100 range, but more importantly, it is
also large enough to overdrive all the BJTs by at least a factor of 10. (Hence,
meeting our requirement of Nmin = 10.)
IB = 10*IB(EOS) = 3.27mA = (5 –VBE(sat))V/RB = (5-0.8)V/Rb
Therefore RB = 4.2V/3.27mA = 1.284k
Question #6:
 I Br  I Bf 
Ts   s * ln 
 and s = f = 100*300 ps = 30ns
 I Br  I B ( EOS ) 
Note: All equations come from the Equation Sheet.
We can assume that VBE(sat) = 0.7V and VCE(sat) = 0.2V.
ViReverse = -10V in Part A and Vir = 0V for Part B.
ViForward = 10V in Part A & Part B.
Now we can calculate:
Calculation
IBr = (ViRev – VBE(sat))/RB
IBf = (ViFor – VBE(sat))/RB
Part A.
(-10-0.7)V/2k =
-5.35mA
(10-0.7)V/2k =
Digital Electronics Solutions #2 page 5
Part B.
(0 – 0.7)V/2k =
-0.35mA
(10-0.7)V/2k =
IB(EOS)=IC(sat)/
=(VCC-VCE(Sat))/RC*
 I Br  I Bf 
Ts   s * ln 

I

I

B ( EOS ) 
 Br
QBE = CjE(av)*VBE
QBC = CjC(av)*VBC
4.65mA
(10–0.2)V/(100*470)=
0.208mA
30ns*ln((-5.35mA4.65mA)/(-5.35mA0.208mA)) =
17.62ns
2.38pF * 10.5V =
25pC
1.41pF * 10.5V =
14.8pC
VBE=VB (with emitter grounded)
-10 – 0.5V =
VB = Vrev – VBE(cut-in)
-10.5V
assume VBE(cut-in) = 0.5V
CjE(av) = 2CjE(0)/(1+K1)1/2
2*4.5pF/(14.33)1/2 =
2.38pF
K1 = VRev1/Vd1
10V/0.75V =
13.33
CjC(av) = 2CjC(0)/(1+K2)1/2
2*3.7pF/(27.67)1/2 =
1.41pF
K2 = VRev2/Vd2
20V/0.75V =
26.67
IBav = (IB init + IB final) /2
(10mA +4.75mA)/2 =
7.375mA
IB init = (Vi For – Vi Rev)/RB
(10--10)V/2k =
10mA
IB fin = (Vi For – VBe cut in)/RB (10 - .5)V/2k =
4.75mA
(25+14.8)pC/7.375mA=
tD =(QBE + QBC)/IBav
5.40 ns
4.65mA
(10 – 0.2)V/(100*470) =
0.208mA
30ns*ln((-0.35mA4.65mA)/(-0.35mA0.208mA)) =
65.79 ns
9 pF * 0.5V =
4.5 pC
1.95 pF * 0.5V =
0.975 pC
0V – 0.5V =
0.5V
2*4.5pF/11/2 =
9pF
0V/Vd1 =
1
2*3.7 pF/(14.33)1/2 =
1.95pF
10V/0.75V =
13.33
(4.75+5)mA/2 =
4.875mA
(10-0)V/2k =
5mA
(10 – 0.5)V/2k =
4.75mA
(4.5pC+.975pC)/4.65mA=
1.12 ns
Now, we need to solve TR where:
TR 
[0.8 * Q A  C jC ( av) * VBC ]
[ I Br  I B ( recombo) ]
Here, QA = BF * IB(EOS) = 30ns * 0.208mA = 6.24 pC (Parts A & B)
We need to calculate IB(recombo), but CjC(av) and IBr have already been calculated.
Digital Electronics Solutions #2 page 6
IB(recombo) = QE(av)/BF = QA/2BF = 6.24pC/(2*30ns) = 0.104mA (Parts A & B)
We need to recalculate VBC & VBE because previously, we assumed that VC
and VE were both constants and only VB changed. Here, VB and VE are held
constant while the output voltage (VC) changes from 10% to 90% of its final
value.
VBC = VC = VC(90%) – VC(10%) = 4.5V –0.5V = 4V (Both Parts)
Part A:
TR 
[0.8 * 6.24 pC  1.95 pF * 4.0V ]
 1.95ns
[5.35mA  0.104mA]
Part B:
TR 
[0.8 * 6.24 pC  1.95 pF * 4.0V ]
 23.4ns
[0.35mA  0.104mA]
And now we must calculate…
TF 
[0.8 * Q A  C jC ( av) * VBC ]
[ I Br  I B ( recombo) ]
Part A:
TF 
[0.8 * 6.24 pC  1.41 pF * 4.0V ]
 2.34ns
[4.65mA  0.104mA]
Part B:
TF 
[0.8 * 6.24 pC  1.95 pF * 4.0V ]
 2.81ns
[4.65mA  0.104mA]
Digital Electronics Solutions #2 page 7
In summary:
Parameter
TS
TD
TR
TF
Part A
(-10V->10V->-10V)
17.62ns
5.40ns
1.95ns
2.34ns
Part B
(0V->10V->0V)
65.79ns
1.12ns
23.4ns
2.81ns
We see that while driving the circuit with a large reverse bias reduces TS, TR
and TF. TD is larger than driving the circuit with a signal which never reverse
biases the base emitter junction.
Digital Electronics Solutions #2 page 8