Download PHYS3610/6610 Electronics I – Final – Thursday December 10th

PHYS3610/6610 Electronics I
– Final –
Thursday December 10th
1) The behavior of the transistor used the circuit shown on the left of the figure is
described by the characteristic curves on the right of the figure. We have V B=5V ,
V CC =10V , R B=220k  and RC =1k  . Take the voltage drop across a
forward biased PN junction to be 0.6V .
a) [16 points] What is the value of the base current
IB ?
The base voltage is 5V−0.6V=4.4V . The base
current I B =4.4V /R B =4.4V/ 22000=20 A
b) [16 points] What are the values of the collector
current I C and voltage V C ?
The load line is set by the points
V CE =10V , I C =0A  and
V CE =0V , I C =10V/1k =0.01A it intersect
the characteristic curve for I B =20  A for
V CE ≈7V , I C ≈0.003A=3mA
c) [16 points] What is the value of the DC current gain  DC ?
I C 3×10−3
=150
By definition  DC = =
I B 20×10−6
d) [17 points] What is the minimal value of R B so the transistor functions as a switch
in this circuit with I C =0.01A when the transistor is on?
From the graph we see the base current I B must be greater than 60  A so we must
have R B5V−0.6V/60×10−6=73.3k 
PHYS3610/6610 Electronics I
– Final –
Thursday December 10th
2) Consider the circuit in the figure with R=60 , C g=8  F and C f =0.22nF .
Note: this exercise does not involve any complicated complex number calculation.
a) [14 points] What is value of the gain at very
low frequencies?
At very low frequencies the impedances of the
4R
capacitors is infinite so A DC =− =−4
R
b) [14 points] What is value of the gain
A=V OUT /V IN at very high frequencies?
At very high frequencies the impedances of the capacitors is negligible in comparison to
4R  // 12R 
3R
=−
=−12
the resistor values so A∞ =−
R //  R /3
R/ 4
c) [14 points] What is the value of the frequency f g for which the input capacitor
impedance ∣Z C ∣=R/3 ? What is the value of the frequency f f for which the
feedback capacitor impedance ∣Z C ∣=12R ?
g
f
1
∣Z C ∣=R/3 ⇒ 2 C
g
g
fg
1
∣Z C ∣=12R ⇒ 2 C
f
f
f
=R / 3⇒ f g =
1
1
=
≈1kHz
2 C g R/ 3 2 8×10−6 60 /3
=12R ⇒ f f =
f
1
1
=
≈1MHz
2 C f 12R 2 0.22×10−9 12×60
d) [14 points] What is the value of the gain
f g and f f ?
A Mid =V OUT /V IN for frequencies between
At a frequency f such that f g ≪ f ≪ f f , the input impedance is
Z g ≈ R //  R/ 3=R/ 4 and the feedback impedance is Z f ≈4R so the gain is
4R
A Mid =−
=−16
R/ 4
e) [14 points] Calculate the dB values of ∣ ADC∣ , ∣ AMid∣ and ∣ A∞∣ and trace a
graph representing the dB gain of the circuit as a function of the frequency on the back of
the page.
∣ ADC∣dB=20 log10 4=12.0dB
∣ AMid∣dB=20 log10 16=24.0dB
∣A∞∣dB=20 log 10 12=21.6dB
PHYS3610/6610 Electronics I
– Final –
Thursday December 10th
PHYS3610/6610 Electronics I
– Final –
Thursday December 10th
3) Consider the three inputs (A,B,C), one output (R) digital circuit in the figure.
a) [16 points] Fill in the truth table of this
circuit.
b) [22 points] Fill in the Karnaugh table of
this circuit.
A
B
C
R
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
1
A B
R
C
0 0 0 1 1 1 1 0
0
0
0
1
0
1
1
1
1
1
c) [16 points] From the Karnaugh table write an expression for R as a sum of minterms.
We can identify two overlaping groups of conditions for which
corresponding expression is R= A⋅BC
R=1 and the
d) [11 points] Manipulate the expression you obtained for R so it can be implemented
with NAND gates only. Provide a schematic of the NAND implementation.
Applying de Morgan relations we have
R=A.B+C = A.B . C