Digital Electronics
Midterm Review
ECE27H Antennas and Propagation, DR. H. Ragheb
1
Problem (1)
• The inverter in Fig. 1 is a member of
the diode-transistor logic (DTL) family
that was used prior to the invention of
the TTL
a) If 𝑣𝐼 = 5 𝑣, find 𝑣𝑜 ,
b) If 𝑣𝐼 = 0.15 𝑣, find 𝑣𝑜 ,
• The transistor has parameters
𝛽𝐹 = 40 , 𝛽𝑅 = 0.25 , 𝑉𝐵𝐸 = 0.7 𝑉 ,
𝑉𝐶𝐸𝑠𝑎𝑡 = 0.15 𝑉 , and the diode has
𝑉𝐷𝑜𝑛 = 0.7 𝑉.
ECE21I Digital Electronics, DR. H. Ragheb
Fig. 1
2
Problem (1)
Solution
• If 𝑣𝐼 = 5 𝑣 𝐷1 is off and 𝐷2 in on also 𝑄2 is on, and
1
1×4
𝑉𝐵𝐵 = 4.3
= 0.86 𝑉
𝑅𝐵𝐵 =
= 0.8 Ω
1+4
1+4
0.86 − 0.7
𝑖𝐵2 =
= 0.2 𝑚𝐴
0.8
• This current is enough to saturate 𝑄2
• Thus
𝑉𝑜 = 0.15 𝑉
• Next if 𝑣𝐼 = 0.15 𝑉 (low), 𝐷1 is on, and
𝑉𝑥 = 0.15 + 0.7 = 0.85 𝑉
• This voltage is not enough to drive 𝐷2 and 𝑄2 is on, thus is 𝑄2 off, and
𝑉𝑜 = 5 𝑉
ECE21I Digital Electronics, DR. H. Ragheb
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Problem (2)
• The circuit shown in Fig. 2 is used as an
inverter. The transistors used are identical
and have 𝛽 = 25 , 𝛽𝑅 = 0.02 and
𝑉𝐶𝐸(𝑠𝑎𝑡) = 0.2 𝑉.
V CC = 5 V
R = 4 k
• Assume 𝑣𝐼𝑛𝑝𝑢𝑡 = 0.2 𝑉 find 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 .
• Assume 𝑣𝐼𝑛𝑝𝑢𝑡 = 5 𝑉 find 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 .
V B1
input
120 
I1
output
Q1
VB2
Answer
• If 𝑣𝐼𝑛𝑝𝑢𝑡 = 0.2 𝑉,
Q2
Q3
V B3
• 𝑄1 is ON and 𝑉𝐵1 = 0.9 𝑉
1 k
5−0.9
• Thus, 𝐼1 =
= 1.025 𝑚𝐴, which is
4
enough to saturate 𝑄1 .
ECE21I Digital Electronics, DR. H. Ragheb
I
Fig. 2
4
Problem (2)
• Accordingly, 𝑉𝐵2 = 0.2 + 0.2 = 0.4 𝑉
• then 𝑄2 and 𝑄3 are off 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 = 5 𝑉.
V CC = 5 V
• 𝑣𝐼𝑛𝑝𝑢𝑡 is low and 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 is high
R = 4 k
• If 𝑣𝐼𝑛𝑝𝑢𝑡 = 5 𝑉, the base-emitter junction of 𝑄1 is reverse biased, while
the base collector junction is forward biased.
• Therefore, the transistor 𝑄1 will work in the inverse active mode.
V B1
input
• In this case 𝑉𝐵1 = 0.7 + 𝑉𝐵2 assuming 𝑄2 is ON, 𝑉𝐵2 = 0.7 + 𝑉𝐵3 if also 𝑄3
is ON, thus 𝑉𝐵1 = 2.1 𝑉.
•
5−2.1
The current 𝐼 can be calculated as 𝐼 = 4 = 0.725 𝑚𝐴.
120 
I1
output
Q1
VB2
Q2
Q3
V B3
1 k
I
• Thus, the current driving 𝑄2 is 𝐼𝐵2 = 1 + 0.02 𝐼 = 0.7395 𝑚𝐴, which is
enough to drive 𝑄2 to saturation. In this case 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 = 0.2 + 0.7 = 0.9 𝑉
• 𝑣𝐼𝑛𝑝𝑢𝑡 is high and 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 is low
ECE21I Digital Electronics, DR. H. Ragheb
5
Problem (3)
• The circuit shown in Fig. Q2 is used as an inverter. The transistors
used are identical and have 𝛽 = 25, 𝛽𝑅 = 0.02 and 𝑉𝐶𝐸(𝑠𝑎𝑡) =
0.2 𝑉.
• Assume 𝑣𝐼𝑛𝑝𝑢𝑡 = 0.2 𝑉 find 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 .
• Assume 𝑣𝐼𝑛𝑝𝑢𝑡 = 5 𝑉 find 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 .
Answer
• If 𝑣𝐼𝑛𝑝𝑢𝑡 = 0.2 𝑉, 𝑄1 is ON and 𝑉𝐵1 = 0.9 𝑉
• Thus, 𝐼1 =
5−0.9
= 1.025 𝑚𝐴, which is enough to saturate 𝑄1 .
4
• Accordingly, 𝑉𝐵3 = 0.2 + 0.2 = 0.4 𝑉
• then 𝑄2 and 𝑄3 are off while 𝑄4 will be on and saturated.
𝑣𝑜𝑢𝑡𝑝𝑢𝑡 = 5 − (0.2 + 0.7) = 4.1 𝑉
Fig. 3
𝑣𝐼𝑛𝑝𝑢𝑡 is low and 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 is high.
ECE21I Digital Electronics, DR. H. Ragheb
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Problem (3)
• If 𝑣𝐼𝑛𝑝𝑢𝑡 = 5 𝑉, the base-emitter junction of 𝑄1 is reverse
biased, while the base collector junction is forward biased.
• Therefore, 𝑄1 will work in the inverse active mode.
• In this case 𝑉𝐵1 = 0.7 + 𝑉𝐵2 assuming 𝑄2 is ON, 𝑉𝐵2 = 0.7 +
𝑉𝐵3 if also 𝑄3 is ON, thus 𝑉𝐵1 = 2.1 𝑉.
• The current 𝐼 can be calculated as 𝐼1 =
5−2.1
= 0.725 𝑚𝐴.
4
• Thus, the current driving 𝑄3 is 𝐼𝐵3 = 1 + 0.02 𝐼 =
0.7395 𝑚𝐴, which is enough to drive 𝑄3 to saturation. In this
case the emitter current of 𝑄3 will go to the base of 𝑄2
which is enough to saturated it
𝑣𝑜𝑢𝑡𝑝𝑢𝑡 = 0. 2 𝑉
• 𝑣𝐼𝑛𝑝𝑢𝑡 is high and 𝑣𝑜𝑢𝑡𝑝𝑢𝑡 is low
ECE21I Digital Electronics, DR. H. Ragheb
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Problem (4)
• Consider the circuit in Fig. 4 where the base
currents of any transistor when it is on is
very small and you can neglect it.
• Determine 𝑅𝐶2 such that 𝑣2 = −1 𝑉 when
𝑄2 is on and 𝑄1 is off.
• For 𝑣𝑖𝑛 = −0.7 𝑉, determine 𝑅𝐶1 such that
𝑣1 = −1 𝑉.
• For 𝑣𝑖𝑛 = −0.7 𝑉, determine 𝑣𝑂1 and 𝑣𝑂2 .
• For 𝑣𝑖𝑛 = −1.7 𝑉, determine 𝑣𝑂1 and 𝑣𝑂2 .
• Assume 𝑉𝐵𝐸𝑜𝑛 = 0.7 𝑉, and 𝛽𝐹 = 50
ECE21I Digital Electronics, DR. H. Ragheb
Fig. 4
8
Problem (4)
Answer
• When 𝑄2 is on and 𝑄1 is off 𝑉𝐸 = −1.2 − 0.7 = −1.9 𝑉
• The current in the emitter resistance is
• Neglecting base current
𝐼=
−1.9− −5.2
2.5
= 1.32 𝑚𝐴
1
𝑅𝐶2 = 1.32 = 0.758 𝑘Ω
• When 𝑣𝑖𝑛 = −0.7 𝑉 𝑄2 is off and 𝑄1 is on 𝑉𝐸 = −0.7 − 0.7 = −1.4 𝑉
−1.4− −5.2
2.5
• The current in the emitter resistance is
𝐼=
• Neglecting base current
𝑅𝐶1 = 1.52 = 0.658 𝑘Ω
= 1.52 𝑚𝐴
1
• When 𝑣𝑖𝑛 = −0.7 𝑉 𝑄2 is off and 𝑄1 is on 𝑉𝑜2 = −1.7 𝑉
• In this case 𝑣2 ≈ 0 and 𝑉𝑜1 = −0.7 𝑉
• When 𝑣𝑖𝑛 = −1.7 𝑉 𝑄2 is on and 𝑄1 is off 𝑉𝑜1 = −1.7 𝑉
• In this case 𝑣1 ≈ 0 and 𝑉𝑜2 = −0.7 𝑉
ECE21I Digital Electronics, DR. H. Ragheb
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Problem (5)
• Which logic gate has the
voltage transfer characteristic
shown in Fig. 5?
• Clearly label 𝑉𝑂𝐿 , 𝑉𝑂𝐻 , 𝑉𝐼𝐿 , and
𝑉𝐼𝐻 in the figure.
• Given that 𝑉𝐷𝐷 = 3.3 𝑉,
𝑉𝑡 = 0.5 𝑉 select 𝑉𝐼𝐿 , and 𝑉𝐼𝐻
such that 𝑁𝑀𝐻 = 𝑁𝑀𝐿 ?
Fig. 5
ECE21I Digital Electronics, DR. H. Ragheb
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Problem (5)
• It is CMOS inverter
• 𝑉𝑂𝐻 = 𝑉𝐷𝐷
• 𝑉𝑂𝐿 = 0
• 𝑁𝑀𝐻 = 𝑉𝑂𝐻 − 𝑉𝐼𝐻
• 𝑁𝑀𝐿 = 𝑉𝐼𝐿 − 𝑉𝑂𝐿
• 𝑉𝐼𝐻 =
1
8
5𝑉𝐷𝐷 − 2𝑉𝑡 =
• 𝑉𝐼𝐿 =
1
8
3𝑉𝐷𝐷 + 2𝑉𝑡 =
• 𝑁𝑀𝐿 =
1
8
1
8
16.5 − 1 =
9.9 + 1 =
15.5
8
10.9
8
10.9
= 1.3625 𝑉
8
• 𝑁𝑀𝐻 = 𝑉𝑂𝐻 − 𝑉𝐼𝐻 = 3.3 −
15.5
= 1.3625
8
ECE21I Digital Electronics, DR. H. Ragheb
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Problem (6)
• An inverter has 𝑉𝐷𝐷 = 2 𝑉 , 𝑘𝑛 =
300 𝜇𝐴/𝑉 2 , 𝑘𝑝 = 200 𝜇𝐴/𝑉 2 , 𝑉𝑡𝑛 =
0.5 𝑉 , |𝑉𝑡𝑝 | = 0.5 𝑉 , (𝑊/𝐿)𝑛 = 2 , and
(𝑊/𝐿)𝑝 = 3.
• Find 𝑉𝑜 for 𝑉𝑖𝑛 = 0 𝑉,
𝑉𝑜
• Find 𝑉𝑜 for 𝑉𝑖𝑛 = 2 𝑉
• Find the values of 𝑉𝑂𝐻 , 𝑉𝑂𝐿 , 𝑉𝐼𝐿 , 𝑉𝐼𝐻
Answer
• For 𝑉𝑖𝑛 = 0 𝑉, 𝑄𝑁 is off and 𝑄𝑝 is on
𝑉𝑜 = 𝑉𝐷𝐷 = 2 𝑉.
ECE21I Digital Electronics, DR. H. Ragheb
Fig. 6
12
Problem (6)
For 𝑉𝑖𝑛 = 2 𝑉, 𝑄𝑁 is on and 𝑄𝑝 is off 𝑉𝑜 = 0 𝑉.
𝑉𝑂𝐻 = 𝑉𝐷𝐷 = 1.3 𝑉
𝑉𝑂𝐿 = 0 𝑉
• To obtain 𝑉𝐼𝐻 we use:
1
1
𝑉𝐼𝐻 = 5𝑉𝐷𝐷 − 2𝑉𝑡 = 5 × 2 − 2 × 0.5 = 1.125 𝑉
8
8
• To obtain 𝑉𝐼𝐿 we use:
1
1
𝑉𝐼𝐿 = 3𝑉𝐷𝐷 + 2𝑉𝑡 = 3 × 2 + 2 × 0.5 = 0.875 𝑉
8
8
ECE21I Digital Electronics, DR. H. Ragheb
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Problem (7)
• Design the complete CMOS logic
circuit using minimum number of
transistors that realize the
function
ሜ
𝑌 = 𝐵(𝐷
+ 𝐴𝐶) .
D
𝐵ത
C
A
B
Y
𝐵ത
(Assume that the available inputs
are 𝐴, 𝐵, 𝐶 and 𝐷).
A
D
• What is total number of
transistors needed?
C
ECE21I Digital Electronics, DR. H. Ragheb
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Thank you
Digital Electronics Dr. H. Ragheb
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