Transistor Circuits XII
DC/AC Load Lines
Formulas for DC Load Line
Saturation Current
• 𝐼𝐶(sat) =
𝑉𝐶𝐶
𝑅𝐶
• 𝐼𝐶(sat) =
𝑉𝐶𝐶 +𝑉𝐸𝐸
𝑅𝐶 +𝑅𝐸
• 𝐼𝐶(sat) =
𝑉𝐶𝐶
𝑅𝐶 +𝑅𝐸
Cutoff Voltage
• 𝑉𝐶𝐸(cutoff) = 𝑉𝐶𝐶
• 𝑉𝐶𝐸(cutoff) = 𝑉𝐶𝐶 + 𝑉𝐸𝐸
Formulas for AC Load Line
• 𝑖𝑐(sat) = 𝐼𝐶 +
𝑉𝐶𝐸
𝑟𝐿
• 𝑣𝑐𝑒(cutoff) = 𝑉𝐶𝐸 + 𝐼𝐶 𝑟𝐿
Formulas for centering Load Line
• 𝑉𝐶𝐸 = 𝐼𝐶 𝑟𝐿 (Due to 50% of source desired)
• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶 + 𝑅𝐸 𝐼𝐶
• 𝑟𝐿 𝐼𝐶 = 𝑉𝐶𝐶 − 𝑅𝐶 + 𝑅𝐸 𝐼𝐶
• 𝐼𝐶 =
𝑉𝐶𝐶
𝑅𝐶 +𝑅𝐸 +𝑟𝐿
• 𝐼𝐶 =
𝑉𝐶𝐶
𝑅𝐶 +𝑟𝐿
(Voltage-divider biased)
(Two-supply and Base-biased)
First example circuit
• Sketch the dc and ac load lines for the circuit
described in the third example of the lecture
from Transistor Circuits VII. Show the
operating point Q and indicate the quiescent
IC to the left of point Q and the VCE below this
point. What is the peak value of the largest
possible unclipped output signal of this
circuit?
18 V
6.8kΩ
RB
1.2MΩ
C1
vs = 2mV (peak)
C2
27kΩ
-vce
Values from previous lecture
• 𝐼𝐶 = 1.16mA
• 𝑉𝐶 = 𝑉𝐶𝐸 = 10.112V
• 𝑟𝐿 = 5.432kΩ
New work
• 𝐼𝐶(sat) =
𝑉𝐶𝐶
𝑅𝐶
=
18V
6.8kΩ
= 2.647mA
• 𝑉𝐶𝐸(cutoff) = 𝑉𝐶𝐶 = 18V
• 𝑖𝑐(sat) = 𝐼𝐶 +
𝑉𝐶𝐸
𝑟𝐿
= 1.16mA +
10.112V
5.432kΩ
=
1.16mA + 1.862mA = 3.022mA
• 𝑣𝑐𝑒(cutoff) = 𝑉𝐶𝐸 + 𝐼𝐶 𝑟𝐿 = 10.112V +
1.16mA 5.432kΩ = 10.112V + 6.301V =
16.413V
Load Line (Scale: 1 line = 0.5)
Q point values
•IC = 1.16 mA
•VCE = 10.1 V
Second example circuit
• Sketch the dc and ac load lines for the circuit
described in the second example of the
lecture from Transistor Circuits VIII. Show the
operating point Q and indicate the quiescent
IC to the left of point Q and the VCE below this
point. What is the peak value of the largest
possible unclipped output signal of this
circuit?
+VCC
RC
C2
C1
RL vo
vs
RB
C3
RE
-VEE
Values from previous lecture
• IC = 1.8mA
• rL = 2.941kΩ
New work
• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 =
18 − 1.8mA 3.3kΩ = 18 − 5.94 =
12.06V
• 𝐼𝐶(sat) =
𝑉𝐶𝐶 +𝑉𝐸𝐸
𝑅𝐶 +𝑅𝐸
=
18+18
3.3kΩ+10kΩ
=
36V
13.3kΩ
=
2.707mA
• 𝑉𝐶𝐸(cutoff) = 𝑉𝐶𝐶 + 𝑉𝐸𝐸 = 18V + 18V = 36V
• 𝑖𝑐(sat) = 𝐼𝐶 +
𝑉𝐶𝐸
𝑟𝐿
= 1.8mA +
12.06V
2.941kΩ
=
1.8mA + 4.101mA = 5.901mA
• 𝑣𝑐𝑒(cutoff) = 𝑉𝐶𝐸 + 𝐼𝐶 𝑟𝐿 = 12.06V +
1.8mA 2.941kΩ = 12.06V + 5.294V =
17.354V
Load Line (Scale: 1 line = 1)
Q point values
•IC = 1.8 mA
•VCE = 12.06 V
Third example circuit
• Sketch the dc and ac load lines for the circuit
described in the first example of the lecture
from Transistor Circuits IX. Show the operating
point Q and indicate the quiescent IC to the
left of point Q and the VCE below this point.
What is the peak value of the largest possible
unclipped output signal of this circuit?
9V
8.2kΩ
C2
82kΩ
C1
vs
10kΩ -vo
12kΩ
1kΩ
C3
Values from previous lecture
• 𝐼𝐶 = 448.936𝜇A
• 𝑟𝐿 = 4.505kΩ
New work
• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 + 𝑅𝐸 =
9 − 448.936µA 8.2kΩ + 1kΩ = 9 −
448.936µA 9.2kΩ = 9 − 4.13 = 4.87V
• 𝐼𝐶(sat) =
𝑉𝐶𝐶
𝑅𝐶 +𝑅𝐸
=
9V
8.2kΩ+1kΩ
978.266µA
• 𝑉𝐶𝐸(cutoff) = 𝑉𝐶𝐶 = 9V
=
9V
9.2kΩ
=
• 𝑖𝑐(sat) = 𝐼𝐶 +
𝑉𝐶𝐸
𝑟𝐿
=
4.87V
448.936µA +
4.505kΩ
448.936µA + 1.081mA = 1.53mA
• 𝑣𝑐𝑒(cutoff) = 𝑉𝐶𝐸 + 𝐼𝐶 𝑟𝐿 = 4.87V +
448.936µA 4.505kΩ = 4.87V +
2.022V = 6.832V
=
Load Line (Scale: 1 line = 0.5)
Q point values
•IC = 449 µA
•VCE = 4.87 V
Fourth circuit example
• Referring to the last example circuit of this
lecture, what value of IC will cause the
operating point Q to be centered on the ac
load line? Replace the 12-kΩ resistor with a
value that will provide this value of IC. Let VBE =
0.7 V and choose from a list of 10% resistors.
Work through
• 𝐼𝐶 =
𝑉𝐶𝐶
𝑅𝐶 +𝑅𝐸 +𝑟𝐿
=
9V
8.2kΩ+1kΩ+4.505kΩ
=
9V
13.705kΩ
=
656.695µA ∴ 𝐼𝐸 = 656.695µA
• 𝑉𝐸 = 𝑅𝐸 𝐼𝐸 = 1kΩ 656.695µA = 656.695mV 0.657V
• 𝑉𝐵 = 𝑉𝐵𝐸 + 𝑉𝐸 = 0.7V + 0.657V = 1.357V
• 𝑉𝑅1 = 𝑉𝐶𝐶 − 𝑉𝐵 = 9V − 1.357V = 7.643V
• 𝐼𝐵 =
𝑉𝑅1
𝑅1
• 𝑅2 =
𝑉𝐵
𝐼𝐵
=
7.643V
82kΩ
=
1.357V
93.211µA
= 93.211µA
= 14.558kΩ
10% Standard Values
Decade multiples are available from 10 Ω through 1 MΩ
10
12
15
18
22
27
33
39
47
56
68
82
http://www.rfcafe.com/references/electrical/resistor-values.htm
• Choosing from 10% standard values. R2 = 15-kΩ
Any questions?
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• Email:
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