University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
List Of examples about High-frequency response of amplifiers amplifier (CE) 1- Example 1: Common-Emitter amplifier
2- Example 2: Design CE amplifier. 3- Example 3: Common-collector amplifier (CC). 4- Example 4: Common-base amplifier (CB). 5- Example 5: Cascode amplifier.
Common-Emitter amplifier (CE) Example (1) For the common-Emitter amplifier circuit shown in Fig. 1.
a.
Calculate the low-cutoff frequency for each coupling and bypass capacitor.
b. Determine the midband gain. c.
Calculate the high-cutoff frequency response.
d. Determine the bandwidth of the amplifier. e. Compare the results to the simulations using a Bode-plot? The transistor parameters: VBE = 0.8 V, β = 153.5, Cπ = 16.7 pF and C µ =10 pF.
5V
33kΩ
4kΩ
Cc2 Cc1 500Ω
1 Vpk 1kHz 0°
20uF 2N2222*
20uF 5kΩ
22kΩ 4kΩ
Cb 20uF
Fig. 1 Solution
First to determine the small-signal parameters rπ and gm we calculate the Q-point: Rth = RB = 22||33 kΩ =13.2 kΩ.
22 ×5 2 ℎ 2233
Page 1 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
20.8 13.2153.614 0.001 0.00199 Ic = IE = 0.29 mA
0.29 11.6 0.025 11.6 13.2 13.2 Ω Mid-band analysis:
∥ 5||41 11.6 25.8 To find Avs, , we first find Zi:
ℎ || 6.6 Ω 6.6 0.93 6.60.5 93 However:
. 25.8∗0.9323.99 20log| 20log|23.99 23.99|| 27.6 27.6 4 kΩ Low frequency response:
The low frequency response is determined using short-circuit method as following: 1) Considering the effect of C c1 c1 . C c2 c2 and C cb cb are short-circuited. The Thevenin resistance of the input coupling capacitor is:
6.2 0.5 6.7 Ω The input coupling circuit has a cutoff frequency of:
1 21 220 6.7 19 Ω 1.19 2) Considering the effect of C c2 c2 . C c1 c1 and C cb cb are short-circuited. The Thevenin resistance of the output coupling capacitor is:
4 5 9 kΩ
Page 2 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
The output-coupling circuit has a cutoff frequency of:
1 21 220 9 Ω 0.88
3) Considering the effect of C cb . C c1 and C c2 are short-circuited. The Thevenin resistance seen by the emitter-bypass capacitor is:
13.2) ∥ 4 ∥ 1 ∥ (0.5 ∥13.2 154.6 ∥ 4 0.088 ∥ 4 0.086 Ω 13.68 154.6 The emitter-bypass circuit has a cutoff frequency of:
1 21 220 0.086 Ω 92.5 4) The results are: f 1 = 0.88 Hz
output-coupling capacitor
f 1 = 1.19 Hz
input-coupling capacitor
f 1 = 92.5 Hz
emitter-bypass capacitor
From these results, the dominant frequency is the emitter-bypass circuit (92.5 Hz). Note the exact lowfrequency response is:
21 0.88 1.19 92.5 94.6 =
High frequency response:
The transistor in the circuit is replaced by its high-signal model. At high frequencies the coupling and bypass capacitors have very low impedances and can be assumed to be short-circuited. This is so called short-circuit method.
1) The input-miller capacitance is found by:
1 10 25.81 268 The total input capacitance
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University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
= 268 16.7 284.7 The resistance seen by the input capacitance:
∥ ∥ 0.5 ∥13.2∥13.2 1 11 1 0.46 Ω The frequency of the input capacitors is
1 21 2284.7 0.46 Ω 1.2 2) The output miller capacitance is found by:
+ 10 .+ . 10.39 The resistance seen by the output capacitance
∥ 4 ∥ 5 2.2 Ω The frequency of the output capacitors is
1 21 22.2 Ω10.39 6.96 3) The results are: f H1 = 1.2 MHz
input capacitor
f H2 = 6.96 MHz
output capacitor
The high 3-dB frequency is 1.2 MHz. The exact high 3-dB frequency is:
1 12 2 1 2284.7 × 0.46 Ω 10.39 × 2.2 Ω = 1.03 Bandwidth of the amplifier: f bW = f H – f L =1.2 MHz – 92.5 Hz = 1.199 MHz ≈ f H
Simulations: The results of a Multisim analysis are given in Fig. 2a. 1.
From Fig. 2a, the simulated midband gain (Av) dB in the bode-plot is 28.37 dB.
2.
From Fig. 2b, the approximate low 3-dB frequency is 103.1 Hz. Page 4 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312 3.
Frequency response
Dr. Amna Elhawil
From Fig. 2c, the approximate high 3-dB frequency is 1.3 MHz.
As you can see these results close to our calculations.
Fig. 2a
Fig. 2b
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University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
Fig. 2c
Fig. 2d Example(2): Redesign the amplifier circuit in example (1) such that the low frequency ( f L) is 200 Hz. Solution:
From example (1), the dominant frequency corresponding to the lowest resistance, which RCcb . However, to get f L = 200 Hz, we need to calculate Cb , since
0.086 Ω
The emitter-bypass circuit has a cutoff frequency of:
1 21 2 0.086 Ω 200 1 21 22000.086 Ω 9.2
The second frequency ( f L2 ) is the next higher frequency corresponding to the next higher resistance, which is the output-coupling circuit. Let f L2 = f L/10 = 200/10 = 20 Hz Page 6 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
21 219 Ω 20
Thus, the output coupling capacitor is:
1 2209 Ω 0.88
The third frequency is the small frequency corresponding to the highest resistance, which is the input-coupling circuit. Let f L3= f L/20 = 200/20 = 10 Hz
21 216.7 Ω 10
Thus, the input coupling capacitor is:
1 2106.7 Ω 2.4 Common-collector amplifier (CC) Example (3) For the common-collector amplifier shown in Fig. 3.1.
f.
Calculate the low-cutoff frequency for each coupling and bypass capacitor.
g.
Determine the midband gain.
h. Calculate the high-cutoff frequency response. i.
Determine the bandwidth of the amplifier.
j.
Compare the results to the simulations using a Bode-plot? The transistor parameters: VBE = 0.8 V, β = 153.5, Cπ = 16.7 pF and C µ =10 pF.
15 V
15kΩ
2N2222 200Ω
2.5 Vpk 1kHz 0°
47uF 11kΩ
47uF 1kΩ
1kΩ
Fig. 3.1 Page 7 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
Solution
First to determine the small-signal parameters rπ and gm we calculate the Q-point: Rth = RB = 15||11 kΩ =6.3 kΩ.
×15 6.3 ℎ 11 1115
6.30.8 0.034 6.3153.511 Ic = IE = 5.2 mA, VCE= 15 – 13.5 = 1.5 V
5.2 208 0.025 0.74 Ω Mid-band analysis: From the circuit:
104.4 || 1 1||1153.51 0.74 || 1 || 1 1104.4 105.4 0.99 104.4 105.4 To find Avs, , we first find Zi:
|| 1 || 6.3||0.73154.5×1| |1 6.3||77.98 5.82 Ω 5.82 0.97 5.820.2 However:
. 0.99∗0.970.96 Page 8 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
20log|0.96| 0.355 The output impedance without load,
0.74 0.2||6.3 0.91 Ω || 1 || 1||154.5 Fig. 3.2 shows the simulated results: the midband gain is -0.375 dB
Fig. 3.2 Low frequency response:
The low frequency response is determined using short-circuit method as following: 5) Considering the effect of the input-coupling capacitor C c1 , we set C c2 as short-circuit. The Thevenin resistance of the input coupling capacitor is:
5.82 0.2 6.02 Ω The input coupling circuit has a cutoff frequency of:
1 21 247 6.02 Ω 0.56 6) Considering the effect of the output-coupling capacitor C c2 . We set C c1 as short-circuit. The Thevenin resistance of the output coupling capacitor is:
0.09111.091 kΩ The output-coupling circuit has a cutoff frequency of:
1 21 247 1.091 Ω 3.1 7) The results are: Page 9 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
f 1 = 0.56 Hz
input-coupling capacitor
f 1 = 3.1 Hz
output-coupling capacitor
Dr. Amna Elhawil
From these results, the dominant frequency is the output-coupling circuit (3.1 Hz). The simulations give about 3.15 Hz as -3dB frequency.
High frequency response:
The transistor in the circuit is replaced by its high-signal model. At high frequencies the coupling and bypass capacitors have very low impedances and can be assumed to be short-circuited. This is so called short-circuit method.
1. Considering the capacitor Cµ .By setting Cπ as open circuit, The resistance see n by Cµ
µ ∥ ∥[ 1 || ] 0.194 ∥ 0.74 154.5 × 0.5 0.19 Ω The frequency of the BC capacitors (C µ ) is
1 21 210 0.19 Ω 83.7 2. Considering the capacitor Cπ .By setting Cµ as open circuit, the resistance seen by Cπ
|| || 1 || || 0.50.194 0.73|| 0.0065 6.5 Ω 0.73|| 1210×0.5 Page 10 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
|| || 1 || || || || 0.5 Ω || 0.194 Ω
Since,
However,
0.50.194 6.5 Ω 0.50.194 1210×0.5 0.73
The frequency of the BE capacitors (C π ) is
1 21 26.5 Ω16.7 1466.2 4) The results are: f H1 = 83.76 MHz
BC capacitor
f H2 = 1466.2 MHz
BE capacitor
The high 3-dB frequency is 83.76 MHz. The result of the simulation is 86.14 MHz.
Common-Base amplifier (CB) Example (4) For the same circuit of example (1), but using common-base configuration shown in Fig. 4.1.
Page 11 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
k.
Calculate the low-cutoff frequency for each coupling and bypass capacitor.
l.
Determine the midband gain.
m. Calculate the high-cutoff frequency response. n. Determine the bandwidth of the amplifier. o. Compare the results to the simulations using a Bode-plot? The transistor parameters: VBE = 0.8 V, β = 153.5, Cπ = 16.7 pF and C µ =10 pF.
5V
33kΩ
4kΩ
Cc2 20uF
Cc1
5kΩ
2N2222 Cc3
10uF 22kΩ
20uF
500Ω
1 Vpk 1kHz 0°
4kΩ
Fig. 4.1 Solution
First to determine the small-signal parameters rπ and gm we calculate the Q-point: Rth = RB = 33||22 kΩ = 13.2 kΩ.
22×5 2 ℎ 2233 −. .++× 0.0031 , Ic = I
E
= 0.31 mA, VCE= 5 – 8*0.31 = 2.52 V
. 12.4 , 8.0 6 Ω, 0.0 8 kΩ . Mid-band analysis: From the circuit:
|| 5||4 ∗12.427.56 To find Avs, , we first find Zi:
|| 0.08||4 0.078 Ω Page 12 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
0.078 0.13 0.0780.5 . 27.56∗0.133.6 20log|3.6| 11.12
However:
The output impedance without load,
4 Ω Fig. 4.2 shows the simulated results: the midband gain is 11.35 dB
Fig. 4.2 Low frequency response:
The low frequency response is determined using short-circuit method as following: 1. The Thevenin resistance of the input coupling capacitor is:
0.078 0.5 0.578 Ω The input coupling circuit has a cutoff frequency of:
1 21 220 0.578 Ω 13.77 8) The Thevenin resistance of the output coupling capacitor is:
4 5 9 kΩ The output-coupling circuit has a cutoff frequency of:
1 21 220 9 Ω 0.88 9) The Thevenin resistance of the bypass capacitor is:
Page 13 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
||[ || ] 13.2 ||[8.06100× 0.5||4] 10.5 kΩ The output-coupling circuit has a cutoff frequency of:
1 21 210 10.5 Ω 1.5
From these results, the dominant frequency is the input-coupling circuit (13.77 Hz). The simulations give about 14.4 Hz as -3dB frequency.
High frequency response:
The transistor in the circuit is replaced by its high-signal model. At high frequencies the coupling and bypass capacitors have very low impedances and can be assumed to be short-circuited. This is so called short-circuit method.
1. The resistance seen by Cµ
µ ∥ 4 ∥ 5 2.22 Ω The frequency of the BC capacitors (Cµ ) is
1 21 210 2.22 Ω 7.17 2. The resistance seen by Cπ
|| || 0.08|4|5 0.73|| 0.0065 0.077 Ω The frequency of the BE capacitors (C π ) is
1 21 20.077 Ω16.7 123.77 Page 14 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
The high 3-dB frequency is 7.17 MHz. The result of the simulation is 10.2 MHz.
The bandwidth is about 7.17 MHz. Note that in example (1) using the same circuit in the common-emitter configuration the bandwidth is 1.2 MHz. That means the common-base gives higher bandwidth than CE because it doesn’t suffer from the miller effect. The terminals of both parasitic capacitors of the
transistor are connected to the ground, which eliminates the feedback or Miller multiplication effect.
Cascode amplifier Example (5) The cascode CE-CB amplifier is shown in Fig. 5.1. The circuit parameters are VBE = 0.8 V, β1 = β2 = 153.5, Cπ1 = Cπ2= 16.7 pF and Cµ1 = Cµ2 =1 pF, The Q-point of both transistors (IC1 = IE1= IC2 =IE2 =0.157
mA , VCE1= VCE2 = 1.87 V) a.
Calculate the low 3-dB frequency.
b. Calculate the high 3-dB frequency. c.
Calculate the mid-band gain.
Page 15 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
5V
R1 22kΩ
CB
Rc 4kΩ C2 20uF Q1
XLV1
RL 5kΩ
2N2222
10uF R2 33kΩ XLV2
Rs 500Ω
C1
Q2 2N2222
20uF R3 22kΩ
RE 4kΩ
CE 20uF
Fig. 5.1 Cascode CE-CB amplifier Solution
The Q-point has been calculated by assuming I B1 = IB2 are v. small. Thus I1 = I2 = I3= Ix
5 0.065 223322 0.8 0.822×0.065 0.157 22
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University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
Fig. 5.2
Also, to find V CE
5 2 58×0.157 1.8 7 2 However, the small-signal parameters are 24.36 Ω, 6.3 , 0.16 Ω. The small signal circuit is shown below. Mid-band analysis: From the circuit:
13.86 || 6.3×4||5
Fig. 5.3
6.3 0.99 1⁄ 6.31/24.36 However,
13.86 × 0.99 13.72 Page 17 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
To find Avs, , we first find Zi:
However:
| 24.36|13.2 8.56 Ω 8.56 0.94 8.560.5 . 13.72∗0.9412.89 20log|12.89| 22.2
Fig. 5.4 shows the simulated results: the midband gain is 24.4 dB
Fig. 6.4
The input impedance without load,
|||| 13.2 ||24.36 8.56 Ω The output impedance without load,
| 4|5 2.2 Ω Low frequency response:
The low frequency response is determined using short-circuit method as following:
Page 18 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
Fig. 5.5
1) The Thevenin resistance of C1 is:
8.56 0.5 9.06 Ω The C1 has a cutoff frequency of:
1 21 220 9.06 Ω 0.88 2) The Thevenin resistance of C2 is:
4 5 9 kΩ The C2 has a cutoff frequency of:
1 21 220 9 Ω 0.88 3) The Thevenin resistance of the bypass capacitor C E1 is:
∥ 0.5||13.224.36 |40.16|4 0.15 Ω ∥ || 1 154.5 1 21 220 0.15 Ω 53.05 4) The Thevenin resistance of the bypass capacitor C B is:
∥ 13.2 Ω 1 21 220 13.2 Ω 0.6 The dominant frequency is (53.05 Hz). The simulations give about 65.88 Hz as -3dB frequency.
Page 19 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
Fig. 6.6
The next formula gives results more closer to the simulation results :
21 53.05 0.8 0.88 0.6 55.33 =
High frequency response:
Fig. 5.7
The first stage has Miller capacitor. The gain of the terminals of C µ1 (B1 and C1)
0.99 1/ 1
The input-Miller capacitance:
1 1 10.99 1.99 The output-Miller capacitance of the first stage:
+ 1 . . 2.0 1
Fig. 5.8 Page 20 of 21
University of Tripoli - Engineering Faculty Computer Engineering Department Spring 2016 - EC312
Frequency response
Dr. Amna Elhawil
1. From the circuit shown in Fig. 5.8, the total input capacitance
= 1.99 16.7 18.69 The resistance seen by the input capacitance:
|||||| 0.5|13.2|24.36 0.47 Ω 1 218.69 0.47 Ω 18.12 1. In the middle part, we have two parallel capacitors (Cπ2 and Cout1M), the resistor seen by these capacitors is re2
1 216.72.01×0.16 Ω 53.16
2. The output part of the circuit has the capacitor Cµ2, Its resistance:
|| 2.2 Ω 2 × 1 1× 2.2 Ω 72.3
The high 3-dB frequency is obtained by
1 12 21 ∑1 21 18.6 9 × 0.47 Ω 18.71 × 0.1 6 Ω 1 × 2.2 Ω = 11.38 The result of the simulation is about 11.93 MHz.
Fig. 5.9
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