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ECs-13122018 - Solution

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: (40 ) One PMOS amplifier of the configuration CS with source resistor 𝑅𝑆 is studied as
π‘Š
shown in . The PMOS parameters are given:
= 20, πœ‡πΆπ‘œπ‘₯ = 100πœ‡π΄/𝑉 , 𝑉𝑑𝑝 = −1.5𝑉 , π‘Ÿπ·π‘† is
𝐿
very large. Note that, if the MOSFET operates in the active region:
πœ‡πΆπ‘œπ‘₯ π‘Š
(𝑉𝑆𝐺 − | 𝑉𝑑𝑝 |)
2𝐿
2𝐼𝐷
π‘Š
βˆ™ π‘”π‘š =
= πœ‡πΆπ‘œπ‘₯ (𝑉𝑆𝐺 − | 𝑉𝑑𝑝 |)
(𝑉𝑆𝐺 − | 𝑉𝑑𝑝 |)
𝐿
βˆ™ 𝐼𝐷 =
𝑉𝑆𝑆 = 10𝑉
𝑅𝐺
10 kΩ
𝑅𝑆
1 kΩ
𝑄
𝐢𝑖𝑛
𝑣𝑖𝑛 (𝑑)
π‘£π‘œπ‘’π‘‘ (𝑑)
𝑅𝐺
πΆπ‘œπ‘’π‘‘
10 kΩ
𝑅𝐷
2.5 kΩ
𝑅𝐿
5 kΩ
Figure 1: Question 1.
Calculate the quiestcent values (𝑉𝑆𝐺𝑄 , 𝑉𝑆𝐷𝑄 , 𝐼𝐷𝑄 ). Deduct the transconductance π‘”π‘š .
βˆ™ Apply the KVL for the SG loop, we obtain:
𝑉𝑆𝐺 = 𝑉𝑆 − 𝑉𝐺 = 𝑉𝑆𝑆 − 𝐼𝑆 𝑅𝑆 − 𝑉𝐺 = 10 − 𝐼𝑆 − 5 = 5 − 𝐼𝑆
βˆ™ Assume the PMOS operates in the saturation mode, we have:
1
20
π‘˜π‘› (𝑉𝑆𝐺 − | 𝑉𝑑𝑝 |) = 100 × 10− × (5 − 𝐼𝐷 − 1.5) = (3.5 − 𝐼𝑆 )
2
2
0 = 𝐼𝐷 − 8𝐼𝐷 + 12.25
{οΈƒ
2.06(π‘šπ΄)
𝐼𝐷 =
5.94(π‘šπ΄)(π‘Ÿπ‘’π‘—π‘’π‘π‘‘π‘’π‘‘)
𝐼𝐷 =
βˆ™ Then
𝑉𝑆𝐺 = 5 − 𝐼𝑆 = 2.94(𝑉 )
𝑉𝑂𝑉 = 𝑉 𝑆𝐺− | 𝑉𝑑𝑝 |= 2.94 − 1.5 = 1.44(𝑉 )
𝑉𝑆𝐷 = 𝑉𝑆𝑆 − 𝐼𝐷 (𝑅𝑆 + 𝑅𝐷 ) = 10 − 2.06(1 + 2.5) = 2.79(𝑉 )
2𝐼𝐷
2 × 2.06
π‘”π‘š =
=
= 2.86(π‘šπ΄/𝑉 )
𝑉𝑆𝐺 − | 𝑉𝑑𝑝 |
2.94 − 1.5
1/6
βˆ™ Therefore 𝑉𝑆𝐺 > 𝑉𝑂𝑉 , PMOS operates in the saturation mode. The assumption is correct.
Draw the AC small signal equivalent circuit of the amplifying stage and calculate the voltage gain
𝐺𝑣 [𝑑𝐡] = π‘£π‘œπ‘’π‘‘ /𝑣𝑖𝑛 ? Suppose that 𝐢𝑖𝑛 and πΆπ‘œπ‘’π‘‘ are of large values.
βˆ™ Small signal equivalent circuit of the amplifier is shown in Figure 2.
𝑖𝑔
𝑖 𝑖 𝑣𝑖
𝑔
π‘–π‘œ
𝑑
π‘£π‘œ 𝑒𝑑
π‘”π‘š 𝑣𝑔𝑠
𝑠
𝑅 ||𝑅
𝑣𝑖𝑛
𝑅𝐷
𝑅𝐿
𝑖𝑠
𝑅𝑆
Figure 2: CaΜ‚u 2
βˆ™ Voltage gain of the amplifier:
𝐺𝑣 [𝑑𝐡] = π‘£π‘œπ‘’π‘‘ /𝑣𝑖𝑛 = −
π‘”π‘š (𝑅𝐷 ||𝑅𝐿 )
2.86π‘šπ΄ × 2.5π‘˜ ||5π‘˜
=−
= −1.23(𝑉 /𝑉 ) = 1.79(𝑑𝐡)
1 + π‘”π‘š 𝑅𝑆
1 + 2.86π‘šπ΄ × 1π‘˜
If 𝐢𝑖𝑛 = 20πœ‡πΉ , πΆπ‘œπ‘’π‘‘ = 10πœ‡πΉ , calculate 𝑓𝑙 and 𝑓𝑙 at the input and output loops. Then calculate
approximately the overall low-band cutoff frequency 𝑓𝐿 .
βˆ™ Input loop: cut-off frequency:
𝑓𝐿
1
≃ 1.59 (𝐻𝑧)
2πœ‹(𝑅 ||𝑅 )𝐢𝑖𝑛
:=
βˆ™ Output loop cut-off frequency:
𝑓𝐿
:=
1
≃ 2.1 (𝐻𝑧)
2πœ‹(𝑅𝐷 + 𝑅 )πΆπ‘œπ‘’π‘‘
βˆ™ Overall cut-off frequency:
𝑓𝐿 :=
√︁
𝑓𝐿 + 𝑓𝐿 ≃ 2.6 (𝐻𝑧)
: (40 ) Given a BJT amplifier as shown in . The BJT 𝑄 has the following
parameters: 𝛽 = 60, 𝑉𝐡𝐸,π‘œπ‘› = 0.7𝑉 , 𝑉𝑇 = 25π‘šπ‘‰ .
Analyze the circuit at DC mode. Calculate the quiescent values (𝐼𝐢𝑄 , 𝑉𝐢𝐸𝑄 )?
βˆ™ Consider the circuit at DC mode as shown in Figure fi
for 𝑉𝐢𝐢 and resistors 𝑅 , 𝑅 , we obtain:
. Apply the Thevenin equivalent circuit
𝑅𝑇 𝐻 = 𝑅 ||𝑅 = 9.89 (kΩ)
𝑅
𝑅 ||𝑅
𝑉𝑇 𝐻 =
𝑉𝐢𝐢 =
𝑉𝐢𝐢 = 2.65 (V)
𝑅 +𝑅
𝑅
2/6
𝑉𝐢𝐢 = 15 V
𝑅
56 kΩ
𝑅
3.9 kΩ
π‘…π‘œπ‘’π‘‘
𝑅𝑖𝑛
π‘£π‘œπ‘’π‘‘
𝐢
𝑅𝑠𝑖𝑔
𝐢
𝑄
𝑅𝐿
1.0 kΩ
𝑣𝑖𝑛
𝑅
𝑅
0.1 kΩ
𝑅
1 kΩ
1 kΩ
12 kΩ
𝐢
Figure 3: Question 2.
βˆ™ Assume Transistor 𝑄 operates in the active region. Apply the KVL for BE loop, we obtain:
0 = −𝑉𝑇 𝐻 + 𝑅𝑇 𝐻 𝐼𝐡𝑄 + 𝑉𝐡𝐸 + (𝑅 + 𝑅 )𝐼𝐸𝑄
0 = −𝑉𝑇 𝐻 + 𝑅𝑇 𝐻 𝐼𝐡𝑄 + 𝑉𝐡𝐸 + (𝛽 + 1)(𝑅 + 𝑅 )𝐼𝐡𝑄
𝑉𝑇 𝐻 − 𝑉𝐡𝐸
𝐼𝐡𝑄 =
= 27.89 (πœ‡A)
𝑅𝑇 𝐻 + (𝛽 + 1)𝑅
𝐼𝐢𝑄 = 𝛽𝐼𝐡𝑄 = 1.67 (mA)
𝑉𝐸 = 𝑅 𝐼𝐸𝑄 = 1.87 (V)
𝑉𝐡 = 𝑉𝐸 + 𝑉𝐡𝐸 = 2.37 (V)
𝑉𝐢 = 𝑉𝐢𝐢 − 𝑅 𝐼𝐢𝑄 = 8.47 (V)
𝑉𝐢𝐸𝑄 = 𝑉𝐢 − 𝑉𝐸 = 6.6 (V)
βˆ™ Transistor 𝑄 operates in the active region. The initial assumption is correct.
βˆ™ The quiescent point: 𝑄 : (𝐼𝐢𝑄 , 𝑉𝐢𝐸𝑄 ) = (1.67 mA, 6.6 V).
𝑉𝐢𝐢 = 15 V
𝑅
56 kΩ
𝑅
𝑉𝐢𝐢 = 15 V
3.9 kΩ
𝑅
3.9 kΩ
𝑅𝑇 𝐻
𝑄
𝑄
𝐼𝐡𝑄
𝐼𝐸𝑄
𝑅
12 kΩ
𝑅
𝑉𝑇 𝐻
𝑅 +𝑅
1.1 kΩ
Figure 4: 𝑅
3/6
Assume that capacitors have very large value. Draw the AC small signal equivalent circuit at
the mid-frequency band and calculate input resistance 𝑅𝑖𝑛 , output resistance π‘…π‘œπ‘’π‘‘ and the voltage gain
𝐺𝑣 = π‘£π‘œπ‘’π‘‘ /𝑣𝑖𝑛 ?
βˆ™ Small signal equivalent circuit of the amplifier is shown in fi .
βˆ™ Resistance π‘Ÿπœ‹ or β„Žπ‘–π‘’ :
𝑉𝑇
= 0.896 (kΩ)
𝐼𝐡𝑄
π‘Ÿπœ‹ = β„Žπ‘–π‘’ :=
βˆ™ Input Impedance:
𝑣𝑖
= π‘Ÿπœ‹ + (𝛽 + 1)𝑅 = 6.99 (kΩ)
𝑖𝑏
𝑣𝑖
:=
= 𝑅 ||𝑅 ||𝑅𝑖𝑏 = 4.1 (kΩ)
𝑖𝑖
𝑅𝑖𝑏 :=
𝑅𝑖𝑛
βˆ™ Output Impedance:
βƒ’
𝑣𝑐 βƒ’βƒ’
−𝑖𝐿 ⃒𝑣𝑠𝑖𝑔
π‘…π‘œπ‘’π‘‘ :=
= 𝑅 = 3.9 (kΩ)
βˆ™ Voltage gain 𝐺𝑣 = π‘£π‘œπ‘’π‘‘ /𝑣𝑖𝑛 :
π‘£π‘œπ‘’π‘‘
𝑣𝑖
π‘£π‘œπ‘’π‘‘
=
×
𝑣𝑖𝑛
𝑣𝑖
𝑣𝑖𝑛
𝑅 ||𝑅𝐿
𝑅𝑖𝑛
= −
×
= − 5.58 (V/V)
𝑅 + π‘Ÿπ‘’ 𝑅𝑖𝑛 + 𝑅𝑠𝑖𝑔
𝐺𝑣 :=
𝑅𝑖𝑛
𝑅𝑠𝑖𝑔
𝑅𝑖𝑏
π‘…π‘œπ‘’π‘‘
𝑣𝑖
𝑣𝑐
𝑐
𝑏
π‘£π‘œπ‘’π‘‘
1.0 kΩ
π‘Ÿπœ‹
π‘”π‘š 𝑣 πœ‹
𝑒
𝑅 ||𝑅
𝑣𝑖𝑛
𝑅
𝑅
3.9 kΩ 𝑅𝐿
1 kΩ
0.1 kΩ
Figure 5: If 𝐢 = 20πœ‡πΉ , 𝐢 = 20πœ‡πΉ and 𝐢 = 100πœ‡πΉ . Draw the small signal equivalent circuit at the low
frequency band and calculate approximately the overall low-band cutoff frequency 𝑓𝐿 of the circuit.
βˆ™ Small signal equivalent circuit at the low frequency band:
βˆ™ Low-band cutoff frequency by 𝐢 :
𝑓𝐢
:=
1
≃ 1.56 (𝐻𝑧)
2πœ‹(𝑅𝑠𝑖𝑔 + 𝑅 ||𝑅 ||(π‘Ÿπœ‹ + (𝛽 + 1)𝑅 ))𝐢
βˆ™ Low-band cutoff frequency by 𝐢 :
𝑓𝐢
:=
1
≃ 1.64 (𝐻𝑧)
2πœ‹(𝑅 ||(𝑅 + π‘Ÿπ‘’ + 𝑅 ||𝑅 ||𝑅𝑠𝑖𝑔 /(𝛽 + 1)))𝐢
βˆ™ Low-band cutoff frequency by 𝐢 :
𝑓𝐢
:=
1
≃ 13.89 (𝐻𝑧)
2πœ‹(𝑅 + 𝑅𝐿 )𝐢
4/6
𝐢
𝑅𝑠𝑖𝑔
𝑣𝑖
𝑣𝑐
𝑐
𝑏
π‘£π‘œπ‘’π‘‘
1.0 kΩ
𝐢
π‘Ÿπœ‹
π‘”π‘š 𝑣 πœ‹
𝑒
𝑅 ||𝑅
𝑣𝑖𝑛
𝑅
𝑅
0.1 kΩ
𝑅
1 kΩ
3.9 kΩ 𝑅𝐿
1 kΩ
𝐢
Figure 6: βˆ™ Overall low-band cutoff frequency 𝑓𝐿 of the circuit:
𝑓𝐿 := 𝑓𝐢 + 𝑓𝐢 + 𝑓𝐢 ≃ 15 (𝐻𝑧)
At high frequency band, the circuit is affected by the parasitic capacitors πΆπœ‹ = πΆπœ‡ = 10𝑝𝐹 . Draw the
small signal equivalent circuit at the high frequency band and calculate approximately the overall high-band
cutoff frequency 𝑓𝐻 of the circuit.
βˆ™ Small signal equivalent circuit at high frequency band:
𝑅𝑠𝑖𝑔
𝑣𝑖
πΆπœ‡
𝑏
𝑣𝑐
𝑐
π‘£π‘œπ‘’π‘‘
1.0 kΩ
πΆπœ‹
π‘Ÿπœ‹
π‘”π‘š π‘£πœ‹
𝑒
𝑅 ||𝑅
𝑣𝑖𝑛
𝑅
𝑅
3.9 kΩ 𝑅𝐿
1 kΩ
0.1 kΩ
Figure 7: βˆ™ Analysis using Open-Circuit Time Constants Method:
Resistance π‘…πœ‹ seen by πΆπœ‹ is
′
π‘…πœ‹
𝑅 + 𝑅𝑠𝑖𝑔
≃ 0.13 (kΩ)
:=
1 + π‘”π‘š 𝑅 + 𝑅 /π‘Ÿπœ‹
Resistance π‘…πœ‡ seen by πΆπœ‡ is
′
π‘…πœ‡ :=
𝑅𝐿
′
′
′
𝑅𝑠𝑖𝑔
𝑅𝑠𝑖𝑔
𝑅𝑠𝑖𝑔
1 + ′ + (π‘”π‘š π‘Ÿπœ‹ + ′ ) ×
′
𝑅𝐿
𝑅𝐿
π‘Ÿπœ‹ + 𝑅 + 𝑅𝑠𝑖𝑔 + 𝑅 π‘”π‘š π‘Ÿπœ‹
′
≃ 7.37 (kΩ)
′
where 𝑅𝑠𝑖𝑔 = 𝑅𝑠𝑖𝑔 ||𝑅 ||𝑅 = 0.908(kΩ) and 𝑅𝐿 = 𝑅𝐿 ||𝑅 = 0.796(kΩ)
The cut-off frequency 𝑓𝐻 is
𝑓𝐻 :=
1
≃ 2.1 (𝑀 𝐻𝑧)
2πœ‹ × (π‘…πœ‹ πΆπœ‹ + π‘…πœ‡ πΆπœ‡ )
5/6
βˆ™ Appoximation method: Analysis using Miller Theorem: assume that 𝑅 is very small and the emitter
is grounded at ac mode.
High-band cutoff frequency by 𝐢𝑖𝑛 :
π‘”π‘š 𝑅 ||𝑅𝐿
) ≃ 89.23 (𝑝𝐹 )
1 + π‘”π‘š 𝑅
𝐢𝑖𝑛 := πΆπœ‹ + πΆπœ‡ × (1 +
1
≃ 2.24 (𝑀 𝐻𝑧)
2πœ‹ × π‘… ||𝑅 ||𝑅𝑠𝑖𝑔 × πΆπ‘–π‘›
𝑓𝐢𝑖𝑛 :=
Low-band cutoff frequency by πΆπ‘œπ‘’π‘‘ :
1 + π‘”π‘š 𝑅
) ≃ 11.4 (𝑝𝐹 )
π‘”π‘š 𝑅 ||𝑅𝐿
πΆπ‘œπ‘’π‘‘ := 𝐢𝑔𝑑 × (1 +
1
≃ 17.54 (𝑀 𝐻𝑧)
2πœ‹ × π‘… ||𝑅𝐿 × πΆπ‘œπ‘’π‘‘
π‘“πΆπ‘œπ‘’π‘‘ :=
Overall high-band cutoff frequency 𝑓𝐻 of the circuit:
𝑓𝐻 := √οΈ‚
1
1
𝑓𝐢𝑖𝑛
+
1
≃ 2.22(𝑀 𝐻𝑧)
π‘“πΆπ‘œπ‘’π‘‘
: (20 ) .
Given the OpAmp circuit as shown in Figure 3. Determine the output voltage π‘£π‘œ as a function of 𝑣
and 𝑣 .
𝑅
𝑅
1 kΩ
3 kΩ
𝑣
𝑅
π‘£π‘œ
𝑣
2 kΩ
𝑅
4 kΩ
Figure 8: Question 3.
βˆ™ The output voltage of the circuit:
π‘£π‘œ = −
𝑅
𝑅
𝑅
8
𝑣 +
(1 +
)𝑣 = 𝑣 − 3𝑣
𝑅
𝑅 +𝑅
𝑅
3
Given voltage source 𝑣 and 𝑣 , use ideal OpAmp(s) and resistor(s) to design circuit that provide the
output π‘£π‘œ = 2𝑣 − 0.5𝑣 .
𝑅
𝑅
𝑅
𝑅
10 kΩ
20 kΩ
10 kΩ
𝑅
10 kΩ
𝑣
𝑣
20 kΩ
π‘£π‘œπ‘’π‘‘
Figure 9: Question 3.2
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