The pMOSFET has e) V₁=-IV, AC CIN R₁ 'k' W=0.5mA/V²,V₁0⁰ "L R₁ 1.6 M V R₂ 820k DD = 10V Rs 2K W f) Overall voltage gain_G₁=1 / RD 1K R. R₁ 1K + a) small signal parameters gro b) Draw the small signal circuit for the amplifier c) Input resistance Rin d) Output Resistance Rout Voltage gain Ay= Vo BAZAART
The pMOSFET has e) V₁=-IV, AC CIN R₁ 'k' W=0.5mA/V²,V₁0⁰ "L R₁ 1.6 M V R₂ 820k DD = 10V Rs 2K W f) Overall voltage gain_G₁=1 / RD 1K R. R₁ 1K + a) small signal parameters gro b) Draw the small signal circuit for the amplifier c) Input resistance Rin d) Output Resistance Rout Voltage gain Ay= Vo BAZAART
Introductory Circuit Analysis (13th Edition)
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Transcribed Image Text:### PMOSFET Amplifier Analysis
**PMOSFET Specifications:**
- Threshold voltage (\( V_t \)): -1V
- Transconductance parameter (\( k_n \)): \(\frac{1}{2}\)
- Width to length ratio (\( \frac{W}{L} \)): \(0.5 \, \text{mA/V}^2\)
- Early voltage (\( V_A \)): \(\infty\)
**Circuit Description:**
The diagram displays a PMOSFET amplifier circuit with the following components and values:
- **\( V_{DD} \)**: 10V supply voltage.
- **Resistors:**
- \( R_1 = 1.6 \, \text{M}\Omega \)
- \( R_2 = 820 \, \text{k}\Omega \)
- \( R_S = 2 \, \text{k}\Omega \)
- \( R_D = 1 \, \text{k}\Omega \)
- \( R_L = 1 \, \text{k}\Omega \)
- **Capacitors:**
- Input coupling capacitor (\( C_{IN} \))
- Output coupling capacitor (\( C_{C} \))
- **MOSFET Configuration:**
- Gate (G)
- Source (S)
- Drain (D)
**Problem Statements:**
a) **Small Signal Parameters**: Determine \( g_m \) (transconductance) and \( r_o \) (output resistance of the MOSFET).
b) **Small Signal Circuit**: Draw the small signal model of the amplifier.
c) **Input Resistance (\( R_{in} \))**: Calculate the input resistance of the amplifier circuit.
d) **Output Resistance (\( R_{out} \))**: Calculate the output resistance seen from the output node.
e) **Voltage Gain (\( A_V \))**: Calculate the voltage gain of the amplifier (\( A_V = \frac{v_o}{v_i} \)), where \( v_o \) is the output voltage and \( v_i \) is the input voltage.
f) **Overall Voltage Gain (\( G_V \))**: Calculate the overall voltage gain (\( G_V = \frac{v_o}{v_s} \)), where \( v_s \) is the source voltage.
![**Transcription for Educational Website:**
**Equations and Calculations:**
1. \( V_{ds} = 5 \times \frac{3}{5} = 3 \, V \)
2. \( V_s = (I_D + x) = 2 \, I_D \)
3. \( I_{D} = \frac{k_n}{2} (V_{gs} - V_T)^2 \)
4. \( I_{D} = 0.5 (3 - 2 \, I_D - 1) \)
5. \( 2 \, I_D = 4 (1 - I_D) \)
6. \( x \, I_D = \frac{1}{2} (I_D^2 + 1 - 2 \, I_D) \)
7. \( I_D^2 + 1 - 2.5 \, I_D = 0 \)
8. \( I_D = \frac{2.5 + \sqrt{6.25 - 4}}{2} \)
9. **Final Result:**
\[ I_{D} = 0.5 \, \mu A \]
**Additional Calculations:**
1. \( g_m = \sqrt{2 \, k_n \, I_D} \)
\[ g_m = \sqrt{2 \times 1 \times 0.5} \]
\[ g_m = 1 \, mA/V \]
2. \( g_o = \frac{V_A}{I_D} \) as \( V_A \to 0 \)
\[ g_o = 6 \]
**Small Signal Equivalent:**
- Capacitors are considered as short circuits.
**Diagram Analysis:**
- The diagram shows a small signal equivalent circuit with resistors and dependent sources:
- **Resistors:** 1kΩ, 2.4kΩ, 4kΩ
- **Voltage-controlled current source:** \( g_m V_{gs} \)
**Additional Calculation:**
1. \( R_{in} = (1 \, k \Omega) + \left( \frac{2 \times 4}{2 + 4} \, k \Omega \right) \)
\[ R_{in} = (1 \, k \Omega) + 1.2 \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93e18a88-13ba-47d2-8c84-ec9d29397d2a%2F03d3072a-e984-4839-a7a9-11e244a4d4f2%2F2m0figg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website:**
**Equations and Calculations:**
1. \( V_{ds} = 5 \times \frac{3}{5} = 3 \, V \)
2. \( V_s = (I_D + x) = 2 \, I_D \)
3. \( I_{D} = \frac{k_n}{2} (V_{gs} - V_T)^2 \)
4. \( I_{D} = 0.5 (3 - 2 \, I_D - 1) \)
5. \( 2 \, I_D = 4 (1 - I_D) \)
6. \( x \, I_D = \frac{1}{2} (I_D^2 + 1 - 2 \, I_D) \)
7. \( I_D^2 + 1 - 2.5 \, I_D = 0 \)
8. \( I_D = \frac{2.5 + \sqrt{6.25 - 4}}{2} \)
9. **Final Result:**
\[ I_{D} = 0.5 \, \mu A \]
**Additional Calculations:**
1. \( g_m = \sqrt{2 \, k_n \, I_D} \)
\[ g_m = \sqrt{2 \times 1 \times 0.5} \]
\[ g_m = 1 \, mA/V \]
2. \( g_o = \frac{V_A}{I_D} \) as \( V_A \to 0 \)
\[ g_o = 6 \]
**Small Signal Equivalent:**
- Capacitors are considered as short circuits.
**Diagram Analysis:**
- The diagram shows a small signal equivalent circuit with resistors and dependent sources:
- **Resistors:** 1kΩ, 2.4kΩ, 4kΩ
- **Voltage-controlled current source:** \( g_m V_{gs} \)
**Additional Calculation:**
1. \( R_{in} = (1 \, k \Omega) + \left( \frac{2 \times 4}{2 + 4} \, k \Omega \right) \)
\[ R_{in} = (1 \, k \Omega) + 1.2 \,
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