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Vin For the circuit shown, assume that Vo = 3V and that the transistor is described as follows: B=100, Va = 100V, VT = .025V. a) Determine the small signal parameters for the transistor. b) Draw the small signal model 5V 5V 300uA Vo www 1||w 4K 625 C

Vin For the circuit shown, assume that Vo = 3V and that the transistor is described as follows: B=100, Va = 100V, VT = .025V. a) Determine the small signal parameters for the transistor. b) Draw the small signal model 5V 5V 300uA Vo www 1||w 4K 625 C

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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### Analyzing a Transistor Circuit For the circuit shown, assume that Vo = 3V and that the transistor is described as follows: - β = 100 - Va = 100V - VT = 0.025V #### Tasks: a) Determine the small signal parameters for the transistor. b) Draw the small signal model. #### Circuit Diagram Description: - The circuit consists of a PNP transistor connected to a voltage source (5V). - A 4KΩ resistor is connected to the collector of the transistor. - A 300µA current source is connected in parallel with the voltage at the collector, indicated as Vo. - The base of the transistor is labeled as Vin, which is the input voltage. - The emitter of the transistor is connected to ground through a 625Ω resistor. - There is also a capacitor (C) connected in parallel with the 625Ω resistor to ground. #### Steps for Solving the Tasks: **For task a:** Determine the small signal parameters for the transistor. 1. **Small signal current gain (β):** The given β = 100. 2. **Transconductance (gm):** \[ I_C = 300\mu A \] \[ V_T = 0.025V \] \[ g_m = \frac{I_C}{V_T} = \frac{300 \times 10^{-6} A}{0.025V} = 0.012 \, S \] 3. **Output resistance (ro):** \[ V_A = 100V \] \[ ro = \frac{V_A}{I_C} = \frac{100V}{300 \times 10^{-6} A} = 333.33 \, k\Omega \] **For task b:** Draw the small signal model The small signal model will include: - **gm*vbe:** A current source in series with the collector-emitter path, representing the small signal current. - **ro:** Resistor ro connected from collector to emitter. - **Small signal input (vin) to base:** The voltage applied to the base. - **Re (625Ω) bypassed by capacitor (C):** In small signal analysis, the capacitor can be seen as a short circuit for AC signals, so the emitter will have
Transcribed Image Text:### Analyzing a Transistor Circuit For the circuit shown, assume that Vo = 3V and that the transistor is described as follows: - β = 100 - Va = 100V - VT = 0.025V #### Tasks: a) Determine the small signal parameters for the transistor. b) Draw the small signal model. #### Circuit Diagram Description: - The circuit consists of a PNP transistor connected to a voltage source (5V). - A 4KΩ resistor is connected to the collector of the transistor. - A 300µA current source is connected in parallel with the voltage at the collector, indicated as Vo. - The base of the transistor is labeled as Vin, which is the input voltage. - The emitter of the transistor is connected to ground through a 625Ω resistor. - There is also a capacitor (C) connected in parallel with the 625Ω resistor to ground. #### Steps for Solving the Tasks: **For task a:** Determine the small signal parameters for the transistor. 1. **Small signal current gain (β):** The given β = 100. 2. **Transconductance (gm):** \[ I_C = 300\mu A \] \[ V_T = 0.025V \] \[ g_m = \frac{I_C}{V_T} = \frac{300 \times 10^{-6} A}{0.025V} = 0.012 \, S \] 3. **Output resistance (ro):** \[ V_A = 100V \] \[ ro = \frac{V_A}{I_C} = \frac{100V}{300 \times 10^{-6} A} = 333.33 \, k\Omega \] **For task b:** Draw the small signal model The small signal model will include: - **gm*vbe:** A current source in series with the collector-emitter path, representing the small signal current. - **ro:** Resistor ro connected from collector to emitter. - **Small signal input (vin) to base:** The voltage applied to the base. - **Re (625Ω) bypassed by capacitor (C):** In small signal analysis, the capacitor can be seen as a short circuit for AC signals, so the emitter will have
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