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Consider a palladium Schottky diode at T = 300 K formed on n-type germanium doped at N = 5 X 1018 cm-3. Determine the electric field at the metal junction for VR = 2 V. O Emax Emax = 3.11 X 106 V/cm Emax = 6.00 X 106 V/cm Emax = 2.01 X 106 V/cm = 1.83 X 106 V/cm

Consider a palladium Schottky diode at T = 300 K formed on n-type germanium doped at N = 5 X 1018 cm-3. Determine the electric field at the metal junction for VR = 2 V. O Emax Emax = 3.11 X 106 V/cm Emax = 6.00 X 106 V/cm Emax = 2.01 X 106 V/cm = 1.83 X 106 V/cm

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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**Title: Calculating the Electric Field in a Palladium Schottky Diode** **Introduction:** Consider a palladium Schottky diode at a temperature of \( T = 300 \, \text{K} \), formed on n-type germanium. The germanium is doped with a donor concentration of \( N_d = 5 \times 10^{18} \, \text{cm}^{-3} \). **Problem:** Determine the maximum electric field (\( E_{\text{max}} \)) at the metal junction when the reverse bias voltage (\( V_R \)) is 2 V. **Options:** - \( E_{\text{max}} = 6.00 \times 10^6 \, \text{V/cm} \) - \( E_{\text{max}} = 3.11 \times 10^6 \, \text{V/cm} \) - \( E_{\text{max}} = 2.01 \times 10^6 \, \text{V/cm} \) - \( E_{\text{max}} = 1.83 \times 10^6 \, \text{V/cm} \) **Instructions:** To solve this problem, use the relevant equations for the electric field in a Schottky diode, considering factors like doping concentration and applied voltage. Compare your calculated \( E_{\text{max}} \) to the given options and select the correct answer.
Transcribed Image Text:**Title: Calculating the Electric Field in a Palladium Schottky Diode** **Introduction:** Consider a palladium Schottky diode at a temperature of \( T = 300 \, \text{K} \), formed on n-type germanium. The germanium is doped with a donor concentration of \( N_d = 5 \times 10^{18} \, \text{cm}^{-3} \). **Problem:** Determine the maximum electric field (\( E_{\text{max}} \)) at the metal junction when the reverse bias voltage (\( V_R \)) is 2 V. **Options:** - \( E_{\text{max}} = 6.00 \times 10^6 \, \text{V/cm} \) - \( E_{\text{max}} = 3.11 \times 10^6 \, \text{V/cm} \) - \( E_{\text{max}} = 2.01 \times 10^6 \, \text{V/cm} \) - \( E_{\text{max}} = 1.83 \times 10^6 \, \text{V/cm} \) **Instructions:** To solve this problem, use the relevant equations for the electric field in a Schottky diode, considering factors like doping concentration and applied voltage. Compare your calculated \( E_{\text{max}} \) to the given options and select the correct answer.
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