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The four-layer diode in Figure 11–52 is biased such that it is in the forward-conduction region. Determine the anode current for VBR(F) = 20 V, VBe = 0.7, and VCE(sat) = 0.2 V. FIGURE 11-52 Rs 1.0 kΩ VBIAS 25 V

The four-layer diode in Figure 11–52 is biased such that it is in the forward-conduction region. Determine the anode current for VBR(F) = 20 V, VBe = 0.7, and VCE(sat) = 0.2 V. FIGURE 11-52 Rs 1.0 kΩ VBIAS 25 V

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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**Question:** 1. The four-layer diode in Figure 11–52 is biased such that it is in the forward-conduction region. Determine the anode current for \(V_{\text{BR(F)}} = 20 \, \text{V}\), \(V_{\text{BE}} = 0.7 \, \text{V}\), and \(V_{\text{CE(sat)}} = 0.2 \, \text{V}\). **Figure 11–52 Description:** The circuit diagram includes: - A four-layer diode connected in series with a resistor \(R_S\) of \(1.0 \, \text{k}\Omega\). - A power supply labeled \(V_{\text{BIAS}} = 25 \, \text{V}\). - The diode is in series with the resistor and then connected to the ground. **Explanation:** The task is to determine the anode current for the given voltages. In the circuit, the total applied voltage is 25 V, and the voltage drop across the forward-biased diode will be \(V_{\text{BE}}\) and \(V_{\text{CE(sat)}}\). The breakdown voltage \(V_{\text{BR(F)}}\) is also given. To find the anode current, use the formula: \[ I_A = \frac{V_{\text{BIAS}} - V_{\text{BE}} - V_{\text{CE(sat)}}}{R_S} \] Substitute the given values: \[ I_A = \frac{25 \, \text{V} - 0.7 \, \text{V} - 0.2 \, \text{V}}{1.0 \, \text{k}\Omega} \] Calculate the current: \[ I_A = \frac{24.1 \, \text{V}}{1.0 \, \text{k}\Omega} = 24.1 \, \text{mA} \] The anode current is 24.1 mA.
Transcribed Image Text:**Question:** 1. The four-layer diode in Figure 11–52 is biased such that it is in the forward-conduction region. Determine the anode current for \(V_{\text{BR(F)}} = 20 \, \text{V}\), \(V_{\text{BE}} = 0.7 \, \text{V}\), and \(V_{\text{CE(sat)}} = 0.2 \, \text{V}\). **Figure 11–52 Description:** The circuit diagram includes: - A four-layer diode connected in series with a resistor \(R_S\) of \(1.0 \, \text{k}\Omega\). - A power supply labeled \(V_{\text{BIAS}} = 25 \, \text{V}\). - The diode is in series with the resistor and then connected to the ground. **Explanation:** The task is to determine the anode current for the given voltages. In the circuit, the total applied voltage is 25 V, and the voltage drop across the forward-biased diode will be \(V_{\text{BE}}\) and \(V_{\text{CE(sat)}}\). The breakdown voltage \(V_{\text{BR(F)}}\) is also given. To find the anode current, use the formula: \[ I_A = \frac{V_{\text{BIAS}} - V_{\text{BE}} - V_{\text{CE(sat)}}}{R_S} \] Substitute the given values: \[ I_A = \frac{25 \, \text{V} - 0.7 \, \text{V} - 0.2 \, \text{V}}{1.0 \, \text{k}\Omega} \] Calculate the current: \[ I_A = \frac{24.1 \, \text{V}}{1.0 \, \text{k}\Omega} = 24.1 \, \text{mA} \] The anode current is 24.1 mA.
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