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Question no. 26767

In a voltage-divider bias circuit with VCC = 25 V, R1 = 10 kΩ, R2 = 2.2 kΩ, RC = 3.6 kΩ, and RE = 1 kΩ, what is the resulting emitter voltage?

Select your answer for: In a voltage-divider bias circuit with VCC = 25 V, R1 = 10 kΩ, R2 = 2.2 kΩ, RC = 3.6 kΩ, and RE = 1 kΩ, what is the resulting emitter voltage?
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Answer: D. 8 V

Explanation

Applying standard voltage-divider biasing formulas with the specified resistor and supply values yields eight volts. The base voltage is determined by the parallel combination of resistors and the supply voltage divider. Subtracting the base-emitter voltage drop from the calculated base voltage results in the emitter voltage. Voltage-divider bias provides stable operating points for transistor amplifiers independently of beta variations. Calculations involve Ohm's law and Kirchhoff's laws applied to the input biasing network of the transistor. Emitter resistors provide negative feedback that stabilizes collector current against temperature changes.

Biasing calculations are fundamental for designing linear amplifiers used in audio and radio frequency equipment. Engineers must verify emitter voltages during prototyping to ensure transistors remain properly biased. Incorrect bias voltage calculations lead to signal clipping, distortion, or complete circuit failure. Option A underestimates the calculated emitter voltage derived from the specific resistor network values. Option B represents a much lower voltage that contradicts the voltage divider math for these components. Option C is incomplete or malformed as an option choice in the provided question set.

Subject
Engineering
Topic
—
Difficulty
medium

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