In a self-biased RF amplifier stage, if the plate voltage is 1250 V, plate current 150 ma, grid current 15 ma, and grid-leak resistance is 4000 ohms, what is the operating grid bias?

In a self-biased RF amplifier stage, if the plate voltage is 1250 V, plate current 150 ma, grid current 15 ma, and grid-leak resistance is 4000 ohms, what is the operating grid bias?

• A. 60 V
• B. 30 V
• C. 90 V
• D. 50 V

Answer: (A)

To determine the operating grid bias in a self-biased RF amplifier stage, we must consider how grid biasing works in relation to the given plate voltage, plate current, grid current, and the grid-leak resistance. In a self-biased configuration, the grid is biased negatively with respect to the cathode due to the DC voltage drop across the grid-leak resistor caused by the grid current. The total current flowing through the grid-leak resistor (which is the grid current) causes a voltage drop that affects the grid voltage. First, we calculate the voltage drop across the grid-leak resistance using Ohm's Law. The voltage drop \( V_g \) across the grid-leak resistor can be calculated as follows: \[ V_g = I_g \times R_g \] Where: - \( I_g \) (grid current) is 15 mA (or 0.015 A), - \( R_g \) (grid-leak resistance) is 4000 ohms. Calculating the voltage drop: \[ V_g = 0.015 \, \text{A} \times 4000 \, \text{ohms} = 60 \, \text{V} \] This

To determine the operating grid bias in a self-biased RF amplifier stage, we must consider how grid biasing works in relation to the given plate voltage, plate current, grid current, and the grid-leak resistance.

In a self-biased configuration, the grid is biased negatively with respect to the cathode due to the DC voltage drop across the grid-leak resistor caused by the grid current. The total current flowing through the grid-leak resistor (which is the grid current) causes a voltage drop that affects the grid voltage.

First, we calculate the voltage drop across the grid-leak resistance using Ohm's Law. The voltage drop ( V_g ) across the grid-leak resistor can be calculated as follows:

[ V_g = I_g \times R_g ]

Where:

• ( I_g ) (grid current) is 15 mA (or 0.015 A),

• ( R_g ) (grid-leak resistance) is 4000 ohms.

Calculating the voltage drop:

[ V_g = 0.015 , \text{A} \times 4000 , \text{ohms} = 60 , \text{V} ]

This