At what frequency will the AC amplitude be halved in this schematic

In summary, the problem involves finding the value for the AC frequency from Vin at which the amplitude at V (CE base) is halved. The suggested approach is to use the impedance formula and set it equal to 2 times the impedance at the other frequency. However, this yields a negative and unreal value due to the large impedance of the capacitor. The alternative is to use the voltage divider equation with the equivalent resistance of 8.246 kΩ and the 0.22μF capacitor in series. The task is to find the frequency that results in an output to input voltage ratio of 1/2.
  • #1
Storm Butler
78
0

Homework Statement



Basically we were given the attached picture and we are supposed to find at what value for the AC frequency from the Vin will the amplitude be halved at V (CE base)

Homework Equations



The Attempt at a Solution



I think i am supposed to use V=Iz where Z is the impedance. Then the expression for the impedance should be sqrt([8.25e3]^2+[1/i*w*.22e-6]^2). Since the value i am looking for is half the voltage then the expression should be equal to 2 the impedance at the other frequency. So i get sqrt([8.25e3]^2+[1/i*w*.22e-6]^2). =2sqrt([8.25e3]^2+[1/i*t*.22e-6]^2). where t is the frequency that it is half at. The problem is that the impedances for the capacitor is too large and negative when i calculate this that i don't get a real answer.

Any suggestions ?
 

Attachments

  • circuit 2.PNG
    circuit 2.PNG
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  • #2
Apparently you've worked out that the equivalent resistance that the resistor divider presents to the AC signal at the junction of the resistors is 8.246 kΩ. Okay.

That means an equivalent circuit that the signal sees would be the 0.22μF capacitor in series with that equivalent resistance (which has its other end tied to ground). So once again you have a voltage divider situation. Write the equation for the output signal voltage using the voltage divider equation. Find the frequency that makes |Vo/Vin| = 1/2.
 

1. What is the significance of halving the AC amplitude in this schematic?

The AC amplitude is a measure of the strength of the alternating current in a circuit. Halving the AC amplitude can have various implications, such as reducing the power or voltage in a circuit, or affecting the performance of electronic components.

2. How is the frequency related to the halving of the AC amplitude?

The frequency of an alternating current is directly proportional to the halving of the AC amplitude. This means that as the frequency increases, the AC amplitude will be halved at a higher frequency, and vice versa.

3. What factors can affect the frequency at which the AC amplitude is halved?

The frequency at which the AC amplitude is halved can be affected by the components in the circuit, such as resistors, capacitors, and inductors. Changes in these components can alter the frequency response of the circuit and thus affect the halving of the AC amplitude.

4. Can we calculate the exact frequency at which the AC amplitude will be halved?

Yes, the frequency at which the AC amplitude will be halved can be calculated using the formula f = 1/2πRC, where f is the frequency, R is the resistance, and C is the capacitance in the circuit. This formula is valid for simple RC circuits.

5. What is the difference between AC and DC amplitude?

The AC (alternating current) amplitude refers to the maximum value of the alternating current, which constantly changes direction and magnitude. DC (direct current) amplitude, on the other hand, refers to the constant value of the direct current, which flows in one direction with a constant magnitude. In general, AC amplitude is higher than DC amplitude.

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