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

1. Feb 7, 2012

### Storm Butler

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

3. 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 dont get a real answer.

Any suggestions ?

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2. Feb 7, 2012

### Staff: Mentor

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.