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# Voltage divider with capacitor

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#### Electronics4you

##### Member
Hi there,

The answer to the exercise may be incorrect, since I get the wrong results every time I do the calculations. It's a simple circuit, and I need the capacitor value, so the output is -3dB at 100Hz.

I get the answer 583 nF, but the answer should be 160 nF

Can someone explain where the calculations goes wrong?

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There should be a "j" in front of the capacitive reactance term. You must use complex arithmetic.

Since above the cutoff frequency, the gain is -1db, at the cutoff frequency the network should be down to -4db (three on top of the one to start).

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A 'j' will give me an imaginary capacitance result

The result of the equation must then be -4dB = 0.636

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A 'j' will give me an imaginary capacitance result...

Yes, so what is your point?

You have to use phasors (complex arithmetic) to account for both the phase and magnitude to solve the problem. I was lazy, and didn't do the complex arithmetic. However, I showed you that the answer given with the exercise is correct.

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MikeMl

You cheated using the computer simulation program. You get zero for not showing your work.

I mean, that when solving the equation, the result for C is on the form C=j*'number', which is false

MikeMl

You cheated using the computer simulation program. You get zero for not showing your work.
We are here to show how to do the work, not do the work.

Could someone then please show the right way to solve the problem?

I mean, that when solving the equation, the result for C is on the form C=j*'number', which is false
Do you know how to do complex calculations? You convert the rectangular function on the bottom to polar form, do the division, and then convert the result back to rectangular.

Yes, it was just the English term "complex arithmetic" (not my language), that confused be shortly. The result to the equation should have been 0.636 + angle. That's why the result was false.

Thanks for the help

I just remembered a short cut. The 3db down frequency of this high pass filter occurs when the magnitude of the capacitive reactance equals the loop resistance, i.e. |1/(2ΠfC)| = 1k + 9K

C = 1/(2Πf*(1k+9k) = 1/6.28318E6 = 159.1549E-9 = 159.1nF

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