Need help with high pass filter

Well here is my guess since impedance for a cap is 1/wc where w =2*PI*f where f is the frequency in Hz. looks like 33,000= 1/(2*PI*130*c) or c= 1/(2*PI*130*33,000) * 1,000,000 in uF = .0362uF How did I do guys (gals) ?

Thank you for your help, Joegio. But how do I incorporate the resistor into the formula (e.g. resistor of 15k ohm)? Is that just added to the amplifier's resistance (33,000 + 15,000), changing the equation to C= 1/(2*Pi*130*48,000)? Or is there another formula to incorporate the resistor? I know that this can be done without the resistor, connecting the ground directly, but based on a Stereophile prototype (Corey Greenberg Vol. 17 No. 1) who incorporated a resistor as well, there "must" be some advantage to that. Does anybody know?

Well I'm not sure about the resistor unless its to create a filter of a higher order. Anyway I took one of the Vandersteen filters apart and found two caps in parallel which sum the capacitance, but no resistor. If the resistor is in series than I think you would add it to the capacitive impedance, if in paralel than its 1/Z = 1/Xc + 1/Xr where Xc is capacative reactance and Xr is resistance reactance. This is where a spreadsheet comes in handy. Again I'm no expert, but I think this is how it goes.

Let's talk circuit topology. For a high pass filter (HPF), you will need the capacitor (C) in series in the signal path from preamp to power amp. The additional resistor (R) can go in one of three places: (1) it could bridge across the preamp outputs, but this will have no effect on the filter characteristics (2) it could parallel C, but this would change your HPF from a true high-pass characteristic to a shelved filter that merely emphasizes high frequency, or (3), it could bridge across the power amplifier input. This is the only logical place, and it serves to better define the actual load seen by the network (the specified input impedance is a "nominal" value. Your mileage may vary). The resistance (R) parallels the power amp input impedance (we'll assume it's all resistive) Rin. The filter resistor R should be chosen to be smaller than Ri, but larger than the lowest load impedance Rlow allowed by the preamp manufacturer. I'd pick R = sqrt(Rin*Rlow) as a compromise. You can compute the effective termination load of the HPF as Reff = 1/((1/R)+(1/Rin)). The cutoff frequency of the HPF will be f=1/(2*pi*Reff*C). Transposing this gives C = 1/(2*pi*f*Reff). Given your values, and assuming Rlow = 10 Kohms gives R=18.2 kohms and Reff = 11.7 kohms. This, in turn gives a required capacitance of 0.10 uF. If you decide to use a 15 kohm resistor, Reff = 10.3 kohm, and the required capacitance is 0.12 uF. If you do not include resistor R in the HPF circuit, then Reff = Rin, and the required capacitance is 0.037 uF (Joegio, check the batteries in your calculator; they're running a little low :-)

Whoops, one more possibility I forgot concerning circuit topologies: the added resistor could be placed in series in the signal path. That would make Reff = R + Rin in the calculation for C in my previous post. The downside of this is that an attenuated signal voltage is presented to the power amp input, limiting the maximum volume available from the system (that may or may not be an issue, depending on the size of R compared to Rin).