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) ?
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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).
Thank you both for you help, Joegio and 1439bhr. After doing some more homework I think I got it. But your- or anyone's feedback would be greatly appreciated... did I get it right? Doing Capacitor in series in the signal path with a Resistor bridging across power amp inputs (choice 3, 1439bhr) I have reached the following conclusions: Using a 0.1 uF Cap and 20 Kohm Resistor: Fx = 1/ 2*Pi*R*C where R = r1*r2/r1+r2 => 1/ 6.283*12.45283*0.0001 => Fx = 127.8Hz. In other words, using 0.1uF Kimber Cap and a 20 Kohm Resistor with my power amp's 33 Kohms input impedance will provide a 127.8Hz High Pass filter. Did I get this right? Now the next step. Any suggestion on how to solder this together... the right way. How do I insulate this (i.e. signal path from ground etc.) and support it in place (such as mounting the RCA jacks on a plate??). Thanks, Arni
Looks good. The thing to realize is that the actual cap value may vary quite a bit from the nominal value. Ordinary commercial caps may have tolerances as large as +50 / -20%. Tolerances on good resistors may be as large as 5%, or less than 1% for "precision" resistors. If you have access to a capacitance bridge, you may want to measure the actual cap value and adjust resistance to suit. That's what "hand picked" components mean. As far as construction technique, take a peek at a couple of electronics hobbyist magazines to get a feel for typical techniques, or visit your local electronics parts store or even Radio shack, nose around the construction hardware sections and ask a few questions. Better yet, ask around and find someone who's into ham radio. Chances are they know something about constructing homebrew electronics. An ugly but effective solution would be a small plastic "project box" drilled for installation of RCA plugs, with the parts mounted on a small piece of perforated board mounted in the box via machine screws and standoffs, or use a "terminal strip", which is a piece of Bakelite plastic with solder lugs mounted on one side and a couple of mounting lugs on the other side (this used to be used a lot in vacuum tube TVs and radios employing "point to point" wiring.) Have fun!