Is it because I am in front of the 10k resistor so it acts like a high pass?, If that is the case, a 1uf gives me 16hz(Below my hearing I would guess) and a 2uf gives me about 8(Highly doubt my computer can go this low, but something in the future may).
I think i need a capacitor value pack or something to play with.
I am still using the op amp inverting. So i would have low impedance(so something like 10k resistor + opamp inverting input) right?
Not exactly. The frequency filter network is determined by the input capacitor and the real input impedance. With the simple input circuit - without resistors referenced to ground -
you have an input impedance of nearly infinite. This is very high and this high impedance does not provide a load for the input path. This is due to the FET input of the OPAs.
FETs in a single device amplifier (1 FET with some external components), FETs can have a lower or a high input impedance - depending on the amplification mode.
The OPA inputs are switched as Source-Followers (Common Drain mode) attached to current sources. Source followers usually have a higher input impedance,
which results in a low output impedance. (Did i mix it up in the earlier posts??) The FET inputs provide an uneven higher impedance compared to Transistors.
OK - i hope i don't mix it up now.. The scheme above should show the actual situation. You have a input capacitor, a 10K serial resistance -
(because it is in serial with the input) and the naked FET input. C1 and R2 are a serial network and the FET is the real impedance**. The FET impedance usually is 1M to 10M (100K worst situation guesstimated).
Now you can count the 0.1uF and the FET impedance (let's take worst situation values - 100K). This provides a 15.9Hz corner frequency - low enough to shake glass.
The serial resistance of 10K (R2) does not join the party - except that it attenuates the incoming signal in conjunction with the feedback resistor. Until here - the corner frequency is only determined by Cin (C1) and the FET -
Changing the corner frequency means changing the capacitance all the time.
To make the input corner frequency adjustable, you can go the simple way and add an optional linear (lin not log) potentiometer* (switched as variable resistor). This allows you to "bend" the frequency curve
like if you would use an equalizer for just the high-pass (low cut) band. This pot (variable resistor) is attached to ground and now works in parallel to the FET input.
The equation (simplified) now is fc = 1 // 2 * pi * Rpot II R_FET * Cin where R_FET is obsolete due to the infinite (best case) impedance.
Now the potentiometer becomes the main input impedance and when you change the resistor value by turning the potentiometer you get the adjustable fc (corner frequency). Puh.
It seems to be simple - but it isn't. As you can see, the potentiometer with the variable impedance is attached to ground. Every turn which cuts the low frequency does result in a lower input impedance and it does load the input.
More cut-off = more load == more load = more loss. It simply also attenuates the signal.
When we stick with the 0.1uF input cap and a 100k potentiometer we get the hum frequency cancellation at about 25k set at the potentiometer (63Hz).
25k is not even bad and the loss is not super high. 25k is still a good impedance value for usual input networks. Some go and mout 10k volume pots (which i never liked - personally one) as input volume regulators.
This can be seen as low. 100k might be OK so. As said - it is the simplest way and only a "Bass cut" (low cut) network.
Unless you don't need an active circuit (with loss compensation) it works. If you want to engage a "multi band" equalizer you don't come around to build an active EQ.
Every single band adjustement does result in heavy losses if you drive the bands without compensation. This is going to be complicated.
Some examples and schematics for further reading:http://www.sentex.ca/~mec1995/tutorial/xtor/xtor5/xtor5.html
Simple tone controls - scroll down to 3/4 of the page - good and useful infoshttp://gilmore2.chem.northwestern.edu/projects/equal_prj.htm
Baxandall controls - lossy unless you put another OPA inbetween - good and useful infoshttp://www.free-circuit.com/3-band-graphic-equalizer-circuit-with-ic-tl072-and-ne5532/
let's get serious (OPA2134 also works)http://www.free-circuit.com/10-band-graphic-stereo-equalizer-circuit-with-tl074/
if someone needs more channels... 20 for stereo