active-filterhigh pass filteroperational-amplifier
I am trying to figure out how this filter circuit works:
The filter is part of a mini color organ from Forrest Mims Engineer's Mini-handbook. I know it is a high pass active filter which will only illuminate the green LED at certain frequencies, but every other high pass active filter I've looked at has the cap and resistor in series both going to the inverting input. Can somebody please help me undertstand how this filter works? Thank you!
I think, if you assume that the internals of the op-amp contain, in effect, a first order low pass filter, you will have created for yourself what is known as a multiple feedback low-pass filter. I used a great simulation tool from Mister Okawa here to produce this: -
If you look closely at the circuit in the picture above and imagine that R2 and C2 are inside the op-amp, your circuit becomes the same. There is some hand-waving here because I'm taking a stab in the dark about what R2 and C2 actually are and "massaging" them numerically to fit close to producing a peak near 800 Hz.
C2 being 5pF is in the right order for the "conventional" stabilizing stage inside an old-fashioned op-amp like the TL081 and I guess R2 would be in the vicinity of a few kohm.
Anyway I'm convinced this is what is happening!!!
I think you'll do just fine with equalization. RG59 (Belden 8212) is pretty good coax and you can find the attenuation figures HERE for virtually any commonly available coax - it's a calculator so just plug the numbers in i.e. frequency, cable length etc. and bingo out pops the attenuation: -
It's really not a hard cable to equalize up to 10MHz (attenuation about 6.5 dB at 10 MHz and I would suggest that pre-emphasis is the best way to go (then your receiver isn't going to be too upset).
So, plot out the counter-response you need and make a peaking filter that roughly matches the emphasis you need to apply. Use (free) LTSpice for this or any PSpise software to hand.
I've had to do this for 660 Mbps over 50m of very good coax but it still benefitted from a bit of emphasis. In the end we decided to use CAT7 but the idea was proven as far as I am concerned.
Best Answer
It's probably best to think of it as a normal differential amplifier, but with a complex impedance on the non-inverting input resistor (R2 in the following diagram):
The derivation of the equation is shown here, but it simplifies down to:
Now replace R2 with the capacitor impedance R2=Xc=1/(2*pi*f*C). We can see that as the frequency increases, this term decreases. This means that the V2 term increases as the frequency decreases, making the output increase and making the circuit act as a high pass filter.