biascouplingfilterhigh pass filter
Hello
I don't have much experience with electronics and I am trying to understand how this circuit works and its effect on AC input signals.
My understanding so far is that the circuit acts as a high pass filter. But I am not sure of the effect of changing the resistor values (different for both resistors).
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!!!
A low pass filter is pretty much defined by the R and the C value: -
\$f_C = \dfrac{1}{2\pi R C}\$
The same applies for a simple RC high pass filter and the steepness of the transition from "pas" to "reject" frequencies is very shallow: -
Even if the cut-off frequency did fluctuate by (say) 1% (and likely very slowly due to warming and cooling effects of the capacitor) the perceived change in tone would be minimal over that long period of time (at least several minutes.
Think about a slow 1dB change in volume - this could occur in a space of 2 seconds and you just wouldn't notice it let alone over a period greater than several minutes.
And there is certainly no dynamic effect of tone shifting due to the way the guitarist picks or strums the strings.
I don't think looking at CR filter changes is going to yield results that you think. I play and record guitar and have done so over many years. I do all sorts of things to the recorded guitar sound from basic equalization (messing with tone controls) to complicated effects and the least clearly obvious things that effect the sound of a guitar is movement of tone unless it is something dramatic like the wah-wah effect.
Best Answer
In a great simplification, ignoring cutoff frequencies and reactance, C can be treated as a short circuit to AC signals and an Open circuit to DC signals. This AC-couples the input node, Vi, to the output node, Vo. In effect, this removes any DC component from the input (when observed from Vo's point of view).
The resistors form a simple biasing network, which creates a voltage divider between the +2.5V rail and the -2.5V rail. Since both Rs have the same resistance value, the DC voltage at node Vo will be 0V, as long as no signal is applied to Vin. When a signal is applied to Vin, the signal at Vout will be centered around the DC bias of 0V.
What does this all mean for you? Well you could change the resistor values to change the DC bias. For example, in a system where you wanted to be able to measure AC signals with some sort of receiver (ADC, input amplifier, etc.), but you didn't have a negative rail, you might set the bias at half of the supply voltage. That way, even if Vin was going from +1V to -1V, you could still read it on your input.
Keep in mind, if you start mucking with the resistor values, your cutoff frequency will change. So, the best solution is to keep your cutoff freqency far enough below your operating band that way it doesn't really matter[1] anymore what the cutoff frequency is.
[1] "doesn't really matter": For example, if you are trying to pass audio, just make sure your cutoff frequency is a couple orders of magnitude below your lowest expected audio frequency, then it won't matter what the attenuation looks like, since there will be almost no attenuation at the low end of your input range.