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Your opamp has an input capacitor, an output capacitor and a feedback to ground capacitor. Each one drops the output 3dB at the cutoff frequency.
Three capacitors cause the output to be -9dB at the cutoff frequency.
-9dB is almost 1/10th the power so if the cutoff frequency is 20Hz then it will be at a very low level and 100Hz will also be reduced.
So select a cutoff frequency of 7Hz or less for good bass.
But I took a look at PRACTICAL ELECTRONICS FOR INVENTORS book and it used the formula I told (What that formula States really?) And when Is help full? What is F-3dB for an input cap which is working at 20Hz to 20Khz?
Remember when I said use the formula and make F (Frequency) equal to the LOWEST frequency you're gonna amplify. That's exactly what they are doing in your attachment. F-3db is the limit of frequency for the amplifier, and is the definition of the amplifiers bandwidth. Generally, there is a -3db maximum and a -3db minimum frequency. However, for your purposes, you need only be concerned about the lowest frequency.
The formula C = 1/2ΠF(-3db)*R is helpful if you want to make the -3db or "cutoff" at a specific frequency. You're orignal question, however, was how to insure the caps behave as closed circuit elements, which does not mean you want specific cutoff frequencies. That's why I posted the formula.
BTW, do you realize that if you take my orignal equation and replace the ">>" with "=", you have the same equation as the one your're asking about?
If the amplifier has three capacitors in the signal path then you must use a cutoff frequency about 1/3rd the desired cutoff frequency of the entire amplifier.
also, having the 10uf cap in the feedback loop, and the one in series with the output, both alter the output impedance. the output cap just adds Xc to the output impedance (Xc is the reactance or "impedance" of the cap), but the one in the feedback loop has an effect on the calculation of Zo=ro/(1+aβ), because it alters the β term to equal:
(R1+Xc)/(R1+R2+Xc), so the output impedance now also depends on Xc. it affects the closed loop gain of the amp because the closed loop gain is determined by Acl=1+(R2/R1), but with the cap becomes Acl=1+(R2/(R1+Xc)). so if we pick a couple of frequencies, such as 20hz and 1khz, the gain of the circuit (temporarily ignoring the input and output coupling cap) is 1+(22k/(1k+796))=13.24 at 20hz and 1+(22k/(1k+16))= 22.65 at 1khz. Xc is 796 ohms at 20hz, and 16 ohms at 1khz. the raw output resistance of a TL071 is about 160 ohms, so the output impedance of a TL071 in your circuit is 160/(1+(200000*((1k+796)/(22k+1k+796))))=~10milliohm. the 200000 figure is the open loop gain of a TL071 at 20hz. at 1khz, the output impedance at 1khz is going change for two reasons, the open loop gain of the TL071 is only about 4000 at 1khz, and Xc is only 16 ohms, so 160/(1+(4000*)(1k+16)/(22k+1k+16))))=0.9 ohms. now, placing the output cap in the circuit, the low output impedance of 0.01 ohms at 20hz, now equals 0.01+796=796.01 ohms, and the output impedance at 1khz is 0.9+16=16.9 ohms. the same effect also happens at the input. since the input cap is in series with the 23.5 k input impedance, so at 20hz it is 24,296 ohms, and at 1khz it is 23,516 ohms. your -3db frequencies will be where Xc is equal to the resistance it is in series with. for the feedback loop the corner frequency will be where Xc=1k, or about 15 hz. this is the frequency where the gain is one half of 1+(R2/R1), because it's equivalent to 1+(R2/(2)R1). if the load impedance on the amp output is 1K (very common), then the corner frequency at the output will also be about 15hz, so you will actually be 6db down at this point. the input corner frequency, however is 0.65hz, so the 10uf cap there isn't a real issue for audio. but the feedback loop and the output cap should be much larger, maybe 100uf each.
btw, the source of the op amp gain figures is the gain vs frequency plot in the data sheet for the TL071, and the source of the raw output resistance figure is the equivalent circuit schematic in the data sheet.
i know this is a long post, and i hope i didn't mangle the math too bad, but i'm trying to illustrate how to plug the variables in to the equations and get some useful results, and to not ignore the "gotchas" like Xc as part of the math. otherwise, i could say that the output impedance was 0.01 ohms at 20 hz, and if someone measured it with the output cap in place someone might say "WRONG! it's almost 800 ohms!!"
btw AG, i saw your post right after i posted this "book". not really, only 2 of the caps might have a "cascade" effect, the input cap turnover frequency is below 1 hz.
Unclejed, you have ignored the fact that capacitive reactance is phase shifted by 90 degrees from resistance. From your example, the magnitude of the impedance of 1k in series with 10µF, at 20Hz, is
Z=√(1k^2+796^2)
Z=1.28k.
I am not sure if this method has already been mentioned or not so i'll be brief...
The output impedance can be viewed as a resistor in series with a perfect source voltage. The idea is to load the output with another resistor that causes the output voltage to fall to some value and then using this information calculate the value of the series resistance and take that as the output resistance.
For the input impedance, we can apply an AC source voltage (appropriate for the amplifier) and knowing the series resistance of the source measure the voltage on the input and calculate the value of the resistance that would cause such a voltage drop and take that as the input resistance.
The only caveat is that some outputs do not accept a load from the output directly to ground, so in some cases you have to use a very very large value capacitor to act as a DC isolator so only the AC signal is measured.
Of course you also can not load the output too heavily.
Unclejed, you have ignored the fact that capacitive reactance is phase shifted by 90 degrees from resistance. From your example, the magnitude of the impedance of 1k in series with 10µF, at 20Hz, is
Z=√(1k^2+796^2)
Z=1.28k.
I am not sure if this method has already been mentioned or not so i'll be brief...
The output impedance can be viewed as a resistor in series with a perfect source voltage. The idea is to load the output with another resistor that causes the output voltage to fall to some value and then using this information calculate the value of the series resistance and take that as the output resistance.
For the input impedance, we can apply an AC source voltage (appropriate for the amplifier) and knowing the series resistance of the source measure the voltage on the input and calculate the value of the resistance that would cause such a voltage drop and take that as the input resistance.
The only caveat is that some outputs do not accept a load from the output directly to ground, so in some cases you have to use a very very large value capacitor to act as a DC isolator so only the AC signal is measured.
Of course you also can not load the output too heavily.
i think that method was mentioned in one form or another. the source impedance and the load impedance act as a voltage divider. i think i mentioned that as the "indirect method" of measuring input or output impedance.
i forgot to mention that with the op amp, even though the output impedance may be 10 milliohms for instance, you still have an effect from the real output resistance, and that is the reduction of output voltage swing across that real resistance. if you loaded the output of the op amp with a 160 ohm resistor, the output voltage swing is to half of the rails. within that range, the op amp will essentially behave as if it has an output impedance of 0.01 ohm, because internally the op amp will do whatever is required to maintain balance across the input terminals, but when heavily loaded, the real output resistance limits the available output current and total voltage swing..
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