Re: Need help with active filters ???
- From: "Dan Coby" <adcoby@xxxxxxxxxxxxx>
- Date: Thu, 31 Jul 2008 12:54:26 -0700
<jalbers@xxxxxxx> wrote in message
news:63046ac3-ad0a-46d8-9488-1667b82a171d@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I have been reading a book about filters both passive and active and
have a few questions about active filters that I need some help on. I
have a link to pages 109-115 containing the information that I have
questions over.
https://ilocker.bsu.edu/users/jalbers/WORLD_SHARED/Electronics/ActiveFilter.PDF
1. For non-inverting op amps the book says that "Typically gain
remains constant up to 10 Khz , then falls steadily to reach 1 at at 1
Mhz." Why is this so?
Real components have some maximum operating frequency. No amplifier
can have a constant gain that goes all the way to infinite frequencies. With
op amps, there are limits due to the technology used to make the IC.
Op amp are also usually (almost always) used with feedback components.
With any feedback circuit, there are also problems with keeping the circuit
stable (i.e. not oscillating). A common example of feedback oscillation is
when a microphone picks up a signal from a speaker, the signal is amplified
and then sent back to the speaker. If the gain (volume) is high enough then
you get a loud squeal as the same signal feedback into the microphone,
Circuits will oscillate anytime that the gain around the feedback path is greater
than one AND there is 180 degrees of phase shift in the feedback path.
(The feedback path includes everything from the inputs through the op amp
and then back through the feedback components - the resistors Rf and Ra
in your example).
At first glance, it seems like thing should be simple. All that you need to do is to
avoid having 180 degree phase shift in the feedback path. After all, the resistors
should not have a phase shift. So no problem - WRONG. If you go to a high
enough frequency, everything has a phase shift. If nothing else, the speed of
light will start to introduce phase shifts. At 1 GHz, you will get a 180 degree
phase shift with about 6 inches of wire.
Since you cannot avoid phase shifts at high frequencies, feedback amplifiers
are usually built with a gain that starts very high at low frequencies and then
the gain is dropped at higher frequencies. This gain control is built into the
typical op amp. Op amps have both a specified maximum gain and a
'gain bandwidth product'. The later indicates the frequency at which the gain
drops to unity. Your book referred to an op amp with a gain bandwith of 1 Mhz.
Typical modern op amps are better than that.
2. In the inverting configuration, why is the resistor connected to
non inverting terminal Rb connected to ground through a resistor
having the resistance value of Ra and Rf in parallel? The book makes
a statement that for less accuracy the non inverting terminal could be
connected to ground.
If the op amp was 'ideal' then it would not matter whether or not a
resistor is connected to the non inverting input. However with real op
amps (i.e. op amps that have a finite input impedance, finite gain, and
non zero offset currents and voltages) then it is better if both inputs to the
op amp have the same source impedance. Having the same source
impedance helps to balance errors caused by having a non ideal op
amp.
3. On the bottom of page 113 a first order active filter circuit is
described. It is basically a low pass RC filter connected to the
input of a non inverting op amp. As far as I can tell from the
formulas the gain of the op amp is 1+Rf/Ra which means that the gain
has to be larger than 1 which means that the op amp is amplifying the
signal instead of attenuating the signal? I am thinking that it would
be better to have a gain of less than 1 to attenuate the signal.
Yes, the circuit is a simple RC low pass filter followed by no inverting
amplifier. The input resistor and capacitor will attenuate the signal before it
gets to the amplifier so the overall result will have a gain which is less than
1 at higher frequency.
4. I have never been to clear as to what output impedance really
means. I can understand why op amps need to have a high input
impeadance so that they do not put too much of a drain on the circuit
that they are trying to measure. The book says that the output
impeadance of an op amp is low (around 75 ohms) for example. What
does the 75 ohms mean?
It means that if you use the op amp without any sort of feedback and you
then compare the output signal size between no load versus a 75 ohm
load then the signal will be 1/2 of the unloaded size when a 75 ohm load
is added.
However if you have feedback then things are different. Due to the feedback,
the amplifier will try to compensate for the voltage drop across the
output impedance. The overall effect is to drop the effective output
impedance by a factor of (1 + GH) where G is the op amp gain and H
is the feedback gain (ratio). (Note; H is 1/gain for a non inverting amplifier).
Thus with an op amp that has a 75 output impedance and a gain of 100,000
and a unity gain (i.e. H = 1) then the effective output impedance is 0.75 milliohms.
However there are a few complications. Do not expect that op amp to drive
a large signal into a 1 ohm load. That 75 ohms is still going to limit the maximum
output current. Likewise, as I mentioned earlier, op amps are intentionally designed
so that their gain drops as the frequency increases. So at higher frequencies,
the effective output impedance of a feedback circuit will go up as the gain
drops.
.
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