Re: Need help with active filters ???
- From: Tim Wescott <tim@xxxxxxxxxxxxxxxx>
- Date: Thu, 31 Jul 2008 16:25:29 -0700
jalbers@xxxxxxx wrote:
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?
Either the author is making too many assumptions or you're not reading things carefully enough.
These days op-amps are nearly (or entirely) self-compensated, which means that you can just implement the "feedback resistor input resistor have a nice day" circuit and it will be stable. In order to accomplish this the op-amp is made so that it (more or less) acts like an integrator. When you wrap the op-amp with resistors to get a finite gain, the resistors only determine the circuit gain as long as the op-amp gain by itself is well in excess of the design circuit gain.
The bandwidth of the circuit that you achieve depends on the gain you designed in and the gain-bandwidth product of the op-amp. So if you have a gain of 100, and an op-amp with a gain-bandwidth product of 1MHz, your circuit power gain will be down by a factor of two at 10kHz, and your circuit voltage gain will be unity at 1MHz.
The reason I say that the author (or you) isn't being careful is that there is a wide variety of op-amps available with a wide variety of gain bandwidth products: you can get fast, expensive, power-hungry op-amps with GBW products into the GHz, you can get slow, expensive, power-miserly op-amps with GBW products down in the 10's of kHz, and you can get jelly-bean, old-technology, inexpensive op-amps with GBW products around 1-10MHz. They are _not_ all 1MHz parts!
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.
Op-amps that have bipolar transistors in their input circuits require a bit of current to operate. This bias current must be supplied by the external circuit, and the bias current that is supplied to the - input terminal through Ra and Rf makes it's voltage a bit lower than it would be otherwise. The resistor to the + terminal will have the same drop if you make it equal to that magic parallel combination, and will minimize the problem (you'll still have some offset voltage due to bias current offsets and resistor mismatch, if accuracy is very important you need to take this into account in your design).
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.
What do you want the filter to do? It'll attenuate higher frequency signals because the capacitor will shunt that current to ground, but maybe you _want_ the DC gain to be higher than one. (Note that in a non-inverting configuration you can make the input resistor -- your Ra -- equal to infinity by leaving it out. Then you'll have a voltage follower with a gain of 1).
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?
In this model the op-amp acts like a 'perfect' voltage source in series with a resistor. The higher the output resistance, the more that changes in the output current will yank the output voltage around.
An op-amp by itself generally has output impedances in the hundreds of ohms; unless you've got a really high-gain circuit the output impedance (at least at DC) should be in the tens or singleton ohms, but that depends a lot on the op-amp and the circuit.
Any help would be greatly appreciated. Thanks
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
.
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