Need help with active filters ???

[jalbers@bsu.edu]

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?

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.

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.

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?

Any help would be greatly appreciated. Thanks

 

[Michael Black]

On Thu, 31 Jul 2008, jalbers@bsu.edu 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?

Check the copyright on the book. Likely it's old. And once upon a time,

op-amps had pretty limited specs, A 1MHz gain-bandwidth was pretty good,
yet once you try to get gain from it you don't get much actual bandwidth.
You could use them fine if you just connected them as voltage followers,
ie no gain, but start trying to get much gain from them and the bandwidth
goes away. It was misleading 35 years ago, for people first playing
with op-amps and seeing only that "1MHz" figure.

I bet that's the limitation, the book is old and the op-amps current
to the book had limited gain-bandwidth.

More recent op-amps have much better gain-bandwidth, and thus can
supply more gain up to a higher frequency.

Michael

 

[John Popelish]

jalbers@bsu.edu 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?

Feedback controls the overall gain by throwing away the
excess gain the opamp has above what the feedback calls for.
Those rules of thumb assume a particular programmed gain
and a maximum opamp frequency capability. Most opamps have
a very high gain at DC and that gain rolls off, above a few
hertz, in proportion to the frequency, reaching a gain of
only 1 at about 1 MHz. Such an opamp can have no more gain
than about 10 at a frequency of 100kHz, no more than 100 at
10kHz, no more than 1000 at 1kHz, etc. So if they are
saying that the gain of the feedback controlled opamp is
constant up to about 10kHz and falls to 1 at 1 MHz, they are
implying a closed loop (programmed) gain of about 100.

The simple way to capture this gain versus frequency is
called gain bandwidth (gain times frequency), since the
product of open loop (not programmed by feedback) gain and
frequency is roughly constant over the whole roll off range
from a fet hertz to the unity gain frequency (the frequency
where the open loop gain has fallen to 1).

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 opamp were perfect (infinite input resistance, which
means the inputs draw zero current) this would make no
difference. But many practical opamps, while drawing low
current through their inputs, do not draw zero. Matching
the total parallel resistance at both inputs produces
matching voltage drops across those two resistances, to the
extent that the two inputs draw similar currents. This
allows the voltage produced across one input's resistance to
be canceled, at least mostly, by the voltage produced by the
current passed by other input passing through its
resistance. This technique improves the DC accuracy of the
closed loop response.

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.

Attenuation is always in comparison to the non attenuated
part of the response (the pass band). Since overall gain
increases both, the attenuation ratio remains unchanged.
Besides, if Rf is zero and Ra is infinity, the amplifier
gain is 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?

That means that, neglecting any effect of an external
feedback network, the opamp output voltage will be altered
by the current passing through the output. A 75 ohm output
resistance means that an open loop output voltage will be
changed by a volt if you draw 1/75th of an ampere through
the output. If you connect negative feedback around the
amplifier, this output voltage change is reduced by the
voltage gain of the amplifier. For instance, if you
connected the opamp as a follower (gain of 1) and loaded the
output with the same 1/75th amp, and the opamp had a DC gain
of 100,000, then the output would have to shift only
100,000th of a volt before that small voltage multiplied by
the DC gain produced the internal 1 volt current caused
swing upstream of the 75 ohm output resistance, shifting the
distortion to inside the closed loop, and reducing the
output voltage sag by a factor of 100,000.

--
Regards,

John Popelish

 

[Tim Wescott]

jalbers@bsu.edu 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