I have a question about the magnitude of the two feedback resistors on opamps. I am familiar with how to determine the ratio of the resistors to get the desired gain, whether inverting or not. What I'm trying to figure out is what determines the magnitude of the resistors involved? I've seen different circuits with resistors of, say thousands, tens or hundered of thousands and even one that used Megohm-range resistors. Why the variation in magnitude of the resistors, assuming a fixed ratio? Does it have anything to do with the overall impedance of the opamp circuit block? I thank you for whatever illumination that can be provided.
** Choosing a particular resistor value for a given spot in a circuit is a major part of electronics design - so your Q is a mile wide and goes way beyond just op-amps.
I think 10k ohm resistors are my favorite. (then 1 k ohm) In general I'd like to choose R's that are smaller. First this tends to reduce the effects of stray capacitance. (Some RC corner that move to higher frequency with smaller R) And then if you care, smaller R has less noise.. though most of the time that doesn't matter much. Now of course you can't make 'em too small. First the opamp mya not have enough poop to drive it. and second if the R's get very small (a few ohms) you then start to worry about the resistance of the traces and contacts.
Thank you George. I've been following that as a general rule of thumb (staying in the huuped singled K and the general 10's of Kohm range), but I was seeing circuits that had other orders of magnatude and I was just starting to wonder why so high, in particular with two of three circuits I saw that just struck me as a wee bid odd..
At reasonable gains (well below the gain-bandwidth product of the OpAmp), input impedance matters not at all. But...
(1) What can matter in bipolar OpAmps is input bias current, which can cause an effective offset voltage if resistors are too large.
(2) Also, with very large feedback resistors, the stray capacitances of the circuit can cause undesirable frequency roll-off.
(3) Too small valued resistors will tax the current output capability of the OpAmp. ...Jim Thompson
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| James E.Thompson | mens |
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I love to cook with wine. Sometimes I even put it in the food.
Capacitance is very important as well, both input and feedback.
Op amps always have at least 1.5 pF of input capacitance, and usually more than that. Say you have a unity gain inverter with 1 meg input and feedback resistors and Cin = 3pF.
Your feedback network will roll off starting at
f_c = 1/(2 pi (3pF) (500k)) = 106 kHz.
That means that the output swing will start to rise there, to keep the input in balance. So you wind up with a gain peak. If you had picked
10k instead of 1 meg, the peak would start 100 times higher in frequency, i.e. 10.6 MHz, which is liable to be outside the amplifier bandwidth, so you wouldn't see it unless the op amp were reasonably fast.
The shunt capacitance of the feedback resistor is also important, though it doesn't really matter for unity gain inverters, since the effect is about the same on each one. It's typically about 0.12 pF for a 1/4 W axial resistor and 0.05 pF for an 0603. (You also have to watch out for the capacitance between pads--a nearby ground plane is your friend here.)
However, if you had an amp with a gain of -100, i.e. 1 meg on the input and 100 meg feedback, that 0.05 pF starts to dominate the feedback at
f_RC = 1/(2 pi (50 fF) 100 meg) = 32 kHz.
Layout problems will make this worse.
Adjusting the feedback capacitance to cancel out the effect of the input capacitance is a typical way of controlling these problems when you can't just use lower impedances.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
Note that with the non-inverting (+) input at 0V, then Vout must assume whatever value is required to force the - input to whatever voltage is on the + input; 0V.
If Vin is at 1V, and if R1 = R2, then we have a voltage divider that looks like this:
1V | [R1] | +---0V | [R2] | -1V
where -1V is the output of the opamp and 0V is the opamp's - input.
Accordingly, as the input voltage swings positive and negative, then the output voltage will swing with the same magnitude as the input, but 180 degrees out of phase.
The circuit, then, has a voltage gain of -1.
Assuming, now, that the signal generator driving R1 has an output impedance of 10kohms and that R1 and R2 are each 10k, our circuit now looks like this: . SIGNAL 10k . GENERATOR +--[R2]--+ . +-----------+ 10k | | . | +--[Rg]--|--[R1]-+--|-\ | . | | 10k | | >--+-->Vout . | OSC | +--|+/ . +-----------+ | . GND
If the signal generator has an open-circuit output of 1 volt, when it's connected to R1 (since the end of R1 connected to the minus input of the opamp will be at 0V) the current through RgR1 will be:
E 1V I = --------- = ------ = 50 microamps Rg + R1 20kR
Now, since Rg, R1, and R2 are in series and the current into the - input of the opamp is miniscule, the current through Rg and R1 also flows through R2.
Then, since R2 = 10kR and the current through it is 50 microamps, the voltage across it must be:
Therefore, even though R1 = R2, the gain of the [entire] circuit won't be -1 because of the effect of R1 loading the source.
Knowing that, and taking the source impedance into consideration, the resistances of R1 and R2 can be selected so that the opamp's output voltage will be whatever's desired.
The resistor selection criteria for the non-inverting amp are arguably less stringent with, roughly, not starving the opamp - input for bias current on one end and minimizing the noise into the opamp on the other.
There's also noise to worry about, of course. Your garden variety op amp has an input noise voltage of very roughly 10 nV/sqrt(Hz), which is about the same as the thermal (Johnson) noise of a 6k resistor. So going too high will cost you voltage noise performance.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
John Fields, thank you very, very much for that thoughtful explanation, it covers exactly what I was wanting to know but didn't quite have the knowledge to ask correctly.
I had thought I understood op amps thoroughly, having used them in some light-weight audio level circuitry in the past. I'm trying to thoroughly understand them as I try to understand everything else, before I embark on designing anything substantial. The most I've managed to design and build from the ground up so far was a power supply with current fold-back, from discrete parts, but it was a seat of the pants design. I want to understand what's really happening in a circuit so I can understand more or less what to expect and understand why I expect it before I ever breadboard anything.
And thank you for treating a basic problem from a first time asker for what it was. Gone Postal
The other nice thing about 10 k ohm, is you can put a 1/4W 10k across the +/- 15V rail and *not* let the magic smoke out. And 10 k is very close the the "quantum" of conductance.
2*e**2/h =~ 1/12.9k ohm) Coincidence? I don't think so. I think God also likes ~10k ohm :^) (please excuse the extreme hubris in the above.)
Maybe that's the Imperial unit of resistance. One attoparsec equals one decifoot, to within observational error (at least mine), which proves once again (if any further proof were needed) that God is an Englishman. ;)
(Of course those sillies have gone all metric on us, but I digress.)
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
They tried metric on all the highway signs around here, but their use has gradually faded away... might have been all the bullet holes they collected >:-} ...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
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