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Opamp gain question.

post #1 of 17
Thread Starter 
I know an opamp can be configured for a gain of 1 or -1 if it's unity gain stable.

But is it possible to configure it for a gain of less than unity?

ie. -0.01 gain in an inverting circuit.
post #2 of 17
Quote:
is it possible to configure it for a gain of less than unity?
It's possible to get inverting attenuation, simply by making the feedback resistor smaller than the input resistor. Beware, though, a 100x division is going to cut your bandwidth by 100x, just as a 100x gain would.

EDIT: Noninverting attenuation isn't possible. Think about it: as the feedback resistor gets smaller, all you do is get closer to the ideal buffer configuration (g=1): a short from OUT to -IN. It doesn't matter what the resistor from -IN to ground is doing once the feedback resistor becomes small enough.

A far simpler method, and one that retains bandwidth and doesn't make you go inverting, is to put a divider in front of an op-amp configured as a buffer.
post #3 of 17
Yes, in an inverting circuit, the gain is Rf/Rin, so pick your resistors appropriately. So, for example, if Rin = 10Kohm and Rf = 1Kohm, G = -0.1.

However...a unity gain stable opamp isn't necessarily going to be stable at less than unity gain. If you're after an attenuation circuit, why not do it with a non-inverting configuration instead. With a unity gain stable opamp, set the feedback loop resistance to zero and use the output of a voltage divider to the noninverting input of the opamp. The gain will be the attenuation factor of the voltage divider.

You can do something similar with an inverting circuit, but it requires twice as many resistors and a little math:



Set the base gain for the opamp where Rf = Rin1 + Rin2. Then choose R where R = Rf(Gain/(2-2(Gain)), where Gain is the desired attenuation.

For example, if you want a gain of 0.01, and Rf = 10K, then R = 10E4*0.01/(2-2(0.01)) = 50 ohms.

-Drew
post #4 of 17
Thread Starter 
Quote:
Originally Posted by tangent
It's possible to get inverting attenuation, simply by making the feedback resistor smaller than the input resistor. Beware, though, a 100x division is going to cut your bandwidth by 100x, just as a 100x gain would.

EDIT: Noninverting attenuation isn't possible. Think about it: as the feedback resistor gets smaller, all you do is get closer to the ideal buffer configuration (g=1): a short from OUT to -IN. It doesn't matter what the resistor from -IN to ground is doing once the feedback resistor becomes small enough.

A far simpler method, and one that retains bandwidth and doesn't make you go inverting, is to put a divider in front of an op-amp configured as a buffer.
Yeah, I understand that non-inverting attenuation isn't possible.. Was interested in whether inverting attenuation is possible..

Thanks for the help. =)
post #5 of 17
you're getting some incorrect advice here, inverting atten will work but stability and bandwidth are determined by the "noise gain" of the circuit

"noise gain" is the gain seen by a V source in series with the +,- inputs; essentially the the gain of the circuit considered as a noninverting amplifier

a -1 inverting amplifier has a noise gain of 2 and only 1/2 the bandwidth of a +1 gain follower

in the "inverting attenuating" configration the noise gain tends toward 1 in the limit as the attenuation increases and bandwidth correspondingly increases to the unity gain bandwidth of the op amp
post #6 of 17
Thread Starter 
Hrmm... ok.. Not that I fully understood what you said.. LOL

I just want to make a 'de-amplifier' for headphones..

At my listening levels (based on calculations it's around 76dB in an air-conditioned room, up to about 82dB if the windows and doors are opened and the noise levels increase), I barely even need to pump 250mV p-p into 300 ohms headphones..

So I was just wondering about the feasibility of making a 2 inverting stage "amplifier" with a total gain of 0.25. That would give a maximum of 500mV rms into headphones with a 2v rms input.
I should be able to hit the sweet spot on my pots. Currently can't even go past 10 o'clock when the minimum is 7 o'clock on my amp with a gain of 3.
post #7 of 17
Quote:
Originally Posted by tangent
...Beware, though, a 100x division is going to cut your bandwidth by 100x, just as a 100x gain would.
Are you sure about that? A Bode plot will claim the opposite.

0.1 in gain will INCREASE the bandwidth compared to gain of 1. Yes?
post #8 of 17
Quote:
Originally Posted by peranders
Are you sure about that?
Nope. SPICE lied to me. Pretty ironic, considering my signature.

Let that be a lesson to y'all: don't believe SPICE unless you know the answer already. I've preached that line before, but I got caught just the same.
post #9 of 17
Quote:
Originally Posted by Dreamslacker
I should be able to hit the sweet spot on my pots. Currently can't even go past 10 o'clock when the minimum is 7 o'clock on my amp with a gain of 3.
wouldnt it be easier to swap in a lower value pot?
post #10 of 17
Thread Starter 
Quote:
Originally Posted by rellik
wouldnt it be easier to swap in a lower value pot?

No.. I actually need the international math community to re-define the log curve so I can push any log-pot up further...
post #11 of 17
How about adding attenuation before or after the pot. Not as much fun, but it should get you closer to the POT setting you're interested in...
post #12 of 17
Quote:
How about adding attenuation before or after the pot. Not as much fun, but it should get you closer to the POT setting you're interested in...
exactly !

why overcomplicate the simple ?

By adding a sinple single resistor inline before the volume control (or even a variable inline pad) you bang down the apparent or "effective' gain while retaining the headroom and bandwidth of the active stage.Passive solutions are your freind and more often than not a simple cap or resistor is all that is required for a seemingly complex job.
post #13 of 17
Quote:
Originally Posted by Dreamslacker
I actually need the international math community to re-define the log curve
If you've read this you know you can define that curve almost any way you like. Rick's given you a pretty good idea of where to start.
post #14 of 17
Thread Starter 
Quote:
Originally Posted by tangent
If you've read this you know you can define that curve almost any way you like. Rick's given you a pretty good idea of where to start.
It's more fun that way.. LOL.. I already know how to use a resistor ladder but I want to try making something different.. Like a de-amplifier...
post #15 of 17
if you want the "fun way" over the simple then check this little program :

http://focus.ti.com/analog/docs/samp...igationId=9742
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