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Help me understand common mode input range
Old 21st September 2012
  #1
Gear Addict
 
🎧 10 years
Help me understand common mode input range

I'm trying to work out how much input signal voltage a piece of equipment will handle. The front end is an OPA1632 differential op amp.

There is a specification listed as "Common mode input range" and is specified as minimum (V-) + 1.5v and maximum (V+) - 1.0v.

Does this refer to the signal voltage presented to each input?

Does it also tell me that the signal could clip asymmetrically? eg if the input signal swung from (V-) + 1.0 to (V+) -1.0, the signal would be unclipped at the top but clipped at the bottom?

I've checked a few online articles, but I'm still unclear. Can any shed some further light?

Paul
Old 21st September 2012
  #2
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tINY's Avatar
 
1 Review written
🎧 15 years

That is the max/min voltage you can present to the inputs of the op-amp...

Remember that most op-amp circuits have circuitry in front of the inputs that limits voltage applied to the inputs. If your Inputs are seeing less than 3v of headroom, you should probably change the circuit topology.

Divided feedback : OPERATIONAL AMPLIFIERS

Read this if it's not already familiar to you.




-tINY

Old 21st September 2012
  #3
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JohnRoberts's Avatar
 
🎧 10 years
Quote:
Originally Posted by PRH ➑️
I'm trying to work out how much input signal voltage a piece of equipment will handle. The front end is an OPA1632 differential op amp.

There is a specification listed as "Common mode input range" and is specified as minimum (V-) + 1.5v and maximum (V+) - 1.0v.

Does this refer to the signal voltage presented to each input?

Does it also tell me that the signal could clip asymmetrically? eg if the input signal swung from (V-) + 1.0 to (V+) -1.0, the signal would be unclipped at the top but clipped at the bottom?

I've checked a few online articles, but I'm still unclear. Can any shed some further light?

Paul
+1


"Common mode" refers to the standard behavior of opamps where they generally respond to the differences between the two inputs, and ignore the same voltage at both (within the valid operating voltage range).

JR
Old 21st September 2012 | Show parent
  #4
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🎧 10 years
Quote:
Originally Posted by tINY ➑️
If your Inputs are seeing less than 3v of headroom, you should probably change the circuit topology.
tINY - I think you misread the OP here. The input range depends on voltage rails so if he has +/- 15v rails his input range is from (-15)+1.5 = -13.5v to (15)-1.0 = +14v so he would have 27.5v, not 2.5v
Old 21st September 2012 | Show parent
  #5
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🎧 10 years
Quote:
Originally Posted by mjrippe ➑️
tINY - I think you misread the OP here. The input range depends on voltage rails so if he has +/- 15v rails his input range is from (-15)+1.5 = -13.5v to (15)-1.0 = +14v so he would have 27.5v, not 2.5v
Or maybe you misread Tiny... Opamps will generally have feedback networks that reduce the actual voltage swing at the opamp inputs.

Tiny's comment may be TMI for the OP but I suspect he is suggesting that good practice keeps input voltage swing to a modest fraction of full rail voltage.

Only for a simple non-inverting unity gain follower will the inputs swing 100% of the signal input.

JR
Old 21st September 2012
  #6
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tINY's Avatar
 
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🎧 15 years

Remember that, when operating in the linear mode, the voltage on one input is the voltage on the other input of an op-amp.

For a single-ended, inverting circuit, you will typically set the DC bias point using the non-inverting input. It is tied to a voltage (ground or virtual ground, usually) though a resistance picked to equalize the bias current and keep the signal accurate. The output of the op-amp drives the feedback so that the inverting input is at your DC bias point.

Common mode range on op-amps is about where you can set your DC bias point when using them in linear circuits.




-tINY

Old 21st September 2012 | Show parent
  #7
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🎧 10 years
Quote:
Originally Posted by tINY ➑️

Remember that, when operating in the linear mode, the voltage on one input is the voltage on the other input of an op-amp.

For a single-ended, inverting circuit, you will typically set the DC bias point using the non-inverting input. It is tied to a voltage (ground or virtual ground, usually) though a resistance picked to equalize the bias current and keep the signal accurate. The output of the op-amp drives the feedback so that the inverting input is at your DC bias point.

Common mode range on op-amps is about where you can set your DC bias point when using them in linear circuits.



-tINY

I know you know this, but to be more precise, it's the opamp output connected through a negative feedback network that makes the - input follow the + input.

CM range defines how close to the rails the two inputs can swing, and the opamp output will still move in the right direction (or move at all) based on the difference between + and - inputs.

JR
Old 21st September 2012
  #8
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tINY's Avatar
 
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🎧 15 years


I always understood "Common mode input range" to refer to the limitations of the front end. However, in most audio circuits, it's a distinction without a real difference. Who would put an output bias anywhere besides the center of the rails... Though I suppose there may be some die-hards trying to eliminate caps that try DC-coupled designs.



-tINY

Old 21st September 2012
  #9
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🎧 15 years
Hi
Both inputs sit at about 4 Volts 'off centre' on some of the transistor assisted, transformerless mic amp designs. As the inputs are essentially the same DC level the output will sit at 0 Volts as the 'difference' is zero.
Common mode range is, as mentioned above the DC voltage that can exist on BOTH inputs which result in the op amp behaving correctly, and is a volt or so lower than the supply rails.
The actual voltages that appear will of course be determined by the overall circuit.
Matt S
Old 22nd September 2012
  #10
Gear Addict
 
🎧 10 years
Thanks all.

The op-amp in question is a fully differential in and out op-amp and there are no + and - inputs per se.

I understand that the common mode voltage is voltage which is common to both sides of the differential signal, and can be DC (as in an offset) or AC (eg noise common to both "sides").

The opamp in question is the analogue front end on an audio AD evulation board. The circuit does signal attenuation and filtering, and passes a differential signal the the ADC.

I know how much attenuation there is, and what what input signal voltage corresponds to full scale input at the ADC, but I am interested to work out how great an input can be applied to the actual op-amp (ignoring, for this discussion, the FS input to the ADC).

Does this clarify my question any further?

Cheers! Paul
Old 22nd September 2012
  #11
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🎧 10 years
Sure... Look at how much the feedback network attenuates the input signal. Then divide the max CM range by that attenuation factor.

JR

PS: It still has + and - inputs
Old 22nd September 2012
  #12
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tINY's Avatar
 
1 Review written
🎧 15 years

Is this a "single op-amp differential amp" circuit?

Building a differential amplifier : OPERATIONAL AMPLIFIERS

You need to look a the voltage on the non-inverting input. Find the "Rrrs" and work out the math for min/max input voltage there.




-tINY

Old 22nd September 2012 | Show parent
  #13
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🎧 10 years
Quote:
Originally Posted by tINY ➑️

Is this a "single op-amp differential amp" circuit?

It is a monolithic differential in and out op-amp. Datasheet: OPA1632

I've located an app note from TI which I couldn't seem to find previously. I think I have a better idea of what is going on.

And yes, JR, the op-amp does of course have + and - inputs, but I initially found the differential architecture confusing

Paul
Old 22nd September 2012
  #14
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tINY's Avatar
 
1 Review written
🎧 15 years

Well, the problem is, TI chose a really bad name for this device.... It's not an op-amp in the traditional sense. Interesting idea.

If you read the app note, starting at page 8, you'll probably find what you need.


-tINY

http://www.ti.com/lit/an/sloa054d/sloa054d.pdf
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