# Op Amp Bode Plot 3db Point

#### SamR

Joined Mar 19, 2019
4,492
Continuing on with the TI Op Amp Handbook, I came across this.

But, I believe it is incorrect. Corrected it should be. Am I correct? Not sure I have a good handle on this 3dB point.

Also, I know what "Unity Gain" is but what is this (XI)?

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#### crutschow

Joined Mar 14, 2008
31,128
Not sure why it's called a 3dB error since that's the normal rolloff of a 1-pole LP filter, but the rolloff is indeed -3dB at the corner frequency.

X1 just appears to be an arbitrary label at the unity gain point.

#### ericgibbs

Joined Jan 29, 2010
16,803
Hi Sam
It is a poorly drawn Graph.
The Lowest 20dB Y scale height is well under the height of the upper scaling interval
Also, the scaling Y scaling interval is too short, compare it with the supposed 20dB scale range point.
E

#### SamR

Joined Mar 19, 2019
4,492
K, thx guys for the input. It wasn't quite making sense and I went and read about the 3dB point but was referring only to filters and not Op Amps so needed some assurance I had the idea right. I looked at the plot for the TL071 I've been using for my breadboarding and have another question about the crossover. This and other plots from the TI Hbk use |A|, absolute gain. But from the TI Hbk and PDFs I start seeing the line at the crossing point continuing on in what would be negative gain. I assume this is for both inverting and noninverting input?

So, this is telling me @ -20dB gain (inverting) the max frequency is higher than Unity Gain which is higher than 20dB gain roll-off (non-inverting)? Since frequency implies a sine wave and unless the zero is shifted then the frequency is actually limited by the 20dB line? I don't understand the line going negative from the crossing point. What are they trying to tell me here?

#### SamR

Joined Mar 19, 2019
4,492
K, I know about the phase shifting near the frequency limit. I was surprised to see the graph going on down to -30dB. Lots to learn still.

#### crutschow

Joined Mar 14, 2008
31,128
I don't understand the line going negative from the crossing point. What are they trying to tell me here?
It's just showing that the gain keeps rolling off below the gain of 1 point.
A negative dB is a gain of <1.

An interesting point is that an inverting gain of one is equal to a non-inverting gain of two for the op amp bandwidth.
That's because the circuit is the same for both circuits, just the location of the circuit signal ground and input are interchanged.

#### SamR

Joined Mar 19, 2019
4,492
just the location of the circuit signal ground and input are interchanged.
Yes, I did find it curious, but not enough to want to dig into the Op Amps internal circuitry to explain why. I'm afraid that is still a bit of a reach for me. Still need to more work on transistors as yet.

#### crutschow

Joined Mar 14, 2008
31,128
but not enough to want to dig into the Op Amps internal circuitry to explain why
The internal op amp circuitry (other than the capacitor that provides the 20dB/decade rolloff) has nothing to do with the change in bandwidth between the inverting and non-inverting configuration.
An ideal op amp would work the same.

#### SamR

Joined Mar 19, 2019
4,492
A negative dB is a gain of <1.
I missed that bit first time through. That is something I had not even considered. I assume that occurs when R1 is greater than R2? Not sure that I can fathom why that would be necessary. Especially considering the low voltages and currents that Op Amps typically deal with. Not only an amplifier but an attenuator as well. Something I should breadboard and think about a bit... Thanks for the heads up!

#### ericgibbs

Joined Jan 29, 2010
16,803
hi Sam,
One common example would be to use the high impedance Non Inverting input of a OPA as say a buffer

eg: R1 =10k and R2 =5k, actual Gain =1

E
As covered by Carl in Post #7
An interesting point is that an inverting gain of one is equal to a non-inverting gain of two for the op amp bandwidth.

#### sparky 1

Joined Nov 3, 2018
732

My current source for single op amp connected to an NE5532 Low Pass filter.
The power supply is set at 13.29 VDC the output of the variable current source is 12VDC a little less than the PS,
the amplifier is sinking 4.14 mA no load. When I adjust the function generator and dummy load,
I will call that the reference point which is 100%, the -3 dB point will be roughly 70% magnitude or 30 dB down.

It may be assumed that you are familiar with TI filterpro.
This one has been very useful. Hopefully that will make it clear.
https://webench.ti.com/filter-design-tool/filter-response

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#### SamR

Joined Mar 19, 2019
4,492
Not actually doing OPA filters as yet, still learning the basics with the TI Handbook. Thx

#### ericgibbs

Joined Jan 29, 2010
16,803
hi Sam.
If you don't have these PDF's in your database, they could be helpful.
E

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#### SamR

Joined Mar 19, 2019
4,492
Yes, I already have them Eric. Planning to run through the Analog Devices one using something other than their obsolete OPAs it uses next. I've used the other one to expand on some of the PDF items I was unsure of. Good stuff, both of them! Lately I've been using the TL081 or TL071 OPAs as a general Op Amp for breadboarding. I started breadboarding to look at attenuation this morning but haven't gotten very far with it yet. So far, I'm not quite halfway with the TI Handbook I'm working with now. Lots on my todo list!

#### SamR

Joined Mar 19, 2019
4,492
Set up a standard noninverting OPA circuit with R1=R2.

EDIT: Then kept increasing R1 to attempt to get attenuation below the input.

Never achieved attenuation below input! So, is this OPA phase shift protected? Is attenuation not possible? The Bode plots show the plot for negative gain so what am I missing here?

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#### bertus

Joined Apr 5, 2008
22,120
Hello,

When R1 is infinite, the circuit is basicaly a buffer.

Bertus

#### ericgibbs

Joined Jan 29, 2010
16,803
hen kept increasing R2 to attempt to get attenuation below the input.
Hi Sam.
R2 is the feedback resistor Gain = Rfb/Rin
E

#### SamR

Joined Mar 19, 2019
4,492
So how do you drive it to attenuation? The OPA already has "Infinite" input impedance, so... Ah crap... I got it backwards again... I was increasing R1, with R2 kept at 10k.

#### eetech00

Joined Jun 8, 2013
3,418
Set up a standard noninverting OPA circuit with R1=R2.
View attachment 257657
View attachment 257658
EDIT: Then kept increasing R1 to attempt to get attenuation below the input.
View attachment 257659View attachment 257660View attachment 257662View attachment 257663View attachment 257664View attachment 257665
Never achieved attenuation below input! So, is this OPA phase shift protected? Is attenuation not possible? The Bode plots show the plot for negative gain so what am I missing here?
Remember the Gain equation for a non-inverting amplifier:

Gain=1+(R2/R1)
Where R2= Feeback resistor

Vin=1Vpp
Gain = 1+(10k/10k)=2
Vout=1Vpp*Gain=2Vpp
Second example:
Gain =1+(10k/30k)=1.333
Vout=1Vpp*Gain=1.333Vpp

And so on....

Or for a non-inverting buffer:
Gain = 1+(0/infinity)=1 100% feeback (or short circuit between -pin and out pin)

If you do this math you will see that the gain can never be less than 1.

Edit: fixed equation above.

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