# Op-Amp design specification for a specific signal condition

#### Junus2012

Hello,

I need to amplify a small sensor signal that have a maximum signal level of 5 mV and bandwidth of up to 100 kHz. The gain accuracy should be 1%.

I want to design an opamp using the 180 nm technology powered by single supply voltage of 1.8 V. The opamp is then driving the fully differential input ADC.

I started like this

To cover the full-scale window of the ADC, the opamp should have a closed loop gain of 1.8/5 mV = 50 dB.
To have a gain accuracy of 1% I should have a loop gain that is 20 db (100) above the 50 dB for the complete bandwidth of the input signal, that is 70 dB at least.

However, since the opamp gain roll off with frequency, the gain-bandwidth product should be more important to consider.

Please help me to determine the required GBW of the opamp to condition this signal and also the required slew rate

Thank you

Regards

Hi,

I don't want to offend you with basic stuff as I know you are an IC designer but maybe one of the links will help to solve the puzzle... The links talk about slew rate formula and also include calculators, and the other two about calculating bandwidth, perhaps Op-Amp Gain Bandwidth Product is a good one to check first - look about halfway down and after this: 'GBP = ACL x fCL', read the third paragraph (it's pretty close to what you are calculating, I believe): 'For example, with a required gain of 40dB (100), if the op-amp needs to be operated at 100 kHz, then the Gain Bandwidth Product of the op-amp should be at least 100 x 10^5 = 10^7 = 10 MHz.'

From my experiences of learning to select op amps and making the typical beginner mistake again and again of thinking advertised GBW of 1 MHz means the OA can be used at 1 MHz, but then the datasheet shows it's actually more like 50 kHz or (much) worse, I have my rough rule-of-thumb that GBW needs to be about at least 50 to 100 times greater than the frequency applied.
I also did an idiotic calculation of 1/100 kHz = 10us, and without factoring in your gain of 70dB, that would mean it could process a maximum of 10 kHz/us signals, which must be too slow for your needs, and really dodgy calculation was (1us as slew rate is usually V/us) 5mV/1us = 5,000 (whatever that might mean) - as neither of those are 'scientific' maths and I'm an ignoramus I'm sure they are numbers that are not much use to consider and your waste mental energy on exploring further.

One more thing, I'm not totally sure what 'constant gm' means, but I bet you do, and I wondered if an OA that will see a varying input frequency would by necessity require a constant gain block, no doubt your design already has that and I'm just showing my utter ignorance of IC design, again...

An Op Amp Gain Bandwidth Product

Slew Rate Calculator (and formula)

Op Amp Slew Rate Calculator and Formula

### Junus2012

Points: 2
Not sure how you calculated your accuracy,
but I would use the following.

You have Acl = 1.8/5e-3 = 360
So, B = 1/360 (feedback factor)

For Accuracy, we can use Acl_err ~ 1/(Ao*B)
so Ao >= 1/(Acl_err*B) >= 1/(.01/360) >= 36000 or 91dB
using 70dB Ao only gives gain 3162 or 11.4% error

If you use f3db = 100kHz, then you can easily solve GBW
(1/B)*f3db = GBW , (360*100e3) = 36MHz

Now, Slew Rate is dV/dt. dt is just Tau or 1/w3db = 1/(2*pi*100e3) = 1.59us
So SR >= 1.8V/1.59us >= 1.13V/us
easy to acheive because low BW.

*These are all theoretical starting points. You can simulate and add margins for practical purposes.

Last edited:

Points: 2

### d123

Points: 2
Hello friends,

Thank you for your help, let me focus on the GBW requirement and pause the SR parameter.

I have concluded the following

Since the ACL=AOL/(1+Beta. AOL),

where Beta.AOL = loop gain=LG

The LG defines the ACL accuracy, and hence if we want 1% accurate gain then LG=100 or 20 dB.

Because we can define the LG alternatively as LG=AOL-ACL and since the AOL frequency dependent and decreases with frequency, therefore the gain error increases at higher frequency.

Therefore f(input signal)*ACL <= 10 to 100 time less that GBW of the amplifier

Thank you once again and I welcome your discussion back to this point

Hello friends,

Thank you for your help, let me focus on the GBW requirement and pause the SR parameter.

I have concluded the following

Since the ACL=AOL/(1+Beta. AOL),

where Beta.AOL = loop gain=LG

The LG defines the ACL accuracy, and hence if we want 1% accurate gain then LG=100 or 20 dB.
I don't understand where some of this is coming from. Could you either show a more clear derivation with equations or references/sources?

For example, LG=100 or 20dB. As far as I know, voltage gain is 20*log10(vo/vi) so 20*log10(100) =40dB not 20 that is likely power gain. And why are you defining accuracy error by loop gain alone? Is there a reference for that? I usually use deltaGain/Actual Gain as the relative error. If you plug it back into Acl = Ao/(1+Ao*Beta), you should be able to derive exactly the %error between closed loop gain ideal and close loop gain actual (regardless of equation form).

### Junus2012

Points: 2
Sorry, I mean for 100 should be 40 dB.

I usually use deltaGain/Actual Gain as the relative error. If you plug it back into Acl = Ao/(1+Ao*Beta), you should be able to derive exactly the %error between closed loop gain ideal and close loop gain actual (regardless of equation form).
yes I have no doupt about this, but you know Ao is frequency dependent, so it it drops as the signal frequency increase which then reduce the loop gain. The loop gain is the distance between the closed loop gain and the Ao.
That is why we need to include the frequency factor here represented by the GBW to insure that we have the required loop gain for certain frequency.

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