OPA858: The open loop gain doesn't meet the data in the datasheet

you might be confusing Aol = 0dB with GBP, normally you would project the Aol gain at 40dB to GBP multiplying that frequency by 100X, 5.4GHz here looks correct

These issues are described in this article, 

https://www.planetanalog.com/why-is-amplifier-gbp-so-confusing-insight-12/

Hi Michael,

Thanks for the feedback. I've read the article, but still a little bit confused.

Which one is the actual GB shall we use, @0dB or @40dB*100?

Do you mean we can trust the simulation result at 0db, the real GB is 3.2G, the reason why the datasheet shows is 5.4G is due to they use 40 dB to estimate? 

Or you mean the simulation model has defect, we can trust the data @40dB*100, It is 5.4G.

Well Ba, the Ti group that did those video's should perhaps answer this one

Hi Ba,

GBW product is defined only when Aol is rolling off at a rate of -20dB/decade. When a devices' Aol starts rolling off at a faster rate (due to additional higher order poles), we cannot use the GBW calculation and any result from that calculation is not valid. This is why Michael used a data point at 40dB, as this will ensure he is in the Aol region where it is rolling off at -20dB/decade. For most higher speed devices, higher order poles will influence the Aol by the time it is around 0dB, therefore we don't take the frequency at 0dB to calculate GBW and instead take the approach Michael took to get GBW for our devices. 

More information can be found in our TI Precision labs video series. The video series specific to this topic is called "Op amps bandwidth theory" and can be found here.

Best regards,

Ignacio

Hi Ignaccio,

Thanks for the feedback, this makes sense. Could you please help answer my second question?

TI Precision labs video series

Stability-->Spice simulation @ 3:04

https://www.ti.com/video/series/precision-labs/ti-precision-labs-op-amps.html

2.Why Aol_loaded=Vo/Vfb and 1/beta=1/Vfb?

For opAMP, (Vin+-Vin-)*Aol=Vo, Vin+=0V,Vin-=Vfb, so Aol=- Vo/Vfb.

And Beta=Vfb/Vo, so 1/beta=Vo/Vfb,

Please help to explain how do you get the equation in the training video.

Hi Ignaccio,

Thanks for the feedback. I understand that Beta will not include Vo as it is now open. But Vfb= the injected Vin*R1/(R1+RF), if Beta= R1/(R1+RF), Vfb= Vin*Beta.

Vfb=Beta only if  Vin=1? Why we consider Vin=1? Please fell free to correct me.

Best Regards,

Eric Ba

Hi Eric,

The beta term is a unitless factor, which would be Beta = Vfb/Vin = R1/(R1+Rf), which happens to be at the node we label Vfb.

Best regards,

Ignacio

Hi Ignacio，

Thanks for the feedback.

I understand that  Beta = Vfb/Vin = R1/(R1+Rf) ,as you said, is a unitless factor,

Now we can get Vfb= Vin*Beta. The question is how do we get Vfb=Beta.

Vfb=Beta only if  Vin=1? Why do we consider Vin=1?  Could you please help to clarify this?

Thanks.

Eric

Hi Eric,

Th ac simulation takes the transfer function with respect to where you are probing relative to the input signal. So, by probing at Vfb the simulation is actually giving you Vfb/Vin which is also the beta term we are calculating. I am not sure if that is the point where it is not clear. The Vfb node can be labeled anything but at that node the simulation is capturing the beta factor we are looking for.

Best Regards,

Ignacio

Hi Ignacio,

Thanks for the feedback.

It is still not clear for me. Let's go back to the equation.

Beta = Vfb/Vin = R1/(R1+Rf), I have no question about this.

But why the we say Beta = Vfb, In the simulation, we can get both Vin and Beta.

Please look into the video below. In the simulation, If it use1/Beta=(R1+Rf)/R1, I can understand. But it use 1/Beta=1/Vfb, I don't understand, I think it shall be 1/Beta=Vin/Vfb. Why Beta=Vfb?

TI Precision labs video series

Stability-->Spice simulation @ 3:04

https://www.ti.com/video/series/precision-labs/ti-precision-labs-op-amps.html

Thanks for this strong support. That makes sense.