Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

op amp high gain vs high accuracy

Status
Not open for further replies.

mpig09

Full Member level 4
Joined
Aug 26, 2005
Messages
232
Helped
8
Reputation
16
Reaction score
2
Trophy points
1,298
Location
Taipei
Activity points
2,810
Hi all:

In analog circuit design, I often get a info
"for high accuracy need high gain".
But I am not understand the Concept in the circuit.

Do you help me to understand
why high gain of opamp will get high accuracy?

Ex: 1% --> need opamp > 40dB.

If I have any mistake, please correct me.

Thanks for your reply.
mpig
 

Here is the closed-loop gain (Acl) formula including the opamp´s open-loop gain Ao:

Acl=Hf*Ao/1+Hr*Ao)=Hf/(1/Ao + Hr)

Hf: Forward transfer, if any (attenuation)
Hr: Feedback transfer factor (example: Hr=Ro/(Ro+Rf)

As you can see, in case of Ao>>1/Hr and Hf=1 (non-inverting operation) the formula reduces to the well known expression

Acl=1/Hr=1+Rf/Ro.

Hence, for very large open-loop gain Ao the closed-loop gain Acl is determined (with a very small error) by the external resistors only.
 

... why high gain of opamp will get high accuracy?

Ex: 1% --> need opamp > 40dB.

If you accept that non-linearity is the opposite of high accuracy:

non-linearity(closed-loop) / non-linearity(open-loop) ≈ Acl/Ao (definitions s. above in LvW's posting),
i.e. the improvement of non-linearity is about equal to Acl/Ao .

With (high) accuracy ~ 1/non-linearity ---> improvement of accuracy ~ Ao/Acl

i.e. for a fixed closed-loop Amplification Acl, accuracy improves ~ Ao , the open-loop gain.
 

Regarding the gain error, a small manipulation of the given formula leads to

Acl=Hf*Ao/1+Hr*Ao)=Hf/(1/Ao + Hr)=...=(1/Hr)*[AoHr/(1+AoHr) ]

The term in brackets gives you the error term with respect to the IDEAL gain Acl=1/Hr.
Now you can solve for Ao*Hr if the term in brackets must not exceed 1%
(Correction: ...must not fall below 100%-1%=99%)
(Note that -Ao*Hr is called "loop gain").
 
Last edited:

Hi LvW and erikl:

Thanks.
I will think it.

mpig
 

mpig09: Watch my correction above.
 

Hi LvW and erikl:
Thanks for your clear reply.

I try to explain by myself:
Acl / Ao = (1/Hr) x (Hr*Ao)/(1+Hr*Ao)
let Hf = Hr = 1
==>Acl / Ao = Ao / (1+Ao)
For 1% accuracy, Acl / Ao > 99%
so the Ao should > 99.
and we select Ao = 100 for design spec.

If I have any mistake, please correct.
mpig
 

Hi Erikl,

Sorry couldn't fully understand your points on non linearity with Acl,. And I'm not sure if we can talk about open loop non linearity. If I understand it correctly, non linearity (tones in band) will be kind of suppressed by the open loop gain and tones and higher harmonics will any way be suppressed by the attenuation of the loop. So, I'm able relate harmonic supression or non linearity supression by open loop gain., but couldn't get it how it is related to the ACl. It would be great if you could help me out.
 

... I'm able relate harmonic supression or non linearity supression by open loop gain
??? How could you get suppression without feedback, i.e. in Ao condition?

... but couldn't get it how it is related to the ACl. It would be great if you could help me out.
It's all given above. Just study it thoroughly!
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top