Opamp based Inverting attenuation circuit and output equation

[skraj123]

Hi,
Am seriously looking on a schematic with output equation for my work.An inverting attenuation circuit with output equation is required. I have find inverting attenuation circuit from TI's SLOA058 document but i am not understand it fig.5 shows the image of schematic. Hope positive results. Its important for me.
I designed TI's SLOA058 inverting attenuation circuit for my requirement vout/vin = 0.100. but my simulation and bread board test not positive.
help me in this.

 

[ebclr]

Everything , and more here

https://en.wikipedia.org/wiki/Operational_amplifier

 

[skraj123]

Everything , and more here

https://en.wikipedia.org/wiki/Operational_amplifier

HI, thanks for your reply but am looking on solution for a complex circuit . Here added image of that schematic please let me the output equation will be ???

 

[FvM]

The circuit can be analyzed bases on the "virtual short" concept. Determine the current through RinB assuming a short across the OP inputs.

 

[skraj123]

Dear FvM,
Hope you could seen the image of the circuit, I tried to analyse as you said "Virtual Short". Please explain in detail and give me output equation alone.
I try to solve based on your ideas and verify with your output equation.

Thanks in Advance

 

[godfreyl]

The circuit is more complicated than needed. If you set RinB = zero ohms (short circuit), then analysis is easy:
Input impedance = RinA
Gain = -Rf / RinA

If the opamp is unity gain stable then you don't need R3 either, just leave it open circuit.

 

[skraj123]

hi godfreyl ,
Am try to produce gain from 0.01 to 50, so that am asking all this. Hopefully expecting reply.

 

[godfreyl]

Am try to produce gain from 0.01 to 50,

Do you mean the gain must be adjustable e.g. with a potentiometer from 0.01 to 50? That's a very wide range for a single control, so it would be difficult to set accurately.

 

[skraj123]

Gain is fixed at any one value. AM not looking variable gain. being Attenuator i said from 0.01, but my requirement is exactly 0.01 . so i want choose values for resistor.
thanks

 

[godfreyl]

Gain is fixed...
requirement is exactly 0.01 . so i want choose values for resistor

OK. As I said in post 6: Gain = -Rf / RinA. So if you want gain = 0.01 then Rf = 0.01 * RinA.

e.g. You can use RinA = 220K, Rf = 2.2K and leave out the other two resistors as mentioned previously.

Similarly, if you want gain = 50 then Rf = 50 * RinA, so you could use RinA = 2K, Rf = 100K

 

[albbg]

Let's start from the pi-network on the input with R3 going to GND (vcc/2 is used because op-amp is supplied from Vcc to GND). By using Thevenin we have:

Voc=R3/(Ra+Rb)*Vi
Rth=Rb+Ra*R3/(Ra+R3)

so all the pi-network can be replaced by a voltage generator Voc followed by a series resistor Rth. The circuit is now a simple inverting-amplifier, thus

Vo=-Rf/Rth*Voc

replacing with complete expression for Voc and Rth we will have:

Vo=-RF*(Ra+R3)/[Rb*(Ra+R3)+Ra*R3] * R3/(Ra+R3) *Vi ==> Vo/Vi=-Rf*R3/[Ra*Rb+(Ra+Rb)*R3]

Since Ra=Rb (let's call this Rs):

Vo/Vi=-Rf*R3/(Rs²+2*R3*Rs) the article says that Rf=2*Rs (have a look at the number written on the components) then:

Vo/Vi=-2*R3/(Rs+2*R3)

normalizing to Rs=1:

Vo/Vi=-2*R3/(1+2*R3)

from this we can now extract R3 normalized (of couse we need tha absolute value of the gain, so the sign "-" have to be eliminated):

(Vo/Vi)+2*R3*(Vo/Vi)=2*R3

from which:

R3=(Vo/Vi)/{2*[1-(Vo/Vi)]}