# Conditioning circuit

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#### gonsays

##### Newbie level 5 Hello guys,

I'm having a problem understanding the values of the solution for the following question amd circuit: The solution is: What are the equations used in order to get those values?

What would be the best strategy to analyze the circuit?

Thank you very much in advance! Any info is appreciated #### barry There's a problem here. Your initial question states that you're trying to convert a current input, and then your final equation is for a voltage input. Which is it? Are you missing something?

• gonsays

### gonsays

Points: 2

#### gonsays

##### Newbie level 5 Hello guys,

Barry, you are right, there must be some problem regarding the question made, but my main objective is to understand the equations used in the analysis of this circuit.

I will present to you guys another example, can you check if it makes sense and the equations used on the analysis? (On the yellow box the solution is presented regarding the dimensions used): Thank you very much in advance! Any link or info is really appreciated #### albbg You have two simple op-amp based circuit that are cascaded. The first one is an inverting amplifier that has gain G1=-Rx/R, the second one is an inverting adder that has an overall gain G2=-Ry/R.

The input/output transfer function can be written as (I changed the input label from o to Vin to avoid confusion):

Vout=(Vin*G1+Vref)*G2

The following analysis only applies to ideal op-amps.

We can apply the virtual ground to the first amplifier (i.e: V+=V-) then applying a current I, the corresponding input voltage will be given by:

Vin=R*I

then:

Vout=(R*I*G1+Vref)*G2

Now, from the specs

0=(R*4mA*G1+Vref)*G2
10=(R*20mA*G1+Vref)*G2

from the first equation -R*4mA*G1=Vref

but G1=-Rx/R, then R*4mA*Rx/R=Vref that means Rx=Vref/4mA
as you can see R is not involved in the calculation. It's correct: if R increases the input voltage increases but the gain lower by the same quantity; the same if R decreases.

from the second equation, using Rx=Vref/4mA and G2=-Ry/R, we will have:

10*R=(20mA/4mA-1)*Vref*Ry from which Ry=10/4*R/Vref

Now choosing arbitrary Vref=2.5V

Rx=2.5V/4mA = 625 ohm
Ry=R

We can also arbitrary choose R; for instance R=100 ohm

Then G1=-625/100=-6.25
G2=-100/100=-1

Substituting in the transfer function:

Vout=(-Vin*6.25+Vref)*(-1) = 6.25*Vin-Vref

The second exercise is very similar to this, but the input is a voltage instead of a current.

#### D.A.(Tony)Stewart you have two requirements by converting in terms of GAIN and OFFSET

The inverting OpAmp (OA), 1st stage is ground referenced on + side, so Rin acts as a current shunt to ground converting current to voltage.

One could also design this with only 1 OA using Shunt on Vin+.

The voltage gain max-min swing OUT/IN
or 10V/Vin_pp (Vin is across shunt R to virtual ground) on (-) or real ground on (+) in my case.

Vin should not too too small or too large and depends on supply voltage at source. Shunt voltages for low current ranging from 50mV to 1V are reasonable.

But Vin smaller means more gain needed so go for around 1V. But you need a precision voltage reference scaled down so what is available?

There are a lot of choices.

But my practice is deal with the GAIN 1st then OFFSET 2nd.

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