# Making sense of non ideal elements on signal conditioning circuit

So I was trying to solve a problem where I would get an input signal of 230V,16A@50Hz and the output had to be the same frequency but in the rage of 0V and 5V. Through the analysis of the circuit below I got to the expressions :

Corresponding to this circuit:

With the values used on MultiSim I should be seeing a sine wave with a slight phase difference between 0V and 5V. Are my calculations wrong? Is there something I am getting wrong in the schematic from the simulation? Are non ideal components causing this much fuss?

Your DC bias conditions are not correct for using the opamp with a single supply. You are saturating the output.

This design shows a single supply opamp with the negative pin to ground. The output can only be driven between zero and 5V (assuming rail to rail opamp). In order for the amplifier to function with a bipolar signal at its input it must be biased so that with zero input voltage the output sits somewhere between the rails, for the maximum output capability it should be at +2.5V.

The potentiometer R3 in circuit 1 is used to adjust this but the amount of bias needed varies with the gain setting of R5. It is not a practical circuit.

You could remove R2 and provide a negative supply to the opamp so the output is centered on ground.

Or calculate the bias provided by R2 to bring the opamp in its linear range. That may be difficult to do and the output will be offset.

A more common way to implement this type of circuit is like this;

The input and output capacitors isolate the DC of the circuit and the gain can be changed without affecting the DC biasing.

Image from Stack Exchange

• could you be more specific where the "DC bias conditions are not correct"? My electronics vocabulary is still quite small... Also, something that sets me off is that if I place a potentiometer on the R3 on multisim just like the schematic I showed, it only takes a 5k potentiometer as 50% to get a sine wave (i.e. the OpAmp working on linear range) and I don't know what I'm getting wrong... Dec 21, 2019 at 19:00
• Just saw the edit. That is true, the circuit is very far from being practical. Unfortunately I have already built and I need to analyze the circuit above. Do you have any tips? I have tried using the superposition principle, but it gives me a wrong answer, I think it's because of the capacitors that turn it into a non linear circuit. But solving it is still a pretty hard task Dec 22, 2019 at 0:39
• @Bidon What do you need to analyze? R3 just adds to R2 with a Thevenin voltage at the other end that's <= 5 V. It's just a DC Thevenin equivalent (no AC.) Vin swings stuff around, but that just means the (+) input moves around. The output can't provide negative feedback of any kind, since you have the output going to a different divider network. So the opamp can't do anything with its output to get the two inputs to equal each other. So it saturates out in just slightly less than 100% of the permutations of your potentiometer values. There is a slight chance left, but it may as well be zero.
– jonk
Dec 22, 2019 at 4:41
• @Bidon Without $R_6$ and $C_3$, your function is:$$V_\text{O}=\frac{R_4+R_5}{R_4+\frac{R_4+R_5}{A_o}}\cdot\frac{V_\text{IN}\,\left(R_2+R_\text{TH}\right)\,R_b+V_\text{TH}\left(R_1\,R_a+R_1\,R_b+R_a\,R_b\right)}{\left(R_1+R_2+R_\text{TH}\right)\left(R_a+R_b\right)+R_a\,R_b}$$where $V_\text{TH}$ and $R_\text{TH}$ are the Thevenin voltage and Thevenin resistance, respectively, presented by $R_3$. $A_o$ is the open loop gain of the opamp and is probably large enough to allow you to simplify the first factor. If $R_\text{TH}$ is made much smaller than $R_2$, it can be neglected.
– jonk
Dec 22, 2019 at 15:01
• @Bidon There's an offset in the above equation. So, while you can get the local slope using derivatives, you can't get a simple $\frac{V_\text{O}}{V_\text{IN}}$ to then apply to $R_6$ and $C_3$ as a low-pass filtering of $V_\text{O}$ in the frequency domain.
– jonk
Dec 22, 2019 at 15:04