resistance purpose in the circuit

Hi guys
I have a circuit.i want to know the purpose of R17 & R18 resistance in the below diagram.

They are the bottom end of a potential divider.

R1, R2, R3, R4 are the top and R18 the bottom in one leg,
R5, R6, R7, R8 are the top and R17 the bottom in the other leg.

I'm not sure what duplicating the differential amp is supposed to do and I suspect a type with higher input impedance would be needed anyway.

Brian.

betwixt said:

They are the bottom end of a potential divider.

R1, R2, R3, R4 are the top and R18 the bottom in one leg,
R5, R6, R7, R8 are the top and R17 the bottom in the other leg.

Click to expand...

I cannot get your point betwixt. What is the relation between top and bottom? R17 & R18 is pulled-up with 5v.

I don't believe it will work accurately with the LM358 anyway and I can't understand why you have two of them in parallel but the point about the potential divider is this:

Input voltage>----- R1-----R2-----R3-----R4--M--R18-----+5V where 'M' is the Measurement point. Its a 2M to 1.2K potential divider, it drops the input voltage by about 1667 times before it reaches the op-amp.

The other half of the circuit is identical using R5, R6,R7,R8 ad R17.

Brian.

I think is a circuit used to measure a floating voltage with an high impedance probing.

If we call Rx=R1+R2+R3+R4 and Ry=R5+R6+R7+R8 and we focus on the meshes Vi-Rx-R18-Vcc(the 5 V voltage) and Vi-Ry-R17-Vcc we can find the voltage
seen at the left side of R9 that is connected also to R16 and R12 that is connected also to R13. Let's call the first node M and the second one N. Applying the superposition we have the voltages:

VM=[R18*(Vcc+Vi)+(Rx+Ry*R17)*Vcc]/(Rx+Ry*R17+R18) and
VN=[R18*(Vcc-Vi)+(Rx+Ry*R17)*Vcc]/(Rx+Ry*R17+R18)

Now the high side op-amp does Vout1=k*(VM-VN) while the second one Vout2=k*(VN-VM) where k is the gain k=R10/R9 (all the series and feedback resistor are identical to these).
Notice that the op-amp are single voltage supplied, then Vout1=k*(VM-VN) if VM>VN, otherwise is 0 and Vout2=k*(VN-VM) if VN>VM otherwise is 0. This means that, depending from the polarity of Vi one of the two Vout1 or Vout2 is 0. Then on the output capacitor you will always have a positive voltage that means the circuit is also a rectifier, this is why two op-amp are used.

Since Rx=Ry (we can call them Rs) and R17=R18 (we can call them Rp) then VM-VN=Rp/(Rs+Rp)*Vi

having Rs>>Rp, then VM-VN=Rp/Rs*Vi or 0 if Vi>0, in an analog way VN-VM

The total gain, from Vi to the output will be H=Rp/Rs*k*0.5. This last 0.5 is due to the voltage divider R19-R20 (don't forget that Vout1 or Vout2 is zero depending from the polarity of Vi)

Numerically H=1.2k/2M*806k/24.9k*0.5≈1/100

It seems to me that the transfer function is Vout=|Vi|/100

R17 and R18 are used to make the single supplied op-amps working with proper common mode voltage, since the inputs are floating.

Please check my calculations I'm not 100% sure I didn't do some mistake.

The circuit dimensioning makes no sense. The pull-up resistor biases LM358 outside its common mode range of 0 to 3.5 V.

FvM said:

The circuit dimensioning makes no sense. The pull-up resistor biases LM358 outside its common mode range of 0 to 3.5 V.

Click to expand...

You are right.
I explained how the circuit works and the meaning of the two resistors R17 and R18. I didn't check (even if it was very easy) if the common mode voltage was in range. However I'm not sure (even if the used symbols suggest so) if the supply of the op-amps is 5V or higher (at least 7.5V).

In a typical amplifier with differential voltage divider, the input would be biased to Vcc/2 or center of common mode range.