What difference will it make on the output between these two circuits?

Question

In the above image there are two figures, Fig(a) consists of force sensor connected to a resistor in a simple voltage divider configuration, and its output is fed to an ADC. Now coming to Fig(b), its the same circuit as Fig(a), however in this case the voltage divider circuit is followed by an op-amp in voltage follower configuration.

The circuit in Fig(a) does not have any output offset voltage, whereas the circuit in Fig(b) has an output offset voltage of 650 mV. I know that the circuit in Fig(b) is mostly preferred over the one in Fig(a) due to the op-amp having high input impedance, but I don't understand what difference it makes in the output fed to the ADC. Now my questions are as follows:

1. What is the difference between the two circuits with respect to the output being fed to the ADC i.e. in Fig(a) and Fig(b). And why is the circuit in Fig(b) more preferred than in the Fig(a)?

2.The OP177 op-amp has pins 1 and 8 for trimming the offset i.e. 650 mV using a POT, however this isn't working. So why isn't the offset nullification working?. The POT has 3 terminals among which, 1st and 3rd terminals are connected to pin 1 and pin 8. And the middle terminal is connected to +Vss(pin7) for offset trimming.

Answer 1

As mentioned in a comment, the second circuit has the output tied to ground so will not work. The ground link at the negative input/output needs removing.

1. Adding the op amp doesn't help much in this situation, ADC's usually have a high input impedance. It will introduce an offset voltage at the output (ADC input), which will need to be compensated for in software.

2. Without knowing what you have done to test the operation of the offset null pin, it is hard to tell you why it is not working. You must ground the inputs to the op amp, then measure the voltage at the output. By adding a voltage to the offset null pin, the output voltage will either reduce/increase depending on this value, showing you what value you must add to the pin to get 0v and no offset at the output.

By the sounds of your description of the POT, I think you are using the wrong pins. The middle pin should be connected to the offset null, with the outer pins connected from VCC to GND.

Answer 2

I think the connection between the negative input of the opamp and the ground is a mistake. If you delete that connection, the opamp has the function of a buffer. V+ and V- are equal, and since the Vout of the opamp is connected to V-, the output is equal to the input voltage. (just like you would have in fig. a) The difference is that no current is leaking from the middle point of your voltage divider, since an opamp doesn't draw current on its input.