0
\$\begingroup\$

I am finding it very strange what is the role of R1 in this circuit, and what would happen if I would just remove R1 along with its connection, i.e., remove the connection inside the pink box:

enter image description here

Wouldn't both leads still get amplified?

No matter how much I read about it, opamps are still not intuitive for me.

Another strange highlight, is that this diagram looks exactly the same as the following, but the author now calls the "non-inverting amplifiers" by "voltage follower":

enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ nemewsys: Like a lot of EE topics, they only become "intuitive" after you study them for years! Half of the problem is the misinformation which is prevalent in the explanations and the poor choice of words used in the descriptions. After all, we are EE's not Nobel Prize Winning authors. For example, in the lower diagram where A1 & A2 are noted as "Voltage Follower"s. Nope, false. These are actually the gain stage of the overall 3-amp configuration. A "voltage follower" by definition has a gain of unity (one, 1.0000000). Wish I had a better quick answer for you. If you removed R1, they would be \$\endgroup\$ – FiddyOhm Oct 2 '17 at 22:48
2
\$\begingroup\$

The Wikipedia Instrumentation amplifier article explains it reasonably well using the theory. Let's use some diagrams instead to make this a bit more intuitive.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. \$ R_{GAIN} \$ removed.

  • With the gain resistor omitted we simply have two voltage followers.
  • In (a) we have a differential input of 1 V and a differential output of 1 V.
  • In (b) we have a common-mode input of 1 V and a differential output of 0 V.

The problem occurs with this circuit that if we want to increase the gain in the normal non-inverting amplifier way we would add resistors from IN- to ground on OA1 and OA2. The problem is that the common mode signal gets amplified too and this may cause overload for the following stage - particularly if the common mode signal is high relative to the signal of interest.

schematic

simulate this circuit

Figure 2. \$ R_{GAIN} \$ reinstalled.

To keep the maths simple I've set Rgain = 20 kΩ.

  • In (a) we know from our non-inverting amplifier theory that the op-amp will settle with V- = V+. Therefore with a 1 V input OA1 will have 1 V on IN-.
  • Similarly OA2 will have 0 V on V-.
  • Now there is 1 V across Rgain so 0.05 mA must flow through R1, Rgain and R2. From this we can work out the op-amp output voltages. We see that we have doubled the differential signal output to 2 V.
  • In (b) we can see that the common mode signal goes through with a gain of one.

So, by adding Rgain = 20 kΩ we have increase the differential gain by a factor of 2 without increasing the common mode gain. As we increase Rgain the benefits improve further.

Read paragraph 3 of the Wikipedia article in the light of this and see if it helps.

\$\endgroup\$
  • \$\begingroup\$ Great explanation! How did you get to those +1.5V and -0.5V? \$\endgroup\$ – nemewsys Oct 2 '17 at 23:49
  • \$\begingroup\$ We can calculate that because we know the current through the 10 k resistors. The op amp input impedance is extremely high so no current goes into the inputs therefore it all flows through the resistor chain and we can work out the voltage drops. The Insight you need Is to realise the voltages at the inverting inputs \$\endgroup\$ – Transistor Oct 3 '17 at 0:03
  • \$\begingroup\$ ... is the same as the voltage at the non-inverting inputs. This is usually the best way to analyse op amp gain circuits for me. \$\endgroup\$ – Transistor Oct 3 '17 at 0:05
1
\$\begingroup\$

R1 in your instrumentation amplifier allows you to adjust the overall gain of the difference using one resistor, without any matching tolerance. Yes, you could remove R1 (and indeed replace both R2's with a short). Then your gain would be fixed, and determined by R3 and R4.

Without R1 and R2, if you wanted to adjust the gain, your accuracy is limited by the matching accuracy of R3 and R3. This would require something like a stereo potentiometer, whose matching (and tracking) tolerance isn't going to be perfect. For high gains, this inaccuracy can swamp the signal of interest.

\$\endgroup\$
1
\$\begingroup\$

R1 sets differential gain. If removed, the circuit will have gain set by R3, R4, R4, R5 (R3 and R4 in the second diagram). Benefit of R1 is that it increases differential gain while leaving common mode gain equal to 1 in both branches.

Another benefit of this topology is that an integrated instrumentation amplifier with high common mode rejection ratio can be made with accurately matching R2, R3, R4 resistors (easy on an IC) while their absolute value accuracy is not so important (not that easy on an IC) and set gain using only one external component.

Regarding naming, "non-inverting amplifier" is similar to "voltage follower". Former can have any positive (as opposed to negative, inverting) gain while later is typically unity gain: output follows input. In this circuit it is more accurate to call it "non-inverting amplifier" to emphasis the fact that both positive and negative input signals preserve their phase and are subtracted in A3. Though if R1 is removed, A1 and A2 becomes both "non-inverting amplifier" and special case of it - "voltage follower".

There is a Wikipedia article about Instrumentation amplifiers with lots of more good reading: https://en.wikipedia.org/wiki/Instrumentation_amplifier

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.