Differential op amp feedback resistors

Question

I understand if you do not have negative feedback in any op amp configuration then the output will be unstable (will oscillate for example) and the feedback resistor and input resistor on the negative input set the gain of the op amp.

What is the purpose of the 'feedback' resistor on the positive input of a differential op amp?

In the below example, there is a differential pair at the input of the op amp and a 2V DC bias being added to the +ve input. The red and blue traces are measured at the +ve and -ve inputs to the op amp and green trace measured at the output. It can be seen that the DC bias has reduced a little from 2V to ~1.98V because VR20(V +ve input)=1k/(10+1k)*2

If we remove R20 as shown below then the V at +ve input is now not split across a voltage divider and we get the full 2V.

So it seems having R20 is actually a bad thing? But I am sure I am missing something because in all textbook examples of differential op amps I have seen, there is always R20 on the +ve input. So what is its purpose?

Answer 1

Referring to the upper schematic in post #1, in a classic differential amplifier there are four resistors that matter: R17, R18, R19, R20. For this analysis, delete R21 and V20.

In the standard circuit, the ratios of the two resistor pairs is exactly the same. Note this does not mean that all four resistors are equal in value; it is the ratios of the values that determines the circuit gain.

So, R17 / R18 equals R20 / R19. For a gain of 5, it might be

R17 = 10K

R18 = 2K

R19 = 200

R20 = 1K

What is the role of R20? R20 is the shunt leg of an attenuator.

In standard inverting opamp circuit, the gain is R17 / R18. But in a non-inverting circuit, the gain is (R17 + R18) / R18, or 1 + (R17 / R18). For the same feedback resistors, the gain is different for the inverting and non-inverting cases by an increment of 1.

Thus, in a differential amplifier circuit where there are both inverting and non-inverting signal paths, the non-inverting input is amplified more than the inverting input. To compensate for this, the non-inverting input needs an attenuator so that the two gains are not just opposite, but equal and opposite. If you crank through the full-form gain equations, it turns out that the attenuator resistance ratio is exactly the same as the feedback resistance ratio.

In the example above, the inverting gain is 5 and the non-inverting gain is 6. The non-inverting input attenuator is 1K / 1200, or 5/6. 6 x 5/6 = 5 for the net non-inverting gain.

In your circuit the four resistors are equal, so the inverting gain is 1, the non-inverting gain is 2, and the non-inverting attenuator is 1/2.

Answer 2

A differential amplifier subtracts two input signals.

For split supply operation an a attempt to perform \$V_b-V_a\$ is made in Figure 1. If the two input signals are equal, the output \$V_o\$ should be zero. Using the configuration gain equations and superposition reveals: $$V_o(V_b)-V_o(V_a)=\left(1+\frac{R_f}{R_i}\right)V_b-\frac{R_f}{R_i}V_a$$

Since both resistors are equal, the result becomes: $$V_o(V_b)-V_o(V_a)=2V_b-V_a\ne0$$ for equal inputs.

\$V_b\$ has twice the gain that \$V_a\$ does.

So a voltage divider is placed between \$V_b\$ and the non-inverting input to attenuate \$V_b\$ by 1/2, as shown in Figure 2. This is the standard reference diagram for a differential amplifier.

So now the output equation is: $$V_o(V_b)-V_o(V_a)=\left(1+\frac{R_f}{R_i}\right)\frac{R_2}{(R_1+R_2)}V_b-\frac{R_f}{R_i}V_a$$ Since all four resistors are equal, the result becomes: $$V_b-V_a=2\frac{1}{2}V_b-V_a=0$$ for equal inputs. The circuit now subtracts correctly.

This is why \$R_2\$ (R20 in the OP) is absolutely necessary to the operation of the differential amplifier.

^{simulate this circuit – Schematic created using CircuitLab}

^{simulate this circuit}

^{simulate this circuit}

For single supply operation the op-amp inputs must operate at 1/2 VCC for equal output swing over the supply voltage. To acheive this R2 (R20 in the OP) is used for an additional purpose. Instead of connecting it to dc 0 volts, it is connected to a low impedance source voltage (\$V_{bias}\$) of 1/2 VCC as shown in Figure 3. If a different operating point is required, then \$V_{bias}\$ can be fixed to any desired value acceptable to the operation of the amplifier.

In the OP, the midpoint bias for VCC=5V would be 2.5V. A 2V bias will restrict the negative swing, but is allowed.

Answer 3

Both diagrams are poor differential amplifiers. They are ruined by this shown in red below: -

R21 and V20 wreck the ability of the diff amp to provide decent common mode rejection. If you need to bias the op-amp then do this (inside lime green box): -

What is the purpose of the 'feedback' resistor on the positive input of a differential op amp?

It's not a feedback resistor.

So it seems having R20 is actually a bad thing?

No, it's vital for ensuring excellent common mode noise/signal/interference rejection.