or each input, Equation 1 defines the effective input resistance as:

Let’s start with the easy part first: the noninverting input. Looking at the diagram in Figure 2, you can see that R3 and R4 are in series. Assuming that no current enters or leaves the op amp input, Equation 2 calculates the effective input resistance as simply:

Now let’s focus on the inverting input. Recall from the ideal op amp rules that the two inputs of the op amp should always be at the same potential (Equation 3):

You can also see that R3 and R4 form a voltage divider at the noninverting input. Equation 4 calculates the voltage at the op amp’s noninverting input (VP) as:

Why is this important? Well, this voltage partly determines the effective input resistance of the inverting input. Consider that the current through R1 (IIN(N)) is the voltage across it divided by its resistance (Equation 5):

Substituting Equation 5 for input current back into Equation 1 gives you Equation 6, a generic equation for the resistance of the inverting input:

Since you know the voltage at the inverting input (VN) will be equal to the voltage at the noninverting input (VP), you can replace VN in Equation 6 with Equation 4 to get Equation 7:

Notice that the effective input resistance at the inverting input actually depends on the ratio of the two input voltages. To understand how this might affect your application, consider an example: the INA134 used in an audio-line-receiver application where the voltages at the two inputs are equal in magnitude but opposite in polarity (Equation 8):

Referring back to Equation 2, the effective input resistance at the noninverting input is fairly straightforward:

However, the inverted relationship of the two input voltages has a drastic effect on the effective input resistance at the inverting input (Equation 9):

The effective input resistance at the inverting input is one-third that of the noninverting input. Thus, you must consider the lower impedance of the inverting input when choosing input coupling capacitors or filter circuits. Furthermore, any amplifier driving the difference amplifier inputs must be capable of driving the lower impedance at the inverting input.

With simple circuits, it’s often the most basic aspects of their operation that cause headaches in the lab. Don’t overlook the obvious!