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Comparator

Comparators compare two voltages and output “high” or “low” based on which voltage is higher. A slightly modified version of the op-amp assumptions apply:

i+=i=0 v+>vvOUT=VDD v+<vvOUT=VSS

Note the following:

Solving comparator problems

When solving comparator problems, you have to essentially solve the circuit for each possible state. Luckily, there are only two possible states:

For a comparator in positive feedback,

Let's look at this specific example:

No matter the state of the circuit, this relation is true due to the voltage divider relation:

v+=R2R1+R2vOUT

And the negative terminal voltage is simply equal to the input voltage:

v=vIN

Let's first look at the case where v<v+, so vOUT=V. In this case,

v+=R2R1+R2V

This means that this circuit will stay in this state as long as v<R2R1+R2V. If v crosses the threshold R2R1+R2V, the output of the circuit will change to vOUT=V.

Let's now look at the other case, where v>v+, so vOUT=V. In this case,

v+=R2R1+R2(V)=R2R1+R2V

This means that this circuit will stay in this state as long as v>R2R1+R2V. If v crosses the threshold R2R1+R2V, the output of the circuit will change to vOUT=V.

To summarize the results, if vOUT=V, for vOUT to transition to V, vin has to become greater than R2R1+R2V. And if vOUT=V, for vOUT to transition to V, vin has to become less than R2R1+R2V. Notice that the threshold for the input voltage changes depending on the current output voltage. This gives us hysteresis.

Let's plot the transfer curve (output voltage as a function of input voltage) of this behavior.

Notice the arrows on the inner rectangle. They signify that to go from a high output to low output, the input must cross R2R1+R2V, whereas to go from a low output to a high output, the input must cross R2R1+R2V.

What if we apply a triangle wave to the input?