# Comparator circuit analysis

#### jay76

Oct 16, 2022
1
comparators are used to measure whether a particular voltage is above or below reference voltage.

- It makes sense for the led to turn on when input voltage (v+ > v-) and turn off for input voltage ( v+ < v-) (please refer to the "Comparator circuit with solution" image)

My question is: what would happen to the led ( turn on/off) if we attach a negative feedback to the comparator circuit ? please explain why

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#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,864
With negative feedback, you would basically be changing the comparator to an operational amplifier that is trying to drive a digital output instead of producing an output that is proportional to the difference in input voltage. How this would affect an LED that is driven by the digital output is uncertain and very likely undesirable. Best case is the output would vary linearly with changes in input voltage.

Comparators, used to create a logic "1" or logic "0" digital output by comparing the polarity and magnitude of two analog inputs, are special forms of high open-loop gain differential amplifiers with digital output driving capability. The differential gain is large enough to cause the output to change states for very small (microvolt) differences between the two inputs. For this reason, any input with sufficient noise will cause the comparator output to oscillate wildly between logic "1" and logic "0" states when the differential input voltage approaches zero. This oscillation is an undesirable condition that can be prevented by adding hysteresis to the comparator circuit with positive feedback.

Negative feedback is never intentionally used in comparator applications because it deceases the closed-loop gain of the comparator, which then increases the allowable difference in compared inputs that will cause a change in logic-level output. You typically want the comparator logic output to change only when the measured input is very nearly the same as the reference input, being a few microvolts more positive or a few microvolts more negative. Differential input voltages between the limits that cause a change in output state should not cause a change in output. This is an attribute of a comparator circuit that has hysteresis.

Hysteresis needs to be as small as practical to prevent changes in comparator output caused by noise on the inputs. If a typical open-collector comparator output is pulled up to logic "1" with a resistor, and the output at logic "0" is near ground potential, a largeish resistor can be inserted between the output and the "positive" input terminal to implement hysteresis. There needs to be a finite resistance to common on the "positive" input terminal for the positive feedback to be effective. Note that the designation of "positive" and "negative" for the two input terminals refers to the direction the output will swing, positive being toward logic "1" and negative being toward logic "0", when the differential input voltage has those polarities.

More information can be found on this thread, which discusses using an LM318 operational amplifier as a comparator.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,448
Mod edit: Moved to the homeworks forum as it looks like such.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,448
Diodes are a one-way, power storing, component.
Please excuse if my language seems to direct, but that is nonsense. Diodes don't "store power".
most comparators(all the ones I made to learn for myself) fail, just from experience,
Most comparator circuits I built worked like a charm. Guess who's to blame?
I don't find my addition to this discussion extreamely important or very understandable
I fully agree to the latter.
they degaussed it so the static would not interfere with the circuit,
Static is afaik considered an electric effect, not a magnetic one. Degaussing minimizes the magnetic field not the electric field, therefore it doesn't help to reduce static.

i have no idea what the rest of your rambling is about

#### crutschow

May 7, 2021
786
i have no idea what the rest of your rambling is about
You get a big 2nd from me on that.

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,864
Diodes are a one-way, power storing, component. You require a ground for electricity to conduct. If you put a volt into a diode, you have to supply a 1 volt of ground to shut it down.
You are so full of nonsense. Diodes do not store either power or energy, except incidentally as a result of a minuscule charge stored on the capacitance of the diode junction when it is reverse biased.

The only requirement for electricity to conduct (between two points in space) is a voltage potential difference between two points and a finite resistance between those two points. Ground has absolutely nothing to do with the conductance of electricity, as any battery-powered flashlight demonstrates.

You don't "put a volt into a diode" or any other component. You apply, or measure, a voltage between the two terminals of a diode. Voltages are always the measure of electrical potential between two non-congruent points in space.

#### davenn

Moderator
Sep 5, 2009
14,213
major tidy up and remove rubbish post

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