# Comparators

A comparator is a device that takes two inputs, one called $V_{in}$, and one called the threshold, $V_{TH}$. If $V_{in}$ is greater than $V_{TH}$, the comparator returns a high signal, otherwise its output is low.


![Figure: A comparator drawn as a triangle. $V_{in}$ connects to the "+" input, $V_{TH}$ connects to the "−" input, and the output is labeled $V_{out}$.](assets/images/comparator1.png)

Essentially, the comparator asks "Is $V_{in} > V_{TH}$?"

- If $V_{in} > V_{TH}$, $V_{out}$ = High.
- If $V_{in} < V_{TH}$, $V_{out}$ = Low.

Note that the case where $V_{in}$ is exactly equal to $V_{TH}$ should also return a low signal, but this situation rarely arises in the real world because all of the input signals will have some noise component.

The orientation of a comparator can also be reversed, so that the question becomes: Is $V_{in} < V_{TH}$?

![Figure: A comparator with the inputs swapped — $V_{in}$ connects to the "−" input and $V_{TH}$ connects to the "+" input, with output $V_{out}$.](assets/images/Comparator2.png)

Although the symbols look similar, do not confuse a comparator with a NOT gate. Also, **no current** flows through an ideal comparator.

:::{admonition} Aside: How does a comparator work?
:class: seealso, dropdown
It is derived from an analog electronics component called an "Operational Amplifier" or more simply an Op-Amp. An Op-Amp uses transistors and feedback to amplify a small difference between two input currents. The ideal output of a differential amplifier is $V_{out} = A(V_{in}^{+} - V_{in}^{-})$ where $A$ is a constant gain factor. However, the amount of amplification is limited by two supply rails ($V_{S+}$ and $V_{S-}$) that supply power to the circuit. To make a comparator, we simply crank up the gain so that any small difference is saturated to the supply limit, and ensure that $V_{S+}$ is equal to our high voltage level and $V_{S-}$ our low level (e.g. $V_{S+} = 5\,\text{V}$ and $V_{S-} = 0\,\text{V}$). In this manner, there is no longer amplification, but a bimodal High or Low output.
:::

## Setting the Threshold Voltage

In order to set the threshold level we can use a voltage divider. $V_{TH}$ is often expressed as some fraction of $V_{cc}$. (Recall that $V_{cc}$ stands for common collector and is equal to our high signal level, typically 5 V.)

![Figure: A circuit with $+V_{cc}$ at the top feeding resistor $R_1$. The node between $R_1$ and $R_2$ is labeled $V_{TH}$ and connects to the "−" input of a comparator; $V_{in}$ connects to the "+" input. $R_2$ connects the $V_{TH}$ node to ground. The comparator output is labeled $V_{out}$.](assets/images/comparator3.png)

From KVL:

$$V_{cc} = I R_1 + I R_2 \quad \text{(Eq 1)}$$

and

$$V_{TH} = I R_2 \;\Leftrightarrow\; I = \frac{V_{TH}}{R_2} \quad \text{(Eq 2)}$$

Rearranging and substituting (2) into (1) gives:

$$V_{cc} = \frac{V_{TH}}{R_2}(R_1 + R_2)$$

$$V_{TH} = \frac{R_2}{R_1 + R_2} V_{cc}$$

Concept Check Orange Box with Circle Exclamation icon
:::{admonition} Concept Check:
:class: attention
What will the threshold voltage of the comparator above be if $R_1 = R_2$?
:::

Green Solution with light bulb icon
:::{admonition} Solution
:class: hint, dropdown
If $R_1 = R_2$, then the threshold level is exactly $\frac{1}{2}V_{cc}$.
:::




## Comparator Output and The Problem with Noise

Let's examine what happens as we raise the input voltage from 0 to $V_{cc}$ and then lower it again for the specific case where $R_1 = R_2$.

![Figure: A timing diagram. $V_{in}$ is a triangle wave crossing a dashed horizontal line labeled $\frac{1}{2}V_{cc}$. Below it, $V_{out}$ is a square pulse that is high while $V_{in}$ is above the threshold, with dashed vertical lines marking the crossing points.](assets/images/comparatortiming1.png)

We can see that the comparator gives us a clear digital signal even though the the voltage level varies continuously. We get a nice, clear transition from a low output to high, and back again. However, we've assumed that the input voltage rises and lowers perfectly smoothly. In a more realistic scenario, the inherent noise in any input signal can lead to very undesireable behaviour.

![Figure: A timing diagram. $V_{in}$ is a noisy, jagged waveform hovering around the threshold line. Below it, $V_{out}$ shows multiple rapid pulses.](assets/images/comparatortiming2.png)

The noise causes very rapid flickers between the on and off states as it makes the input quickly vacillate above and below the threshold. This behaviour could very easily lead to circuits not functioning correctly. How can we design a circuit that can correct this noisy behaviour?

## Schmitt Triggers

To deal with noisy signals (which all real-world signals are) we use what is known as a Schmitt Trigger. The comparator is combined with feedback to create two different trigger levels depending on the current output state. This concept is known as a hysteresis loop: To go high has one trigger level, but once that high threshold is reached it switches to a different lower threshold to go low once again. As long as the two thresholds are far enough apart, noise should not cause undesired transitions, as shown in the timing diagram below.

![Figure: A timing diagram. $V_{in}$ is a noisy waveform rising and falling across two dashed horizontal lines: a red "High Threshold" and a green "Low Threshold" below it. $V_{out}$ is a single clean pulse: it goes high when $V_{in}$ crosses the high threshold and returns low when $V_{in}$ falls below the low threshold.](assets/images/comparatortiming3.png)

The hysteresis loop can be represented on a graph like so:

![Figure: A graph of $V_{out}$ (y-axis, marked High and Low) versus $V_{in}$ (x-axis, marked Low Threshold and High Threshold). A rectangular loop with arrows shows the output switching high at the high threshold and low at the low threshold.](assets/images/hysteresis.png)

Any logic gate that uses a Schmitt Trigger at its input is symbolized using this loop. For example: a Schmitt NOT Gate aka a Schmitt Inverter.

![Figure: An inverter symbol (triangle with a bubble on the output) with a small hysteresis-loop symbol drawn inside the triangle.](assets/images/SchmittInverter.png)

## Schmitt Inverter Construction

A Schmitt Inverter is made using a comparator and feedback.

![Figure: A circuit with $+V_{cc}$ at the top feeding $R_1$. The node below $R_1$ is labeled $V_{TH}$ and connects to the "+" input of a comparator; $V_{in}$ connects to the "−" input. A feedback resistor $R_2$ connects the output $V_{out}$ back to the $V_{TH}$ node, and $R_3$ connects the $V_{TH}$ node to ground. An arrow notes: \*Note the orientation of the comparator!](assets/images/SchmittInverter2.png)

So how does this work? Let's start by assuming $V_{in} < V_{TH}$.

This means that $V_{out} = \text{High} = V_{cc}$.

What is the threshold level that we would need to reach to change the output to low?

First, we note that since $V_{out}$ is High, current will flow from the output, through $R_2$ and then to ground. From KCL then we have:

$$I_1 + I_2 = I_3 \quad \text{(Eq. 1)}$$

where the subscript denotes current flowing through the corresponding resistor.

From KVL we have:

$$V_{cc} = I_1 R_1 + V_{TH} \quad \text{(Eq. 2)}$$

$$V_{out} = I_2 R_2 + V_{TH} = V_{cc} \quad \text{(Eq. 3)} \quad \text{(because we are starting with } V_{in} < V_{TH}\text{)}$$

Rearranging (2) gives:

$$I_1 = \frac{V_{cc} - V_{TH}}{R_1} \quad \text{(Eq. 4)}$$

and doing the same for (3) yields:

$$I_2 = \frac{V_{cc} - V_{TH}}{R_2} \quad \text{(Eq. 5)}$$

From Ohm's Law,

$$V_{TH} = I_3 R_3$$

$$V_{TH} = (I_1 + I_2) R_3 \quad \text{(from Eq. 1)}$$

$$V_{TH} = \left( \frac{V_{cc} - V_{TH}}{R_1} + \frac{V_{cc} - V_{TH}}{R_2} \right) R_3 \quad \text{(from Eqs. 4 \& 5)}$$

$$R_1 R_2 V_{TH} = R_2 R_3 (V_{cc} - V_{TH}) + R_1 R_3 (V_{cc} - V_{TH})$$

$$R_1 R_2 V_{TH} + R_2 R_3 V_{TH} + R_1 R_3 V_{TH} = R_2 R_3 V_{cc} + R_1 R_3 V_{cc}$$

$$V_{TH} (R_1 R_2 + R_2 R_3 + R_1 R_3) = (R_1 + R_2) R_3 V_{cc}$$

$$\boxed{V_{TH} = \frac{(R_2 + R_1) R_3}{R_1 R_2 + R_2 R_3 + R_1 R_3} \, V_{cc}}$$

This is the high threshold level. For the case where $R_1 = R_2 = R_3$, then $V_{TH} = \frac{2}{3}V_{cc}$.

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Once $V_{in}$ is raised past that high threshold level, the output of the comparator switches to low. What does our new threshold level become when $V_{out} = 0$???

Since $V_{out}$ is now 0 V, current will flow the other way through $R_2$, and so our KCL equation becomes:

$$I_1 = I_2 + I_3 \quad \text{(Eq. 1)}$$

Ohm's Law gives

$$V_{TH} = I_3 R_3 \quad \text{(Eq. 2)}$$

$$V_{TH} = I_2 R_2 \quad \text{(Eq. 3)}$$

And KVL gives

$$V_{cc} = I_1 R_1 + V_{TH}$$

$$V_{cc} - V_{TH} = (I_2 + I_3) R_1$$

$$V_{cc} - V_{TH} = \left( \frac{V_{TH}}{R_2} + \frac{V_{TH}}{R_3} \right) R_1$$

$$R_2 R_3 V_{cc} = R_2 R_3 V_{TH} + R_1 R_3 V_{TH} + R_1 R_2 V_{TH}$$

$$\boxed{V_{TH} = \frac{R_2 R_3}{R_1 R_2 + R_2 R_3 + R_1 R_3} \, V_{cc}}$$

This is our new low threshold level. If $R_1 = R_2 = R_3$, this is $\frac{1}{3}V_{cc}$.

Therefore, the inverter won't change from low to high until the input voltage falls below the new threshold.