# Schmitt Trigger Here’s another simple, common, and powerful circuit. The Schmitt trigger, named after its inventor (Otto H. Schmitt), is a system with one output and one input that uses positive feedback. We implemented our Schmitt trigger with an op amp and two resistors, but numerous other implementations exist.

## Behavior

Let’s find out how the Schmitt trigger behaves by analyzing it with the 3-state-based ideal op amp model. Recall the regions:

• Region 1: $V_+ > V_-$ in which $V_{out} = V_{dd}$
• Region 2: $% $ in which $V_{out} = V_{ss}$
• Region 3: $V_+ = V_-$

And note from the circuit diagram that $V_- = V_{in}$ and $V_+ = V_{out}$.

#### Region 1: $V_+ > V_-$

In this region, $V_{out} = V_{dd}$. The circuit enforces $V_+ = aV_{out}$, so

$V_+ = aV_{dd}$.

Plugging this back into the region 1 boundary conditions, we find that region 1 holds if

$% $.

#### Region 2: $% $

Following the identical procedure for region 1 we arrive at the boundary conditions

$V_{in} > aV_{ss}$.

#### Region 3: $V_+ = V_-$

Plugging in $V_+$ and $V_-$:

## Input/Output Plot The plot above shows how these conditions manifest when $a=0.5$. Our input/output curve is no longer a function! For a given $V_{in}$, there could be up to three values of $V_{out}$.

How do we know which value $V_{out}$ will take? The current value of $V_{out}$ will depend on its past value. Say $V_{in}$ starts at $V_{dd}$. As $V_{in}$ decreases, it passes $aV_{dd}$, at which point there are three possible values of $V_{out}$ – three possible equilibrium states of the circuit. But the circuit is stable at its current equilibrium, and has no reason to change the output. So the output remains at $V_{ss}$ until $V_{in}$ passes $aV_{dd}$. At this point, there is only one possible value of $V_{out}$$V_{dd}$.

The vertical lines on the plot show where the output voltages will transition from low to high or high to low, with the arrows indicating the direction. The area between the lines is called the hysteresis band, as the value of $V_{out}$ in this band is determined by its value before $V_{in}$ entered the band. The value of $a$ changes the width of the hysteresis band – a larger $a$ will make it wider and a smaller $a$ will make it thinner. At $a=0$, this circuit essentially becomes an inverting comparator with the non-inverting input tied to Ground.

This explanation doesn’t cover the case where $V_{in}$ and $V_{out}$ both start at $0V$. Following our earlier reasoning, it might make sense to think that if the circuit starts in equilibrium 3, it will stay there for as long as it can (as long as $V_{in}$ is in the hysteresis band). However, because of the positive feedback, equilibrium 3 is unstable. Any noise that disturbs the inputs of the op amp will cause $V_{out}$ to diverge, bringing the circuit to either equilibrium 1 or 2. In the real world, there will always be some noise, so equilibrium 3 is never seen in practice. In many Schmitt trigger input/output curves, this diagonal line is not even drawn (but it does exist, and it will show up in simulations!).