Schmitt Trigger - OpAmp based switches exhibiting hysteresis

A Schmitt trigger is a comparator with intentional positive feedback. Instead of a single switching threshold, it has two: one for rising inputs and one for falling inputs. This creates hysteresis. This is often needed when a signal is noisy or slowly varying. Without the hysteresis the output can chatter rapidly around a single threshold and cause digital inputs to flap.

In this post we derive the model behaviour of the two common operational amplifier (OpAmp) configurations (inverting and non-inverting) using the idealized amplifier model. We assume infinite input impedance of the operational amplifier, i.e. zero bias currents, as well as an infinite open loop gain. The amplifier is assumed to swing to it’s saturation voltage - which is assumed to be the rail voltage in this example. For a practical circuit you will have to check which output levels you really need. The output will only change between the upper and lower saturation voltage. In addition you will have to verify if the input bias current is negligible.

In this blog article we are going to look into:

• The inverting Schmitt trigger and it’s characteristics
• And the non inverting Schmitt trigger
• A short conclusion

Inverting Schmitt Trigger

First let’s recall the output of an optimal operational amplifier. The open loop gain is assumed to be $G \to \infty$. This means that the operational amplifier will swing to it’s maximum or minimum output voltage (i.e. close to the supply rail) depending on the inbalance between the inverting input $U_{-}$ and the non inverting input $U_{+}$:

[ \begin{aligned} U_{out} &= \begin{cases} U_{vcc} & \text{if } U_{+} \gt U_{-} \\ -U_{vcc} & \text{if } U_{+} \lt U_{-} \\ \end{cases} \end{aligned} ]

The voltage on the inverting input is given by the input voltage $U_{-} = U_{in}$. The voltage on the non inverting input can be described via a simple voltage divider between the output voltage $U_{out}$ and the bias voltage $U_{b}$:

[ \begin{aligned} I_{g} &= \frac{U_{out} - U_b}{R_1 + R_2} \\ U_{+} &= R_2 I_g + U_b \\ &= \frac{R_2}{R_1 + R_2} (U_{out} - U_b) + U_b \\ &= \frac{R_2}{R_1 + R_2} U_{out} + \left( 1 - \frac{R_2}{R_1 + R_2} \right) U_b \end{aligned} ]

Taking a look at the two possible output cases one can determine the switchpoints of the circuit.

In the first case case $U_{out} = U_{vcc}$ one gets:

[ \begin{aligned} U_{out} &= U_{vcc} \\ U_{+} &= \frac{R_2}{R_1 + R_2} U_{vcc} + U_b \left( 1 - \frac{R_2}{R_1 + R_2} \right) = U_A \end{aligned} ]

Switching happens when $U_{in} > U_A$.

In the second case $U_{out} = -U_{vcc}$:

[ \begin{aligned} U_{out} &= -U_{vcc} \\ U_{+} &= -\frac{R_2}{R_1 + R_2} U_{vcc} + U_b \left( 1 - \frac{R_2}{R_1 + R_2} \right) = U_B \end{aligned} ]

Under assumption of a symmetric supply voltage a look at the width of the hysteresis curve yields

[ \begin{aligned} \Delta U &= (U_A - U_B) \\ &= 2 \frac{R_2}{R_1 + R_2} U_{vcc} \end{aligned} ]

The center of the hysteresis is given by $U_b$. As one can see the width only depends on the resistor values, not on the bias voltage.

Non Inverting Schmitt Trigger

The output of the operational amplifier is again determined by the inbalance between the inverting input $U_{-}$ and the non inverting input $U_{+}$. Again assuming infinite open loop gain $G \to \infty$ the amplifier simply swings to it’s maximum or minimum output voltage:

[ \begin{aligned} U_{out} &= \begin{cases} U_{vcc} & \text{if } U_{+} \gt U_{-} \\ -U_{vcc} & \text{if } U_{+} \lt U_{-} \\ \end{cases} \end{aligned} ]

Applying the voltage divider like in the inverting case again:

[ \begin{aligned} I_{g} &= \frac{U_{out} - U_{in}}{R_1 + R_2} \\ U_{+} &= R_2 I_{g} + U_{in} \\ &= \frac{R_2}{R_1 + R_2} U_{out} + \left(1 - \frac{R_2}{R_1 + R_2} \right) U_{in} \end{aligned} ]

The switch points between the two states are now a bit more complex than in the inverting case due to the contribution of the input signal.

In the first case $U_{out} = U_{vcc}$ we get

[ \begin{aligned} U_{out} &= U_{vcc} \\ U_{+} &= \frac{R_2}{R_1 + R_2} U_{vcc} + \left(1 - \frac{R_2}{R_1 + R_2} \right) U_{in} \end{aligned} ]

The switch happens when $U_{+} \lt U_b$:

[ \begin{aligned} U_{+} &\lt U_b \\ \frac{R_2}{R_1 + R_2} U_{vcc} + \left(1 - \frac{R_2}{R_1 + R_2} \right) U_{in} &\lt U_b \\ U_{in} &\lt \underbrace{\frac{U_b - \frac{R_2}{R_1 + R_2} U_{vcc}}{\left(1 - \frac{R_2}{R_1 + R_2} \right)}}_{U_A} \end{aligned} ]

In the second case $U_{out} = -U_{vcc}$ we get

[ \begin{aligned} U_{out} &= -U_{Vcc} \\ U_{+} &= -\frac{R_2}{R_1 + R_2} U_{Vcc} + \left(1 - \frac{R_2}{R_1 + R_2} \right) U_{in} \end{aligned} ]

The switch happens when $U_{+} \gt U_b$:

[ \begin{aligned} -\frac{R_2}{R_1 + R_2} U_{Vcc} + \left(1 - \frac{R_2}{R_1 + R_2} \right) U_{in} &\gt U_b \\ U_{in} &\gt \underbrace{\frac{U_b + \frac{R_2}{R_1 + R_2} U_{vcc}}{\left(1 - \frac{R_2}{R_1 + R_2} \right)}}_{U_B} \end{aligned} ]

Again we get the characteristic shape exhibiting the hysteresis as we have already seen from the inverting Schmitt Trigger.

Under assumption of a symmetric supply voltage a look at the width of the hysteresis curve yields

[ \begin{aligned} \Delta U &= (U_A - U_B) \\ &= 2 \frac{R_2}{R_1} U_{vcc} \end{aligned} ]

The center of the hysteresis is given by $U_b$. As one can see the width again only depends on the resistor values, not on the bias voltage.

Conclusion

Both Schmitt trigger configurations implement the same idea. The positive feedback creates two distinct switching points. In the inverting configuration the input goes straight into an op-amp input, which in the ideal case is assumed to have infinite input impedance. Practically, the input current is the bias current. There is no intentional resistive path from output to the input source and thus no direct back-action of the circuit to the input.

In the non inverting configuration the input is connected through a resistive divider into a summing node that is also connected to the output. This means the output can source or sink current through the resistors causing a back action of the circuit onto the input. The input impedance seen by the source is finite and depends on the present output state.

For practical applications keep in mind that dedicated components implementing Schmitt triggers exist. These are usually faster than OpAmp based circuits and contain polysilicon resistors already on the chip die, which makes the circuit simpler and more compact.

This article is tagged: Tutorial, Electronics, Basics, OpAmp