Analog

Schmitt Trigger Non Inverting

By CourseTron Team ·  8 min read ·  Updated Oct 2026 💬 𝕏 in f
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    In short: Both circuits wrap positive feedback around a comparator to create two switching thresholds; the inverting form sets them at ±β·Vsat with β = R2/(R1+R2) and flips the output, while the non-inverting form sets them at ±(R1/R2)·Vsat and keeps the output in phase, so one ±4 V window can be built with either topology.

    Why a bare comparator is not enough

    A comparator flips its output whenever the input crosses a single reference. That is fine for a clean, fast edge, but real signals carry noise. Imagine a sensor slowly rising through 2.5 V with 40 mV of ripple on top. Each time the ripple pushes the signal back across the reference, the output toggles, so one slow crossing becomes a burst of edges. A downstream counter, interrupt line or motor driver then sees dozens of events instead of one.

    A Schmitt trigger cures this by giving the circuit memory. Once the output has switched, the reference moves away from the input, so the signal must travel a deliberate distance (the hysteresis window) before the output can switch back. If your window is wider than the peak-to-peak noise, chatter disappears. The classic circuits for this are the inverting Schmitt trigger and its non-inverting cousin, and interviewers love asking you to derive both on a whiteboard.

    Inverting Vin − + R2 Vout R1 β = R2 / (R1 + R2) VUT = +β·Vsat VLT = −β·Vsat Non-inverting Vref = 0 − + Vin R1 Vout R2 k = R1 / R2 VUT = +k·|Vsat| VLT = −k·|Vsat| CourseTron · coursetron.com
    Figure 1. Same op-amp, different feedback wiring. The inverting circuit feeds the signal to the minus pin; the non-inverting circuit feeds it through R1 into the same node that carries the feedback.

    Inverting Schmitt trigger: thresholds from a divider

    Here the signal drives the inverting pin and the output is divided down by R1 and R2 onto the non-inverting pin. The reference seen by the comparator is therefore V+ = β·Vout, where β = R2/(R1+R2). With the output saturated at +Vsat the reference is +β·Vsat; when the input rises above it, the output falls to −Vsat and the reference jumps to −β·Vsat. The input must now fall below that lower value before the output returns high.

    Take R1 = 20 kΩ, R2 = 10 kΩ and ±12 V output swing. β = 1/3, so VUT = +4 V, VLT = −4 V and the hysteresis width is 8 V. The output is the logical inverse of the input, and the transfer plot loops clockwise.

    Key fact: hysteresis width equals β·(Vsat+ − Vsat−). Raise R1 or lower R2 and the window narrows; a very large R1 makes β tiny and the circuit degenerates into a plain comparator.

    Non-inverting Schmitt trigger: thresholds from a balance point

    Now the input enters through R1 into the non-inverting pin, and R2 returns the output to the same node, while the inverting pin sits at a reference (0 V here). The output flips exactly when the node voltage equals the reference. Superposition gives V+ = Vin·R2/(R1+R2) + Vout·R1/(R1+R2). Setting V+ = 0 yields Vin = −(R1/R2)·Vout.

    So with the output at −Vsat the circuit waits until Vin reaches +(R1/R2)·|Vsat|, then jumps high; it falls again only when Vin drops to −(R1/R2)·|Vsat|. Choose R1 = 10 kΩ and R2 = 30 kΩ with the same ±12 V swing and you again get ±4 V. The output is in phase with the input, and the loop runs counter-clockwise.

    If the inverting pin is tied to Vref instead of ground, the whole window slides: VUT = Vref·(1 + R1/R2) − (R1/R2)·Vsat−, and the width stays (R1/R2)·(Vsat+ − Vsat−). The inverting version offers the same trick by returning R2 to Vref, which gives V+ = β·Vout + (1−β)·Vref.

    Inverting transfer curve Vin Vout +Vsat −Vsat VLT −4 V VUT +4 V Clockwise loop, output inverted Non-inverting transfer curve Vin Vout +Vsat −Vsat VLT −4 V VUT +4 V Counter-clockwise loop, output in phase CourseTron · coursetron.com
    Figure 2. Orange arrows show the direction of travel. Identical thresholds, opposite output polarity: the only visible difference is which rail the output sits on while the input is low.

    Designing from a specification

    Work backwards from three numbers: the window centre, the window width and the output swing. For a single 5 V supply with a rail-to-rail comparator (0 V and 5 V outputs), suppose you need to switch up at 3.0 V and down at 2.0 V. The width is 1.0 V, so R1/R2 = 1.0/5 = 0.2. The upper threshold is Vref·(1 + 0.2) − 0 = 3.0 V, hence Vref = 2.5 V. Pick R1 = 10 kΩ and R2 = 50 kΩ and the divider on the supply that creates Vref should be much stiffer than the input network, or the reference will wander.

    PropertyInvertingNon-inverting
    Signal enters atInverting pinThrough R1 to the + node
    Feedback ratioβ = R2/(R1+R2) < 1k = R1/R2, any positive value
    Thresholds (±12 V, ratio 1/3)±4 V±4 V
    Output polarityInvertedIn phase
    Input impedanceVery high (pin only)About R1 plus feedback loading
    Input common-mode swingFull input rangeSmall, pins sit near Vref

    Key fact: the non-inverting version can have a window narrower or wider than the supply, because k is not limited to values below one, while the inverting version is capped by β < 1.

    What the waveforms actually do

    Feed a triangle wave of ±5 V amplitude into the non-inverting circuit above. While it climbs, nothing happens until it crosses +4 V, then the output slams to +12 V. The input keeps rising to +5 V, turns around, and the output stays high through 0 V, and falls only when the input reaches −4 V. A noisy input with 1 V peak-to-peak ripple near 0 V would be ignored completely, because the output has no reason to move until the signal travels 8 V from its last switching point. The same waveform through the inverting circuit produces a mirror-image output.

    Non-inverting Schmitt in motion: ±4 V thresholds VUT +4 V VLT −4 V HIGH LOW Vin Vout Output is HIGH Output is LOW CourseTron · coursetron.com
    Figure 3. The vertical cursor scans the triangle input. The output rises only at the upper threshold on the way up and falls only at the lower threshold on the way down, never at 0 V.

    Practical pitfalls and verification

    • Real saturation levels. A general-purpose op-amp such as the LM358 saturates roughly 1.5 V inside its rails, and the two rails are rarely symmetric, so measured thresholds differ from the ideal ±Vsat figures. Always compute with the actual high and low output levels.
    • Open-collector comparators. Parts like the LM393 need a pull-up. That pull-up forms a divider with the feedback resistor, so make it small compared with the feedback path or include it in the maths.
    • Speed. Op-amps used as comparators recover slowly from saturation. A dedicated comparator gives cleaner edges, and a few picofarads across the feedback resistor speeds the regenerative switch.
    • Input impedance. The non-inverting form loads the source with roughly R1, so drive it from a low-impedance stage or the thresholds will shift.
    • Digital cousins. Gates like the 74HC14 build the same hysteresis inside the input stage, which is why they clean up slow clocks and push-buttons.

    Before tape-out or a PCB spin, verify with simulation: run a transient sweep with a slow ramp plus injected noise in Cadence Spectre inside Virtuoso ADE, Synopsys HSPICE or ngspice, and plot Vout against Vin to confirm both threshold values and the loop direction. Siemens EDA's Eldo and Questa ADMS offer equivalent flows. Corner and Monte Carlo runs reveal how resistor tolerance moves the window; 1 % parts typically shift a 4 V threshold by well under 100 mV, but offset voltage of a cheap op-amp can easily dominate.

    For further depth on the trade-offs between amplifier and comparator topologies, see op-amp comparators and hysteresis design. If you want hands-on lab practice and placement support, the analog track at ChipXpert VLSI training pairs well with these notes.

    Ready to design comparators, oscillators and mixed-signal blocks with confidence? Explore the structured analog modules in our CourseTron courses and practise every circuit in this article with guided simulations.

    Frequently asked questions

    Which Schmitt trigger should I choose?

    Choose the inverting form when you want a very high input impedance and do not mind an inverted output. Choose the non-inverting form when you need the output in phase, thresholds set by a ratio above one, or a window shifted by a reference voltage.

    Can both circuits give identical thresholds?

    Yes. With the same output swing, an inverting circuit with β = 1/3 and a non-inverting circuit with R1/R2 = 1/3 both switch at ±Vsat/3. Only the output polarity and input loading differ.

    How wide should the hysteresis window be?

    Make it comfortably larger than the worst-case peak-to-peak noise at the comparator input, plus the offset and resistor tolerance error, but small enough that the signal still crosses it. Two to three times the noise is a common starting rule.

    Why does my Schmitt trigger oscillate on the output?

    Usually the supply or reference is coupled to the output edge, or the feedback is too strong for a slow op-amp. Add supply decoupling, keep the reference node stiff, and prefer a real comparator for fast signals.

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