Download EE 4429 Practice 3

Practice 3: Schmitt Triggers
[Background]
1. A Schmitt Trigger is essentially a comparator with hysteresis. It can be implemented by
connecting positive feedback to the non-inverting input of a comparator or differential amplifier.
In the non-inverting configuration, when the input voltage is > a chosen threshold (VTH), the
output voltage is high. When the input voltage is < another chosen threshold (VTL), the output
voltage is low. When the input voltage is between VTL and VTH, the output voltage maintains its
present value. This dual-threshold behavior is called hysteresis.
In fact, we need to consider 2 output voltage levels, VOH and VOL, when we look at VTL and VTH.
Furthermore, we should also consider the inverting configuration. The VTC for both
configurations are shown in the figures below:
The left VTC is for an inverting Schmitt Trigger; the right VTC is for a non-inverting trigger.
The VTC of Schmitt Trigger dictates that its operation must “cross” the hysteresis region. We
define the hysteresis voltage VH ≡ VTH – VTL and mid-hysteresis voltage VM ≡ (VTH + VTL)/2.
The positive feedback increases the switching speed, i.e. a slowly changing input wave form will
become a fast (steep) changing output wave form. The hysteresis region reduces impact by input
noise (as determined by VH).
The major difference between a simple comparator and the Schmitt Trigger is how they respond
to input noise. If the input signal Vin shows noise near VT, then the output signal Vout will show
some corresponding chattering (i.e. bouncing between low and high), for the simple comparator
in (a). The Schmitt Trigger, as in (b), shows more robustness.
(a)
(b)
2. There are many ways to implement the Schmitt Trigger. Let’s look at some op-amp
implementations, starting with the non-inverting one.
In this circuit, R1 and R2 form a parallel voltage summer. This parallel positive feedback creates
the hysteresis controlled by the proportion of R1 and R2. Since Vout is referenced to ground, this
circuit does not need an amplifier with a differential input. Because conventional op-amps have
a differential input, the inverting input is grounded to ensure the reference point is 0 V.
Vout will always follow the sign of the op-amp input voltage, but not necessarily the circuit input
voltage (i.e. the 2 input voltages can have different signs!) When Vin is > VTH or < VTL, Vout has
the same sign has Vin (since this circuit is non-inverting). When VTL < Vin < VTH, Vout is
undefined and depends on its last state – i.e. here this circuit works like a latch.
From V+ = V- we know that:
0 = Vin/R1 + Vout/R2 ⇒ Vin = - Vout (R1/R2) which gives
(1) When Vout = VOH, Vin = - VOH (R1/R2) = VTL
(2) When Vout = VOL, Vin = - VOL (R1/R2) = VTH
For this circuit, we obtain VH = VTH – VTL = (R1/R2)(VOH – VOL).
We also get VM = (1/2)( VTH + VTL) = -(1/2) (R1/R2)(VOH – VOL).
We also know that VOH = +VS, the positive power supply voltage, and that VOL = -VS, because
these output voltages are obtained directly from the op-amp output. Thus VTL = -VS(R1/R2) and
VTH = +VS(R1/R2).
From this derivation, we see that for this Schmitt Trigger, T = VTH and –T = VTL, which defines
the horizontal boundaries of the Trigger’s switching band. M, on the other hand, is VS, the
positive power supply voltage itself. The band is centered around the origin. We can shift the
band to left and right by applying a bias voltage to the op-amp’s inverting input, creating a nonzero reference point.
3. The other commonly seen implementation is the inverting op-amp configuration. The
following figure shows it.
In this case, R1 and R2 work entirely like a voltage divider. The input loop acts as a simple series
voltage summer that adds a part of Vout in series to the circuit input voltage Vin. This series
positive feedback creates the needed hysteresis controlled by the proportion between R1 and
(R1+R2). Because the effective applied to the op-amp input is floating, the op-amp must have a
differential input in this configuration.
This circuit is inverting since Vout always has an opposite sign to Vin when it is out of the
hysteresis (i.e. its switching band), when Vin is > VTH or < VTL. When VTL < Vin < VTH, Vout is
undefined and depends on its last state – i.e. here this circuit works like a latch. The circuit can
be inverting as well as non-inverting (why?)
Similar to the non-inverting configuration, from V+ = V- we know that:
Vin = Vout (R1/(R1+R2)) which gives
(1) When Vout = VOH, Vin = VOH (R1/(R1+R2)) = VTH
(2) When Vout = VOL, Vin = VOL (R1/(R1+R2)) = VTL
For this circuit, we obtain VH = VTH – VTL = (R1/(R1+R2))(VOH – VOL).
We also get VM = (1/2)( VTH + VTL) = (1/2) (R1/(R1+R2))(VOH + VOL).
We also know that VOH = +VS, the positive power supply voltage, and that VOL = -VS, because
these output voltages are obtained directly from the op-amp output. Thus VTL = -VS(R1/(R1+R2))
and VTH = +VS(R1/(R1+R2)). Note that the circuit will switch when Vin = V+.
The T and M remain the same in quantities. But the VTC for inverting Schmitt Trigger is the
mirror image (w.r.t. the vertical axis) of that for the non-inverting trigger.
What happens if R1 = 0 and R2 → ∞?
Compared to the non-inverting trigger, this circuit does not have impact on the input Vin, as Vin
is separated from the voltage divider output by the high op-amp input differential impedance.