Page 308 - Mechatronics with Experiments

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where V = V sat or V =−V , the output of the amplifier is essentially always saturated
o
sat
o
depending on the signals on its input terminals (+) and (−).
R
+ 2
v = ⋅ V sat during V = V sat (5.212)
o
R + R
1 2
R 2
+
v =− ⋅ V sat during V =−V sat (5.213)
o
R + R
1 2
Let us trace the operation on the op-amp along the input–output relationship curve that has
the hysteresis loop. Assume that initially we start on the curve from the left hand side,
R 2
+
V = V , v = ⋅ V , v < 0 (5.214)
o
sat
i
sat
R + R 2
1
For the op-amp output to change state, V =−V (move to the C part of the curve), the
o sat
−
+
inverting input must slightly exceed (v = v ) > (v ). In other words, we move along B
i
−
+
on the input–output curve, where v must be larger than v = R 2 V sat > 0. Here, we
R 1 +R 2
neglect the transient response details of the change in the output state of op-amp. At that
+
point, v = R 2 V sat and V =−V . In order to return to the previous state, now the
o
sat
R 1 +R 2
inverting input must be slightly more negative than the noninverting input which is at the
+
value of v =− R 2 V . This can happen on the curve to the left side of region D. Then,
sat
R 1 +R 2
the state of the op-amp output is switched back to ON (D line along the hysteresis loop).
The noninverting Schmitt trigger circuit works with the same principle except the output
polarity is different.
Let us consider the noninverting Schmitt trigger circuit shown in Figure 5.28a. The
−
voltage on inverting terminal v is connected to ground zero. The voltage on the noninverting
+
terminal, v , is a value between V and V as follows (assuming that there is no current flow
i o
through the op-amp). The current through the circuit between V and V , through resistors
i o
R and R is as follows,
i f
V − V o
i
i = (5.215)
R + R f
i
where a positive direction of the current is assumed as the direction from V to V . Then,
o
i
+
the voltage at v , which is the point between the resistors is
+
v = V − R ⋅ i (5.216)
i
i
V − V o
i
= V − R ⋅ (5.217)
i i
R + R
i f
1
= (R V + R V ) (5.218)
i
i
f
o
R + R f
i
−
+
Now, let us consider the case starting with V ≪ 0.0, then V =−V . Then, v < 0.0 = V ,
o
sat
i
and then V =−V sat will continue to be in negative saturation output values. The only way
o
+
−
the output will switch state, that is V = V , is when v > v = 0.0, which can happen
o
sat
+
(see above equation for v )if
R V + R V > 0.0 (5.219)
i
f
o
i
R V − R V sat > 0.0 (5.220)
i
i
f
R i
V > V (5.221)
i sat
R
f
At this point and above values of V , the output V will switch to a V = V sat value. In order
i
o
o
for the op-amp output to return to the previous state, V =−V , the following condition
sat
o