SchmittTriggerAll circuits we have discussed so far are having an op-amp inopen loop operation at its core. Since open loop gain is very high, even for asmall Vd , output will change. This means that a steady signalhovering around (just above or below) threshold may trigger the circuit manytimes due to noise. For example, in a zero-crossing detector, input sine wavemay have HF noise superimposed on it.
So, during the expected zero crossing ofinput, due to noise, actual amplitude may cross zero more than once. This willtrigger the circuit more than once, something that we don’t wish for as we areinterested only in the number of timesthe signal cross zero. A solution to this problem is to implement a circuitwhich has some kind of memory. Such a zero- crossing detector circuit willchange output only if input is sufficiently high (more than some positivevoltage, say V) while increasing and only if input is sufficiently small ( lessthan a negative voltage, say -V) while decreasing. If this V is more than theamplitude of noise (which is the usual case), only actual zero crossings by thesignal trigger the circuit.
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Schmitt trigger is one such circuit which is a comparatorwith hysteresis introduced due to positive (regenerative) feedback. InvertingSchmit triggerConsider the basic Inverting Schmitt trigger circuitgiven in the left page for explanation purpose.
e., Vout = L- . Thiscauses the VP to change to ?L- . This is obviously anegative voltage. So for further rise ininput voltage, output will remain at L- . Thisvoltage at which output switches its state while input is rising may be definedas VTH (Higher Threshold) . Here, VTH = ?L+ .
Nowconsider the other case. Let the input be largely positive. Obviously theoutput will be L- . VP= ?L- .
So when the input is higher than VP , output willremain at L- . Now, when input reduces to a value less than ?L-, output suddenly switches to L+ making VP = ?L+, which is positive. Further increase of VN in the negativedirection won’t change output at VP is a positive value.Thisvoltage at which output changes its state when input is decreasing, may bedefined as VTL (Lower Threshold). Here, VTL = ?L-.As we cansee, this device show some kind of memory.
For e.g., once input is above VTH, even if it gets below VTH , output won’t change.
It must get lowerthan VTL to switch the output. Generally, |L+| = |L-| which implies |VTH| = |VTL|.Forsolving this issue, we will use external voltage to fix the value of VPas is done in the next circuit.
Operation is straightforward as we have seen in the case ofsimple circuit discussed above. Only difference is in the design procedurewhich is mentioned below:We canwrite VP in terms of VCC and Vout usingsuperposition as followsFor getting the desiredVTH and VTL , replace VP with VTH ( or VTL) and Voutwith L+ ( or L- ) and solve for m and n. Choose resistorby assuming a suitable value of R, say 3.3 k? .Non – Inverting Schmitt TriggerThis also similar tothe previous case but the main difference is that input is given to invertingterminal. Consider Fig 5 a .
Here also, we use the same notations weused in the inverting case. The main thing to note here is that VPis not fixed. Using superposition, we can write VP as follows:It should be noted thatoutput switches its state when VP cross zero in either direction.Consider the input tobe largely negative. This means thatoutput will be at L- .
When we increase Vin , VPalso strart to become less negative. At one particular input voltage, VPcross zero. Corresponding input voltage is denoted as VTH . Once VPgoes positive, Vout becomes L+ , increasing VP further. This means that for furtherrise in Vin , there is no change in Vout ( Even if Vinis kept as 0 just after output has changed, Vout won’t trip back toL- as VP is still positive due to feedback) . Putting VP =0 , Vout = L- and Vin = VTH in theabove eqn, we getWhen we go in the otherdirection, Vout will remain as L+ until VPcross zero.
This crossing occur at an input voltage of VTL . Once Videcrease below VTL , output becomes L- , which furtherdecrease VP which ensure output won’t change for further decrease inVin. ( Even if Vin is kept as 0 just after output haschanged, Vout won’t trip back to L+ as VP isstill negative due to feedback)Using equation x , weget
Design can be done using the following equations.For switching, VPshould cross VN wherei.e., (Usingsuperposition)OrSubstitute Vin= VTH ( or VTL ) and Vout = L- ( orL+ ) to obtain expressionsfor threshold voltages as given below :AndSolve for R1, R2 ,Ra and R4using above expressions ( Assume some parameters to get remaining )