 on smith trigger using transistor. Learning outcomes are at the end of session students will be able to explain operation of smith trigger circuit as well as they can identify the difference between bistable multivibrator circuit and smith trigger circuit. These are the contents smith trigger circuit can be obtained with the help of transistor or operational amplifier. Here we will concentrate on smith trigger circuit using transistor. It consists of 2 stages of amplifier and 2 stable state. As I explained you, bistable multivibrator in the previous session, we have the same thing that is 2 stage amplifier with 2 stable state but here we do not have coupling and we do not apply any kind of trigger pulse over here. Here also we have the 2 stable state and the duration or the switching between the stable state will be determined by the amplitude of the input voltage. So, we are determines the stable state either logic 0 or logic 1. Coupling is not provided over here and positive feedback is provided through RE that is emitter bypass resistor. We have the 2 transistors and emitter circuit or emitter terminal of both transistors are short circuited to each other. So, it is called as emitter coupled and it oscillates between the logic 1 and logic 0. So, circuit is also called as binary oscillator. Figure 1 shows the circuit diagram for smith trigger using transistor. Emitter terminal of Q1 and Q2 is short circuited and R3 is your emitter bypass resistor. Here resistor R1, R2 and R4 forms the voltage divider circuit which controls the biasing or the input voltage at the base of transistor Q2. Supply voltage is applied to R1 and R out. R out can be selected based on the voltage level you want at the output and you can also use R in or base resistor at base of transistor Q1. So, how it generates the square waveform that we will see. Now, assume input voltage is 0 and supply voltage is provided to the R1 and R out. As input voltage is 0, transistor Q1 will not conduct. Now, see the base voltage. Base voltage will be nothing but the voltage I cross resistor R4. So, it can be calculated with voltage division rule like V R4 equal to V supply nothing but total input voltage into R4 divided by total resistances in the circuit. So, these are series resistances. So, voltage appear at the base of Q2 will be R4 into V supply divided by R1 plus R2 plus R4. Now, sufficient voltage will be there at base of Q2 and therefore Q2 will be on and that will be remain in the on state. Now, as you know this is on nothing but Q2 goes to the saturation level. As Q2 goes to the saturation level, it will act as a short circuit, output voltage will be reduced nothing but logic 0 will be obtained. This is the stable state where Q2 will be in the saturation and Q1 will be in the cutoff condition which gives the logic 0 stable state of the smith trigger. 4, IC will flows through the circuit and same flows to the immediate so logic 0 is obtained. Now, this is what one stable state. Now, if the input voltage is applied to the base of transistor Q1 and that input voltage is greater than the emitter voltage, then and then it is possible to drive the transistor Q1 into saturation region. Now, when input voltage is greater than VBE1 plus voltage at the emitter terminal that is VE. So, when input voltage is greater than VBE1 plus VE transistor Q1 will be now in the saturation condition that is on state. When it is in on state, its collector voltage will goes on reducing. This reduced collector voltage is applied to the base terminal of Q2 and due to that Q2 goes to the cutoff region nothing but transistor Q2 will be cutoff. So, when transistor Q2 is in the cutoff region or in the off condition it will give the increase in the collector voltage and that is approximately equal to VCC. So, we have observed previous stable state giving you the logic 0 state and now when Q2 is off and Q1 is on you will obtain the VCC that is logic 1 level. So, in this way you can switch or you can get the switching of the signal from logic 0 to logic 1 nothing but square wave at the output. But difference between the Smith-Riegel circuit and the bi-stable multivibrator is that output time period nothing but on and off time is not symmetrical. Because here we are not doing a thing with the negative half cycle that we will see in the waveform but here very important thing is that base terminal is free from the switching action. So, wherever we want the base terminal free in that case we use the Smith-Riegel circuit rather than bi-stable multivibrator circuit. Figure 2 shows the output waveforms for the Smith-Riegel circuit you can see as I told you output switches between logic 0 and logic 1 but it is not exactly logic 0 and logic 1 but it is VCE sat nothing but 0.3 volt practically based on the transistor used for the circuit and logic 1 state is VCC. As you can see output waveform nothing but on and off time are not same that is why it is the asymmetrical or it is not giving the same time for on and off time. Because we are doing nothing with the negative half cycle nothing but there is no change or no alteration in the complete negative half cycle. Output switches between VCC and VCE sat and this can be obtained with the help of voltage divider socket that is with the help of R1, R2 and R4. Here upper threshold level and lower threshold level can be calculated with the help of formula that we will see but keep in mind that transistor switching action only gives you the switching of stable state from logic 0 to logic 1 and logic 1 to again logic 0. So, whenever input voltage is greater than UTP level nothing but upper threshold level that gives the VCC signal. So, UTP is nothing but VBE plus VE and LTP is nothing but VCC minus IC2 into RE. IC2 is nothing but current flowing through the second transistor and RE is nothing but emitter bypass resistor. So, that is approximately equal to VCE sat. Now, see the formulas to calculate the higher threshold level of the circuit and the lower threshold level it can be calculated like this supply voltage into the R4 R4 is nothing but the third resistor in the voltage divider circuit divided by total resistance minus that is this complete term minus 0.62 where this is nothing but typical cutting voltage of transistor. It can be 0.7 also depends on the transistor used and lower threshold level can be calculated with the help of supply voltage into R4 total divided by R1 plus R2 plus R4 plus R1 into R4 divided by R3 plus 0.61. So, this is the lower threshold level nothing but as I told you it is VCC minus IC2 into RE but for the calculation of the resistance you can use this formula. Now, just recall the concept of hysteresis and identify or just select the correct option. So, hysteresis prevents the false triggering and that is associated with which of the following. Is it with sinusoidal input? Is it with noise voltage? Stake capacities are three points. Now, just recall the concept of hysteresis. What are the hysteresis? Hysteresis is nothing but output voltage direction changes and that gives the change in the direction of the signal. So, here it is associated with the noise voltage and that is why Smit trigger is very important to filter out the noise signal. So, correct option is noise voltage. It can be used as a window detector nothing but comparator. It could be used with the thermistor for heating control or to control the light with the help of LDR. Smit trigger convert any kind of input signal into the square wave. So, it is also called as a square or squaring circuit which also removes the noise. So, also called as filtering circuit. So, these are the applications of Smit trigger references are like this. Thank you.