in this video we will learn how to design our really good and how come experiment described here after doing this experiment we should be able to design and test a voltage control with a circuitous pitch control contribute or a half-hour control rectifier should be able to identify the major features issues and suggest modifications for different situation like what if we make it a full wave control to differ and so on this triggering circuit is used for obtaining a triggering pulse at a decide phase shift angle or time proportional to a control voltage we control that means there is a weak control and this week control is going to roll the alpha so here is your triggering circuit again crepitus as a black box concept for the circuit when I changed the control voltage this alpha the position of the Turing purse is also changed cute that we intend to apply to during circuity is a half a controlled rectifier then AC source and a CR and a simple R o our Lord we are going to apply this pulse to the gate with respect to the cathode in this experiment we are not worried about the isolation requirements here because we will be focusing on generating the or Spallone will concentrate our attention on the circuit here this complete circuit the partition this acute in two major stages you can see that you have a current source here made up off a try dear transistor being busy one and seven this is the current source okay so this current source is basically charging this capacitor here in charge so that is where you get the RAM voltage then we have another section which is a synchronizing acute is used to reset the ramp which comprises of this secured here so this is the cute okay please use to reset the ramp while we need a synchronizing so good because I'd if this is the input AC voltage our getting pulse at every I could should be exactly the same position with respect to the previous zero crossings the chrono synchronizing our get pushed to the zero crossings of the input power supplies very essential okay the final and third portion here is the comparator circuit here is a comparator circuit which produces the gating purse ok so these are the three important stages in this circuit a current source is synchronizing circuit and a comparator let us understand how this works with the basic concept piece to have a ramp like this ok this is synchronized to the zero crossings are synchronized to the zero crossings of the input image is having a height a ramp case to be compared with a control voltage control and the comparison point we get a pulse the position of the pearls is from this 0 which is synchronised to the zero crossing of the sine wave so if you superimpose a sign right here and see that respect to the zero we controlled rectifier we don't want pulse in the entity her sacred so during this entire period the RAM will be reset and again at the zero crossing the RAM will start so the start of the ramp is always synchronized to the zero crossing of the input sine-wave and when I in the control voltage up and see that the crossing point and the comparison operation is moving up for that the edge of the pulse is also moving right so you increase the alpha when I increase the control voltage will increase our decrease the control voltage I will decrease the alpha okay so this is the basic principle used in most of the PWM Isis used in power electronics okay so first we need a ram generator I'm genrika current source capacitor okay when the capacitor voltage reaches the required Ram voltage via ramp what we'll do is we'll simply reset this capacitor voltage so you produce a ramp perfect ramp so for that we need a constant current source and a constant current source can be made using a transistor because you know that when you bias the transistor in this constant current region you can make it work as a constant current source - re are chosen such a way that we transistor BC one sensor and produce a required current but how do we fix that current before that we need to look at the ramp required ramp we know that the a AC waveform is having a cycle of T by 2 and we want our clamp to start at the beginning of this zero crossing at this little crossing and then and exactly T wait okay to the ramp time time period is T by T and the ramp required height is say V ramp okay so let us say V ramp is equal to 5 volts we need charge a capacitor see with I see as the charging current the voltage across a capacitor is equal to 1 by C integral you see Katie and if I keep I see as a constant then you see of t is equal to ABC in DT plus the integration constant represents the ramp so at the end of time T by to ram voltages iran so at the end of at t is equal to t by 2 we have we see of t reaching v ramp which is equal to our 5 volt this is equal to I by E into T by 2 okay so from this if we assumed see you can get current or if he assumes I can get capacitance usually we will assume a capacitor because that is available in only certain standard values let us take 1 micro farad here so the current required for this is C I I is equal to that 5 volt into C divided by T by 2 oh this is 5 into 1 microfarad / people - is 10 milliseconds so this will give us point 5 milliampere this is the current required so later if you observe in your ramp if you are not getting exact 5-volt in exit T by two who may have situation like this or situation like this we can adjust the biasing of the transistor to adjust the current so that it reaches exact old voltage required is a exactly now ready is to bias the transistor in such a way that it operates with a constant current high he is equal Familia if you look back could hear this I see is going to charge the capacitor the electric current approximately to the collector current that is missing pay and here see here the metal current flows through the reducing re here drop Gary we need to assume that voltage assume that the voltage across VR II to be around on fourth of the VCC which is equal to approximately three volts so because VCC we have chosen of 12 volt if we Ares 3 volt VR is equal to ie into re that is equal to 3 would so this means our it can be calculated risk reward divided by 0.5 milliampere which is 6 kilo ohms you know that there is no 6 chrome standard values so we will assume F 5 at that point 6 K ohms here we can calculate back back what will be the voltage across the emitter resistance which is equal to 5 point 6 K into point 5 milliampere which will be run to 0.8 volt Ahri's fixed us trying to be I point 6 K okay now we need to design our own and r2 for that we assume a current I do which is approximately 1 point on milliampere this is much larger than the base current so with respect to the base current with respect to this I to base current is negligible so we may approximate this current i to us the supply voltage divided by the sum of resistance actually there is a current I'm branching out from this yangshin but for the design approximation we will neglect that base current and IO R 1 plus R 2 is the total resistance across VCC so I 2 we will assume as point 1 milliampere and it that means I 2 is equal to VCC divided by total resistance a path which means R 1 plus R 2 is equal to VCC divided by point on milliamp ampere which is equal to 12 divided by point 1 milliampere this is 120 kick you need to look at the base voltage here maybe from in this loop if you write the equation you see see - VRE - the be a drop point seven EB is equal to VB so VB can be written as here VB can be written as ECC - a re - a B which is equal to 12 minus or 2.8 volt ER e is he which is points on what which is equal to 8 point 5 volt now because we have the potential divider here r1 r2 you see see here and this point is connected to the base we have a potential divider principle applying here so VB from that principle VB is equal to VCC into r2 divided by told us around us are two which means that which means that our to can be calculated at EB into r1 plus r2 divided by EC which is equal to VB is 8 point 5 volt - 120k very big 12 oh this is to fight kilo-ohms who are two can be two saunas the standard value nearest to 82 is 85 is 82 K Oh artist useless 82 case with an r1 is equal to 120 km minus 82 K which is 38 K standard values r1 is equal to 39 K or the lower value 33 K can also be used only thing is that both these values you make sure that the base emitter Junction is forward biased and the collector base Junction is reverse by nothing proper transistor biasing here so this r2 is 82 K and this is 33 K or bettin in case there's a work now we have to come to the resetting circuit five and alpha can we used design for making you see or not seven act as a switch you can assume some current here I see is equal to say 1 or 2 milli ampere and IB is 1/10 of that or hard saturation you can design we can choose our success if our success in k then this can be 1/10 of that that is 1 K should be enough or also choose double this 2.2 key okay and this barrier is one and 4001 or four thousand seven in that series again our four is because we have a pulse coming here this should drive this transistor into saturation so you can put on kilohms should be enough to drive enough base current here and this resistance is to limit the discharge current of the arteries used to limit the discharge current of a capacitor and if artery is large then it will cause the ramp to look like this okay if r2 is r3 small then it will quickly discharge so this has to be a very small resistance input 10 ohm here coming to the comparator section we need to change the control voltage VCC in the range of the ramp height if the ramp height is 5 volt then the control voltage should go up to 5 volt because a bow 5 volt you won't get any comparison operations you need to limit the control voltage to 5 volt so or if you want to limit the voltage to a lower voltage that also can be done so for that you need to put a divider arrangement here one potentiometer in the bottom side and a fixer resistance on the top side so let us take that's fishing here this is our rate this is e1 this is VCC which is 12 volt so this voltage has to be a maximum so this is the control voltage okay control voltage maximum should be equal to V ramp or full 180 degree this means we control maximum is equal to e CC into e1 at its maximum point divided by our write plus e1 max this we can get right to be approximately 1.3 K I choose p1 max a 4.7 K port we are ready to be 1.3 case again 26.8 case tattered value this can be 6.8 K and finally the here which is a must for LM 311 this can be 1k or any value up to 10k do you know 1k to 10k you can choose okay but this is this complete citizen don't forget to give the - apply voltage here suppose we want to limit the maximum firing angle to some point then I let us say maximum firing angle to this point then you can come calculate the corresponding voltage control voltage maximum here and then you can really design our rate in such a way that even if you vary the control voltage here to the maximum it will not exceed the maximum specified here so that your maximum firing angle can be controlled