 Welcome myself Giridhar Jain, Assistant Professor in Electronics and Telecommunication Engineering, Vulture Institute of Technology, Sallapur. Now today, I am going to explain Holtage to Holtage Converters. Learning outcomes of this session are, at the end of this session, students will be able to draw, circuit and derive expressions for output voltage for fee-to-fee converter. And second output, students will be able to design Holtage to Holtage converter. Contents of this session are, fee-to-fee converter circuit and transfer characteristics, derivation of output Holtage and design of fee-to-fee converter. So, this figure shows the circuit diagram for fee-to-fee converter. This is first op-amp and this is a second op-amp. So, 741 is used as an operational amplifier. If we look at the circuit, the first operational amplifier is connected as inverting summing amplifier, where there are two inputs, one is fee-in and second is fee-os. Fee-os is the offset Holtage, which is obtained from a potentiometer RP, where variable point is connected, where there is a fee-os and plus fee and minus fee are the power supplies. Second operational amplifier is connected as a inverting amplifier with RF is equal to R and R1 is equal to R. Therefore, gain of the second stage is minus 1 and this is the transfer curve. So, output versus input. When input changes from fee-a to fee-b, there is a corresponding change in output is fee-o-a to fee-o-b as shown in figure. Derivation for output Holtage, first operational amplifier is a inverting summing amplifier. Therefore, fee-o-1 is given by fee-in minus RF by RI minus fee-o-s RF by ROS. Therefore, taking this minus sign outside fee-o-n equal to fee-in RF by RI plus fee-o-s RF by ROS. Second op-amp is inverting amplifier with gain of minus 1. Therefore, fee-o is equal to fee-in RF by RI plus fee-o-s RF by ROS. Now, if RF is equal to ROS, then fee-o becomes equal to fee-in RF by RI plus fee-o-s. Therefore, this is equation of straight line Y is equal to MX plus C, where M is the slope of straight line, which is M is equal to RF by RI and C equal to fee-o-s. So, C is the intercept on Y axis. So, here Y is fee-o and X is fee-in. Now, slope of the transfer characteristics can be obtained by this formulae. So, this is transfer curve. Input varies from V-A to V-B and output varies from fee-o-A to fee-o-B. Therefore, slope M is given by delta fee-o upon delta fee-in that is fee-o-B minus fee-o-A divided by fee-B minus fee-A. Now, pause this video and think on following question. What is maximum output voltage of this fee-to-fee converter circuit? Maximum output voltage of any operational amplifier circuit is limited by plus fee-sat and minus fee-sat of the op-ohm used in the circuit. So, in this circuit, 741 is used as op-ohm and plus fee-sat and minus fee-sat depends on the plus minus fee-c-c. So, it is approximately 1 volt less than the fee-c-c. So, design of fee-to-fee converter. So, let us solve this design example. So, design interfacing circuit for conversion of 2 to 6 volt input voltage into a equivalent of 0 to 5 volt. Now, solution, the circuit diagram for which we have made the derivation earlier, same circuit is used and for the circuit from the given data. So, it is given that the input changes from 2 to 6 volt and output changes from 0 to 5 volt. So, from this, a slope m is obtained by the formulae m equal to delta fee-o divided by delta fee-in that is fee-o-B minus fee-o-A divided by fee-B minus fee-A. Substituting the values, fee minus 0 divided by 6 minus 2 equal to 5 by 4 that is 1.25. Therefore, m is equal to RF by RI 1.25. Therefore, RF is equal to 1.25 RI. Now, select RI is equal to 1 kilo ohm. Therefore, RF is equal to 1.25 kilo ohm. To find the offset voltage fee-o-S, we use the equation fee-o-B equal to fee-in-B into RF by RI plus fee-o-S. Substituting the values in this equation. Fee-o-B is 5 volt. At that time, fee-in-B is 6 volt. RF by RI already we have obtained as a 1.25. So, plus fee-o-S. Solving this equation, we get fee-o-S equal to minus 2.5 volt. Now, this is the transfer curve of the design. So, in this transfer curve, input voltage varies from 2 to 6 volt and corresponding output voltage is 0 to 5 volt. So, this is transfer curve of the design. Now, complete circuit diagram and the component list is given as below. So, this is the complete circuit diagram. So, in this circuit diagram, the component values which we have calculated earlier are RF is equal to ROS that is equal to 1.25 kilo ohm. Now, all the resistance used in the design are quarter watt. Now, after selecting RF and ROS, let us calculate RI. Now, RF is 1.25 RI. Therefore, RI is equal to 1 kilo ohm. Now, in the second design, this is resistance R. This is resistance R. So, both are equal. Therefore, RI is equal to R is equal to 1 kilo ohm. Now, this is R compensation. So, R compensation is obtained by taking the parallel combination of RF and RI. So, it is parallel combination of 1.25 K and 1 K which comes approximately to 560 ohms. Now, RP. RP is a potentiometer. So, 10 kilo ohm linear potentiometer is used as a RP. Now, select power supply voltages as plus phi is equal to 12 volt and minus phi is equal to minus 12 volt. So, these are the power supply voltages plus 12 and minus 12. So, now this is the complete circuit and these are the component values. All resistors are quarter watt and this power supply is plus 12 volt and minus 12 volt is a regulated power supply. These are references, operational amplifier and linear integrated circuits by Ramakan Gaiquat, PHI publisher and second is electronic system design by Vaibhav Tarate, Electrotech Publications Satara. Thank you for watching this video.