 Welcome to this session. Myself, Mr. Giridhar Jain, Assistant Professor in Electronics and Telecommunication Engineering, Walchen Institute of Technology, Sholapur. Now, today, I am going to explain you Current-to-Holtage Converters. Now, Learning Outcomes. At the end of this session, students will be able to draw circuits and design I2V and V2I converters. Contents of this session are I2V converter with offset, design of I2V and V2I converters. First is I2V converter with offset. This is circuit. Figure shows the circuit for I2V converter with offset. Two operational amplifiers are used. Lower operational amplifier is used as a buffer. Input is applied at non-enoting terminal, which is derived from a dual power supply and a potentiometer as shown in figure. Output of the buffer is Vz. Now, that goes to this I2V converter circuit. Now, input is I in and the resistances are R1, RF, R1, RF as shown in figure. Designed formally for this I2V converter R, output voltage V0 is given by RF by R1 in the bracket I in into R span plus Vz. Means output voltage is proportional to the input current I in plus offset Vz. R span is given by Vb minus Va divided by RF by R1 into Ib minus Ia and Vz is given by Va minus RF by R1 into Ia into R span. So, these are the design equations of I2V converter with offset and this is the circuit. Let us proceed further for design of I2V converter with offset problems. Now, this is problem one. This problem states that design offset I2V converter to convert a current of 4 milliampere to 20 milliampere into a voltage of 0 to 10 volt solution. Select I2V converter with offset as a basic circuit. From the given data, the transfer curve can be plotted as shown in figure. Input is 4 milliampere to 20 milliampere and the output is 0 to 10 volt. For the given circuit, select RF by R1 is equal to 10 and next is calculation of R span. R span is calculated by the formulae, R span is equal to Vb minus Va divided by RF by R1 in the bracket Ib minus Ia. That is equal to substituting values of Va, Ia, Vb, Ib and RF by R1 we get. R span is equal to 10 minus 0 divided by 10 in the bracket 20 minus 4 bracket close into 10 to the power minus 3. This is equal to 62.5 ohm. Offset voltage Vz is given by Vz is equal to Va minus RF by R1 into Ia into R span. Substituting the values, Va is 0, RF by R1 is 10, Ia is 4 milliampere and R span is 62.5. Solving further, we get Vz equal to minus 2.5 volt. Now, select R1 is equal to 1 kilo ohm. Therefore, RF is equal to 10 times R1. Therefore, RF is equal to 10 kilo ohm. Now, as the R span is calculated as a 62.5 ohm, this is adjusted by using a potentiometer of 100 ohm, which is adjusted to 62.5 ohm. Now, let us proceed further for design problem 2. A transducer circuit produces 2.5 volt to 7.5 volt, which is to be converted into a current of 4 to 20 milliampere. Now, this current is transmitted over a wire. At a receiver end, this current is to be converted to 0 to 5 volt design circuit for above requirement. Think on solution of this design. Now, in this design, the input voltage produced by, is 2.5 to 7.5 volt produced by a transducer, which is to be converted into a 4 to 20 milliampere. Means, on the input side, V to I converter is needed. Now, this current is to be converted to 0 to 5 volt at the receiver end. Means, at receiver end, I to V converter is required. Now, this is the complete circuit, which consists of V to I converter on input side and I to V converter on the output side. Now, first op-amp is configured as I to V converter, V to I converter with offset and second circuit is I to V converter with offset. Now, this is fair reference derived from the 10 K potentiometer. Two resistances of one make. This is R-span for this circuit. Output is a current. Now, this current is transmitted through wire. This is R-span and here that current is converted into a equivalent voltage. Now, design of V to I converter. R-span is given by V B minus V A divided by 2 in the bracket I B minus I A, substituting the values of V A, I A, V B, I B. 7.5 minus 2.5 divided by 2 in the bracket 20 minus 4 bracket close into 10 raise to minus 3. This comes to be 156.25 ohm. And V reference is given by 2 into R-span into I B minus V B. This is equal to substitute the values. R-span is 156.25 ohm, I B is 20 milliampere and V B is minus here 7.5. Therefore, V reference equal to minus 1.25 volt. Now, design of I to V converter. For design of I to V converter, R-span is given by V B minus V A divided by R F by R 1 in the bracket I B minus I A. Substituting the values, R-span equal to 5 minus 0 divided by 10 in the bracket 20 minus 4 into 10 raise to minus 3. Simplifying further, R-span comes as 31.25 ohm. And V Z can be calculated by the formulae V A minus R F by R 1 into I A into R-span. Substituting the values, V A is 0, R F by R 1 is 10 and I A is 4 milliampere and R-span 31.25 volt. Therefore, V Z comes to be minus 1.25 volt. Now, select R F by R 1 is equal to 10, R 1 is equal to 1 kilo ohm and R F comes as a 10 K. Now, this is the complete circuit for the solution. On the input side, this is your V to I converter. Then output of heat to I, there is a transistor for boosting the current. And this current is transmitted over a cable and then this is I to V converter with offset. Now, in a summary for complete circuit, all the operational amplifiers are 741 operational amplifiers. Then, second is selection of the power supply. V CC is selected as a plus 12 volt and V EE as a minus 12 volt means dual power supply of plus minus 12 volt is required for the circuit. And the R-span is used as a 500 ohm potentiometer which can be adjusted to the R-span of this V to I converter and R-span of I to V converter. These two potentiometer 500 ohm here and 500 ohm for the second circuit are used. And both are adjusted separately for the required value of R-span for the respective circuits. These are references. Thank you for watching the video.