 Welcome myself Mr. Giridhar Jain, Assistant Professor in Electronics and Telecommunication Engineering, WIT, Sholapur. Now, today I am going to explain, Whole Test to Current Converters. Now, learning outcomes of this session are, at the end of this session, students will be able to draw and derive expressions for output current of Whole Test to Current Converters. Now, content, these are the contents of the session, Necessity of V2I Converters, Second V2I Converter with Floating Load, Third is V2I Converter with Grounded Load and Differential V2I Converter. Now, think on the Whole Test transmission over a long wire, what problems will occur during the transmission. Now, during transmission of voltage over a long wire, the potential drop takes place across the resistance of wire and due to this, there is a reduction in voltage during transmission. Therefore, at the receiver end, there is a voltage drop due to potential drop across the resistance. So, this is the problem in transmission of voltage over a wire. Now, to solve this problem, what is done at the transmission end? The voltage is converted into current by using V2I Converter. Current is transmitted over a cable and at the receiver end, the current is again converted into a voltage. So, this will solve the problem for transmission of voltage over a long cable or wire. Now, first we will understand V2I Converter with Floating Load. So, this is circuit. Now, if we look at the circuit, circuit make use of the operational amplifier, input is applied at non-inverting terminal and for inverting terminal, they are at inverting terminal, there are two resistances RL and R1 as shown in the figure and load is floating. Means, load is connected between the output and inverting terminal, load is not connected to the ground. That is why it is called as floating load. Now, considering ideal op-amp, voltage at inverting and non-inverting terminal should be equal, that is V plus equal to V minus and voltage at V plus is V in. Therefore, this is equal to V in. Now, as the voltage at inverting terminal is V in, current I naught through resistance R1 is given by voltage across R1 divided by resistance R1. So, voltage across R1 is nothing but V in. Therefore, I naught equal to V in divided by R1. So, output current I naught will flow through resistance RL because current flowing into the inverting terminal of op-amp is 0. So, this I naught will flow through RL as shown in the figure in this direction. Therefore, current through load I naught is V in by R1 means output current or current through RL is directly proportional to the input voltage. Thus, input voltage V in is converted into current I naught. Now, second type of V2I converter is V2I converter with grounded load. So, this is the circuit for V2I converter with grounded load. So, op-amp, this is plus terminal, this is load which is connected to the ground. These are the two resistances. It is applied here and these are the feedback resistance RF and here there is R1. Now, by looking at the circuit, here load is connected to the ground. Now, this is point X. Now, applying KCL at point X, we get I1 plus I2 is equal to IL and I1 is given by voltage across R divided by resistance R that is V in minus VX divided by R and I2 is given by V0 minus VX divided by R. Substituting this values of I1 and I2. So, this is I1, this is I2, I1 plus I2 is equal to IL. Going further, we get IL is equal to V in plus V0 minus 2 V1 divided by R. Now, looking at the circuit, basic amplifier is a non-inverting type. Now, voltage at non-inverting terminal is V1. Now, gain of non-inverting amplifier is 1 plus RF by R1. So, output voltage V0 is given by gain of amplifier multiplied by V1. Now, select RF is equal to R1 is equal to R. Therefore, gain of amplifier becomes 1 plus R by R. R by R is 1. So, gain of amplifier becomes 2. Therefore, V0 becomes 2 V1. So, here V0 becomes 2 V1 and IL is given by V in, V0 is replaced by 2 V1 in the basic formulae. So, 2 V1 minus 2 V1 divided by R. So, this gets cancelled. Therefore, IL equal to V in divided by R. So, IL is the current flowing through load and it is given by V in divided by R means load current is directly proportional to the input voltage and proportionality constant is 1 by R. So, in this way the input voltage V in is converted into the proportional current through grounded load as shown in the figure. And here point to be noted is that the value of load current does not depend on the load. Whatever is, whatever is the value of load the IL will flow through that load. Now, we will study the last type of converter that is V2 I, differential V2 I converter. So, this is circuit. Now, in some applications there is a requirement of converting differential input voltage to a proportional current. So, here this is circuit. If we look at the circuit E1 and E2 are the two inputs. So, E2 minus E1 is converted into the proportional current IL through the load. This is load. Now, select R1 is equal to RF is equal to R2 is equal to R3 is equal to R means one these four resistances are taken to be R. Now, let us make the derivation for the load current. Now, as far as the circuit is concerned there are three voltages E1, E2 and VL acting simultaneously. Therefore, output voltage at the output of op-amp VO is obtained by using superposition theorem. So, here VO1 is the output voltage due to E1 acting alone. VO2 is the output voltage due to E2 acting alone. And similarly VOL is the VO due to VL acting alone and the output voltage is sum of all the three voltages. Now, let us obtain VO1, VO2 and VOL separately and then make sum of all the three. This is circuit. Now, by superposition theorem VO1 means output due to E1 acting alone means E2 and VL is connected to the ground. So, this amplifier will act as inverting amplifier and gain is minus RF by R1 which is minus R by R into E1. So, that is equal to minus E1. Thus, VO1 is minus E1. Now, for VO2 means E2 acting alone, E1 is grounded, VL is grounded. So, basic amplifier will act as a non-inverting amplifier. Therefore, VO2 is 1 plus R by R into V1. But V1 is given by R upon R plus R that is R upon 2R into E2 because this is grounded. Now, substituting value of V1 here we get VO2 is equal to this RR gets cancelled 2 gets cancelled VO2 is E2. And similarly, VOL is given by 1 plus R by R into V1 and V1 is again R upon 2R into VL. So, VOL is VL. Now, VO is sum of all the three voltages. Now, IS is given by VO minus VL upon RS. Substituting value of VO this VL gets cancelled and IS becomes E2 minus E1 upon RS. Now, this is also equal to the load current because current flowing by seeing current is 0. Therefore, load current is given by E2 minus E1 upon RS. Thus, load current is directly proportional to the differential input voltage. So, these are the references. Thank you.