 Hello everyone, welcome to lecture on differential amplifier with one opamp. At the end of this session students will be able to identify differential amplifier and also able to calculate different parameters of it like input resistance, voltage gain, etc. Now before starting with the actual session let's recall that what is the total output offset voltage with feedback of the opamp. So let's pause the video and think about it. So it is nothing but total output offset voltage with feedback of the opamp is nothing but equals to total output offset voltage without feedback divided by 1 plus a b equals to plus or minus v sat divided by 1 plus a b where v sat is nothing but the saturation voltage of the opamp a is nothing but the open loop voltage gain and b is nothing but the gain of the feedback circuit of the opamp. Now let's start with the differential amplifier with one opamp. If you see figure one shows the differential amplifier with one opamp. So it has two input signals basically it is a comparator you can see that it is having two input signal that is vx and vy it amplifies actual difference between these two input signals also it is better able to reject common mode voltages that is noise which is common to both input voltage sources. So because of this it is also present balanced input impedance. So you can see this figure is nothing but the having combination of both inverting and non-inverting amplifier in single circuit here you can see that when vx is reduced this vx is reduced to 0. So this combination becomes non-inverting amplifier why because when you reduce this to 0 this one is grounded so there is only single source that is vy so that connected to non-inverting terminal of the opamp so that is why this combination becomes non-inverting amplifier. Similarly when vy reduced to 0 that terminal is grounded so only single source becomes vx which is connected to inverting terminal of the opamp so this circuit or combination becomes inverting amplifier. So that is why you can say that it has both the combination of inverting and non-inverting amplifiers of the in single circuit. So mostly these amplifiers are used in an instrumentation and industrial applications because it amplifies the difference between two input signals. So in the industry or there are some application where it requires to amplify such a small difference between two input signals or small difference of the signal which is not able to amplify using normal amplifiers so that is why their differential amplifiers are used. Now let us see the different parameters of this differential amplifier with one opamp. So first voltage gain so from figure 1 you can see that it has two input signals that is vx and vy so here by using superposition theorem we can calculate the relationship between the input and output so when vy is 0 circuit becomes an inverting amplifier as we already seen in the previous slide. So by for that circuit we have a equation one for voltage output that is vo when vy is 0 the circuit is having inverting amplifier so we have the equation for inverting amplifiers output voltage that is vox because of the voltage source vx equals to minus rf upon r1 into vx where rf is nothing but the feedback resistor which is present in the circuit r1 is the resistor which is in series with the vx and vx is nothing but the voltage source. So minus sign indicates that it is a inverting amplifier means whatever you are having input you get the output with the this gain and negative sign. So this becomes equation one so this voltage vox is nothing but the output voltage because of the vox. Similarly when vx is 0 the circuit becomes non-inverting amplifier so for non-inverting amplifier we have equation for output voltage so for that we can write the output voltage is nothing but the vo y. Vo y is the output voltage because of the input voltage now in the previous figure you can see that the input voltage to the non-inverting amplifier is a v1 and for v1 input v1 is nothing but the voltage which is derived from the voltage divider circuit so v1 is nothing but the r3 divided by r2 plus r3 into vy vy is nothing but the input voltage connected to non-inverting terminal. So v1 is nothing but the voltage across r3 which is derived by this equation so that v1 is nothing but input to the actual input to the non-inverting terminal of the opam. So final output of the vo because of the vy is given by 1 plus rf upon r1 into v1 because we already know that for non-inverting amplifier equation is 1 plus rf upon r1 into v1. So now in this case v1 is a actual input to the non-inverting amplifier. So from equation 2 you can see that we have value of v1 so put that value of v1 in equation 3 so we can get the final equation of vo y that is output voltage because of the voltage source vy equals to r3 upon r2 plus r3 into bracket r1 plus rf upon r1 so into vy. So since r1 equals to r2 and r3 equals to rf so this two terms get cancel out each other so final equation becomes vo y equals to rf upon r1 into vy. So this we have two equations that is a vo x that is the output voltage because of the voltage source vx that is minus rf upon r1 plus v into vx and we have output voltage because of the voltage source vy that is rf upon r1 into vy. So from these two equations we can get the total output voltage of the differential amplifier with one opamp that is vo equals to vo x plus vo y. So from the previous two equation put the value of vo x and vo y we can get vo x is nothing but the minus rf upon r1 into vx plus rf upon r1 into vy. So rf upon r1 become take common so minus rf upon r1 minus sign common so vx minus vy. So we know that that is vx minus vy is nothing but the vxy that is a difference input voltage of the opamp. So final equation becomes vo equals to minus rf into vx divided by r1. So from this equation we can say that vo upon this vxy comes over here. So vo upon vxy is nothing but the gain equals to minus rf upon r1. So gain of the differential amplifier that is why it is treated as a d, d is for differential. Gain of the differential amplifier is given by the minus rf upon r1. So means you can say that the gain of the differential amplifier is same as that of inverting amplifier. Means that the gain of the differential amplifier is controlled or controlled by feedback resistor rf and resistor which is in series with the vx. Now next parameter is a input resistance. Input resistance rf of the differential amplifier is a resistance determined looking into either one of the two input terminals with the other grounded as we done for the voltage gain keeping other source grounded and finding the voltage and then doing for the other one. Similarly over here therefore when vy is 0 the circuit which is from figure 1 is an inverting amplifier we already seen. So the input resistance of inverting amplifier we have equation rifx nearly equals to r1. r1 is nothing but the resistance in series with the input source vx. So that similarly when vx equals to 0 that circuit becomes a non-inverting amplifier. So for non-inverting amplifier we have input resistance equation from the previous sessions we already seen that. So rify nearly equals to r2 plus r3. So from these two equation you can see that the input resistance seen by the different sources input sources vx and vi are not the same. So here you can see that because of the vx input voltage source vx it is nearly equals to r1 and for because of the vy it is nearly equals to r2 plus r3. So these are two not same it is different because of the these two different input signal sources. So this inequality can be corrected and both input resistance can be made equal if we modify the differential amplifier which is we seen in the figure 1. So this figure 2 shows the modification done in the differential amplifier to make a equal input resistance. So you can see the figure 2 is nothing but the differential amplifier with variable gain. So to perform properly both the input resistance rifx and rify can be made larger than the source resistance to avoid loading of the signal sources. So it has a variable gain because if you see the modification is nothing but the one pot is connected in series with the r4 which is nothing but the load resistance that pot is rp and the wiper of that pot is connected to the feedback resistor means as the wiper changes its position the feedback resistance total resistance value also changes. So because of that so gain of the op-amp also changes. So final condition is that if r4 is equals to rp means if the this two value is equal then the maximum voltage you are getting from this circuit is nothing but the vo equals to minus 2 rf upon r1 into vxy. So as I said depending on the position of the wiper in the rp voltage gain can be controlled. These are the references. Thank you.