 Hello everyone, welcome to lecture on voltage shunt feedback amplifier. At the end of this session, students will be able to identify voltage shunt feedback amplifier and also able to calculate different parameters of it. Now before starting with the actual session, let's recall what is the voltage for how. So pause the video and think about it. Voltage follower is nothing but the non-inverting amplifier. It has input connected to non-inverting terminal and the feedback path is connected to the inverting terminal of the opamp. So that feedback path having a gain of 1. So that's why it follows whatever the input you are applying that the same input is taken as output. So that's why it follows voltage follower. Now let's see what is mean by voltage shunt feedback amplifier. Figure 1 shows the voltage shunt feedback amplifier. You can see there is one opamp which is denoted by A. It is also having a feedback which is connected to non-inverting terminal and non-inverting terminal is connected to ground. So it is also known as an inverting amplifier with feedback. Input signal drives the inverting terminal of the opamp and amplified as well as inverted output signal is also applied to inverting terminal through RF and non-inverting terminal is connected to ground. So due to this arrangement it forms a negative feedback. So because of that any increasing the output signal results in a feedback signal into the inverting input. So because of that causing a decrease in the output signal. The feedback circuit has only one component that is nothing but the RF. After resistor R1 is also used which is connected in series with the input voltage source VIN. Now let's see the parameters of voltage shunt feedback amplifier one by one. First parameter is nothing but the close-loop voltage gain that is AF. So it can be calculated by the following procedure. AF can be obtained by writing Kirchhoff's current equation at the node V2. You can refer the figure 1 in the previous slide. So at that node V2. IN is equals to IF plus IBU IN is nothing but the current flowing through R1. IF is nothing but the current flowing through the RF resistor and IB is nothing but the input bias current which is flowing through the non-inverting terminal of the opamp. Since RI that is input resistance of the opamp is much larger. So that's why the input bias current is negligibly small. So that equation above equation becomes IN near approximately equals to IF that is VIN minus V2 divided by R1 equals to V2 minus V0 divided by RF. So IN can be given by the equation VIN minus V2 divided by R1 and IF is given by the equation V2 minus V0 divided by RF. But we know that V1 minus VT equals to V0 upon A. A is nothing but the gain of the opamp and V0 is nothing but the output voltage. Since V1 is approximately 0 voltage because it is connected to ground so V2 becomes V2 equals to minus V0 upon A. So substituting this value of V2 in the equation 2 we get VIN plus V0 upon A into bracket divided by R1 equals to minus V0 upon A minus V0 divided by RF. So by rearranging this we get the equation for AF AF equals to V0 upon VIN equals to minus ARF divided by R1 plus RF plus AR1 that this one is exact equation. Negative sign indicates that in the above equation input and output signals are out of phase by 180 degree. Because of this this configuration is called inverting amplifier with a feedback. Since the gain of the opamp A is much greater value ideally it is infinite. So AR1 term is much greater than R1 plus RF. So from equation 3 R1 plus RF plus AR1 that becomes AR1. So you can rewrite that equation that final equation of AF becomes AF equals to V0 upon VIN equals to minus RF upon R1 that is the ideal equation for the closed loop voltage gain of voltage and feedback amplifier. Means that the gain of the inverting amplifier is said by the selecting a ratio of feedback resistance to the input resistance. Next parameter is the inverting input terminal at virtual ground. Now this is one interesting concept about the voltage and feedback amplifier. From figure you can see that non-inverting terminal is grounded and input signal is at inverting terminal via resistor R1. But we know that difference input voltage is ideally 0 for the opamp. Means that voltage at the inverting terminal V2 is approximately equal to the voltage at non-inverting terminal V1. In other words you can say that inverting terminal voltage V2 is approximately equal at ground potential because the V1 is connected to ground. Therefore the inverting terminal is said to be at the virtual ground. Next parameter is input resistance with feedback. Figure 2 shows the input resistance with feedback for inverting amplifier. For to find the input resistance the easiest method is that RI is to millerized the feedback resistor RF. This means split RF into two miller components as shown in the figure 2. Therefore the input resistance with feedback that is RIF is then given by the equation equals to R1 plus RF divided by 1 plus A which is parallel with the R1. So you can from the figure it can easily see that RI divided by 1 plus A is parallel with the input resistance RI. So R1 plus RF that equation becomes RI and A are very large. So final equation RF divided by 1 plus A parallel with RI is nearly equals to 0. Hence RIF equals to R1. That one is the ideal equation for the input resistance with feedback for voltage and feedback amplifier. Next parameter is output resistance with feedback. Figure 3 shows the diagram for the output resistance with feedback. Output resistance ROF can be obtained by using Thevenin's theorem. Thevenin's equivalent circuit for ROF is shown in figure 3. So from that figure you can write equation IO equals to IA plus IB that is current flowing through node equals to current flowing out from the node. This RF plus R1 parallel with RI is having value much more larger than RO. That's why current flowing through that branch that is IA is much greater than current flowing through feedback path IB. Therefore IO nearly equals to IA. So we have equation VO minus ROI minus AVID equals to 0. But we know that input voltage difference is nothing but the V1 minus V2. V1 is connected to ground so that becomes 0. So 0 minus beta VO equals to final equation minus beta VO. So therefore for ROF final equation you can write equals to VO upon IO equals to RO upon 1 plus A beta. Next parameter is bandwidth with feedback. We know that the bandwidth is nothing but the band of frequency over which the gain remains constant. For opamp the gain at 5 hertz is having value is larger that is a 2 lakhs and at 1 mega hertz it is having a value gain is 1. So gain bandwidth product if you see that is a 1 mega hertz at 5 hertz and at 1 mega hertz it is a 1 mega hertz. So it is a constant throughout the frequency. So for 741 CIC it has 2 frequencies one is a break frequency and second one is a unity gain bandwidth. Unity gain bandwidth is nothing but the gain having value 1 that frequency is called as a unity gain bandwidth. So for open loop you can write a unicode to get bandwidth equation is A into FO. A is nothing but the gain of the opamp FO is nothing but the break frequency. Similarly you can write for the closed loop opamp that is unity gain bandwidth is equal to AF into FF. So from these 2 equation you can write A into FO equals to AF into FF. So by rearranging in this we will get final equation for the break frequency FF equals to FO into 1 plus A beta. This is the equation if you know the FO frequency and if you know the unity gain bandwidth you can use the equation IF equals to UGB into bracket 1 plus A beta divided by A. So from this FO equals to nothing but unity gain bandwidth divided by A. These are the references. Thank you.