1. The problem statement, all variables and given/known data Find the input impedance. 2. Relevant equations 3. The attempt at a solution Input impedance= R1. I don't get why we can ignore the other resistors. Problem 2 : 1. The problem statement, all variables and given/known data Find the gain of the system. 2. Relevant equations 3. The attempt at a solution Gain = 2 My professor wrote vout= (R3/R2) Vin + vin = 2Vin, so gain =2. I understand that (R3/R2) Vin is from the output of the op amp but I do not know where the other Vin came from. Can someone explain this please? Thank you so much!
Which opamp input is the inverting input and which is the non-inverting. Sure, I can guess and probably be right -- but engineering isn't about guessing. You need to try to communicate your ideas in a clear, unambiguous manner (not always easy and you'll never be perfect at it, but it is a learned skill that must be practiced). Why do you think that the input impedance is R1. I'm not saying it isn't, but just stating that hardly qualifies as an attempt at a solution. Show you WORK (or your line of reasoning) so that we can see where your thinking is leading you and, perhaps, leading you astray. The same with the gain. Stating "Gain = 2" doesn't count as an attempt. Your professor's equation, even if right, seems to assume that R3=R2. Upon what is that based? Your figure has four independent and unrelated resistances with no hint of constraints between them. DON'T MAKE US GUESS! You say that you understand where the (R3/R2)Vin comes from. Okay. So present some analysis that shows where that comes from! Here's a key point to consider: How much current is flowing in R2 if the opamp is in its linear (i.e., active) region?
My apologies. I'm a bit new at this. Input 2 is the inverting input and input 3 is the non inverting input for both problems. For the first problem, I am clueless. I know that the input impedance is the resistance seen at the input. R1 is at the input but so is R2 and R3. Are they ignoring it because they are not directly at the inputs? Or am I completely off? For the second problem, R2= R3= 3kOhms. The diagram is a bit messy. (R3/R2)Vin comes from the fact that it is an inverting op amp and the gain is (R3/R2)Vin. Is the current through R2, Vin/ R2? So if that is the case, the current across R3 is also Vin/ R2 and the voltage across R3 is Vin? Thanks for the reply Wbahn.
Where are you getting that the gain is (R3/R2)? You just keep stating that, but on what is this claim based? Why would the current through R2 be Vin/R2? Remember, to apply Ohm's Law you need the voltage ACROSS the resistor. What is the basis for claiming that the voltage Vin appears across R2? If the opamp is in its active regain, what do you know about the voltage at the non-inverting input relative to the voltage at the inverting input?
That is not possible. Why? Because before you get to R2, you have a voltage drop across R1, let us call it VR1. So at best the voltage across R2 is some portion of Vin-VR1. I say some portion of Vin-VR1 because it looks to me like R2 and R3 for a voltage divider, which means that only a portion of Vin-VR1 will appear across R2, another portion will appear across R3. It might be easier for you to understand if you draw the currents. I think there is one exiting the Vin, another exiting the op-amp and goes through R4. The two currents would combine at Pin 2 and flow through R2 then R3 to ground.
Wait the wires at the 2 and 3 input are not connected. I thought that the gain would be (R3/R2)Vin since it is an ideal inverting op amp, but I realize that is should be -(R3/R2)Vin.
For the third and final time. What will the voltage at pin 2 be compared to the voltage at pin 3. If you can't answer this question, they you do not understand the most basic concept of opamp behavior and have no hope of answering any opamp circuit question (except by regurgitating memorized formulas without any idea of whether those formulas even apply).
And if there is a current flowing through R2, what does this mean for the differential input voltage of the opamp?
Well, if I recall correctly, it will indicate a voltage drop between Pin 2 and Pin 3, which means that Pin 2 is at higher potential than Pin 3, which in turn means that differential voltage is not zero.
The voltage at input 3 is Vin to make the current flowing in zero. In an ideal op amp, the voltage difference between input 3 and 2 is zero.
Ok, I am a bit fuzzy about it so bear with me. The "op-amp" will try to bring the Pin 2 potential down to match the potential at Pin 3, so that the result is that their difference is zero. To do this, the voltage across R4 must be equal to voltage Vin-VR1.
But what says that the current flowing in is zero? Okay, So if the voltage difference between pins 2 and 3 is zero, what does this mean the voltage across R2 is? And, from there, what is the current in R2?
For the point here, it doesn't matter how the circuit makes the Pin 2 voltage equal to the Pin 3 voltage, just that this is what is going to happen (as long as the opamp is in its active region). So, what will the voltage across R2 be?
Yes, I see what you are driving at. Voltage across R2 should be zero. Which means what? It means there is no current though R2, 0 volts=0 amperes*R2, whatever value R2 has. Since we know for a fact that R2 has a value, the only way for voltage across R2 to be zero is for current though R2 to be zero. So all the questions about current though R2 are basically irrelevant.
And if the current through R2 is zero, what is the current through R3? Given the current through R3, what is the voltage at the non-inverting input? From that, what is the voltage at the inverting input? From that, what is the current in R1? Where does the current in R1 have to go? Given that, what is the voltage drop across R4? Given that, what is the output voltage? Given that, what is the gain?