Need help with Circuit Analysis basics, op amps basics. Please help. (10 Points!)?

Question

Hello,
I am having some major trouble with understanding how to analyze op amp circuits. I am familiar with circuit analysis very basic concepts like ohms law, kvl, kcl, nodal and loop analysis, the general approach, etc....

I was wondering if anyone can help me understand how to solve this problem step by step, I am more concerned with understanding how to solve this type of problem then the final answer.

Here is a picture of the circuit problem: http://tinypic.com/view.php?pic=9ghqmq&s=7

Thank you

Answer 1

The way to think of opamp circuits with negative feedback is that the opamp has so much gain, the only way the output will not be saturated is if the two inputs are essentially at the same voltage, because the opamp amplifies the difference between those two input voltages with all that gain. That means, that, if it is possible for the negative feedback to keep the output between the supply rails, you can just assume, for the simplified analysis, that the output does whatever it can to force the two inputs to the same voltage. In this circuit, the + input is held at +4volts. So the output must drive current through the two resistors so that the junction of those two resistors settles at +4volts, also. This reduces the problem to a simple voltage divider between +5volts and the output voltage.



Because of the assumed equality of the voltage on the two inputs, you could write an equation with whatever formula or constant defines the voltage on one input on side of the equation and whatever formula or constant determines the voltage on the other input on the other side of the equation (because the two input must be equal to each other), as long as the output voltage appears somewhere in this equation. Then you solve that equation for the output voltage. This simple analog between an opamps operation and an equation is why they are called opamps (mathematical operation amplifiers) and analog computers solve arbitrary equations that depend only on feedback components, are made with opamps. The feedback components act as the constants in equations and input voltages act as the variables.



Note that this circuit is not very practical for most integrated opamps, because they have a very low current limited output. Something like 0.02 amps maximum, though large current, high power opamps do exist. But that is a separate limitation, not covered in this problem.



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Regards,



John Popelish

Answer 2

the trick is to assume the op amp is ideal, this makes the analysis simple (but it is still possible with finite gain!)

Remember the opamp is a differential amplifier with very high gain, so the output voltage is A*(V+ - V-)

where V+ is the voltage at the non-inverting input, and V- is the inverting input, A is the gain.



If we have a small 'reasonable' voltage at the output, and the gain is very very large, then the difference between V+ and V- must be very small. It could be so small we can ignore it, and say the voltages at the + and - inputs are equal. So now we can easily do some nodal analysis:



from the above argument, the voltage at the - input must be 4 volts (or really close to it.). Now write a node equation for that node:



(4-5)/1 + (4-Vo)/4 = 0



Now you can easily solve this!

Answer 3

The best shortcut you can use is to assume that the inputs of the opamp (+ and -) always have the same voltage. Don't worry about input polarity or how it happens for a moment; you can always inversigate these details later. Also, assume that the inputs act as open circuits with infinitely large impedance that does not affect the circuit at all. In the real circuit, you have to worry about all these and many other issues, but for the purposes of solving an academic problem, these assumptions are good enough.



Your voltage of the "-" terminal is the same as the "+" terminal, or 4V. The voltage across R1 is 5V-4V = 1V, so the current through R1 is 1V/1Ohms = 1A. SInce no current goes into the input of the opamp, all of it goes into the 4Ohms resistor producing a 4V drop across it. Since the voltage on the "-" input is 4V, the output voltage is 4V-4V = 0V.



You can do the same calculation in many other ways as long as you understand how to handle the opamp.

Answer 4

It's been a while...



As I recall, op amps are assumed to have infinite gain, or at least large compared to what you are doing, and circuits are arranged with negative feedback. I think the gain is given by R2/R1, i.e. 4,

so with the 4V offset you will get -(5-1) * 4 + 4 i.e. 0V on the output\



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Yes, the input impedance is very high, treated as infinite. I seem to remember the -ve input being referred to as a "virtual ground" - usually the +ve input is grounded, and negative feedback drives the -ve to 0V. So if no current goes in the input, all the current must go through the feedback resistor. This approach works for integrator circuits, too, where there is a capacitor in the feedback line.

Answer 5

2 regulations approximately op amps: a million.) no modern enters the terminals 2.) voltage at the two terminals is an identical To perform a little circuit prognosis, set a floor or reference element, ultimate to do on the + terminal hence. We now know that the two + and - terminals are at 0V Write 2 equations for i1 leaving the battery; i1 = (1V - 0V)/1k = (1V - Vo)/6k I have been given -5V it incredibly is clever i think of via fact it fairly is an inverting op amp with a earnings of 5.