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The Current rule states that there is no flow of current toward the inputs of an op-amp whereas the voltage rule states that the op-amp voltage tries to ensure that the voltage disparity between the two op-amp inputs is zero. From the above non-inverting op-amp circuit, once the voltage rule is applied to that circuit, the voltage at the inverting input will be the same as the non-inverting input. So the applied voltage will be Vin. So the voltage gain can be calculated as,.
Therefore the non-inverting op-amp will generate an amplified signal that is in phase through the input. In a non-inverting operational amplifier circuit, the input impedance Zin can be calculated by using the following formula. So, for a non-inverting operational amplifier circuit, the input impedance Zin can be calculated as.
The voltage gain is dependent on two resistances R1 and Rf. By changing the values of the two resistances required gain can be adjusted. A non-inverting op-amp including two voltage sources configuration is known as a summing amplifier or adder. So this is one of the most essential applications of an op-amp. In the summing amplifier circuit, multiple voltage sources are used.
The non-inverting summing amplifier circuit uses the configuration of a non-inverting op-amp circuit. The main benefit of the non-inverting summing amplifier circuit is there is no effective earth condition across the input terminals; its input impedance is much higher than that of the standard inverting amplifier configuration. So the flow of current in the non-inverting op-amp with two voltage sources can be defined as:. Op-amp gain mainly depends on its configuration. The inverting op-amp gain is negative because the output of the op-amp is out of phase with the input.
There is a direct tradeoff between amplifier performance in terms of amplification, and performance in terms of bandwidth. This is not merely theoretical. You are likely to run into this problem in real-world op-amp design! For example, if you need a gain of , and you simultaneously need to handle signals of 10 5 Hz , you have a few options:. The limited frequency response also manifests as a slower step response in the time domain.
Simulate the circuit above and see how long it takes to settle to its final value after an input step for different gain configurations. This is actually a simple case of a common but confusing concept in feedback systems: a modification in the feedback path such as multiplication by f generally causes the inverse or reciprocal effect such as multiplication by 1 f to the whole system after closed-loop feedback is applied.
For readers familiar with transfer functions: this is equivalent to saying that the feedback transfer function ends up in the denominator of the closed-loop response. In a general way, we can look at a feedback system with a forward transfer function G and a feedback transfer function H as depicted here:. For simplicity, consider these multipliers G and H to be constants, performing multiplicative scalings of their input. The three block diagram elements one subtraction and two transfer function multiplications let us build a system of three equations :.
We can combine the above equations, substituting V fb and V err to find:. This last equation is the closed-loop transfer function , and it relates the input to the output, after considering the effects of the feedback loop. This is a remarkable result: if the magnitude of the loop gain G H is large compared to 1, then the foward transfer function G actually cancels out of the closed-loop result, and the closed-loop response is determined only by the reciprocal of the feedback transfer function, 1 H.
So the closed-loop gain is just:. When we care about the response of systems with frequency-dependent behavior, such as when we analyzed the gain-bandwidth tradeoff above, we can still apply the Laplace-domain to the same general closed-loop result:.
We can even use a potentiometer to make an adjustable-gain amplifier. But how should we choose the absolute resistor values? The answers are similar to the tradeoffs discussed in the Voltage Dividers section. There are concerns and drawbacks on either extreme:. What does the resulting signal look like?
What happens if you change R1 and R2 to both be 2x smaller or larger? As an exercise, add a load resistance to the output and see how the signal changes. These problems cause nonlinear clipping , which destroys information and causes distortion for all later signal stages.
What happens if there is some unintentional but unavoidable parasitic capacitance in the feedback path? Conceptually, we can follow the ideal op-amp adjusting its output up or down based on the immediate difference in its inputs:. How does even a few picofarads of parasitic capacitance affect the step response?
Does anything change if C1 is connected between the two op-amp inputs, rather than from the inverting input to ground? Why or why not? Parasitic capacitance is a real problem in high-speed amplifiers, and issues with feedback loop stability is one of them. You can approach this in the simulator by making C1 very large.
In the simulator it will eventually settle out, but in the real world, it may actually oscillate forever because of additional delay imposed by the speed of light, if nothing else! The last option for mitigating our stability problem is worth demonstrating briefly. We can add a small feed-forward compensation capacitor C2 in parallel with R1. Remember that C1 is not a capacitor component that we buy and install in our circuit, but is instead just a parasitic capacitance that arises from the circuit layout.