Adding & Subtracting Fractions Calculator

Adding & Subtracting Fractions Calculator – An op-amp voltage buffer mirrors the voltage from a high-impedance input to a low-impedance output. 8 min read

A voltage buffer, also known as a voltage-following or unity-gain amplifier, is an amplifier with a gain of 1. This is one of the simplest op-amp circuits possible with closed-loop feedback.

Adding & Subtracting Fractions Calculator

Adding & Subtracting Fractions Calculator

Although the 1 gain does not amplify the voltage, the buffer is useful because it prevents the input impedance of one stage from loading the impedance of the previous stage, causing unwanted loss of signal transmission. We covered this concept in the section on Maximum Signal Transfer and Minimum Interstate Load.

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A voltage gain of 1 means that when the input voltage increases by ΔV, the output voltage is designed to increase by the same ΔV.

We mentioned in the Ideal Op-Amp section that an op-amp will vary the output voltage until the two inputs are the same. Now we have our first chance to see how this works, as this circuit is switched from the output of the op-amp back to one of the inputs. Before we do the math, let’s build a qualitative, intuitive model.

In this case, we can slow down time and imagine what would happen if we assumed a steady state and suddenly changed the input voltage:

This is an example of a connected source feedback because there is a connection from the output to one of the feedback inputs.

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We can substitute V + into V to find and algebraically solve the equation that connects V to V above:

V out = A OL (V + – V -) V out = A OL (V in – V out) V out = A OL V in – A OL V out V out (1 + A OL) = V OL V out = (A OL 1 + A OL) V.

In a truly ideal op-amp, with infinite gain, bandwidth, and slew rate, the process described in the intuitive model is straightforward.

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In the real world, op-amps have a finite gain-bandwidth product, so the intuitive modeling process is immediately time-limited. We can simulate this using an op-amp with a finite product bandwidth of 1 GHz and a 100 MHz square wave input signal:

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Exercise Click to open and copy the diagram above. How long does it take to respond after a change in input?

With an ideal op-amp, the voltage buffer will often have a flat frequency response, with an output from 1 to an infinite frequency.

In a real op-amp with an extremely wide product, the voltage buffer configuration has a closed loop of 1, so the bandwidth is equal to the product bandwidth. Try this simulation with a 10 MHz GBV op-amp and observe that the gain flattens out until you reach the 10 MHz corner:

Exercise Click to open and copy the diagram above. What is the frequency -3dB? Double-click OA1, adjust A_OL, and run the simulation again: Does the Bode plot change? Next, do the same for GBW.

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A simulation of this circuit shows that the -3 dB knee in the frequency response curve occurs in the gain width product of the op-amp (GBW).

For practical purposes, we can assume that a real-world op-amp voltage buffer will perform well for much lower frequency signals than an op-amp GBV. . from input to output. GBW is listed in the op-amp datasheet, so you can solve this problem by simply buying a faster op-amp.

As shown in the Ideal Op-Amp section, we can model the open-loop function of an ideal op-amp (with finite GBW and A_OL) in the Laplace domain:

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In fact, CircuitLab makes it easy to simulate this Laplace transform in a closed inverse configuration by removing the op-amp OA1 from our circuit and replacing it with a voltage gain and Laplace transfer function:

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Exercise Click to open and copy the diagram above. Note that the frequency response of this Laplace block model is similar to the frequency response shown for the op-amp circuit above. . 1 and phase 0. See the CircuitLab documentation on frequency-domain simulation for more details.)

The simulation shows that with a wire that reverses the loop and the loop, the large open-loop gain is changed to a closed-loop gain of 1 until the GBW limit is reached.

And as at the beginning of this section, we can substitute V + (s) = V (s) and V – (s) = V to get:

V out (s) = (V + (s) – V – (s)) H (s) V out (s) = (V in (s) – V out (s)) H (s) (1 + H ) ) (s) V from (s) = H (s) V in (s) V from (s) V in (s) = H (s) (1 + H (s))

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This fraction of (V) V gives us the total closed loop of the op-amp voltage buffer circuit. Let’s simplify algebraically by connecting H(s):

V out (s) V in (s) = H (s) 1 + H (s) V out (s) V in (s) = A OL 1 + s (A OL 2 π GBW) 1 + A OL 1 + s (A OL 2 π GBW) V from (lar) V in (s) = A OL 1 + s (A OL 2 π GBW) + A OL V from (s) V in (s) = 1 (1 + 1 ) ) A OL) + s (1 2 π GBW)

As expected, since we configured the op-amp in a closed configuration with a gain of 1, this closed transfer function is f c = the transfer function of the cutoff frequency filter with frequency GBW.

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Let’s look at our LED current controller example from the voltage divider section. In this example, we used a voltage divider (R1 and R2) to set the base terminal voltage of the NPN BJT:

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Exercise Click to open and copy the diagram above. How do V and I L change as the resistance parameter X is increased?

The problem we examined in this section was how to choose the value of the voltage divider resistors, since we know we want a ratio of 100:52 between R1 and R2. If we made the distance too small, they consumed a lot of power, even more than the LED we wanted to control. If we made the distance too big, they will not be able to handle the increased load of main current to I quarter, and the voltage will drop much more than we predicted. In this section, we used a simulator to find the sweet spot between these two effects.

But now that we have op-amps, a simple use of an op-amp voltage buffer is to buffer a voltage divider by placing a buffer between the voltage divider and the base of Q1:

Exercise Click to open and copy the diagram above. Now that we have added a buffer, how do V and I L change as we increase the resistance parameter?

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As you can quickly see from the smooth DC sweep simulations, this allows us to use much larger resistance values ​​for R1 and R2 (thus reducing current consumption in the voltage divider) without changing the LED current!

In the real world, there is a trade-off with this method. By adding an op-amp voltage buffer, we overcome the power loss of R1 and R2, but lose a bit by adding new current consumption from the op-amp’s own current. We also increased the cost and space required by adding a component. But in many cases math is a winning trade!

More importantly, the buffer allows us to separate different sections of the design: R1 + R2 design choices are now much more independent of design choices.

Adding & Subtracting Fractions Calculator

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