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Опубликовано 2013-05-03 14:06:12 автором MRS Operational amplifieroperational amplifier - this sort of thing , which can be used in different ways to change the analog signal ( amplify, attenuate , invert , etc.). On the concept denoted as follows U out = (U inp1 -U inp2 ) * K , here U inp1 - the voltage at the non-inverting input , U inp2 - the inverting input voltage , U out - the output voltage , K - gain without feedback. Let us model the option to include the OS without feedback in Proteus and see what will be the output. scheme
:Gain is calculated by the formula: K = 1 + (R2/R1)
As we remember, the OS operating principle is to make the output voltage such that the input voltage difference is zero. Here we are on the line input is 5V , and inverted through a resistor divider half of the output voltage. DU remains nothing how to raise the output voltage twice relative to the direct input. With this integration , you can multiply the voltage op amp to a number greater than 1. But where it can come in handy ? For example, we need to measure the current . To do this, take a small shunt resistance of 0.01 ohm, the voltage drop across it at a current of 2 A is U = I * R, U = 2 * 0.01 = 0.02 . Amplified by the OS 100 times , it will be 2B, and fed to an ADC of the microcontroller.
inverting amplifierGain is calculated by the formula: K = - (R2/R1)
Direct entry to us to put on the ground, the OS will try to do so and at the inverse input was zero. For this he needs to lower the output voltage is below zero, and that you see in the diagram .
repeaterOutput voltage equal to the voltage on the direct input , but you may ask why this is necessary because you can just throw a straight wire . Imagine the situation: there is a resistive divider and to the outlet you need to connect a light bulb . But the light bulb has its own resistance, and it will affect our resistive delitel.Chtoby remove this influence bufferiziruetsya output operational amplifier inputs , then he has a very high resistance and influence he will be on the lowest resistive divider and the output current can provide dozens or even hundreds miliamper , which is enough for the bulb. Комментарии - (2)
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