Suppose we want to multiply this input by a number

to produce another voltage that we will refer to as the output.

Or, suppose that we have several input voltages,

and we wish to add them together,

and for the output voltage to represent their sum.

Operations such as these, as well as many far more advanced calculations

can be performed with circuits that have what we call an

operational amplifier, typically referred to as an “op amp.”

In this video, we will discuss circuits that have an “ideal op amp.”

An op amp has two input terminals and one output terminal.

The op amp also has two terminals for providing power to the device.

Current can never flow into or out of the input terminals.

Current can flow into and out of the output terminal.

The current that flows into or out of the output is supplied

by the two terminals providing power to the op amp.

The two terminals providing power to the op amp are typically not shown.

But, it is important to keep in mind that the op amp

only has the ability to produce output voltages that are

in between the voltages of the two power terminals.

One of the two inputs has a plus sign next to it,

and the other input has a minus sign next to it.

The op amp takes the voltage value of the “plus” input,

and subtracts from it the voltage value of the “minus” input.

The op amp then takes this difference between the two input voltages,

and multiplies it by a very large number.

The op amp then tries to make the output voltage

equal to the value of this result.

This means that if the “plus” input is even slightly lower than the “minus” input,

the op amp will try to make the output voltage equal to

the largest negative number it is capable of producing.

And if the “plus” input is slightly higher than the “minus” input,

the op amp will try to make the output voltage equal to

the largest positive number it is capable of producing.

Now, suppose that we do something that we call providing negative feedback.

These two resistors cause a portion of the output voltage to be added

to the value of the op amp's “minus” input terminal.

Suppose that the voltage of the op amp's “plus” input terminal

is lower than the value of the op amp's “minus” input terminal.

This will cause the op amp to want to decrease the voltage value

of the output terminal, which will then also end up decreasing

the voltage of the op amp's “minus” input terminal.

The output voltage will stop decreasing when the op amp's “minus” input terminal

is almost exactly equal to the value of the op amp's “plus” input terminal.

Now, let us consider a new scenario.

Suppose that the voltage of the op amp's “plus” input terminal

is higher than the value of the op amp's “minus” input terminal.

This will cause the op amp to increase the voltage value of

the output terminal, which will then also end up increasing

the voltage of the op amp's “minus” input terminal.

The output voltage will stop increasing when the op amp's “minus” input terminal is

almost exactly equal to the value of the op amp's “plus” input terminal.

Therefore, the presence of negative feedback always forces

the op amp's two input terminals to always be at almost the same voltage value.

Once the voltages of the op amp's two input terminals are almost at the same value,

the difference between them is a very small number.

The output voltage is then this very small number

multiplied by a very large number.

For this example, let us suppose that the value of the

op amp's “plus” input terminal is always set to zero volts.

The negative feedback will force the op amp's “minus” input terminal

to also always be set to close to zero volts.

Let us define this point here as the input of the circuit.

If we apply a voltage to this input, a current will flow.

Since current is unable to flow into or out of the op amp's input terminals,

the value of current that flows through the first resistor

must also be the exact same value of current that flow through the second resistor.

The voltage drop across each resistor is the value of

this current multiplied by the resistor's resistance.

Now, let us suppose that one of the resistors has a larger value,

which we will represent as two resistors in series.

Now, the magnitude of output voltage is changed as shown.

If we want the output voltage to be positive when the input voltage is positive,

then we can change the location of the input voltage as shown.

Our circuit can have several different inputs.

All resistance values are the same.

Since current can't flow into the op amp's input terminals,

this means that the current through the resistor on the right

is equal to the sum of the current flowing through the other three resistors.

The voltage drop of the resistor on the right is the sum of all these currents

multiplied by the resistor's resistance.

The output voltage therefore represents the sum of all the input voltages.

Suppose we use a capacitor in the circuit as shown.

The output voltage represents

the integral of the input voltage waveform with respect to time.

Suppose we swap the positions of the resistor and the capacitor.

The output voltage represents

derivative of the input voltage waveform with respect to time.

Another important use of op amps is to act as a buffer.

In this case, the output voltage is exactly equal to the input voltage.

Although no current is ever drawn from the input,

the output terminal supplies a current coming from the op amp's power terminals.

Much more information about electric circuits

is available in the other videos on this channel.

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