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The topic of this problem is Operational Amplifiers, and

Â we're going to work with a circuit that is an inverting amplifier.

Â The problem is to find V sub out in terms of the input voltages.

Â In our case, we have one input voltage, it's V sub n,

Â that's attached to a branch with Rs of 1,

Â and ultimately, to the inverting input of our op-amp.

Â We know that, for op-amps, we have a number of properties,

Â which help us solve our circuits.

Â So, if we look at, just a general op-amp configuration,

Â and this is the circuit symbol for it, we have currents associated

Â with the inverting input and with the non-inverting input.

Â We also have voltages associated with those inverting and

Â non-inverting inputs as well.

Â And we know that some of the properties of ideal op amps

Â are that the current into the op amp are equal to 0.

Â They're equal to each other and they're equal 0.

Â We also know that another important property for us when we're solving for

Â these op amps is that the voltage at the inverting input

Â of the op am is equal to the voltage at the non inverting input of the op amp.

Â So V minus is equal to V plus.

Â So we're going to use these two properties when we solve circuits

Â with op amp in them for linear circuit applications.

Â So let's look and see how we can solve the problem that we have in this example,

Â using our the two properties of the op amp.

Â If we use those properties, one thing that we know is that the voltage at this point

Â is equal to the voltage at ground, which is zero volts.

Â So we're at 0 volts at this non-inverting input.

Â So if it's 0 volts at the non-inverting input, using our second

Â characteristic of op amps, then we're also at 0 volts at the inverting inputs.

Â So that's 0 volts as well at that point.

Â So once we have that,

Â we can use tools that we've already learned in the past to solve this problem.

Â Again, we're looking for V out in terms of the input voltages.

Â In our case the one input voltage which is tied to our inverting input.

Â So if we use Kirchhoff's current law at that node, we'll call it node one.

Â This is Kirchhoff's current law at node one,

Â then we can sum the currents into that node.

Â We have a current through R1, we have a current through R sub F and

Â we have the zero current associated with the non-inverting or the inverting input.

Â So let's first look at the current through R sub 1.

Â It's going to be the N minus 0 over R1.

Â 3:35

And we know that the other current, which is associated with the inverting input,

Â is equal to 0.

Â And the sum of those currents using Kirchhoff's current law is equal to 0.

Â So we're summing currents into node 1.

Â So we have an equation which relates V out to Vn along with our

Â resistor values which we've associated with our circuit.

Â So if we solve that equation for the V out,

Â we end up with a V out which is equal to

Â minus R sub F divided by R1 times V sub n.

Â So if we look at that relationship,

Â the output voltage is equal to an inversion of our input voltage,

Â as amplified by the factor Rf/R1.

Â So we can use our external components to control properties of our op amps.

Â In this case, the amplification factor for

Â an input voltage, with respect to its output.

Â So the negative sign gives us this inverting.

Â Characteristic that we call inverting operational amplifiers.

Â And so V out is equal to V in times a factor which is a negative number.

Â It could be a number greater than one, it can be a number less than one, or

Â it can be a number equal to one where R sub f is equal to R1.

Â Depends on the application.

Â