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Why does the input voltage v_in only appear over R_1, hence making i_1 = v_in / R_1?

Cilcuit diagram

I understand that a positive voltage on the inverting input results in a large negative output. Some of the output is redirected through R_2 back making v_i = 0 (ideally). Since the input current is also equal to 0 (due to ideally infinite input impedance), in my eyes, the voltage from v_in should appear both over R_1 and R_2, and perhaps even over R_L as well?

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  • \$\begingroup\$ btw Vo is - , +(gnd), not +, -(gnd) as shown - its an inverting amplifier. \$\endgroup\$ – JIm Dearden Jan 12 '17 at 12:50
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Because the theoretical op-amp has infinite gain AND the non-inverting input is tied to 0 volts, the effect of negative feedback (R2) forces the inverting input to be at 0 volts.

Given that we assume gain to be infinite (in simple op-amp analysis) we can say (without much error) that the inverting input IS at 0 volts hence R1 connects to 0 volts hence all the input voltage is across R1.

Since the input current is also equal to 0

The current into the op-amp inputs can be reasonably assumed to be zero but not the current taken through R1 - that current is Vin/R1.

You should think of the op-amp as a control system - the demand is set on the non-inverting input and the output HAS to swing to a voltage that forces the inverting input to produce a zero error. Maybe think of it like a servo system like this: -

enter image description here

Imagine the motor and position sensor were inside the op-amp. The op-amp output would be the output from the position sensor. This voltage is forced (due to negative feedback) to be the same voltage as the position demand (0 volts in reality). It makes no difference that there is an extra input (Vin); the inverting input voltage will be 0 volts and all the input voltage (Vin) appears across R1.

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Some op amp basics:

(1) The input is 'differential' - it amplifies the difference in voltage between the inverting (-) and non-inverting (+) input.

The minus signifies that the output will be 'inverted' or 180 degrees out of phase with the input.

The plus signifies the output will be in phase with that input.

(2) The inputs take only a very small current, so small it can be taken as effectively zero.

(3) The voltage gain of the op amp is very large.

enter image description here

Consider an op amp without feedback. One input is connected to 0V. How much voltage at the other input would be needed to swing the output to its maximum value? The supply is +/- 20V so the maximum output is 20V. If we take the gain to be 1 million then an input of 20/1000000 V or 20uV is all that is needed.

For all practical purposes Va is virtually ground voltage.

Why does the input voltage v_in only appear over R_1, hence making i_1 = v_in / R_1?

Lets add some feedback resistors (as in your example)

enter image description here

By kirchoff I1 + I2 + I3 = 0 (what goes in must come out) we know that I3 can be taken as 0 so we get the simple result that I1 = -I2

We also know that Va = Vb = 0V (virtual ground) so that to get the value of I1 we have a simple Ohm's law calculation. I1 = Vin/R1

Similarly we can easily calculate I2 = Vout/R2

But we know that I1 = - I2

So Vin/R1 = - Vout/R2

The voltage amplification of the overall circuit is Vout/Vin

this is equal to - R2/R1 (the minus sign shows the phase inversion)

= 1

The effect of RL - Provided that the op amp is capable of supplying sufficient current at its output the effect of Rl is negligible. However in real op amps there would be an effective output resistance. Too great a load (to small a load resistance) would pull the output voltage down.

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You may find that you'll develop perspectives and views on circuits that work best for you. For some, the purely theoretical view is clearest. For others, they need to see how all the gears and wheels work to get it clear in their head.

So when an op-amp is used with feedback, a simple principle appears: the op-amp will drive its output to whatever voltage is necessary to make its inputs be equal.

This principle applies to correctly-designed feedback circuits, such as you have, and within practical supply voltage limits and a good few other things. But it can simplify one's view of it when thumbing through a circuit.

Looking at your circuit...

The ideal op-amp, on its own, will subtract the Vi- voltage from the Vi+ voltage and multiple the result by its huge open-loop gain. Odds on, this will make the op-amp output go as high as its positive supply or as low as its positive supply. It can't go beyond them, no more volts available.

With the feedback resistors added, it's all a bit more laid-back. The op-amp still does exactly the same thing but the overall circuit sees a more limited behaviour from it.

If Vin is driven with 2 V, the op-amp will do (0-2)*huge = -huge and its Vo output will take a swift trip towards its negative supply rail, trying to get to Vo = -huge V.

However, the div2 potential divider of R1:R2 is connected across Vin and the op-amp's Vo...and the op-amp's Vi- input is actually driven by that potential divider. So Vi- will be driven with the voltage halfway between Vin and Vo, as per simple potential divider behaviour. And as the op-amp's is driving towards its negative supply, Vo will soon travel down to -2 V on its way.

Now we have Vo = -2V driving the bottom of the potential divider and Vin = 2 V driving the top end of it. So the voltage out of the div2 is 0 V.

If the op-amp output carried on swinging negative to, say, Vo = -3 V, the R1:R2 potential divider would present the op-amp's Vi- input with ( ((2 - -3)/2) + -3) = -0.5 V. The op-amp would do it's gain thing and conclude (0- -0.5)*huge = +huge and start driving towards the positive rail. On the way, it would cross -2 V and start driving to -huge again.

But it doesn't go too far in your circuit, it stops nicely when Vi+ = Vi-.

As you know that your Vi+ is tied to 0 V, you also know that the op-amp will do its best in this circuit to keep its Vi- input at the same voltage. Hence the name 'virtual earth': it isn't one, but the op-amp will do it's best to keep Vi- pulled to that by the feedback network of R1 and R2.

From the op-amp's point of view, Rload is outside of the feedback divider so it just gets driven to Vo all the time.

All the stuff about being able to take the R1 current as Vin/R1, that comes from the knowledge that the op-amp output will pull the R1:R2 potential divider about until Vi- is at the same voltage level as Vi+. So it lets you simplify your calculations for Ir1 a little.

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What's happening there is the effect of the negative feedback provided by \$R_2\$ and when working in the linear region. That forces \$V^+\approx V^-\$. That's what's called virtual ground.

Take a look at this link

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