Circuits, inverting amp: Why is Ground have a Resistor in it that is equal to R1 and R2 in parallel?

Question

I have a question about an inverting amplifier.

|||--R3---------|.\
. . . . . . . . . . . .|...\_____V(out)
V(in)--R1-----|.../. . . . | .
. . . . . . . .. |. . .|./. . . . . | .
.................|_ __R2__| .

R1 is at V(in) of the amp, R2 is connected to V(in) and V(out) and R3 is attached to the ground and the (+) input, just like a normal inverting amp.
ok this is suppose to be an inverting amplifier, with
R1 = R2 =10kOhms
What should R3 be and why? I figured out that it is suppose to be the sum of R1//R2 = 5kOhms but I'm not sure why.

Any help explaining the reasoning behind it.

Answer 1

Bi-polar transistor op-amp front ends have a bias current, the current necessary to forward bias the base-emitter junction of each input transistor pair. The bias current from the +In terminal flows through R3, and the bias current from the -In terminal flows through R1 in parallel with R2. Assuming the bias currents are equal (sometimes this can be trimmed), any difference in impedance at the two input terminals creates a differential input offset voltage.

Bias current doubles about every 10 degrees C, so it is important to compensate for it by matching input impedances seen at the +In and -In terminals.

All this becomes less important with FET front-ends, which have very little bias current to begin with.

Answer 2

it facilitates to think of of the present circulate. modern-day = voltage / resistance. think of if there replaced into in basic terms one resistor, the present could be plenty. in case you place yet another resistor in parralel, then there is yet another direction for the present to bypass. so extra modern-day flows. so the comprehensive resistance ought to be decrease than for in basic terms one resistor. so the comprehensive resistance Rt ought to be a decrease variety than the two of R1 or R2. lower back look at it from the attitude of the present. modern-day for R1 could be voltage / R1 modern-day for R2 could be voltage / R2 the comprehensive modern-day could be voltage / R1 + voltage / R2. see how this is such as the equation for Rt? so finally in case you generalize the equations, a million/ R is the comparable as a million/R1 + a million/R2 or a million/Rt = a million/R1+ a million/R2 then you definately do an algabreic circulate to place Rt on right and Rt = a million / (a million/R1 + a million/R2) wish the reason facilitates

Answer 3

The combination of R1 and R2 controls the gain of the amplifier. Because they are equal, the gain is unity --- you get out what you put in, upside down. If you doubled R2, you would also double the gain.

Answer 4

If R1 = R2 there is no amplification, just a signal inversion.

In my opino R3 should be 10X R1, to keep the impedence high!