Help understanding simple OP-Amp

operational-amplifier

Im trying to understand this basic circuit from a book.

The book said:
"The analysis is simple, if you remember your golden rules:

(The golden rules:

I. The output attempts to do whatever is
necessary to make the voltage difference
between the inputs zero.

II. The inputs draw no current. )
enter image description here

  1. Point B is at ground, so rule I implies that point A is also.
  2. This means that (a) the voltage across R2 is VOut and (b) the voltage across R1 is Vin"

  3. So, using rule 11, we have
    In other words,
    voltage gain = Vout/Vin = -R2/R1

I cant understand why voltage across R1 is Vin and in R2 is Vout

Best Answer

Because the inverting (-) input is at 0V (a virtual ground), the voltage across R1 is:

$$V_{in} - 0 = V_{in}$$

By the same reasoning, the voltage across R2 is:

$$V_{out} - 0 = V_{out}$$

The thing that might not be so obvious is that if \$V_{in}\$ is positive, \$V_{out}\$ will be negative. In that case the current flows straight from \$V_{in}\$ to \$V_{out}\$, passing through R1 and R2. The inverting input draws no current, so all current must flow from in to out, and through the resistors.

So as an example, say \$V_{in}\$ is 1V, R1 is 100Ω and R2 is 200Ω. Then \$V_{out}\$ will be -2V by your formula, the difference between \$V_{in}\$ and \$V_{out}\$ is 3V, dropped across a total of 300Ω, gives a current of 10mA from in to out through R1 and R2.