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

  • \$\begingroup\$ Because of point 1. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 1 '15 at 3:27
  • \$\begingroup\$ ... and because non-inverting input (point B) is at a ground potential (i.e. zero volts). \$\endgroup\$ – Nick Alexeev Oct 1 '15 at 3:29
  • \$\begingroup\$ Remember rule II. All the current flowing through R1 has to flow through R2. \$\endgroup\$ – George Herold Oct 1 '15 at 12:59
  • \$\begingroup\$ Yeah, I got a little confuse because (Vin-A)/R1 = Current = (A-Vout)/R2, and A = 0v so Vin/R1=-Vout/R2 Right? is that the right way to do it? \$\endgroup\$ – Pulse9 Oct 1 '15 at 13:03

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.

  • \$\begingroup\$ Just one more question the voltage acorss R2 should be 0-Vout? \$\endgroup\$ – Pulse9 Oct 1 '15 at 4:26
  • \$\begingroup\$ @Pulse9 it would be -Vout since this is an inverting config. \$\endgroup\$ – efox29 Oct 1 '15 at 4:36
  • \$\begingroup\$ The magnitude of the voltage across R2 would be \$V_{out}\$. The sign of the voltage depends on which end you assign the positive sign to, or in practical terms, which side you put the red probe of the DMM on. It's somewhat conventional to assign the positive sign to the end current is flowing into. In that case, the voltage would be \$-V_{out}\$ as @efox29 says, or \$-(-2V)\$ in my example, or 2V. It's fairly typical in my experience to state voltage drops in terms of magnitude (which is always positive of course), because of this arbitrariness in applying a sign. \$\endgroup\$ – scanny Oct 1 '15 at 4:44
  • \$\begingroup\$ Is this the right way to think it? (Vin-A)/R1 = Current = (A-Vout)/R2, and A = 0v so Vin/R1=-Vout/R2 ? \$\endgroup\$ – Pulse9 Oct 1 '15 at 13:03
  • \$\begingroup\$ Yes, I think that works fine. Although I would avoid the choice of \$A\$ as the variable symbol here since that's used to indicate op amp gain. \$V_{inverting}\$ or perhaps \$V_{inv}\$ or V- would be better choices. \$\endgroup\$ – scanny Oct 1 '15 at 13:15

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