I have the following problem. I have to fill in the blank spots in the picture below.

enter image description here

I'm not sure how to answer this question.

My book just says:

If \$V_+ > V_-\$ then \$ V_{out}=max\$

If \$V_- < V_-\$ then \$V_{out}=min\$

Is there an equation of sorts that will help me solve this problem?


For this question, all you need to know is that opamp inputs don't draw current. This turns it into a straightforward resistor divider.


simulate this circuit – Schematic created using CircuitLab

You can calculate it in a few different ways, but the way it probably wants you to do it is using current.

In the case where V2 = 0:

  • What is the total current through the resistors?

  • Knowing the total current, you can calculate the voltage drop across R1

  • Then you just add this voltage drop to V1 to get V+

The same thing happens for V2 = 5V

  • \$\begingroup\$ So when V2=0 the voltage drop is 2.5V. So the current \$i=2.5V/10 kohm = 0,25 mA\$. The voltage drop across R1 must then be \$V=0,25 \$. So \$V_+=2,75 V\$ Is this correct? \$\endgroup\$ – Carl Nov 24 '19 at 20:12
  • \$\begingroup\$ i=0.25mA, not 0.25A \$\endgroup\$ – BeB00 Nov 24 '19 at 20:13
  • 1
    \$\begingroup\$ And also in this case, if V2 is 0V, you would subtract 0.25V from 2.5V, so the answer is 2.25V. If you keep all of your signes consistent you will get this result, but you can also see that you have to subtract it because the answer will always be between V1 and V2. \$\endgroup\$ – BeB00 Nov 25 '19 at 2:12

BeB00 is right mentioning min and max is not relevant to the question.


  • the output of the opamp (\$V_{out}\$ is given in the question)
  • the inputs of an ideal opamp don't draw current
  • Vref = 2.5V

you can calculate \$V_+\$ using both resistor values...

  • \$\begingroup\$ Wait, is the answer just \$V_+=0 V\$ and \$V_+=5 V\$? \$\endgroup\$ – Carl Nov 24 '19 at 19:57
  • \$\begingroup\$ The min and max voltage are unrelated to what this question is asking \$\endgroup\$ – BeB00 Nov 24 '19 at 20:02
  • \$\begingroup\$ @Carl No, please check the update \$\endgroup\$ – Huisman Nov 24 '19 at 20:02
  • \$\begingroup\$ @BeB00 You're right. I misread the question at first. \$\endgroup\$ – Huisman Nov 24 '19 at 20:07

This is a simple basic op amp question. Assuming an ideal op amp, where the input pins draw no current, you can work out an equation for node V+ by simply writing KCL (I assume you know what a KCL is, if not google it) at that node. Drive the equation and then plug in the values for Vout to get the voltage at the non-inverting pin.


As formulated, the question is not related to the operational amplifier. It is a question about the simple 2-resistor summing network R1-R2 driven by two voltage sources - Vref = 2.5 V and Vout = 0/5 V. In the first case it is a voltage divider driven from the left; in the second case it is a complete voltage summer driven from both sides (inputs).

Generally, its output voltage can be found by applying the superposition principle: V(+) = Vref.R2/(R1 + R2) + Vout.R1/(R1 + R2)... i.e., by summing the two partial output voltages of the voltage divider R1-R2 driven from the left and the voltage divider R2-R1 driven from the right.

For those who think visually, I offer two geometric interpretations:

...by a triangle...

Voltage divider visualized

... and trapezoid:

Voltage summer visualized

As for the full circuit, it is an op-amp inverting comparator with hysteresis. The answers to the question give its two thresholds.

  • \$\begingroup\$ In contrast to the 'serial voltage summer' where the input voltage sources are directly connected in series (according to KVL), in this 'parallel voltage summer' the input sources are connected through resistors in parallel to the summing point. As a result, their voltages are scaled with coefficients R2/(R1+R2) and R1/(R1+R2). That is why they have weighting inputs. \$\endgroup\$ – Circuit fantasist Nov 24 '19 at 22:25

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