I have managed to confuse myself when it comes to calculating the voltage drop over a series resistor connected to a potential divider (See below):

op amp connected to a potential divider with extra series resistor

The op amp non-inverting input is connected to a 20K resistor which is then connected to the output of a voltage divider (Top resistor is 56K, bottom is 7K). The input voltage to the potential divider is 12V.

I would like to calculate the voltage "seen" by the op amp (set as a voltage buffer) on its non-inverting input (in this case pin 3).

The input resistance for an ideal op amp should be infinite, however in practise there is a very minute input leakage current. This gives me the following questions:

1) Is the resistance "seen" on the potential divider output branch, equal to the sum of the op amp input resistance and 20k resistor? Or would it be equal to the equivalent parallel resistance of 7k||20k+op amp input resistance?

2) As the input resistance of the comparator is very high, (and I don't have specific values for it) could I somehow calculate the potential divider output voltage drops over the 20k resistor and op amp without calculating current draw?

3) Related to the previous questions, can I "split" the problem into two parts? i.e Calculate the potential divider output voltage and then calculate the voltage drop over the 20k resistor and op amp?

4) Would an equivalent circuit without the op amp (i.e 20k resistor is left floating on one side) help me calculate the voltage drop over the 20k resistor?

Thank you for your help- I've been driving myself mad for days with this problem...

  • \$\begingroup\$ How can you say that 7k and 20k are parallel? Have a look at the datasheet. You may find some information regarding the input impedance. \$\endgroup\$
    – G-aura-V
    Commented Apr 28, 2020 at 11:44
  • 1
    \$\begingroup\$ Explain why R6 (20k) is needed. In my opinion there is no need to have this resistor as the opamp inputs draw no current to no voltage will develop across the resistor so you can just use a wire, but feel free to prove me wrong. \$\endgroup\$ Commented Apr 28, 2020 at 11:52
  • 3
    \$\begingroup\$ It's not a comparator circuit; it's an op-amp circuit; specifically a unity gain buffer. \$\endgroup\$
    – Andy aka
    Commented Apr 28, 2020 at 11:52
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    \$\begingroup\$ I don't see the point why R6 is there. If it drives you mad, why can't you remove it? \$\endgroup\$
    – Justme
    Commented Apr 28, 2020 at 11:54
  • \$\begingroup\$ Sorry, I got the names mixed up when looking at the comparators connected to the output of the op amp (not shown in this drawing). The 20K resistor is internal to the LM3914 LED driver shown here ti.com/lit/ds/symlink/lm3914.pdf I'm just trying to understand the voltage drop across the 20k resistor in this scenario. \$\endgroup\$
    – dyode254
    Commented Apr 28, 2020 at 12:08

2 Answers 2

  1. The resistance is "seen", but its effect is so small that unless you're working in precision circuitry you will not be able to measure it. (the op amp input is really high impedance)

  2. You can refer to the bias input current if you want to know the voltage drop as this will dominate that input resistance in most op amps, e.g. 5pA * 20K = 100nV,

  3. As the effect is about 7 significant figures away from the the divider voltage, in this case your safe to ignore it, if you had say 1 Giga-ohm resistors for your divider then it would be a different story (noise dominates there, but the DC value would be affected)

  4. refer to the input bias currents if you want the 20K drop, it will dominate the input impedance.

  1. The op-amp input current is typically modeled as a constant current, meaning that it does not behave like a resistance at all (an ideal current source has infinite resistance). Rather, it would increase or decrease the input voltage by the effective source resistance of the actual resistor network multiplied by the input bias current. This is the principle of superposition.

  2. It's connected as an op-amp, not a comparator. See above.

  3. Yes. Calculate the Thevenin equivalent source and resistance first. That involves only the three resistors and the voltage source.

  4. Without the op-amp there is no bias current so the 20K resistor won't make any difference, it will, however, affect the Thevenin equivalent source resistance by quite a bit (you can see from inspection that the source resistance of the 56K/7K divider alone is less than 7K, so 20K will increase it quite a bit).


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