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This question already has an answer here:

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Above schematics is just a representation of an OpAmp circuit coupled to a measuring device. Imagine we know the values of the following parameters: Vcc, Vout, R_feedback and R_measuring device. R_measuring_device can be the input impedance of a scope or a DAQ hardware. V_out is the output voltage of the circuit which is coupled to a voltage divider in the schematics.

If one wants to couple V_out to the measuring device, one may need to scale V_out in many cases(for example to lower the voltage). In this case the ratio of R1 and R2 matters and by selecting the right ratio we can scale the output voltage. So far I understand..

But we don't choose in real life R1 = 22 ohm and R2 = 10 ohm for instance. I mean the ratio is easy to figure out when we aim to scale the voltage, but how about the optimum resistor values for R1 and R2?

What is the criteria here? Okay I might guess if R1 and R2 is too small, the OpAmp could be loaded with too much current? But how to quantify the R1 and R2 to prevent it to load. Should we check the max current output of this OpAmp from the schematics?

Same confusion with the relation of R1 and R2 with input impedance of the measuring device(R_measuring_device). If R1 and R2 are too big or too small it could be problem?

And there is also a relation or limitation to R1 and R2 coming from R_feedback I guess. Is that right? R_feedback is connected to the output of many opAmp circuits? Does R_feedback have effect on choosing R1 and R2 values?

These all three things are in relation with R1 and R2 here. How can one quantify R1 and R2 considering surrounding resistors are known(R_feedbcak, R_measuring_device) as well as Vcc and V_out are also known? Is there a rule of thumb or a practical easy way?

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marked as duplicate by Null, Michael Karas, The Photon, Daniel Grillo, brhans Mar 2 '16 at 22:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ Also related: electronics.stackexchange.com/q/129759/51760. Your op amp inputs are swapped, too -- you're using positive feedback. \$\endgroup\$ – Null Feb 29 '16 at 18:32
  • \$\begingroup\$ Not duplicate. Im asking about a very specific situation where it is surrounded by three resistors, where it acts as an interface between an opAmp and a DAQ. \$\endgroup\$ – user16307 Feb 29 '16 at 18:32
  • \$\begingroup\$ @user16307: You have a positive feedback circuit. Your output will latch-up or down at power-up. Swap the + and - inputs. If you are just creating a voltage follower then connect out to '-' without a resistor. \$\endgroup\$ – Transistor Feb 29 '16 at 18:32
  • \$\begingroup\$ @user16307 There's no difference between your question and the duplicate -- in both cases it's the load current you need to consider. \$\endgroup\$ – Null Feb 29 '16 at 18:33
  • \$\begingroup\$ no I dont agree I also ask about the relation between R_feedback. \$\endgroup\$ – user16307 Feb 29 '16 at 18:34
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Let's say we have a generator with a 10 volt output (let's call it "E1") and zero output resistance (source impedance) and we want to connect that to an amplifier with an input which can take 1 volt max (let's call it "E2"), and which has an infinite input resistance. (load impedance)

In that case we're pretty much free to do whatever we want to, so let's say we're comfortable taking 1 milliampere from the generator and using a voltage divider to drop the generator's 10 volt output down to 1 volt for the amplifier.

Since, from Ohm's law, \$ R = \frac{E}{I}\$, we can write:

$$ R2= \frac{E2}{I} = \frac{1V}{0.001A} = 1000\text { ohms.} $$

Now, since we started with 10 volts and we've gotten our one volt by passing 1 milliampere through a 1000 ohm resistor (let's call it "R2"), we must now get rid of the other 9 volts.

We can do that by connecting a resistor (R1) in series with the 1000 ohm resistor, and since current in a series circuit is everywhere the same, we can use Ohm's law again to find the value of R1, like this:

$$ R1 =\frac {E1-E2}{I} = \frac{9V}{0.001A} = 9000\text { ohms}$$

Schematically, then, we have:

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Now let's assume that our load impedance (let's call it "\$Z_L\$") is no longer infinite but is, instead, 1000 ohms, like this:

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In that case, It'll be in parallel with R2 which will make the total resistance from E2 to GND

$$ R_T = \frac{R2 \times Z_L}{R2 + Z_L} = 500 \text { ohms}$$

and we can use the voltage divider formula to find the voltage at E2:

$$ E2 = \frac{E1\times R_T}{R1 + R_T}= \frac{10V\times 500\Omega}{9000\Omega + 500\Omega} = 0.526\text{ volts} $$

A gross error, but easy to fix by just removing R2 from the divider and letting \$Z_L \$ do the work.

At the other end of the divider, the generator's output impedance will affect the value of V1 and, in the end, E2.

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  • \$\begingroup\$ So are you explaining how to quantify R1 and R2 here? I didnt get it.. How bout the R_feedback. Does it have any effect in choosing R1 R2 values? \$\endgroup\$ – user16307 Feb 29 '16 at 18:39
  • \$\begingroup\$ @user16307: Yes, more or less. Sorry, but I've got some errands to run so I can't attend to this right now. I'll edit my answer to make it a little less cryptic later on this afternoon or tomorrow. \$\endgroup\$ – EM Fields Feb 29 '16 at 19:57

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