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I am trying to create simple variable voltage divider circuit, with several constraints that are causing me to look for alternative solution.


  • The input voltage is 1.5 V
  • Maximum R1 is 5 M\$\Omega\$
  • Minimum R1 above 5 k\$\Omega\$
  • \$V_{out}(V_{R1})\$ is 0.1 mV upto 500 mV

The R1 max and min window is too small to allow for a 0.1 mV to 500 mV output.

My attempt and why I think this will not work:

  • To satisfy 5 k\$\Omega\$ or above resistance at output of 0.1 mV I used the equation as below:

$$0.1mV = \frac{1.5 \times 5k}{5k + R2}$$

$$R2 = 75 M\Omega$$

$$\text{This value gives me the value of R2 and hence }R_{total}$$

  • To find if 75 M\$\Omega\$ total will satisfy the 5 M\$\Omega\$ max at 0.5 V I continue from above:

$$0.5V = \frac{1.5 \times R1}{75M}$$

$$R1 = 25 M\Omega$$

$$\text{This value is way above the 5 }M\Omega \text{ limit}$$

It is clear that I can not satisfy the constraints with a voltage divider. My question is are there other solutions that may help me solve this issue?

Some background on why the constraints are there:

These constraints are due to available material and circuit requirements, i.e 5 M\$\Omega\$ limit on R1 is simply because I can't find a trim pot at any larger values. And the minimum R1 is because this circuit is simulating a electrochemical sensor that has an impedance above that value. If the voltage source has less of an impedance, the analyzer circuit will reject the source. Also the reason for using 1.5 V is because the analyzer uses 1.2 V detection signals.

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It would help if instead of starting from your solution, you gave the exact problem you're trying to solve. – sbell Aug 8 '14 at 14:16
It came across to me that he wants to use a voltage divider with an output frmo 100uV to 500mV over the range of the potentiometer. – sherrellbc Aug 8 '14 at 14:19
up vote 4 down vote accepted

A rather convoluted question, but I think what you need is an adjustable voltage source with a minimum output impedance of 5k.

In that case, why don't you design your voltage divider to give you the correct output voltage and then buffer it like so:

enter image description here

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Hi I tried this today and it was what I needed, I am looking into OpAMPS in more detail, thank you very much. – EcEng Aug 11 '14 at 21:44

This is not really an answer, but to help a bit your numbers are not correct unless I missed something. Assuming R1 is the "top" resistor of the divder (the potentiometer), and the bottom resistor is R2 then it appears as if you have done the calculations backwards.

enter image description here

$$V_{OUT} = V_{IN}\frac{R_2}{R_1+R_2}$$

So, when R1 is minimum (5k), the output is going to be at its maximum value.

$$500mV = 1.5V\frac{R_2}{5k+R_2}$$ $$R_2 = \frac{0.5*R_1}{1.5 - 0.5} = 5000*0.5 = 2500 \Omega$$

Now since R2 has been chosen, we go back to the original equation above and determine the output voltage when the potentimeter is much higher (i.e. the output voltage is much lower).

$$100uV = 1.5V\frac{2500}{R_1+2500}$$ $$R_1 = \frac{1.5*2500}{0.0001} = 37,497,500 \Omega$$

Or about 37.5M \$\Omega\$, although still out of the range of your potentiomter.

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is it not a fact that the less resistance means smaller Vout? therefore at 0.1mV i should have the minimum resistance, i.e. 5K ohms. at 0.5 V I should expect a larger R value. I think you have it flipped around. – EcEng Aug 11 '14 at 21:52
@EcEng, it depends on which resistor in the divider you have designated as your variable. In the case of the derivation above, the top resistor (R1) was chosen to be variable. I could not make heads or tails of what you did in the OP so I made some assumptions and worked though it in this configuration. The output behavior will be opposite (i.e. higher resistance, lower/higher output) depending on which resistor is chosen as the variable. For the variable resistor chosen as the "top" component in the divider, what I have done above is correct. – sherrellbc Aug 12 '14 at 1:52

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