As a followup to my previous question it turns out there was more to it than I thought. It seems that between my voltage divider and the ADC on the ESP there is a second voltage divider to step down the maximum of 3.3V from the development board to a maximum of 1V at the ESP unit.

So it actually looks like this:


simulate this circuit – Schematic created using CircuitLab

I'm told R2 in parallel with the the R3/R4 (100K/220K) divider makes the R1/R2 divider act like it was 100K and 76K at Vout such that 3.3V at the first divider yields 1.42V at Vout. That is consistent with what I measure with a DMM.

Can the two voltage dividers as diagrammed above be expressed mathematically? IOW, mathematically, if one voltage divider is Vout=Vin*(R2/(R1+R2)) is there an equation for two voltage dividers such as the one above?


So I've been struggling with the math and I have come the conclusion that my 50 year-old brain is definitely not what my teenage brain was when it comes to algebra. It's not something I've needed since learning it in high school and it's gotten really dusty. :-( Let me apologise profusely for my rusty alegbra in advance. But...

If R3 and R4 are fixed at 220kΩ and 100kΩ respectively and R2 will be a fixed value also (I'm not sure what yet, but it will be a value that will give me good readings on R1 for the range of values on R1 that I am interested in) and R1 will be a sensor with variable resistance (i.e. a thermistor), what is the equation to figure out the resistance of R1 given a value for R2 and a value for Vout?

For argument's sake, let's say R2 is 5.1kΩ since that's a value that has been giving me good readings in the range of temperatures on R1 that I am interested in at the moment. But I'd like to be able to change in my equation in the future if my range of interest in R1 changes.

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    \$\begingroup\$ There is an equation for just about everything.. and there is even a method to generate them, it is called Algebra. Figure out the equation for the voltage on top of R2, then multiply that by the resistor divider R3/R4 equation. \$\endgroup\$
    – Trevor_G
    Dec 21 '17 at 20:00
  • \$\begingroup\$ Current from source is i=v/r or 18.7uA=3.3/176.19k. Voltage at top node is v_src-v_drop_across_R1 or 3.3 - 18.7u*100k=1.43V. Current through left leg is i=v/r or 4.47uA. V_out=v_top_node-v_drop_across_R3 or 1.43-4.47u*220k=0.45V. I'll let you work out the Vo/Vi equation. \$\endgroup\$ Dec 21 '17 at 20:52

For one voltage divider there is the following rule:

  • A voltage divider with a constant voltage input behaves like:
  • A constant voltage source with a resistor \$R_p\$ in series while
  • \$R_p\$ has the value of \$R_1||R_2\$ (the resistance of resistors \$R_1\$ and \$R_2\$ in parallel)

Let's look at the right voltage divider (\$R_1\$ and \$R_2\$) first:

Without \$R_3\$ connected the voltage between the resistors is 1.65V (\$3.3V *\frac{100k\Omega}{100k\Omega+100k\Omega}\$). \$R_p\$ is the value that we had if \$R_1\$ and \$R_2\$ were in parallel: \$R_p = R_1||R_2 = 50k\Omega\$.

So we can replace the right voltage divider by a constant voltage source of 1.65V with a 50 kOhms resistor in series.

Now the left voltage divider consists of a voltage divider with three resistors: \$R_p\$, \$R_3\$ and \$R_4\$ and an input voltage of 1.65V.

Calculating the voltage over \$R_3\$ and \$R_4\$ in this replacement we get: \$1.65V*\frac{R_3+R_4}{R_p+R_3+R_4} \approx 1.43V\$. This is nearly the value you measured.

Now we can calculate the output voltage of the left voltage divider in the case that there is no current flowing through the output:

\$V_{out} = 1.65V*\frac{R_4}{R_p+R_3+R_4} \approx 0.446V\$

... and we can calculate the "\$R_p\$" value of the left voltage divider; let's call this value \$R_q\$ because \$R_p\$ is already the value for the right voltage divider:

\$R_q = (R_p+R_3)||R_4 \approx 73k\Omega\$

This means the complex voltage divider shown in your schematic behaves like a constant voltage source with a voltage of 0.446 Volts with a resistor of 73 kOhms in series.

(However if you have no current flow at the \$V_{out}\$ pin the \$R_q\$ value is not of interest.)


\$R_1||R_2 = \frac 1 {\frac 1 {R_1} + \frac 1 {R_2}}\$

  • \$\begingroup\$ Good explanation. Consider replacing the \$\Omega\$ units to \$k\Omega\$ to match OP's values. \$\endgroup\$
    – altai
    Dec 21 '17 at 21:01

A glance at the schematic should allow you to immediately write:

\$\frac {V_O}{V_{IN}} = \frac{R_2||(R_3+R_4)}{R_1+R_2||(R_3|R_4)}\cdot\frac{R_4}{R_3+R_4}\$

which is easily simplified to the following:

\$\frac {V_O}{V_{IN}} = \frac{R_2 R_4}{R_1(R_2+R_3+R_4) + R_2 (R_3+R_4)}\$

  • \$\begingroup\$ I guess this formula with it's laws have a name? If I want to read up more on the subject? I've been struggling to find key words to even get me started. If there is such a thing, that would be very helpful in terms of digging deeper. Thanks in advance! \$\endgroup\$
    – Torxed
    Apr 21 '20 at 19:53
  • \$\begingroup\$ Top equation, right hand part is just the 2-resistor voltage divider equation. The left hand part of the product is also a voltage divider equation but calculates the voltage at the junction of R1 and R2. So the output voltage is a fraction of that voltage. The left hand part is more complex because we have to calculate the loading effect of R3+R4, but you can think of R2||(R3+R4) = R2*(R3+R4)/(R2+R3+R4) as a single resistance. \$\endgroup\$ Apr 21 '20 at 19:59
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    \$\begingroup\$ Thanks, I'll dig deeper and see if I can make sense of this. Really impressed with how easy it's made out to be once you understand it and hopefully I'll come to this level of understanding one day :) Thanks for the quick answer. \$\endgroup\$
    – Torxed
    Apr 21 '20 at 20:57

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