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I have 900k--100k resistor voltage divider, which works ok - scales down input voltage 10 times. Now I want to offset attenuated value by 1.65, i.e. I apply bias voltage at the bottom of voltage divider. The schematics is easy:

schematic

simulate this circuit – Schematic created using CircuitLab

Now if I apply 10 V on Vin I expect to see 2.65 V on the output, but in practice I measure 2.45 V, why is it so?

At first I thought maybe I have my probes set to x1, but no, on x10 I read the same voltage. Have no idea where are my 200 mV lost.

Other thought was that that 220 Ohm resistor is basically in the divider chain, so it is 3 resistor divider 900k, 100k and 200 Ohms, but I checked - since it is orders of magnitude less than other it should take 0.002 Volts, not 0.2.

The question is - where are 200 mV lost and how to make it "right" i.e. give output of 2.65 V?
Just to clarify without offset I get exactly 0.9999999 volts with the divider and x10 probe, I checked resistors multiple times, all of them are practically up to specs, <0.5% off.

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    \$\begingroup\$ You have added more resistors to the impedance network without calculating what the value of those resistors should be in order to balance the rest of them. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 28 '16 at 22:50
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You have forgotten that you are no longer dividing to ground. You are dividing to 1.65 V.

\$ V_O = \frac {(V_I - 1.65)}{10} + 1.65 \$.

At \$ V_I = 10~V \$ you get \$ V_O = \frac {(10 - 1.65)}{10} + 1.65 = 2.485~V \$.

For a simple (in)sanity check think what happens when \$ V_I = 1.65~V \$ and then when \$ V_I = 0~V \$.

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  • \$\begingroup\$ hm... makes sense. But what should I do to maintain "division by 10" and also biasing by 1.65? I can of course change division ration to 8.3 for example, but that would work only for 10V, so that is not the solution. Changing bias voltage to 1.83 makes everything work, but it is not easy since I would need 3.66 power supply, which is unusual unless I make some ugly workaround. What to do in such case? Thanks for answer anyway. \$\endgroup\$ – ScienceSamovar Aug 28 '16 at 23:12
  • \$\begingroup\$ damn I am stupid, just use different divider for bias voltage :D Thanks!!! \$\endgroup\$ – ScienceSamovar Aug 28 '16 at 23:15
  • \$\begingroup\$ Of course, given the current entering the bias node, that's not an exact solution \$\endgroup\$ – Scott Seidman Aug 29 '16 at 2:01
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You can save a heck of a lot of power, and one part, with minimal design effort as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

Design equations are

R3||R2 = R1 (so output voltage at 0V in is 1.65V)

and

(3.3/R1 + 10/R3)*(R1||R2||R3) = 2.65

I arbitrarily picked R3 to be 1M. If you find you need lower output impedance (for example to go into an MCU analog-to-digital converter that will only tolerate a certain maximum source impedance), you can just scale the whole thing.

Of course the output impedance is just R1||R2||R3, or 100K in this case, so if you needed 10K you could use 20K/25K/100K (24.9K is the closest E48 value to R2, the rest are exact nominal).

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