pulldownvoltage divider
Hello I am currently a beginner in the world of electrical engineering and would like some help on voltage-divider circuits.
My question mainly concerns the "loading down effect". Can someone please explain what this effect is and what effect it has on a simple voltage divider circuit?
Please try to keep it as simple as possible without going into too much detail, I want to know the main idea behind it.
In this answer I showed that the optimal series resistor value is
\$ R_S = \sqrt{R_{MIN} \cdot R_{MAX}} \$
where \$R_{MIN}\$ and \$R_{MAX}\$ are the minimum and maximum resistances of your sensor. The Sparkfun page speaks of between infinity and 300 kΩ. If we assume 10 MΩ as maximum value then your series resistor should be 1.8 MΩ (rounded to nearest E12 value).
These are pretty high values, too high for an ADC, which, like Mathieu says, likes a impedance of less than 10 kΩ. So you'll need to buffer the divider with an unity gain buffer opamp:
Note that a common opamp may have an input bias current as high as 1 µA, and this could distort your reading. A CMOS opamp like the MCP600x has a much lower input bias current, 1 pA typical for the MCP600x, which won't deteriorate your reading.
Your calculations are then incorrect.
The ATmega644 works fine with 5 V (which is also close enough to the output voltage of the microcontroller which itself is rated at minimal 4.2 V on page 316 of the datasheet) and at that voltage using a \$ 330 \mbox{ } \Omega \$ and \$ 75 \mbox{ } \Omega\$ voltage divider, you get 0.926 V which is close enough to 1 V for most purposes.
Using the \$1 \mbox{ } k\Omega\$ and \$ 75 \mbox{ } \Omega\$ voltage divider, you get around 0.349 V which is for most purposes close enough to 0.3 V. And when both pins are off, you get 0 V.
Best Answer
simulate this circuit – Schematic created using CircuitLab
Here's an example, see if this makes sense.
Consider the circuit on the left. Assume the meter VM1 is perfect- it has infinite input resistance. The voltage VM1 measures is \$ 10V \cdot \$ \$R2 \over (R1 + R2) \$ = 5V
Consider the circuit on the right. The meter VM2 has an input resistance of 1M represented by an ideal meter in parallel with 1M ohms. The equivalent resistance of R4 in parallel with R5 is R4||R5 = \$ R4\cdot R5 \over (R4 + R5) \$ = 90.909K ohms. The voltage VM2 reads is then \$ 10V \cdot \$ \$R4 || R5 \over (R3 + R4 || R5) \$ = 4.76V
The loading down of the 1M resistance reduced the reading by about 5%.
If you've been doing Thévenin equivalents, this should come as no surprise, since the unloaded divider could be replaced by a 5V source with 50K in series, and 1M is 20 times the source resistance.