hey, I've got 5 volts here, I can just do another 50% voltage divider to get 2.5 volts. So a couple of 10k resistors and now I had what I thought should be 2.5 volts heading into that 3v3 pin.
But that's not what I measure there.
Yeah. Welcome to Ohm's Law.
If you measure the voltage at the midpoint there with nothing else connected to the midpoint, it will read exactly half of the input voltage. But only because the voltmeter has a high impedance (resistance here), and doesn't suck more than a microamp or two.A voltage divider (two resistors) works only for a tiny current draw. Because of Ohm's Law, a tiny current draw will still yield close to the desired output voltage. But draw more current (like.. powering a bluetooth chip with it) and E=IR says that the resulting voltage drop through the resistor will be high, leaving nearly nothing left.
My sleep-impaired brain says there is a simpler (meaning, more correct) way to explain it:
The WT32i board is also a "resistor", or at least so from the point of view of its power supply. So that resistor, or "internal resistance", needs to be included in the calculation for the resistor ladder you are building. Except the WT32i is a variable resistor, having different internal resistance depending upon what it is doing at any given point.
This makes it nigh impossible to power with a resistor ladder, unless the resistors you choose for the ladder are significantly lower than the internal resistance of the WT43i, so that its resistance doesn't throw off the ladder's value too much.
Same result though: do NOT attempt this. it will likely draw too much current and fry the resistors (literally: smoke). Eg, if you used two 100ohm resistors, then even with nothing connected to their midpoint that circuit will draw (5V / 200ohms) = (0.0925amps, or 25 milliamps. The heat generated would be (5V * 0.025amps) = 1/8watt. Which is just enough to fry the typical small 1/8W resistors used with Arduinos and such.
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