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So I was creating my own circuit based closely off the minty boost charging diagram for iDevices and I had slightly messed up on my calculations. If all my math was correct, I accidentally bought resistors 1000 times stronger than the ones I actually needed. What caused me to make that mistake? The original diagram I had based my circuit off of was in the thousands.

My question is, would it be possible to create a voltage regulator following the same diagram and using the same values for the resistors except 1000 times less (eg 43k is only a 43)? If so, then would would be the reason to use a 43k instead of a 43 ohm? What about using a 43k ohm if the specified voltage regulator schematic asked for a 43 ohm? Are there any advantages/disadvantages to either?

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  • \$\begingroup\$ Looks like your smaller divider on D- will turn about a quarter of a watt into heat, a bit less for the D+. That may be enough to get noticeably warm. \$\endgroup\$ Aug 11, 2013 at 15:04

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These resistors are used simply to provide very specific voltages in order to notify the connected iDevice that it is able to draw a lot of current for charging. The voltage you'll get will still be the same as long as all of your resistors are off by the same factor, so you can get away with your resistors off by a few orders of magnitude.

The reason such high values are used is because resistors waste power. The data line connections D- and D+ are high-impedance, meaning they require very little current, and high resistor values are acceptable. The original 43K, 51K, and 75K resistors will waste around 93μA (plus a tiny bit of current into the data lines), which is quite small. Using 43, 51, and 75 ohm resistors will waste around 93mA, which starts to be significant. To make matters worse, that wasted power has to go somewhere, and physically small resistors can have trouble dissipating it.

You don't want to increase your resistor values too high either, because the current leaked into (or out of) the D+ and D- lines will start to become significant, and will throw off your desired voltage. In the end, deviating in resistor values by a few orders of magnitude in this case won't change much, but there is definitely a good reason for the specific values given.

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  • \$\begingroup\$ Finely (and accurately) stated for a first answer. Only thing I would add is that Voltage Dividers work on Ratios, not just specific values, and tolerances mean that a resistor step up or down is probably not enough to throw the voltage sensing that the iDevice does off. \$\endgroup\$
    – Passerby
    Aug 11, 2013 at 13:11
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A voltage divider which uses smaller resistances draws more current from the source, but appears more "stiff" to the load. That is all: the trade off is between energy use and stability of the voltage.

As a rule of thumb, a voltage divider is firm if the load which uses it is ten times more resistive. If the load is a hundred times more resistive, then the divider is stiff.

Voltage dividers do not regulate voltage; they just scale down an input voltage. Any noise or fluctuation in the voltage is scaled down by the same amount and so is still present in the output voltage in the same proportion. (A voltage divider's stiffness w.r.t load variations does not amount to regulation.)

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