It seems like it's always the simple projects that are confusing. Here is a simple voltage divider circuit to power a device from a vehicle.

Details

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12-14v DC power source

5-8v DC power requirements

1 watt power draw

Requirements

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1. Components must fit inside the 12v plug

2. Minimize current draw on the battery when the device is powered on or off

Calculations

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Load current

1 watt @ 5v = 200ma

1 watt @ 6v = 166ma (used as a reference)

1 watt @ 8v = 125ma

14v power source used for the calculations below

Option1: Eliminate R1 and connect the load in series with R2.

Problem: When the load is powered off, there is infinite resistance across the open circuit which will have 14v potential to the load. There is a risk of damaging the device when it is first powered on.

Option2: To minimize the current draw on the battery, choose R1 so that almost all of the current is going through the load ... like 10K. Then, choose R2 to handle the appropriate voltage and current.

a) if V(load) = 5v, then R(load) = 5 / .200 or 25 ohms

if R1 is 10K, then R(parallel) = 1 / ( (1/R1) + (1/Rload) ) or 1 / ( (1/10K) + (1/25) ) or 24.93 ohms

I(parallel) = I(R1) + I(load) or .5ma + 200ma or 200.5ma

if V(parallel) = 5v, then V(R2) = 9v (14v used for calculations)

R2 = V(R2) / I(load) or 9 / .2005 or 44.88 ohms

P(R2) = V(R2) * I(load) or 9 * .2005 or 1.8 watt

b) if V(load) = 6v, then R(load) = 6 / .166 or 36 ohms

if R1 is 10K, then R(parallel) = 1 / ( (1/R1) + (1/Rload) ) or 1 / ( (1/10K) + (1/36) ) or 35.87 ohms

I(parallel) = I(R1) + I(load) or .6ma + 166ma or 166.6ma

if V(parallel) = 6v, then V(R2) = 8v

R2 = 8 / .1666 or 48.02 ohms

P(R2) = 8 * .1666 or 1.33 watt

c) if V(load) = 8v, then R(load) = 8 / .125 or 64 ohms

if R1 is 10K, then R(parallel) = 1 / ( (1/R1) + (1/Rload) ) or 1 / ( (1/10K) + (1/64) ) or 63.59 ohms

I(parallel) = I(R1) + I(load) or .8ma + 125ma or 125.8ma

if V(parallel) = 8v, then V(R2) = 6v

R2 = 6 / .1258 or 47.69 ohms

P(R2) = 6 * .1258 or .75 watt

Problem: The value of R1 is so high that there is still 14v potential across R1. Again, there is a risk of damaging the device when it is first powered on.

Option3: Choose R1 and R2 to divide the voltage with the device powered off.

a) if V(load) = 6v, then R(load) = V(load) / I(load) or 6v / .166 or 36 ohms

if R1 is also 36 ohms, then I(R1) = 166ma

P(R1) = 6 * .166 or 1 watt

R(parallel) = 1 / ( (1/R1) + (1/Rload) ) or 1 / ( (1/36) + (1/36) ) or 18 ohms

I(parallel) = I(R1) + I(Rload) or 166ma + 166ma or 333ma

if V(parallel) = 6v, then V(R2) = 8v

R2 = V(R2) / I(parallel) or 8 / .333 or 24 ohms

P(R2) = 8 * .333 or 2.6 watt

When the load is powered off, V(R1) = V(total) * R1 / (R1 + R2) or 14 * 36 / 60 or 8.4v

Problem: The current draw is twice the load. 50% efficiency

There is still 166ma of draw when the device is powered off.

I am writing a computer program to go through all of the possible combinations of R1 and R2 to minimize the current draw, and still provide the proper voltages whether the device is powered on or off.

I am also looking into a 6v regulator if I can find one small enough. Any other ideas? Many thanks.