A boost regulator has to transform low voltage into higher voltage. This can only be done by using more current on the input (battery) side. The amount of additional voltage needed is expressed as:
(Volts out - Volts in) / Volts out = Percentage of voltage increase, also know as the switch duty cycle.
Now that we know the duty cycle, we can figure the additional input current required to obtain the desired output voltage.
Amps out / ( 1 - Duty Cycle) = Amps in
Example...
Known factors:
3.7v in
8 watts out
3.0 ohm carto
Ohms Law tells us that we'll need 4.9v and 1.63 amps output to achieve 8 watts.
(4.9-3.7)/4.9 = .24 = 24% increase in voltage = 24% switch duty cycle
1.63 / (1-.24) = 2.16 amps input drawn from the battery.
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Now for validation.
As deemed by the Law of Conservation of Energy, power (watts) in must equal power out. (The true statement of the law is power in equals power out + efficiency losses. But for simplicity sake, we'll get to efficiency below.)
Power = Volts * Amps
Input:
3.7 * 2.16 = 8 watts input
Output:
4.9 * 1.93 = 8 watts output
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But what about efficiency?
Typical efficiency for a boost converter is in the 75-90% range. Efficiency is not constant. It varies with the desired outputs. You can find the efficiency for certain [manufacturer chosen] situations in the regulators data sheet. Using an optimistic value of 90% efficiency we can figure our adjusted input current.
Power out / efficiency = adjusted power in
8 / .9 = 8.89 watts input.
Adjusted power in / Volts in = adjusted amps in
8.89 / 3.7 =
2.4 amps input.
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And comparison:
A fixed voltage
device @ 3.7v will achieve an 8 watt output using 2.16 amps (Ohms Law)
8 / 3.7 =
2.16 amps input
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The above calculations explain why I say that boost regulators will get less battery life than a same size fixed volt, and also that you will not achieve better battery life by using a higher resistance coil.
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