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I am driving an OLED display via two AAA batteries plus a charge pump to step up the voltage to 9V. I was wondering if there is likely to be any significant difference in efficiency between (a) using the two batteries in parallel and using a charge pump to step up from 1.5V to 9V, or (b) connecting the batteries in series and stepping up from 3V to 9V.

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  • \$\begingroup\$ That depends on your chargepump: how does the efficiency varies with output/input voltage ratio. \$\endgroup\$ – Cornelius May 2 '14 at 20:37
  • \$\begingroup\$ Yes, sorry that the question is vague in that respect. I'm currently just using a very simple two-stage Dickson charge pump clocked by a microcontroller. I am not sure how to go about measuring the efficiency as I only have a multimeter to work with. With a small load it gives about 8.8V from a 3V input (which is fine, the OLED needs 7V+). \$\endgroup\$ – foldl May 2 '14 at 20:41
  • \$\begingroup\$ See also this answer \$\endgroup\$ – Dave Tweed May 3 '14 at 0:53
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You can calculate the efficiency by using \$n = 1 - \dfrac{p_{in}}{p_{out}}\$ and get the power \$p_{in}\$ by measuring the current from the batteries times the batteries voltage and the power \$p_{out}\$ by multiplying the current to the LED times the voltage across the LED(s) (at the output of the converter).

This assumes the converter has a filter capacitor. Use a scope and a shunt resistor of about 0.1 Ohms if not. Although trickier, you can estimate the current consumption by averaging the pulses height times the duty cycle.

$$I_{avg} = I_{peak} \times \frac{t_{on}}{t_{off}}$$

From my experience doing a lot of this sort of thing, using a higher voltage into a boost converter will improve efficiency.

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As a rule of thumb, the closer your input and output voltages of smps's the more efficient it will be. With a Dickson, that rule is the same as you've got diode (or mosfet) voltage drops to overcome. And you'll need more stages to get to higher voltage with a 1.5V input. With 3V you have less stages and therefore less voltage drops wasting your energy.

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As a generic answer, you need to pick some values, and do the math. So we have 3V or 1.5V, to 9V. Let's assume 100mA load, though your OLED probably uses less. And let's assume 85% efficiency, a good average switching regulator efficiency.

So first we need to know the output power usage.

$$P = V \times I$$

$$9V \times 0.1A = 0.9W$$

Next, we can calculate the input power usage.

$$N \times 0.85 = 0.9W$$

$$N = \frac{0.9W}{0.85}$$

$$N = ~1.06W$$

So at 85% efficiency, you need 1.06W at the input. So we can figure out the current pull at the input.

Reverse the power formula.

$$I = \frac{P}{V}$$

$$\frac{1.06W}{3V} = 0.350A $$

while

$$\frac{1.06W}{1.5V} = 0.7A$$

With the typical AAA having 1500mAh, you have 3V 1500mAh, or 1.5V 3000mAh. You would have 4 hours with either.

BUT your charge pump might not be 85% efficient with both setups. And battery drain may not be the same in parallel. You need to figure out the efficiency by measuring the current draw at the battery in both setups, as well as the current draw of the regulator + load. Any multimeter with a current/ammeter setting will help. This is not including varying loads, as they are more difficult to figure out.

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  • \$\begingroup\$ Right, I understand that in theory the power consumption will be equal for both methods. I guess I will just have to try both and measure the values, as you say. Thanks. \$\endgroup\$ – foldl May 2 '14 at 21:17

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