simulate this circuit – Schematic created using CircuitLab

I have a 12v 14amp hour lithium ion battery. Connected to a dc-dc buck step down voltage regulator with a potentiometer. Then this is connected to a set of heated goggles.

I have the potentiometer set to output 9.8v, as I was hoping to make the battery last longer than a direct connection, as well as run the goggles on a lower heat setting (they get very hot).

However it seems as though the battery lasts less time. I expected to get 12 hours of run time and only got 5.

Will the step down regulator actually draw more power from the battery over time? What is the best way to supply 10v and preserve the battery for the longest period of time?

I took these measurements with my multimeter. Amps was in-line on the positive connection, and volts was measured with the circuit complete and metering off the positive and negative.

Battery Direct to Goggles: 1.13 Amps 12.75 Volts

DC-DC Buck Step down Regulator installed between battery and goggles:

Between Battery and Regulator: .74 Amps 13.06 Volts

Between Regulator and Goggles: .9 Amps 9.8 Volts

I don't understand how it would be more amps after the regulator vs. before?

In any case, is there a better way for me to reduce the draw on the battery and make it last longer?

  • \$\begingroup\$ Please draw a diagram or schematic with the tool \$\endgroup\$
    – Voltage Spike
    Feb 8 at 22:27
  • \$\begingroup\$ The DC-DC converter passes power (voltage times current), not just voltage or just current. so, if the output voltage is lower than the input voltage, the output current will be higher than the input current. \$\endgroup\$ Feb 8 at 22:31

1 Answer 1


First are you sure your battery specs are actual or are they boosted? Typically the capacity is the pre-boosted capacity of the raw battery in a battery pack. You may not have actual 12V 14aH.

Second, a dc-dc converter is another word for a switching regulator. This could be a buck or step down switching regulator. The specifics of it's energy efficiency will depend on its components and the voltages involved. This could be a high 95% efficiency or it could be a low 80%. That means 5 to 20% power penalty wasted as heat. If it's on the low end, you may not be saving any power by reducing the voltage.

You measure the input at 13.06V .74 Amps. That's 9.6 Watts of power. The step down is putting out 9.8V .9 Amps. That's 8.82 Watts. That's about 92% efficient. With a switching regulator, the input and output current doesn't matter, what matters it the actual power (in watts). You can freely change the current if you change the voltage. That's the benefit of a switching regulator.

But based on your own measurements, the bucked power is 9.6 W vs the unbucked power at 14.4 W. Unless your measurement tools are adding some large resistance skewing your numbers, you are saving almost 5W of power. it should be lasting longer than not regulating it down.

The only other way to save power is a more efficient regulator or... bypassing or modifying what I assume is the internal step up or boost regulator of your battery pack. A third option is replace your googles with something less draining.

  • \$\begingroup\$ Thank you for clarifying the efficiency of the Step Down Regulator. After running some tests, I have figured out that the battery was quickly losing quality, and the charge it was holding was dropping. I am going to get a bigger and newer battery and hopefully that will solve that issue. I had someone else suggest I ditch the switching regulator for PWN dimmer. Would that be more efficient than the Step Down Regulator? \$\endgroup\$ Feb 10 at 14:37
  • \$\begingroup\$ Depends on how much you are dimming it. You are at 66% of the unbucked current. Unless you are going to run it at lower than that then you aren't getting any extra benefit. \$\endgroup\$
    – Passerby
    Feb 10 at 15:26

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