Currently I’m in the progress of designing a PCB (hardware circuit) in order to control a parking brake and signal lights. I am using 36V, 16,5 Ah lithium iron phosphate battery pack to power the whole circuit as well as the driver unit. Since the whole circuit is powered by the battery pack itself I need to use as little current (power) as I can. In other words I need to design the circuit in more efficient way.

In order to power all the circuits I need regulated 12 V and 10 A of current in total. This 10 A will not be used completely at a time but based on each circuit functionality the required current will be used.

Since my battery pack is 36 V 16,5 A/H capacity, I planned to use a DC to DC buck converter to get 12V 10 A. When I calculate the current consumption of the DC / DC buck converter using below formula

Rπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ π‘œπ‘’π‘‘π‘π‘’π‘‘ π‘π‘œπ‘€π‘’π‘Ÿ π‘€π‘–π‘‘β„Ž π‘™π‘œπ‘ π‘ π‘’π‘  = π‘…π‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ π‘π‘œπ‘€π‘’π‘Ÿ (π‘Š) / Efficiency (%) = 120 Watts/ 80% = 150 W

Current consumption with losses = π‘…π‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ π‘π‘œπ‘€π‘’π‘Ÿ with losses (π‘Š) / Input Voltage = 150 W/36 V = 4.1 A

So the DC-DC converter consumes 4.1 A which is a lot for a 16.5 Ah battery pack.

Is there a better method or circuit or PCB device (semiconductor) to convert 36 V 16.5 A to 12 V 10 A in more efficient way by consuming less current?

  • 4
    \$\begingroup\$ ”16,5 Amp/H” No such thing. Do you mean 16.5 Ah? \$\endgroup\$
    – winny
    Dec 16, 2021 at 11:50
  • 2
    \$\begingroup\$ The battery is not rated in amperes per hour (A/h) but ampere-hours. This means that if you draw 4 A from a 16 A-h battery you will get about 4 hours of life. \$\endgroup\$ Dec 16, 2021 at 11:50
  • 2
    \$\begingroup\$ Power in = power out plus losses. To get from 36 V to 12 V (1/3rd of voltage) with a 100% efficient converter, you'd need to provide 1/3rd of output current, 10/3 = 3.33A, as an absolute, thermodynamic, laws of physics, minimum. With losses, it would be more than that, maybe 4 A. Maybe less than 4 if you've got a really efficient converter, a bit more than 4 if it's a rubbish one. 16.5 Ah says nothing about what current the battery can provide, only how long it can last, you need to check the data sheet. If it can provide 4 A, then that would drain it in about 4 hours. \$\endgroup\$
    – Neil_UK
    Dec 16, 2021 at 11:57
  • \$\begingroup\$ LED signal lights should use very little current, which leaves 10A for the parking brake. Is it a motor? If it is a motor, you don't need a buck, just a MOSFET. Just check the insulation can withstand 36V, then run it on 36V with 1/3 PWM ratio. It will get 12V average. \$\endgroup\$
    – bobflux
    Dec 16, 2021 at 12:04

2 Answers 2


This appears to be the question: -

Is there a better method or circuit or PCB device (semiconductor) to convert 36 V 16.5 amperes to 12V 10 amperes in more efficient way by consuming less current?

This part (36 V 16.5 amperes) is wrong because the OP is confused about the meaning of ampere-hours (the OP also refers to them as A/h and that is wrong). So, if you meant this: -

  • 36 V 4.1 amperes instead of this:
  • 36 V 16.5 amperes then...

There are certainly better methods.

  • 36 volts and 4.1 amps is a power in of 147.6 watts whereas,

  • 12 volt out at 10 amps is 120 watts.

  • This is a power efficiency of 81.3% and certainly, with care (and using synchronous buck converters), I would expect a power efficiency of greater than 90%.

  • So, with 10% losses the input power is 132 watts or, 36 volts at 3.67 amps.

  • With 5% losses, the input power is 126 watts or, 36 volts at 3.5 amps.

And, to answer the question, 3.67 amps (or 3.5 amps) is less than 4.1 amps.

Maybe one of these devices (shown wired as a dual parallel synchronous regulator) might be a consideration: -

enter image description here

It can produce 12 volts at 30 amps at efficiencies greater than 90% by the looks of it. Or, maybe one of these: -

enter image description here

It should be good for about 90% efficiency according to the data sheet.


Is there a better or [more] efficient method to convert 36 V / 48V , 16,5 amp/ H battery supply to 12 V 10 A than a DC/DC buck convertor chip?

Probably not. Depending on the buck used, efficiency could be as high as 95% or so.

Also consider that if the brake is a motor, it can likely be PWM'ed at the bus voltage but reduced duty cycle to approximate a lower supply voltage, foregoing any conversion at all. Of course there will be some small losses with either choice.

There are other things to consider for braking applications, like:

  • How often/long does the brake need to be applied? Is it something that could benefit from a static servo control (mechanical means to adjusting braking action; something that can be set once and stays that way until changed again.)
  • Does it need some fail-safe state? (i.e. Roller-coaster brake, defaults to ON.)
  • Is it something that could be regenerative? ("Brake" is really a generator, the power of which is the braking torque.)

battery pack is 36 V 16,5 A/H

Batteries are rated in Amperes * hours, called "Ah" or Ampere-hours. A 16.5Ah battery will provide 16.5A*1h for 1h, or 8.25A*2h, 5.5A*3h, etc.


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