0
\$\begingroup\$

What is the most efficient way to step down 48V to 12V? I have a power supply that can output 48V at 40 amps. Assuming the power supply is perfect, that gives me 1920 watts. However, I'm powering some RC equipment such as ESCs and brushless DC motors that need 12V. The perfect step down would be able to push 12V at 160 amps. Is there anything more efficient than the regulators I have found which at most can push 12V at 12 amps?

The ones I have found:

\$\endgroup\$
  • \$\begingroup\$ Are you looking to design this, just understand which would be most efficient or buy one? \$\endgroup\$ – David Aug 24 '14 at 21:42
  • \$\begingroup\$ I would prefer to buy a pre-made one but if I have to make my own I'll do it. \$\endgroup\$ – Chromey Aug 24 '14 at 21:46
  • \$\begingroup\$ Is the 48v/40 amps power supply something you are trying to use for the project because it seems like a good idea, or is it the only power source available? \$\endgroup\$ – gbulmer Aug 24 '14 at 21:53
  • \$\begingroup\$ Unfortunately it is the only power source available otherwise I would just use a 12v native power source. \$\endgroup\$ – Chromey Aug 24 '14 at 21:57
  • 1
    \$\begingroup\$ I just thought of this. It could be possible to use multiple regulators (1 for each component) because each component shouldn't pull more than 20 amps. \$\endgroup\$ – Chromey Aug 24 '14 at 22:04
3
\$\begingroup\$

The most efficient way to create a lower voltage at higher current from a higher voltage at lower current is a type of switching power supply called a buck converter. For a buck converter, (watts out) = (watts in) - losses. For a linear regulator (current out) = (current in) - losses.

Buck converters up to 85% or so efficient are relatively easy to make yourself. You have to wake up and take it seriously to get over 90%. Getting 95% requires someone that knows what they are doing really applying themselves to the problem.

There is much written about buck converters out there, and the term "buck converter" should be a useful search term. Therefore I'll only explain the general concept briefly.

When the switch is closed, current builds up in the inductor. When the switch is opened, the instantaneous inductor current must continue to flow. D1 provides a path for this current. Since the voltage accross the inductor is now negative, the current in it decreases. The switch opens and closes rapidly to add current to the inductor when closed and causes the inductor current to ramp down when open. The fraction of the time the switch is closed regulates the overall output current. This fraction is usually modulated by a feedback loop to regulate the output voltage.

Due to the current path thru D1, the output current is higher than the input current. If all the components are ideal, no power can be dissipated, and all input power is transferred to the output.

\$\endgroup\$
2
\$\begingroup\$

I know Olin has answered this question and you have accepted it but I would recommend using a synchronous buck converter - it uses two MOSFETs and is more efficient and, surprisingly, easier to understand and control.

Imagine your 48 volts is fed into a circuit that chopped it up into a square wave of a certain mark-space ratio - this is what a sync buck converter does and, the average voltage of that square wave (when fed through a series inductor and capacitor to gnd), is a dc level that corresponds to the output level you need so, for an output of 12 volts you need to use two FETs fed from a PWM source that creates a 25% mark-space ratio.

This converts 48 volts into 12 volts.

If you have really low on resistance fets and a really low resistance inductor all you need to do is set the mark space ratio to 25% and, if the input voltage varies a bit, make this mark-space ratio modifiable by the input voltage changing.

Inevitably it is usually a bit more complex than this because even the best fets in the world drop a little voltage and so does the inductor and under heavy load conditions the voltage will sag - this can be countered by slightly increasing the mark-space ratio.

A non-sync buck converter isn't as efficient and is prone to more instabilities so I'd urge you to consider this route. A PWM circuit such as the LTC6992 is very useful as the heart of this type of switcher - it's a voltage controlled PWM oscillator.

\$\endgroup\$
  • \$\begingroup\$ Thanks. I would upvote your answer but I'm a noob. :( \$\endgroup\$ – Chromey Aug 25 '14 at 0:04
  • \$\begingroup\$ Okay I'm doing my calculations and now I'm stuck. I'm using simonthenerd.com/files/smps/SMPSBuckDesign_031809.pdf as a tutorial. The highest amp rated inductor that I found and can afford is 65 amps and has an inductance of 500 uH. Also, as Olin pointed out that making it 85% efficient is easy, I used 136 amps as current load. Doing the calculations this gives me a switching frequency of -0.000211 KHz. This seems impossible to me. Can you steer me in the right direction? The inductor is digikey.com/product-detail/en/RD8137-64-0M5/817-1844-ND/1997813 \$\endgroup\$ – Chromey Aug 25 '14 at 3:15
  • \$\begingroup\$ Listen to my words - don't use this type of buck regulator - use a synchronous buck regulator AND, most importantly get LTSpice (free from linear technology) so you can simulate stuff. It's also likely that the inductor you'll need is going to be in the tens of micro henry range and will need to be hand-wound - your operating frequency is going to be in the tens of kHz range also. \$\endgroup\$ – Andy aka Aug 25 '14 at 11:17
  • \$\begingroup\$ Can you tell me or link me to a site that has the formulas required to design a synchronous buck regulator? \$\endgroup\$ – Chromey Aug 25 '14 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.