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I'm currently building a power module to power a few raspberry 3 pis and a set of microcontrollers. For the 5V power source, I'm using a buck converter for efficiency to deliver 5 Amps to power the pis.

However, I'm not too sure if using a simple linear regulator to source a 3.3V rail would be better than using another buck converter, as the load would only be within the mA range?

I'm using a set of batteries to power the power module too, so I'm trying to be efficient as possible.

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    \$\begingroup\$ The Pi's should have an onboard 3V3 buck converter, I think (didn't check). Could you just use this 3V3? \$\endgroup\$ – peufeu Nov 9 '17 at 21:01
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    \$\begingroup\$ well, the loss of the Linear Regulator would mainly be the product of your 1.7V drop and your microcontroller current, whereas the loss for the buck would be the MCU current times 3.3V times (1-buck efficiency) + quiescent current (which usually is higher for switch mode converters than for linear ones). So, this really boils down to how efficient a buck converter you choose/build, and can't be answered in general without knowing how efficient that converter would work at 5V->3.3V for your requested current. \$\endgroup\$ – Marcus Müller Nov 9 '17 at 21:01
  • \$\begingroup\$ Thanks. I'm looking to be drawing about 125mA from the 3.3V rail \$\endgroup\$ – Nasher Nov 9 '17 at 21:10
  • \$\begingroup\$ 1/4 watt does not seem like much if you use a 3.3V LDO, if it's a battery operated bob though that might be a lot. \$\endgroup\$ – Trevor_G Nov 9 '17 at 21:19
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At 125 mA, a linear regulator from 5 V to 3.3 V wastes (5 V - 3.3 V) · 125 mA = 0.22 W, which is not much compared to the 25 W being delivered to the Pi cluster.

If you used a buck regulator instead, you could probably get 90% efficiency without too much effort, which corresponds to a waste of (3.3 V · 125 mA · 10%) = 0.04 W.

However, a buck converter will be somewhat more complicated, somewhat more expensive, and will likely have a noisier output. It really depends on your specific application whether saving 0.18 W is worth the extra complexity.

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  • \$\begingroup\$ This has been my thought process exactly. I think going down the linear regulator route maybe the best way, as there isn't much need in over complicating things when dealing with such small currents \$\endgroup\$ – Nasher Nov 9 '17 at 21:58
  • \$\begingroup\$ On second thoughts, I've just came across the LM3671 which doesn't seem too complicated at all. 3.3V @ 600mA would be enough. \$\endgroup\$ – Nasher Nov 9 '17 at 22:04
  • \$\begingroup\$ @Nasher Looks like a pretty typical (but good) buck regulator. You're still going to have pay a lot of attention to component selection and board layout. \$\endgroup\$ – Abe Karplus Nov 9 '17 at 23:04
  • \$\begingroup\$ Thanks. About the board layout, I know the datasheets for most buck converters have a layout example, but in some cases where you cannot follow them, is there a general rule to follow or is this really chip specific? \$\endgroup\$ – Nasher Nov 10 '17 at 7:26
  • \$\begingroup\$ @Nasher The basic rule is to make the current loops as small as possible: there'll be a lot of (high-frequency) current flowing through the capacitors, switch, and inductor, so you need to make the loop area as low as possible. There are other considerations too: search for "switching regulator layout" and you'll find plenty of app notes. \$\endgroup\$ – Abe Karplus Nov 10 '17 at 9:02

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