0
\$\begingroup\$

Currently i'm bulding a circuit that need two voltages.

  • 5V to feed the microprocessor, sensor, leds, LCD and others
  • 12V to temporary feed small water pump (12V 4.8W) and a Electric Solenoid Water Valve (250mAh) [Activated by a MOSFET from MCU]

5V Will always run, but with a timeout that disable lot of sensors, leds, lcd, etc to save power, when user click any button it awakes. From x to x seconds it will auto awake to update sensors and values and sleep again.

12V will only be needed to turn on pump and valve by user interaction (Click a button) to fill a glass of water for example.

I have a 5V 5A brick power supply with stabilization that i run some of my projects. The main question here is if i use this power supply with some or one boost convert 5V->12V will be better than run 12V and buck converter to 5V to main system

Note: 5V Circuit can reach max 2A when its all ON and at Max Brightness

\$\endgroup\$
6
  • \$\begingroup\$ It really depends which convertor you will use. Some has better efficiency and some worse. Did you try to compare some convertors? \$\endgroup\$ Commented Aug 21, 2016 at 19:21
  • \$\begingroup\$ Not yet, since i can't find a proper 5V to 12V Convertor with good current supply (1A or more) \$\endgroup\$ Commented Aug 21, 2016 at 19:33
  • \$\begingroup\$ There is plenty of DC-DC convertors capable of 1A or more. I just looked at digikey and one of many is MC34166. There are lots of similar ones. It is necessary to read datasheets and pick one that is best for you. \$\endgroup\$ Commented Aug 21, 2016 at 19:53
  • \$\begingroup\$ input says 7.5V, can it work with 5V input? \$\endgroup\$ Commented Aug 22, 2016 at 0:26
  • \$\begingroup\$ you use "mAh" and "A" interchangeably in the question and in comments in a way that is confusing. "A"/Amperes is a measure of current and "mAh" is a measure of energy. \$\endgroup\$
    – Wesley Lee
    Commented Aug 22, 2016 at 4:32

3 Answers 3

2
\$\begingroup\$

There are lots of power supplies that deliver +12V, 2A and +5V, 2A; that is a common power requirement for external hard disk drives. I'd use one of those, they're assembled and tested and available whenever an external drive dies...

\$\endgroup\$
3
  • \$\begingroup\$ Keeping mind you should find a quality one. Some are cheap transformer ones without much stabilization, the voltage will vary by load. \$\endgroup\$
    – Passerby
    Commented Aug 22, 2016 at 4:20
  • \$\begingroup\$ Good advice, can you recommend any or show me one? \$\endgroup\$ Commented Aug 22, 2016 at 13:16
  • \$\begingroup\$ By popular demand, here's a kit which includes such a supply: <banggood.com/…> \$\endgroup\$
    – Whit3rd
    Commented Aug 23, 2016 at 2:51
0
\$\begingroup\$

Your 5V drain appears to be significantly more (over an extended period), which would imply that it is more efficient to take this rail directly from the primary supply, and generate the 12V rail on demand. If you start at 12V, you will have an additional penalty from the 5V switcher efficiency.

However, without comparing all of the components you are planning to use, it is hard to be sure. For example, it is possible that the low power state is dominated by leakage in the primary supply if it is rated for the peak load.

\$\endgroup\$
3
  • \$\begingroup\$ components at 5v are just sensors, lcd, lcd, no more than 2A load. Some sensors and components will be disabled with mosfet at sleep time At 12v are 500mAh. 5V Power supply is stabilized 5000mAh. \$\endgroup\$ Commented Aug 22, 2016 at 0:06
  • \$\begingroup\$ Telling me the current says nothing about the energy. How many joules over a 24 hour period? You need to know this to calculate the impact of switcher efficiency on the wall-socket average power. I suspect you actually care more about practicality than electrical efficiency though. \$\endgroup\$ Commented Aug 22, 2016 at 6:56
  • \$\begingroup\$ I still designing the circuit and build the system. Never tested the whole system running, i only know the draw current of each part. I'm not an electrical engineer, my knowledge is basic in the electric maths, sorry. If it help here the schemantic of the system: dl.dropboxusercontent.com/u/18602178/waterdispensor.pdf \$\endgroup\$ Commented Aug 22, 2016 at 14:34
0
\$\begingroup\$

What's 'Better' depends on the situation. I prefer to use a separate power supply for each device I make, unless it is only used occasionally and I don't need my bench power supply for something else.

In general a buck converter is more efficient than a boost converter. Whichever way you go the primary power supply must be sized to handle both loads (including the loss in the converter).

A 12V supply stepped down provides an extra level of stability on the 5V supply, so the 12V supply could be run hard out to operate the pump and valve without glitching the MPU. If you boost from 5V to 12V then the 5V supply must maintain stability over a wide power range. This may be quite difficult to achieve, particularly if your pump has a high startup current.

\$\endgroup\$
3
  • \$\begingroup\$ So you use two power supply if you circuit use 2 voltages? I don't know if it's clear but i'm using dedicated power supply too. Pump are weak 4.8W, 12V rail need no more than 500mAh or 600mAh peak \$\endgroup\$ Commented Aug 22, 2016 at 0:09
  • \$\begingroup\$ Not two power supplies, 1 power supply, 1 step-down converter. You said you already use the 5V 5A PSU to run 'some of' your projects, so I assumed you wouldn't be dedicating it to this one. 600mA at 12V = ~1.7A into the booster at 5V, + 2A = 3.7A total. That's within the PSU's rating so you will probably be OK, but it's generally not a good idea to run high power intermittent loads directly off the microprocessor's +5V rail. \$\endgroup\$ Commented Aug 22, 2016 at 1:19
  • \$\begingroup\$ Ah my bad, i want to say that i use that power supply with other projects and run fine at high loads, not sharing with multiple projects. Sorry. So what you recommend then? I was thinking about use a mosfet to feed boost converter on the go, any way to isolate from MCU rail? \$\endgroup\$ Commented Aug 22, 2016 at 2:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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