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I have 12v power source Mean well PSC-60A with UPS function and acid buttery. In datasheet says, that output can be regulated from 12-15V - but it's not true, because tested results for output was:

  • from 11v(when not connected to 220v and buttery discharged)
  • to 13.8v(when connected to 220v and battery charged)
  • impossible to regulate output to 12v on charging(min about 13v)
  • regulating output voltage affects to battery charging voltage (

How to get a output with voltage of not more than <=12 volts (+/- 0.5V) with max 2A load on peak (avg load 500-800mA) from this 11-13.8v source ?

Output should be less or equal to 12v.

Which option is better, how right:

Help me please with this. enter image description here

Part of used scheme enter image description here

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    \$\begingroup\$ Who knows how you have connected this lot together. Any mind-readers in the house? \$\endgroup\$
    – Andy aka
    Mar 9, 2018 at 11:55
  • \$\begingroup\$ added scheme to question \$\endgroup\$
    – Alexandr Kramarenko
    Mar 9, 2018 at 12:14
  • \$\begingroup\$ A shunt regulator can't increase voltage. It also can't (reasonably) supply a lot of current. And, it wastes a lot of power when used. \$\endgroup\$
    – JRE
    Mar 9, 2018 at 12:19
  • \$\begingroup\$ I do not need to increase the voltage, it is necessary to prevent excess of more than 12v \$\endgroup\$
    – Alexandr Kramarenko
    Mar 9, 2018 at 12:23
  • \$\begingroup\$ I mean, output should be less or equal to 12v \$\endgroup\$
    – Alexandr Kramarenko
    Mar 9, 2018 at 12:24

2 Answers 2

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If all you need is to ensure your load doesn't ever see more than 12V, placing a good and hefty LDO like the MIC29302-12WT will do it. It will need some small heatsink, but especially if those 2A are only short bursts, even the smallest ones will ensure it won't overheat. Take care to look into the datasheet on the capacitor requirements for the input and output of the LDO and place those capacitors physically close to its pins.

On the other hand, if your load requires 12V and cannot live with less than 11V, you may want to place a step-up converter that outputs 12.5V, then follow it with the LDO. If your power supply gives off 13.5V, the step-up converter won't try to do anything, and it is almost the same as if it weren't there.

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  • \$\begingroup\$ ty for MIC29302-12WT \$\endgroup\$
    – Alexandr Kramarenko
    Mar 10, 2018 at 7:01
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As you do not put a schematic diagram I think you have an external board that you have no control over. In view of this it makes sense to add a Buck converter, since it allows you to control the regulation of output voltage and have a high efficiency. The voltage regulator (zener) can do it, but it is not completely efficient. The only detail you will have is that when you add the Buck converter it is possible that you introduce noise to the board since it may not contemplate its inclusion. But since the application is charging a battery, this may not affect the PCB you use.

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  • \$\begingroup\$ placed part of my scheme to question description. and ty for comments \$\endgroup\$
    – Alexandr Kramarenko
    Mar 9, 2018 at 12:40
  • \$\begingroup\$ I understand, in theory it must work. The only thing I do not understand is because the 12v are connected to the Vout (-). I do not find in the datasheet a negative voltage. Another consideration to take into account is the type of capacitor that you will choose, you must be very careful in that, since that is where inexplicable problems arise. \$\endgroup\$
    – Everq
    Mar 9, 2018 at 13:09
  • \$\begingroup\$ ty, where you find -12v ? \$\endgroup\$
    – Alexandr Kramarenko
    Mar 9, 2018 at 13:18
  • \$\begingroup\$ Ok, you have the following problem: + IN is the positive that comes from the power supply, while the -IN is the negative coming from the power supply. I do not understand why you put the 12V to the negative terminal, there you will not have an answer. The buck converter will deliver a single voltage or is 5V or 12V not both at the same time. I recommend you have 12V of the Buck converter and then use an LDO to carry the 5V you need for the Raspberri pi board. \$\endgroup\$
    – Everq
    Mar 9, 2018 at 13:27
  • \$\begingroup\$ oh, understood, it's simple scheme mistake \$\endgroup\$
    – Alexandr Kramarenko
    Mar 9, 2018 at 13:44

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