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I am working on a project in which there is an existing enclosure, an existing power supply board and a bought board that I need to interface with.

The board I need to interface with needs to receive a supply voltage with the following demand:

+3.3 VDC ±5% with less than 100mV ripple

I don't that much experience in board design, so what I need to know is how to supply that 3.3V with the "less than 100mV ripple". I have 5V and 10V from that power supply board that I have.

The power supply board that I have uses the NCS6S1205C for 5V and NCS6S1212C.

  1. Which components do I need? an LDO? a DC-DC converter? I suppose I also need components like inductors and capacitors to meet the <100mV ripple?
  2. Can I do this somehow without a board? maybe put the voltage converter on the enclosure itself?
  3. If I do need a board, I would love if someone can provide a reference schematic I can use, or provide me with a good component that has such a reference schematic in it's datasheet.

I did search and find some stuff like:

http://www.ti.com/product/TPS60201

http://www.ti.com/product/TPS60205

https://www.rohm.com/documents/11486/47257/BD99010_BD99011_Datasheet_Rev+003_EN.PDF

But I really prefer to ask here, since I need the simplest design that will use the least components as possible.

Thank you!

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    \$\begingroup\$ I have a couple of questions: Amps to drive? Do your board have any sensitive-to-noise components like ADC? If so, LDO. If not, use a dc/dc. Mainly because efficiency. A LDO has ~60% instead of a DC/DC (~80%-95%), contains the ripple you mention. There are nanoboards like the LM317 (TO 220 package) which handles small currents (up to 1A) using buck converters. \$\endgroup\$
    – José Manuel Ramos
    Commented Nov 20, 2018 at 10:19
  • \$\begingroup\$ In the datasheet of the board, it is mentioned that the power consumption is 2.5W, so I assume the current is 0.75A? In that case, maybe it would be best to think of a current of 1A. But it is also said that In-Rush Power Consumption 6.6 A for less than 60 us (typical) \$\endgroup\$
    – nettek
    Commented Nov 20, 2018 at 10:31
  • \$\begingroup\$ About sensitive-to-noise components like ADC, I'm not sure. It is a board that has GPS functions and lots of BGA components, so I think it would be a safe bet to say the answer is yes? \$\endgroup\$
    – nettek
    Commented Nov 20, 2018 at 10:33
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    \$\begingroup\$ But it is also said that In-Rush Power Consumption 6.6 A for less than 60 us (typical) That would normally be taken care of by decoupling capacitors. I guess that the 6.6 A only occurs when you connect the board to a 3.3 V lab supply or such. When using a DCDC converter to supply the board, everything is always connected, then the board must be kept in reset while the supply switches on. Only when the supply is stable can the reset be released. That's the most common way to do this, the board might be doing this already by itself. \$\endgroup\$
    – Bimpelrekkie
    Commented Nov 20, 2018 at 10:47

1 Answer 1

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I would simply use the 3.3 V version of the DCDC converter modules you mention ( NCS6S1205C and NCS6S1212C), they have a 3.3 V version also: NCS6S1203C.

Then feed that 3.3 V module from the same input voltage that the NCS6S1205C and NCS6S1212C modules are using already, that way the two existing modules are not loaded more than they are now.

The NCS6S1203C has a 1.52 A maximum output current and 30 mVpp ripple which satisfies your needs.

The TPS60201 and others you mention are only "bare" chips, to use them you need to design a PCB for it and the required support components. That's much more work (and more risk that it doesn't work, especially if you have not done this before) than using a ready made module like the NCS6S1203C.

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  • \$\begingroup\$ I thought about this as well. What you suggest it take the power supply board I have, and replace the NCS6S1205C with the NCS6S1203C? \$\endgroup\$
    – nettek
    Commented Nov 20, 2018 at 10:44
  • \$\begingroup\$ If so, the problem with this is that I need to think about all the peripheral components that exist around that module (probably for filtering), and I really prefer not to do that. \$\endgroup\$
    – nettek
    Commented Nov 20, 2018 at 10:45
  • \$\begingroup\$ If you do not need the output voltages from the 5 V module then sure, you can replace it with the 3.3 V version. My bet is that the filtering components can remain the same, often supply filtering designed for 5 V will also work fine for 3.3 V. But to be 100% sure I would need to see a schematic of the power supply board. \$\endgroup\$
    – Bimpelrekkie
    Commented Nov 20, 2018 at 10:49

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