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Problem:

I'm attempting to develop a power supply for a system that needs 1.2V, 3.3V, and 5V from mains 120V 60Hz. I have never ran into needing all three at once and am curious if anyone knows of a more simple/smaller solution then what I'm proposing.

Attempt:

I'm considering using a buck regulator SMPS to obtain the 5V from 120V and 2 linear regulators. First regulator to step the 5V down to 3.3V and then the next one to step down the 3.3V to 1.2V. Being that size is a big issue I'm up to hear some ideas.

Side note: The total current draw of 3.3V is no more then 1.0A and total current draw of 1.2V is no more then 1.0A.

Using some of the comments and answers, I went out and found this IC BA3259HFP. Any thoughts? It appears to solve the issue with 5 pins, 3 caps and 2 resistors.

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  • \$\begingroup\$ What's your input voltage? \$\endgroup\$ – Gustavo Litovsky Oct 28 '13 at 14:30
  • \$\begingroup\$ Main 120V ill update question! Thanks. \$\endgroup\$ – tman Oct 28 '13 at 14:31
  • \$\begingroup\$ You mentioned 5.5V in the title and 5V in the body, which one do you need? \$\endgroup\$ – Gustavo Litovsky Oct 28 '13 at 14:32
  • \$\begingroup\$ 5V...aHHH. Sorry about that. \$\endgroup\$ – tman Oct 28 '13 at 14:33
  • \$\begingroup\$ Important information is missing. What is the current consumption for the 3.3V and 1.2V separately. The total draw or 5V current draw is not important at all. \$\endgroup\$ – johnfound Oct 28 '13 at 15:24
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Since you need 1 Amp at both 3.3 and 1.2v, there will be high power dissipation with an LDO.

Your suggested BA3259 has a max power dissipation of 2300mW or 2.3W. With a 5v input, that is ((5 - 3.3) x 1 ) + (( 5 - 1.2) x 1) or 5.5W of power dissipation. Not recommended, unless you can keep the chip cooled properly and have the recommended board copper layout.

Since you haven't mentioned the current draw of the 3.3 and 1.2v rails, here are some 6 pin 1.5~2mm dual ldos with 200mA per channel.

  1. http://www.ti.com/product/tlv7111333d (1.3 and 3.3v)
  2. http://www.ti.com/product/tlv7111233
  3. http://www.ti.com/product/tps71812-33

Slightly larger 10 pin adjustable dual 250mA per channel

  1. http://www.ti.com/product/tps71202-ep
  2. http://www.ti.com/product/tps71202 (cheaper)
  3. http://www.ti.com/product/tps71334 (1 fixed 3.3, 1 adjustable)
  4. http://www.ti.com/product/lp5996 (1 at 300mA, 1 at 150mA)

Unless I glossed over it, all these should accept 5v in.

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  • \$\begingroup\$ Sorry for not explaining in detail. Its 1A per output \$\endgroup\$ – tman Oct 28 '13 at 18:32
  • \$\begingroup\$ Then you will have to bite the bullet and get large single output ldos that can do 3.3 and 1.8. Or go for switching regulators if efficiency is concerned. Remember, for LDOs, you are wasting (Vin - Vout) * Iout in heat. That's 3.4 Watts for 5 to 3.3v (2 Amps) and another 2.1 Watts from 3.3 to 1.2 (1 Amp). \$\endgroup\$ – Passerby Oct 28 '13 at 18:36
  • \$\begingroup\$ @tman updated answer \$\endgroup\$ – Passerby Oct 28 '13 at 19:55
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You haven't mentioned whether you need a fully enclosed solution, but if this is for a prototype I would go the following route:

Use a 110V/220V to 5V adapter. These are very common with USB output and their current output is typically 1A or sometimes more.

Unless you have very sensitive circuitry, use a Buck regulator to go 5V to 3.3, and then use the 3.3V to go to 1.2V. Avoid using LDOs when possible because they're inefficient and will dissipate a low of power. You're doing good in attempting to reduce the input to output voltage difference (it is a big factor in LDO inefficiency if you go that route).

TI, Linear and others make combo devices that will give you a small solution.

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  • \$\begingroup\$ Can you suggest any combo IC? \$\endgroup\$ – tman Oct 28 '13 at 14:40
  • \$\begingroup\$ That's the issue when I look at the Dual output LDO's your talking about 20-28 pins where as with using 2 single Liner regulators like the TLV1117LV Im at 8 pins total. PCB real estate is limited so I'm trying to get this fairly small. Thought? \$\endgroup\$ – tman Oct 28 '13 at 14:57
  • \$\begingroup\$ @tman: can you be specific on the output currents for each of the 3 rails? \$\endgroup\$ – Gustavo Litovsky Oct 28 '13 at 15:25
  • \$\begingroup\$ I corrected this above...Its less then 1A for each \$\endgroup\$ – tman Oct 28 '13 at 18:31
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Depends on the current draw of the smaller rails; if they're a few hundred miliamps then you can just add an LDO in SOT-23 packaging and you're done.

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  • \$\begingroup\$ That's the issue when I look at the Dual output LDO's your talking about 20-28 pins where as with using 2 single Liner regulators like the TLV1117LV Im at 8 pins total. PCB real estate is limited so I'm trying to get this fairly small. Thought? \$\endgroup\$ – tman Oct 28 '13 at 14:57
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I wouldn't chain the ldos. 1 A at 1.2V and 1A at 3.3V gives you 2A through the 3.3V ldo.

There are finished switching regulator components that fir in a to220 footprint. I'd look at some of those for efficiency and expediency. Otherwise, use a multi-tap transformer with fly-back regulation and regulate only one of the rails directly.

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Regardless of the structure - chained - 5V -> 3.3 -> 1.2 or unchained 5V->3.3V + 5V->1.2V the power dissipation of the linear regulators will be the same total: 5.5W

Unchained: (5V-3.3V)*1A + (5V-1.2V)*1A = 1.7W + 3.8W = 5.5W

On the chained circuit, the power will be distributed a little bit different (5V-3.3V)*2A + (3.3V-1.2A)*1A = 3.4W + 2.1W = 5.5W

It is not great power, but will need heat sinks on the regulators.

The efficiency will be 45% - 10W input on 4.5W output power which is actually pretty good for linear regulators.

So, if the simplicity of the design is most important - go with this solution.

If the efficiency and the thermal mode is primary - use switching regulators.

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I would first use a regular tranfo, able to output around 15VA with 6 volts. Then I would use a 4 diode bridge, a big capacitor, and then 3 separate buck regulators from TI or Linear or whatever...

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