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I am designing a 21" tablet using a development board, the development board takes 5V DC whereas the display takes 12V DC. The display will take around 800mA and the development board around 500mA.

I am wondering should I use a 12V adapter and Step-Down transformer to convert to 5V, or a 5V and Step-Up to convert to 12V?

What is the standard technique?

And how do you decided which to choose? Will the power consumption be same in both?

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If you have a 5Vdc or 12Vdc input, then you will not be stepping it up or down using a transformer (transformers don't work with dc). But then why use a transformer, when you can use a switch-mode power supply. See one of the answers. – richard Mar 4 '14 at 12:29
Use "mA" for milliamps, not "mamp." – JYelton Mar 4 '14 at 21:42
up vote 2 down vote accepted

Since you mentioned transformers and \$12\$V adapters, I assume you are stuck in the old ways of linear power supplies, which typically only have a power efficiency of between \$50\$%-\$60\$%

So if you go from \$120\$V AC at the wall to \$12\$VDC to \$5\$VDC there is a \$50\$% worst case power conversion loss at each stage. It won't matter if you step up or step down, you can assume that the up-conversion or down-conversion is equally efficient, and the losses are more from the linear regulators than anything else.

In this day and age, you should really only be using switching power supplies, especially for digital electronics like a tablet. There are numerous designs available and numerous chipsets with efficiencies approaching \$98\$% these days.

If you go the switching regulator route, then the path will be \$120\$VAC -> \$12\$VDC -> \$5\$VDC and you can expect \$80\$%-\$90\$% efficiencies at each stage. This is the path I would use if you were coming from the wall, because this is two buck regulators in series and a buck regulator is inherently more efficient than a boost regulator. This is because a buck regulator is either on or off, and in the on cycle it is delivering current to the load, and in the off cycle the inductor is doing it. However, a boost regulator first must deliver current to ground through the inductor to build up the inductor's magnetic field, then in the off cycle (and during the collapsing field) the inductor delivers current to the load. So the resistive losses of the inductor during the buildup phase of a boost regulator are wasteful (however, modern boost controllers go through extra hoops to minimize this wasted energy)

enter image description here

However, if you are working off a battery source (you did say "tablet", so I assume portable), then your choice of buck vs boost is often directed by your source voltage available. If your source is a couple of lithium-polymer cells, then you only have a few volts to start with, and so you will need a battery -> boost \$5\$V -> boost \$12\$V configuration.

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The 12V adapter i am talking about are common adapters that come with external displays. Which can supply 12V and 3.3A. – rajat Mar 4 '14 at 8:14
Then it seems like you already have the 12V adapter. My final comment in my answer is true.. the choice of what you use depends on your source voltage. In this case, your source is the 12VDC from the adapter so assuming that the adapter has enough excess power to also supply the development board, you can just use a buck converter to go from 12VDC to 5VDC. It will be the most efficient solution. – Brian Onn Mar 4 '14 at 8:24

You can get one chip to do both. Here are a couple of examples: -

enter image description here

At the top are a couple of examples of using the LT3988. Below is the LT3992.

You could also consider designing an off-line power supply with multiple tapped outputs to feed 5V and 12V. Power Integrations springs to mind with their portfolio of devices.

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LT3992 looks like a nice device.. I'm going to remember that one for the future if I need 12V and 5V. OP may be using an offline 12VDC switcher that came with a display he's using.. I too thought he was designing from scratch but it appears not – Brian Onn Mar 4 '14 at 8:29

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