power supplytransformer
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?
You are correct. The transformer will only reduce the voltage (and increase the available current), so you need to add additional circuitry to rectify, smooth and regulate your 12\$V_{AC}\$ transformer output to 5\$V_{DC}\$.
This is the type of circuit you should be looking to build:
The smallest travel chargers / adapters available these days do not contain a magnetic transformer.
Instead, isolation and AC coupling is achieved by using piezoelectric transformers: These are really tiny, and flat, in comparison to the conventional magnetic coil based transformer, and work at greater efficiency as well. The frequencies involved are not AC power line frequencies, they are much higher.
See the image below:
The design is the same as described in the answer by Nick Alexeev (switched mode), with power line AC being rectified to DC, then high frequency AC generated, isolated via the Piezoelectric transformer, and rectified again for DC.
The use of high frequencies also reduces the required size of capacitors for smoothing the DC voltage, another key factor contributing to the bulk of older chargers / wall warts.
For additional information about such tiny travel chargers, see this answer.
Best Answer
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)
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.