I'm planning to power devices with different ratings from a single supply. Some of the devices have ratings such as 5V/2A, 5V/4A, 12V/2A etc., Currently they're being powered with seperate adapters. I plan to use a single supply to power all of them, and from other questions/answers, I've gathered that the best way to do it would be to use different DC-DC converters.
My question is, since they have different ratings, would the power supply be chosen such that the max. current rating is the sum of all of the device's current rating? Would it pose a problem then, if I'm connecting the converters to that single power supply since it's likely that the converters would be having different input current limitation.
Best Answer
First I want to answer your second question about the different ratings. I think it's easier to explain with an example. Let's say we have 4 devices which have the following power requirements:
To connect these devices we need 2 power "rails", 5V and 12V. The 5V rail must supply \$4A + 2A = 6A\$. This doesn't work for the 12V devices because we have a device with a wattage rating. Divide the wattage and the voltage to get the current.
$$ P = U \cdot I $$ $$ I = \frac{P}{U} = \frac{18W}{12V} = 1.5A $$
Now we can sum the ampere ratings again to get the current on the 12V rail: \$2A + 1.5A = 3.5A\$.
So the end result is 12V / 3.5A and 5V / 6A
There are several good options if you want to reduce the number of power adapters, these are my favorite:
I choose option 2 whenever possible (price, availability). Something like the Traco Power TXL 100-0512DI comes to mind (details here). You can use the calculated ratings from above to see if a power supply is suitable.
If you want to go for option 1 you need to pay attention to the efficiency of the converters and the total power. I'll extend the example from above and calculate some sample ratings.
Let's say we have a 12V power supply and use a DC-DC converter to get the 5V rail.
The DC-DC converter has an efficiency of 80%. The maximum power consumption on the 5V rail is \$5V \cdot 6A = 30W\$. Because the converter has an efficiency of 80% it draws \$\frac{30W}{0.8} = 37.5W\$.
The 12V devices combined consume \$12V \cdot 3.5A = 42W\$.
This means the 12V must have at least \$37.5W + 42W = 79.5W\$ output power.
Power supplies and dc-dc converters can be found at Farnell, Digi-Key, eBay or if you don't need quality material on chinese sites like Dealextreme, Banggood or AliExpress.