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.

  • \$\begingroup\$ Take the power ratings of the 3 adapters and add them together to get a combined power rating for the new adapter. \$\endgroup\$
    – Andy aka
    Jul 6, 2015 at 9:39
  • \$\begingroup\$ from what input voltage you are deriving all these voltages. Any topology you are using Buck, Boost etc. Once you have fixed these things, Calculate input currents of each regulator using Output power, Efficiency and Input Voltage. Sum all the currents..... \$\endgroup\$
    – user19579
    Jul 6, 2015 at 14:26

2 Answers 2


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:

  1. 5V 4A
  2. 5V 2A
  3. 12V 2A
  4. 12V 18W

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:

  1. Use a power supply with an output voltage equivalent to your highest input voltage (12V in your example ratings) and generate the other voltages with DC-DC converters
  2. Use a power supply with multiple outputs that match your needs

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.

Diagram: Arrangement of supply and DC-DC converter

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.

  • \$\begingroup\$ Thank you. This was very informative. I was unaware of power supplies with multiple outputs. I think I might consider going for option 2 and modify some of the outputs accordingly. \$\endgroup\$
    – Koushik S
    Jul 7, 2015 at 12:09

Basically, the power supply you use will need to rated for a total power equal to the sum of: 1. Max power demanded by each load (For your example 5v/2A (10W) + 5V/4A (20W) + 12V/2A (24W) = 54W) 2. The losses in the three DC/DC converters you use for each output 3. The losses in the power supply itself (assuming you're drawing power from a wall socket).

You should make your power budget with conservative estimates of efficiency, along with allowances for heating etc.

  • \$\begingroup\$ So, assuming that the converters can take in 12V as input, would a 12V/5A (60W) power supply be a good choice? I haven't done this before so excuse my inexperience with these. \$\endgroup\$
    – Koushik S
    Jul 7, 2015 at 6:20
  • \$\begingroup\$ Depends on the losses. You'll need to clarify in your original question as to what the loads are, what are these three Dc/DC converters - are you making them yourself? Are they all off the shelf parts? \$\endgroup\$
    – kabZX
    Jul 7, 2015 at 18:46

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