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I have two separate components, one of which needs a 5V 500mA-1A power supply, the other of which needs a 12V 5A power supply. Is there any way to combine a single power supply somehow so that I only need to plug one thing into the wall in order to power both components?

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Yes, use a 12VDC input (>6.5A or more), then regulate it down to 5V for the low voltage section.
You can use a simple fixed linear regulator like an LM7085 for this, or for better efficiency use a buck regulator like the TL2575 (lots more options here).

EDIT - more detail on the difference between linear and switching regulator options.

With a linear regulator, it drops and regulates the voltage by simply dissipating the difference as heat.
For example in your case of 12V in and 5V out at 1A, the dissipation will be:
(12 - 5) * 1A = 7W (as Steven also calculated in his answer)

So do we need a heatsink? how hot will the regulator get?

In the datasheet for the LM7805, the thermal resistance for junction to air (Rθja) is given as 65°C/W. This means that for every watt dissipated, the junction temperature will rise by 65°C above ambient.
So at 7W, we get 7 * 65°C = 455°C = way too hot! (the regulator will actually shut down at ~150°C to protect itself)
The absolute maximum temperature before damage is given as 150°C, and maximum operating temperature is given as 125°C, so you would need a reasonably sized heatsink to keep the temperature within limits.
To calculate the heatsink needed, you take the junction to case rating (Rθjc), add this to the case to heatsink rating (Rθc-hs) and heatsink to air rating (Rθhs). Here is an introduction to heatsink selection: Heatsink Basics

A switching regulator is a different story. Switching regulators regulate by transforming the power rather than dissipating it (much like a transformer)
If we ignore the small inefficiencies for a moment and assume 100%, we know that power in must equal power out.
So if voltage in = 12V, and voltage out equals 5V and needs to be 1A, we know that 5W of power is needed at the output.
With this information we can calculate the current in:
5W / 12V = 417mA.

Now lets add the inefficiency (say 88%, from the TL2575 datasheet linked to above) and calculate:

5W / 0.88 = 5.68W at the input
5.88W / 12V = 473mA input current
5.68W - 5W = 680mW dissipated as heat.

How about the heatsink?
The TL2575 has a thermal resistance of 31.8°C/W, so:

31.8 * 0.68 = 21.62°C rise above ambient.
If the ambient is 20°C, the temperature will be 41.62°C.
Even at 50°C ambient, we would still be well within operating limits.

So you can see using the switching regulator in this case makes sense. You can go for a "ready rolled" module like in Steven's answer, or build your own using something like the TL2575 IC linked to above. The datasheet will have a few example circuits to help you along (note that a "buck regulator" is a type of switching regulator that lowers the voltage. A "boost regulator" is a switching regulator that raises the voltage)

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  • \$\begingroup\$ So I would connect a normal power supply that outputs 12VDC 6.5A to the LM7085, for example, and that would regulate it for me? As long as my power supply is greater than 6A, would it work? Or is there an upper limit to the current output of the power supply? (There shouldn't be, right? The components will draw as much current as they need). Can you go into a little more detail about how exactly that would work? And how much of a difference would Steven's improvement make practically? Would I need a heat sink if I didn't use a switching regulator? \$\endgroup\$ – Mason Aug 27 '12 at 0:59
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    \$\begingroup\$ @Mason - see edits for more detail. You are correct, the power supply current rating can be anything over 6A (I would go for at least 6.5A to give headroom) as the circuit will only draw what it needs. Stevens module and the TL2575 are both switching regulators, just that the module has the peripheral components added so is ready to use (it uses a similar IC to the TL2575 as its main IC, the LM2596) \$\endgroup\$ – Oli Glaser Aug 27 '12 at 2:26
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Oli mentions a 12 V, 6.5 A supply. The 6.5 A is the 5 A you need for the 12 V + 1 A for the 5 V + 500 mA headroom. The 1 A for the 5 V is required if you use a linear regulator like the LM7805 Oli mentions. That will supply 5 V x 1 A = 5 W to its load, but also dissipate (12 V - 5 V) x 1 A = 7 W itself.

A switching regulator will cause much less power dissipation.

enter image description here

Instead of losing 7 W in the LM7805 you may lose for instance 0.88 W if the efficiency is 85 %. It won't do very much in your total power consumption; it will go down from 72 W to 66 W, or a gain of only 9 %, but you won't need a heatsink for the 5 V regulator.

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  • \$\begingroup\$ Would this particular switching regulator work for my components? And will I need two of them (one for each component) or can one do the job for both components. Thanks! \$\endgroup\$ – Mason Aug 27 '12 at 17:20
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    \$\begingroup\$ It is in any case suitable for going from 12 V to 5 V. What's the 12 V coming from? \$\endgroup\$ – stevenvh Aug 27 '12 at 18:03
  • \$\begingroup\$ It would be from a DC power supply that plugs into a wall socket. \$\endgroup\$ – Mason Aug 28 '12 at 2:08
  • \$\begingroup\$ Probably something like this. \$\endgroup\$ – Mason Aug 28 '12 at 2:40
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    \$\begingroup\$ @Mason - That seems to take care of the 12 V part, so you'll only need the buck converter for the 5 V. \$\endgroup\$ – stevenvh Aug 28 '12 at 15:17

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