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I need to power my device with three different voltages: 3.3V, 5V and 12V. I already have a 12V 1A power supply. So my questions are:

  1. Can I use the circuit below to get these three voltages?
  2. Is it safe when ground of LM317's is shared?
  3. Do I need any additional diodes or anything else?


simulate this circuit – Schematic created using CircuitLab

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Maybe recalculate R4 and R2 - there is a problem here. What current are you supplying on each rail? – Andy aka May 7 '14 at 12:44
Why are all the caps in the schematic and weird angles? – Matt Young May 7 '14 at 12:54
@Andyaka What problem? R1 and R3 want 1,25V across them, so the values look fine to me. – Lundin May 7 '14 at 12:55
Not what the OP is asking, but just a note. You can get what you're asking for very easily with almost any standard PC ATX power supply. All you need to do is short the green wire on the main connector to ground (black) to have it turn on automatically. I say almost as some supplies require a specified minimum load on each rail for proper operation. – DoxyLover May 7 '14 at 15:31
@MattYoung I used CircuitLab – user37741 May 7 '14 at 19:19

What you suggest will work, although I didn't look at your specific resistor values to see if you will get the voltages that you intend.

LM317 is designed so that the output voltage can be set by a resistor divider. That is useful when you want unusual voltages. However, 5 V and 3.3 V are very common, so you could simplify your circuit and use fixed regulators. For example, the very common and cheap 7805 regulator could be used to make the 5 V output.

Another issue is that these are linear regulators and will therefore get quite warm at significant output currents. You didn't say how much current you need at 5 V and 3.3 V, but if more than a few 100 mA, a switcher is probably a better choice. For example, if you draw 300 mA from your 5 V supply, the regulator will drop 7 V and therefore dissipate (7 V)(300 mA) = 2.1 W. A regulator in a TO-220 case will probably need a heatsink at that power.

I'd look into using a switcher to make the 5 V supply. Then you may be able to get away with linearly regulating that down to make the 3.3 V supply. Let's say you are drawing 500 mA at 3.3 V. That would only cause 850 mW dissipation in the 3.3 V regulator, which a TO-220 in free air can probably handle. If you do this, make sure that whatever 3.3 V regulator you use will work with the 1.7 V headroom provided by 5 V in.

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Planned load for 3.3v: stm32f1; for 5v: two l298, two l293n, ina125p. I'm afraid of soldering a switcher because of poor skills and more complex board configuration (I'm going to make pcb by myself and it will be 1 layer only). There's also a cost value — switcher will cost at least $12 in the local shops. – user37741 May 7 '14 at 20:23

Looks okay to me.

You're throwing away the majority of the power on both rails. Your maximum output current (total of 5V and 3.3V lines) will be about 1A (minus 10mA for the regulators).

But even 250mA on the 3.3V rail will result in more than 2W of dissipation in the 3.3V regulator, requiring a heat sink. If you had 500mA on each, you'd be dumping 4.4W + 3.5W or almost 8W of waste heat.

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Thanks, can you suggest a way of more power-efficient solution? – user37741 May 7 '14 at 19:22
One way would be to use a single switching supply to reduce the 12V to 5V, then an LDO regulator to reduce the 5V to 3.3V (66% efficient). Or two switching regulators (a bit better but more complex). Eg. LM2596. – Spehro Pefhany May 7 '14 at 19:45
  1. There is nothing wrong if LM317 share the same ground
  2. Use two separate heatsinks, not electrically connected
  3. It is recommended to add a diode across each OUT-IN of LM317, because you are using a capacitor across the output.
share|improve this answer
No need of the diodes- they're only recommended for high output voltage (25V or more) and high output capacitance (10uF or more). Neither is true in this case. – Spehro Pefhany May 7 '14 at 13:11

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