I too want both supplies (3.3v and 5v) in my circuit and made these schematics:
On 5v supply typical load current is ~8mA.
On 3.3v it is ~30mA.
Regulators are lm1117 in both cases,sot-23 packages and no timing issues Which one is better?
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1\$\begingroup\$ What are the rated current and power dissipation capabilities of the two regulators? What sort of packages are they in and do they have heatsinks? \$\endgroup\$ – Joe Hass Jan 3 '14 at 20:09
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\$\begingroup\$ Are there any timing issues related to startup/shutdown? When power is first applied does one rail need to be stable before the other turns on, or vice versa at power-down? \$\endgroup\$ – Matt B. Jan 3 '14 at 21:20
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\$\begingroup\$ Also note that you said typical load current on the 5V supply is 8mA and 30mA on the 3.3V, but, for the top configuration, the load current from the 3.3V supply will also be a load on the 5V regulator since you have them in series. \$\endgroup\$ – sherrellbc Jun 22 '14 at 21:13
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Depending on your input voltage, but at that low current, it makes no real difference. You would simply be changing which one gets the bigger energy wastage (in watts).
The LM1117 shown in your picture has a 5mA quiescent current, a minimum load current of 1.7mA, and typically up to 800mA regardless of the package. 15v Max input. You are using 9v, through a diode.
This is simple Ohm's Law calculations for Power (P = I * V)
Since they are Linear regulators, Current in is Current Out. For the first schematic, that means 9v - 0.7v (Diode Drop), 8.3v - 5v = 3.3v voltage drop at the 5v regulator. Then 5v - 3.3v = 1.7v voltage drop at the 3.3v regulator.
Then we can calculate heat dissipation. 1.7v * (30mA Load + 5mA Quiescent) = 0.06W for the 3.3v reg. For the 5v reg, we have to add the 3.3v's current load to the calculation. 3.3v * (8mA Load + 35mA 3.3v's Reg + 5mA Quiescent) = 0.159W. Grand total wasted power of 0.219W, with 75% of it on the 5v regulator.
For the second schematic, its almost the same. 3.3v drop for the 5v regulator, but 8.3v - 3.3v = 5v drop for the 3.3v regulator. Since the 5v regulator doesn't carry the 3.3v's load, it's just 3.3v * (8mA Load + 5mA Quiescent) = 0.043W. The 3.3v regulator though, 5v * (30mA Load + 5mA Quiescent) = 0.175W. Added together, it's 0.218W, with 75% on the 3.3v regulator.
In any case, the 0.218W in heat is nothing to be concern about. The regulators won't even get warm, let alone hot enough for a heatsink.
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\$\begingroup\$ FWIW (not much :-) ) if they do have 1.7 mA minimum load current then the series arrangement draws the 3v3 1.7 mA min (and 5mA quiescent) to the 5V regulator so true 5V load can drop to 0 mA so for currents under 1.7 mA at 5V you draw less current overall. fwiw :-). That's a shocking spec, all else being equal. \$\endgroup\$ – Russell McMahon Jan 4 '14 at 6:22
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The first circuit requires the 5 V regulator to dissipate more power than the second does, because its current load is much higher.
The second circuit requires the 3.3 V regulator to dissipate more power than the first does, because its voltage drop is much higher.
If you can replace one of your regulators with a smaller/cheaper part by using one or the other version, that should drive your decision.
For example, in the second circuit you might be able to use a tiny SOT-23 regulator for the 5 V.
Or in the first circuit you might be able to use a smaller package or avoid using a heat sink on the 3.3 V regulator.
answered Jan 3 '14 at 20:29
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\$\begingroup\$ I should add, the first circuit could have some advantage for supressing ripple on the 3.3 V rail, because any noise coming in at 5 V will go through 2 stages of regulation. But it would be a rare case where that matters. \$\endgroup\$ – The Photon Jan 3 '14 at 20:41
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Which is better? is a too broad question. Better in what sense? Current source capability? Voltage drop? You need to provide the judgement factor. The second one has follow advantages over first:
- Immunity wrt fault in 5volt regulator. If 5v regulator goes off for any reason then input to 3.3volt regulator might cut-off.
- Better current branching. In the first design the current supplied to the 3.3 regulator is coming from 5v regulator. In certain cases the 5v regulator may bottleneck(limit) the 3.3volt regulator's current output.
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\$\begingroup\$ Aren't both regulators sourcing current instead of sinking current? Can you explain what the word "bottleneck" means in this context, for the benefit of readers who might not be native English speakers? \$\endgroup\$ – Joe Hass Jan 3 '14 at 20:35
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\$\begingroup\$ You're correct! The regulators must be sourcing the current. As for the bottleneck I'll try to clarify more. \$\endgroup\$ – vvy Jan 3 '14 at 20:37
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\$\begingroup\$ @vvy Thanx Which is better? Better in the sense the one which is more reliable \$\endgroup\$ – Sanjay Jan 4 '14 at 8:30
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\$\begingroup\$ @Sanjay I'd say the second one is more reliable. Reason being independent operation of the regulators. \$\endgroup\$ – vvy Jan 4 '14 at 13:38
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The specified LM1117 "dropout voltage" is 1.2V max at low current so the 3V3 regulator with 5-3.3 = 1.7V "headroom" will be OK.
The 1st circuit does not allow 3V3 out with 5V off. This may or may not matter in your case.
The 1st circuit has potentially better power input noise rejection due to having two regulators in series BUT construction has to be done with due care to achieve this in reality.
At the power levels mentioned max dissipations are small and either arrangement is OK heat wise in almost any situation. At higher power levels the 1st circuit limits the amount of power you can get at 3V3 due to heating of the 5V regulator by the 3V3 current. Not an issue here.
answered Jan 3 '14 at 22:43
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\$\begingroup\$ No, the OP specified LM1117 parts which are available in fixed outputs of 3.3V and 5.0V in addition to the adjustable version. They are LDO regulators. \$\endgroup\$ – Joe Hass Jan 4 '14 at 2:22
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\$\begingroup\$ @JoeHass THanks - I'll change my answer. Brain fade / Old eyes/new year/other excuses. \$\endgroup\$ – Russell McMahon Jan 4 '14 at 4:07
