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I would like to know what type of regulator is better to use in this scenario: (What is best way to convert 15V to 5V 200mAmax with lowest ripple)

  • Vin: 15 V from isolated DC-DC converter (RS6-0515S ripple: 75 mV)
  • Vout: 5 V
  • Iout(max): 250 mA

Linear will be preferred because I need low ripple and a simple circuit, but it needs a heat sink, and it also requires bigger packages. Switching needs more components, a more complicated layout, and it has a ripple problem.

Linear regulator option: EX: P/n: L7805ACD2T-TR => RthJA = 62.5 °C/W, RthJC = 3 °C/W

P = (15 V - 5 V) * 0.25 A = 2.5 W

Tj = 62.5 * 2.5 = 156.25 °C; TOO HOT, needs heat sink

Heat sink TR = 20 to keep temperature at 80 °C (formula used: Tj - TA = (THJC + THCS + THSA)P)

I could not find any other linear regulator with a lower junction temperature.

I am not providing an example for switching since there is no need for a heatsink.

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    \$\begingroup\$ Can you clarify what level of ripple you require? \$\endgroup\$
    – Frog
    Commented Dec 29, 2021 at 20:19
  • \$\begingroup\$ @Frog Thats good question, this will run DAC, RS485, and ADC. so lower ripple much better. since DC-DC convertor has it own ripple I would like to not add any other ripple. Top of my head under 5-10mV ripple should be fine. \$\endgroup\$
    – Shahreza
    Commented Dec 29, 2021 at 20:22
  • \$\begingroup\$ RS6-0515S are boost DC-DC... which are boosting from 5V with 75mV ripple already. Why are you boosting 5VDC to 15VDC and later would like to covert it back to 5VDC with linear reg? \$\endgroup\$
    – NStorm
    Commented Dec 29, 2021 at 20:27
  • \$\begingroup\$ Analog Devices has a white paper on paralleling multiple LDOs to share current and thermal load. But I've got a feeling it'll end up being no less complicated in terms of additional components than a switching regulator with filtering to reach your ripple requirements. \$\endgroup\$
    – TypeIA
    Commented Dec 29, 2021 at 20:29
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    \$\begingroup\$ @Shahreza Look for switching regulator ICs instead of "DC-DCs" (different category). These can be had wayyyyyy cheaper than $40 even in today's supply crisis. \$\endgroup\$
    – TypeIA
    Commented Dec 29, 2021 at 20:50

4 Answers 4

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You can use a hybrid approach: use a DC-DC step-down (buck) regulator to make 5.2V, then use an LDO (low drop out) linear regulator to make the 'clean' 5V for your analog section. Then you get efficiency and low noise.

What's an LDO? It's an improved linear regulator that allows a much smaller overhead voltage than earlier regulators like the LM7805. Overhead voltage is the same as dropout, which is defined as the minimum input vs. output voltage (Vin-Vout).

Why an LDO? Many LDOs have overhead voltages of less than 100mV, compared to 2V or more for the LM7805. The reduced LDO overhead makes it possible to substantially reduce linear regulator losses by taking advantage of its much smaller Vin-Vout requirement.

Let's compare:

  • LM7805: The LM7805 overhead is at least 2V, so Vin needs to be at least 7V; let's make it 7.5V to be safe. At 250mA, the LM7805 will dissipate about 0.625W. Do you need a heatsink for it? Probably not, depending on your worst-case ambient.

  • Typical LDO: to make 5V we can skate by with just 5.2V input. Power used in the LDO will be just 50mW at 250mA, well below any need for a heatsink.

That's not the whole story. We also have losses in the DC-DC regulator, which will be about 10 - 20% of the output power. Let's compare using LM7805 vs. LDO post-regulators:

  • LM7805 @ 250mA, DC-DC out 7.5V, 85% efficient (15% loss), loss = 281mW
  • LDO @ 250mA, DC-DC out 5.2V, 85% efficient (15% loss), loss = 195mW

So using an LDO helps on the DC-DC side too, as the DC-DC is not making excess power that just gets burned up in the regulator.

Adding up all the losses for 5V out @250mA:

  • DC-DC + LM7805 losses: 0.281W + 0.625 = 0.906W
  • DC-DC + LDO losses: 0.195 + 0.050W = 0.245W

Which LDO? I personally like the 'cap-free' NMOS type that's offered by TI and others. They're stable without an output capacitor, and offer low noise and good transient response.

Here's one from MPS: https://www.monolithicpower.com/en/mp20048.html

Here's one from TI: https://www.ti.com/product/TPS73601-EP

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  • \$\begingroup\$ Let me ask questions what difference is between using DC-DC convertor and switching regulator. is DC-DC convertor use switching regulator to bring the voltage to 5.2V \$\endgroup\$
    – Shahreza
    Commented Dec 29, 2021 at 23:54
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    \$\begingroup\$ DC-DC converter is the same thing as a switching regulator. There are types of regulators: step-down/buck (which you'd be using in this case), step-up/boost, buck-boost, and others. What I'm proposing is that you step down 15V to 5.2V, then post-regulate with an LDO. \$\endgroup\$ Commented Dec 30, 2021 at 0:26
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    \$\begingroup\$ What I think you're mixing up is the difference between a regulator and a regulator module (what you're calling a "DC-DC converter".) The former is an IC, the latter is a complete power supply unit with all the support stuff it needs to work. \$\endgroup\$ Commented Dec 30, 2021 at 1:39
  • \$\begingroup\$ You are correct, I thought I was able to use a linear regulator with small heat sink to convert 15V (DC-DC convertor module) to 5V, but I like your idea to use regulator module to get 5.2V and then use 5V LDO regulator. \$\endgroup\$
    – Shahreza
    Commented Dec 30, 2021 at 2:22
  • \$\begingroup\$ The 5.2V DC-DC output is a cool idea except it won't happen. It's a fantasy. If you're extremely lucky, you may end up getting away with 5.5V. You need to consider: 1. LDO worst-case dropout under the expected worst-case load. 2. LDO worst-case static error due to reference voltage. 3. DC-DC worst-case static error due to reference voltage. 4. DC-DC output worst case dynamic glitch due to line and load changes. 5. LDO worst-case dynamic glitch due to load dump, and then back-propagate that as a further limit on minimum drop-out voltage. Even trimming static errors won't help with dynamic stuff \$\endgroup\$ Commented Dec 30, 2021 at 6:49
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  1. A synchronous switcher to drop 15V down to 6V. Shoot for 250kHz+.
  2. A "pi" C-L-C filter consisting of switcher's output capacitor, LDO's input capacitor, and a series inductor, plus a small ferrite bead in series with the inductor. Set cutoff to get rid of some of the switcher's ripple & harmonics.
  3. An LDO like LP2989 - low noise, decent ripple rejection, and excellent line and load regulation with the 10nF bypass capacitor - for relatively cheap.

The 78xx series has decent noise but its line and load regulation are bad, as is the supply rejection. You want something that will suppress some of the switcher's ripple as well as react switfly to load and line changes (those that passed through the DC-DC).

There are significantly better LDOs in terms of high frequency ripple rejection, but they cost more. 2989 is quite a decent starting point. In my experience, it's quite suitable for directly powering strain gages and such sensitive circuitry.

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You are doing it wrong way. Based on the discussion in comments you expect input power to be 5-12V which you are trying to covert to 15V with DC-DC module, which only accepts 4.5-9V.

Next you want to have isolated 5V & 15V supply. That's ok, but the DC-DC module you've picked (not to mention it doesn't works with +12V supply, as noted above) are extra chain in 5V supply. According to that your are looking for low ripple & low noise supply for your analog part, you don't need to feed your +15V converted power to get 5V back again. First and simple way - you can use 5V to 5V isolated DC-DC module. Something like 0505S.

But ideally (if you really looking for good power supply solution with low ripple & noise) you need to design your own SMPS starting from DC-DC converter or controller IC. As already noted in some people comments it's a less integrated than your DC-DC ready made module and you have to add some external parts. But these are pretty cheap and comes in a lot of varieties & topologies. You can find some sort of flyback DC-DC controller with 2 outputs and design your complete power solution accepting 5-12V input & giving you 2 nice & clean outputs of 5V and 15V. Some vendors offer a nice tools which helps to design those circuits based on their ICs.

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If you cascade two liner regulators, one to drop the 15V to 12V (which can be used by your 12V circuits) and from 12V to 5V, you will spread the heat and not need a heat sink. If you don't have problems with size, you can use the TI LM340K liner regulator in a 2-pin TO-3 package for the conversion of 12V to 5V, and the LM7812/LM340 in a 3-pin TO-220 package for the conversion of 15V to 12V.

The Power dissipation at 0.25A for the 15V-12V will be 0.75W, and for the 12V-5V will be 1.75W.

Looking at the Power Dissipation charts of the two packages, they will run within the No Heat Sink lines. enter image description here

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  • \$\begingroup\$ Let me ask question if total power is same then does temp rise same amount just divided by two. (So ambient temp rise will same at end) \$\endgroup\$
    – Shahreza
    Commented Dec 30, 2021 at 23:28
  • \$\begingroup\$ The configuration proposed distribute the total power dissipation of 2.5W into two lower values (0.75W and 1,75W) that can be handled by the regulators within their power dissipation capacity without the need for a heat sink, at the same ambient temperature (which should be below 50 degrees as per charts.) \$\endgroup\$
    – VictorTito
    Commented Dec 31, 2021 at 15:34

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