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I am trying to design an ultra low noise power supply for my sensitive circuit. The supply will provide 5V and 600mA. (Worst case, 200mA typical). My input voltage is 2V to 4V. My ripple target under load is less than 5mV.

The signals I am trying to measure are very small, nA level, and the opamps need to be on a very clean supply hence the very low noise DC/DC converter requirement.

What are the precautions I can take and what are the parameters I should look at while I am selecting the DC/DC converter. I know the forum doesn't like people to provide specific requirements but in this case I appreciate if you could indicate parts that meets the parameters indicated.

Also, what other circuit level precautions I can take? Decoupling caps for once but are there others?

UPDATE My source is a simple two AA batteries, alkaline, but people may use other stuff. Anything that fits to AA battery compartment.

My signal, the opamp input, is 1Mhz with 75% duty cycle

UPDATE 2 I went back and completely re-look at the design, change the Opamps so that I can turn off the power to them (and they can work at 4.5V), arranged GPIOs etc so that we can control the power flow, changed some of the modulation schemes etc. At last, I was able to bring down the total requirement of power to 400mA where 200mA for 5V and 200mA for 3.3V.

I have chosen the high efficiency TI part for the design on the switcher side. TPS661100. This has an LDO which I will use for 3.3V. (interestingly, this LDO has a great PSRR but I am afraid to use it for the 5V supply since I cannot pass it through a pi network before feeding into the LDO) Now I am trying to choose an external LDO to for 5V. This switcher has 500Khz switching frequency but finding an LDO that has a good rejection at 500Khz proved to be challenging. I can find parts that has a low drop out and 30dB PSRR but nothing like you guys recommended such as 60-70dB. I am now open to suggestions and feedback.

Final note, when my design is complete, I will extend this question so that the next guy can follow or learn from my experience. Thanks to you all.

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    \$\begingroup\$ It's going to be pretty difficult to measure picoampere currents at 1 mHz, considering the equation \$ I=\frac{dQ}{dT}\$. Plugging in I = 1 pA and integrating to t = 1/f, with f = 1 mHz, you get Q to be \$10^{−18}\$, which is two orders of magnitude less than the charge on a single electron! \$\endgroup\$ – Bitrex Sep 25 '11 at 8:21
  • \$\begingroup\$ Ok.. this is an interesting input. My nominal currents will be around 10nA. pA was a mistake. \$\endgroup\$ – Ktc Sep 25 '11 at 8:21
  • \$\begingroup\$ @user3685, on the note of the forum does not like people asking for specifics. A great answer can always have parts given as examples, but ideally they teach you during the process of how to find what you need and the examples are just a simple time saver. Also, in general information similar to what noise level is acceptable and such can help users find solutions. \$\endgroup\$ – Kortuk Sep 25 '11 at 13:00
  • \$\begingroup\$ @user3685 - See added "power supply" section at end of my answer. 4 hour battery life OK? / Actual opamp part number would be useful. What else apart from opamps is taking supply current? (200-600 mA is a lot for opamps only). What voltage will opamp and other equipment tolerate as Vmin. \$\endgroup\$ – Russell McMahon Sep 25 '11 at 13:08
  • \$\begingroup\$ 2 x 14500 = AA LiIon would fit into battery compartment :-). Not much seen out of captivity. = 6V+ - 8.5V. Wh energy capacity is about equal to AA NimH. \$\endgroup\$ – Russell McMahon Sep 25 '11 at 13:11
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You'll need a switcher to pump up the input voltage, and have it followed by a linear regulator, like Russell proposes. His TPS717xx, however, while having a high PSRR doesn't have a high output current (150mA). High current and high PSRR don't go together well on LDOs. Nevertheless, an LDO may be required, since the 2-4V input volatge suggests that you're powering from batteries, and then you don't want to lose too much power in regulators.
A solution for the PSRR may be to use 2 LDOs in cascade, so that the PSRRs add.
The LP3878-ADJ may be a good choice for your second stage; it's low-noise, can supply 800mA and has an adjustable output which can be set to 5V.
For the preceding stage you could use an LP38690-ADJ. The datasheet shows how you can improve ripple rejection at higher frequencies if you place a 100\$\mu\$F output cap. Caps between each stage (100\$\mu\$F + 10\$\mu\$F + 1\$\mu\$F) are a must, but \$\pi\$-filters are a better idea: caps + ferrite bead + caps. Those should suppress the higher frequencies generated by the switcher.

edit (re your update)
Dave rightly has second thoughts on your using AA batteries, but it's even worse than he calculated. I estimate that you'll need 7V from the switcher to feed the cascaded LDOs, so 600mA @ 7V is 4.2W, at a conversion efficiency of 80% that's 5.25W from the batteries. Two AA cells deliver 3V, so that's 1.75A from the batteries! Not only won't they last long, they won't like it either. I suggest you use a LiPo cell, which delivers 3.7V with a boost regulator, or even two cells in series, giving 7.4V, and use just the LP3878-ADJ. (If you don't use a switcher you don't need the second LDO to suppress its ripple.)

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  • \$\begingroup\$ @stevenh, the use case almost dictates user buys new batteries, if we make it rechargable, the product will loose certain convenience. I am looking at optimizing power now, we have never considered this but now it is a must. I will try to get running consumption as low as 30mA and worst case to 400mA. There are ways but require serious software work. \$\endgroup\$ – Ktc Sep 26 '11 at 6:40
  • \$\begingroup\$ @user3685 - D type batteries have a capacity 5x that of an AA cell, so they will last longer. Customer may call fail if she needs to replace the batteries after a few hours. Anyway, alkaline power is damn expensive as well (one to two thousand euro per kWh), and as a customer I would prefer rechargeables, though I appreciate that you have your reasons. \$\endgroup\$ – stevenvh Sep 26 '11 at 6:58
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Summary:

  • Avoid using a switched mode power supply (SMPS) if at all possible.

  • Use a high PSRR regulator either without a SMPS or following one.

  • Look at all the usual design advice on PCB design etc - see below and othes

  • Look at all the PSRR design references below


You say input is 2V - 4V. Why is this? What is the energy source? If batteries,
1 x LiIon ~= 3V- 4.2V 1 x LiFePO4 ~= 2V - 3.5V
2 x NimH ~= 2V - 2.8V ...?

You mention "opamps". Are they 5V single supply ?

eg 2 x LiIon will give you 6V - 8.4V with most operation in the 6.5V - 7.5V range. Using a linear regulator gives you about 70% efficiency typically BUT a vast potential reduction in power supply noise. 5 x NiMh gives you say 5.x V to 6.5V and somewhat higher average efficiency.

If you MUST use a Switched Mode Power Supply (SMPS), which will add to your woes with its noise compared to a good quality purely linear supply, then after you follow all the good practices that everyone else will tell you about you are going to also want a superb linear regulator to follow the smps.

For the smps you'll

  • Consider using a sinusoidal waveform or resonant design (Royer maybe).
  • You'll want to look at a balance L-C-L input and output filter to the smps
  • with balanced bifilar wound common mode filtering choke.
  • All the usual care with return currents stuff,
  • No loops in PCB go & return paths to pick up em noise
  • No slots intersecting ground plane return paths
  • Decoupling caps AT IC pins with possibly several caps of a range of values
  • Ceramic caps and tantalum maybe (on output)(Tantalum on input OKish if utterly spike free operation above Cap_Vmax is guaranteed.
  • With a smps pay attention to any oscillatory portions of waveforms, leakage inductance that isn't properly snubbed, diodes not to hard recovery, switching waveforms somewhat bandwidth limited, ... .
  • A switched capacitor smps will have the advantage of no magnetic field interference - still has E field (minor) and power supply noise (can be significant.
  • And ...

BUT

You will then want a linear regulator with an immensely good PSRR (Power Supply Rejection Ratio).

Ye olde standard LM317 claims up to around 80 dB PSRR BUT results can vary vastly across frequency range and with implementation.

If you search eg Digikey for PSRR you'll get devices that claim high PSRR as it doesn't get mentioned unless they are trying to make it better than usual. I got 206 voltage regulators with PSRR mentioned at Digikey - a small number compared to most searches.

As a disturbing example, (not enough current for your application but indicative of the PSRR performance you may expect) here's the datasheet for the TI TPS717xx which mentions low noise and "high bandwidth PSRR" in the title. BUT claoms 70/67/45 dB PSRR at 1/100/1000 kHz. That's actually good BUT that may not be obvious looking at apparent datasheet figures for standard parts. From the graph below it's clear that it would be a REALLY good idea to get rod of as much HF noise before the regulator as possible. This is reasonably easy using a mains sourced linear supply, and "not so easy" using a SMPS ahead of the high PSRR regulator.

I looked at this datasheet based on it being the dearest in 1000 quantity at Digikey that mentions PSRR It's far worse than the TPS717xx - but much higher current rated.


enter image description here


Real world design advice:

Here's a useful Maxim application note on good PSRR design.

Here's an Omicron PSRR testing application note - they want to sell you test gear but it's a useful guide to what you are trying to achieve.

If serious or desperate enough this for $ paper may be of use 96dB rejection ratio PSU design.

Here's a useful TI design note) making the good point that performance may vary "strangely" across frequency.

As ever Wikipedia have something to say


Power supply

Output: 5V and 600mA. (Worst case, 200mA typical).
Supply: 2VDC - 3VDC (2 x Alkaline)

Assume 250 mA mean. At say Vbattery = 2.5V and say 80% all up converter efficiency:

Iin = 5V x 0.25 A / 2.5V x 1/80% = 625 mA A 2500 mAh capacity AA cell will last nominally 2.5/0.625 ~= 4 hours.

For NimH the voltage of 2 cells under that order of load will be 2.4V typical so slightly lower life nominal but they achieve closer to their rated capacity than Alkalines as loads start to approach 1C (here load = C/4).
Only the very very best AA NimH actually achieve 2500 mAh and then only when new so real world lifetime will be 4 hours or less.

Is a 4 hour battery life acceptable?
Can you increase battery voltage to allow a linear only supply? What is the lowest opamp etc supply rail that you can really tolerate? (eg 5.0V, 4.9V, 4.5V ...?)

Overall real world result with NimH is liable to be better than using Alkalines.

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  • \$\begingroup\$ opamps are 5V single supply \$\endgroup\$ – Ktc Sep 25 '11 at 8:23
  • \$\begingroup\$ Overall real world result with...what? Great answer, nonetheless. \$\endgroup\$ – Kevin Vermeer Sep 26 '11 at 10:16
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You say you have a ripple target of 5mV (P2P?) under load, which is 60dB. Even a funky-junky LM317 should be able to accomplish that. However, I believe you may need a much tighter specification, if you want to amplify signals any greater than a couple Hz. Even using an op-amp with a PSRR of 90dB, if you have 5mV P2P of ripple/noise on the power supply that means you'll have 156 nV of noise on the first stage amplifier output.

If you then use the amplifier as a transimpedance amp to amplify a 1 pA P2P signal, it means that you'll need a first stage gain well in excess of 156,000 to get it above the power supply noise. It's probably possible to build a circuit that can do that, since there are opamps with fA input bias currents and only a couple \$nV/\sqrt{Hz}\$ of noise, but the circuit will be useless for anything but the very lowest frequencies because of gain-bandwidth product limitations.

Edit: I apologize, I didn't read the question carefully enough! I see now that it is perhaps implied that you will follow up your switcher with a post-regulator to further clean up the output. I was going to delete this answer, but since I went to the trouble of doing the math I'll leave it here for now. :)

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  • \$\begingroup\$ +1. All good stuff. Post regulator arose as part of answer process AFAIK. \$\endgroup\$ – Russell McMahon Sep 25 '11 at 13:09
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Two points, which others have also made:

  1. 5 mV on 5V isn't all that ultra low noise.

  2. The best way to achieve this within your stated parameters is to boost switcher followed by a linear post regulator.

To conserve power, you want to use a LDO (but see point 3). However, 600 mA is a lot for a LDO. And as Steven mentioned, the nature of LDOs usually gives them poorer input ripple rejection, at least at higher frequencies. So, a few points:

  1. Break up your 600 mA load into multiple pieces and power each piece from its own LDO. Once you get down to 250 mA per regulator, you will have a lot more choices.

  2. Put a separate L-C filter in front of every LDO. LDOs in general have poor capability to adjust to high frequency input voltage changes. I've seen this in action and in one case had to retrofit a chip inductor and capacitor to a existing circuit because too much swithing noise was getting onto the supposedly clean linearly regulated output. Just a 0805 chip inductor followed by 22 µF to ground will help a lot. These chip inductors will have a few 100 mΩ resistance too, but that actually helps a little here. You do have to watch the current rating and possibly use a larger package to get the current you want. You can use two such filters in series if neccessary to really squash the high frequencies the LDO can't handle well.

  3. You don't have to put all the low ripple burden on the power supply. The switching ripple can be easily filtered out when it is high frequency enough. You could use a bare off the shelf boost chip that runs at 500 kHz or more, then add L-C filters as above where needed. These won't be needed for powering digital circuitry like a microcontroller. You do want to put the filters on any analog part. Most likely the sensitive circuitry isn't where most of the 600 mA is going. This solution is the most effecient because you don't have the extra voltage drop of a linear regulator in there.

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  • \$\begingroup\$ thanks.. We will try to utilize your approach on 3. It is indeed a good suggestion. What is ultra low noise? Do you think it is possible to achieve <1mV? \$\endgroup\$ – Ktc Sep 26 '11 at 7:27
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If you are lucky, you will achieve 75% efficiency in the power conversion circuitry. To deliver 600 mA at the output will require 600/0.75 = 800 mA from the batteries. This is a pretty substantial load for random AA batteries. Are other batteries possible? Even C Cells would work out better.

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  • \$\begingroup\$ Actually the battery current will be more than you show because you forgot to account for the fact that he is also boosting the voltage. \$\endgroup\$ – Olin Lathrop Sep 25 '11 at 20:36
  • \$\begingroup\$ Note to self-Coffee First/Internet Second. That's a good point. A real power budget accounting for conversion losses at each stage will be very instructive, and show why Switch mode regulator are good and LDO's are bad. \$\endgroup\$ – rfdave Sep 25 '11 at 21:21
  • \$\begingroup\$ Actually LDOs aren't necessarily all that bad. For example, 5.5V in and 5.0V out is 91% efficient. \$\endgroup\$ – Olin Lathrop Sep 25 '11 at 21:38
  • \$\begingroup\$ @dave We got 4aa but 2 sets are in parallel. Do you think this would be ok? \$\endgroup\$ – Ktc Sep 26 '11 at 4:34
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    \$\begingroup\$ @stevenh ok.. hell of a comment, didn't think of this.. I guess need to go back to drawing board for sure. \$\endgroup\$ – Ktc Sep 26 '11 at 8:19

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