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What I am doing: I am making a wearable device which will use Li-po battery/ies to operate. I am measuring small currents (meaning I want to avoid buck/boost converters). My design draws a current of ~56mA.

What I want to achieve: I need 5V on the device, which should be as voltage-ripple-free as possible. so the way I think of it, using Li-po batteries as power source, I have two options:

  1. Use two Li-po batteries in series, so I always have a voltage of at least 7V, and use Op-amp to output always 5V to my circuit. This way, I do not need to use a buck converter. Like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The above option has the disadvantage that I need two batteries to operate the whole circuit.

  1. The option I do not want to use, is to have only one battery and use a boost converter to boost the voltage to 5V. I want to avoid that since it will introduce switching to my circuit.

schematic

simulate this circuit

And my question is: is there a way to pump/boost voltage to a higher voltage, without introducing noise like a boost converter would? (The reverse with downscaling using op-amp in my first option). I think there might be a way using Op-amps, which I do not know of.

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    \$\begingroup\$ You can preregulate with a boost up to, say, 7-8V and then regulate to 5V with a low noise LDO (depending on your budget); lots of capacitors help too! \$\endgroup\$ Apr 21 at 12:18
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    \$\begingroup\$ You must be really afraid of noise if you have solution one on the table. That will be wasting 30% of the power most of the time. Are you sure that you cannot just filter the noise? Can you add some information on the noise requirements of your load? \$\endgroup\$ Apr 21 at 12:19
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    \$\begingroup\$ HINT: You can design and achieve ANYTHING you need with good specs. The only specs you have are at DC with 5V / 56 mA. That's not enough. What about at 100kHz? 1 MHz. I appreciate you may want to avoid ingress/egress EMI but it must be defined , measurable and verifiable. \$\endgroup\$ Apr 21 at 12:28
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    \$\begingroup\$ If you use a boost converter, if you're careful about not having excess loops of conductor to your filter capacitors, and a sensible layout, there is no need to be afraid of switchers. The last big product built where I worked was a synthesiser, an instrument grade well-specified one. This had several switchers sprinkled about between the RF sections. We were careful with layout, and the design was fine. \$\endgroup\$
    – Neil_UK
    Apr 21 at 16:21
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    \$\begingroup\$ There is very high probability that biggest noise source in your circuit will be the MCU :) because it shares power source with analog part. If there is low current, then highly likely that signals are changing quite slowly, so adequate low-pass filter(s) will filter out all noises (remember also about 50 Hz). Also, many modern opamps, ADCs and MCUs can operate directly from single Li cell. \$\endgroup\$
    – Vladimir
    Apr 21 at 21:15

6 Answers 6

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is there a way to pump/boost voltage to a higher voltage, without introducing noise like a boost converter would?

No, it's not possible to boost voltage with reasonable efficiency without a switching converter. Basically you need something to "pump" energy to a higher voltage, and that has to switch. It can be a boost converter or a charge pump, but it will switch.

That said, some boost chips make less noise than others. Layout is very important, and a synchronous boost will usually give a better layout than one with a diode. Using a high switching frequency makes ripple easier to filter (the LC components become smaller, and ripple amplitude is smaller) but it produces more HF harmonics, so it's a tradeoff.

You should define what kind of noise your circuit can tolerate on its supply first.

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    \$\begingroup\$ re: "with reasonable efficiency"; I would say it is not possible at all, period. \$\endgroup\$ Apr 21 at 11:31
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    \$\begingroup\$ Oh yes, you can stick a LED in front of some solar cells and get higher voltage, but the efficiency will definitely not be "reasonable" XD \$\endgroup\$
    – bobflux
    Apr 21 at 11:41
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    \$\begingroup\$ uhm… rotary converter, a motor in front of a generator :D well, it switches anyway, in a sense. Passable efficiency, power limited only by the available space \$\endgroup\$ Apr 21 at 12:16
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    \$\begingroup\$ There is a way to boost DCDC voltages using sine waves, can you think of one? \$\endgroup\$ Apr 21 at 12:23
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    \$\begingroup\$ Those still charge with peak current pulses @rdtsc \$\endgroup\$ Apr 21 at 12:24
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Just for completeness: Yes, there is a way, but the efficiency is horribly bad.

Using photovoltaic isolators, you send current through an infrared diode (~1.2 V) and obtain a typical output voltage from a photodiode string of around 8-10 V without load for typical integrated parts.

Efficiency is less than 1% usually.

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    \$\begingroup\$ Good to know though! \$\endgroup\$ Apr 21 at 13:42
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    \$\begingroup\$ With careful design you could probably get this idea up to 20 or perhaps even 30% efficiency, or more. 10% should be pretty easy to hit without too much trouble. Cheap LEDs are maybe 25% efficient, and photodetector QEs can be 80-90%. The challenges would be selecting a good wavelength range to optimize, and designing the optical system to collect as much of the light as possible. The costs to take it to that level would surely be prohibitive, of course. On the plus side, you would get complete electrical isolation. Interesting idea. \$\endgroup\$
    – J...
    Apr 21 at 20:40
  • \$\begingroup\$ @J... Interesting consideration. I am wondering though, why integrated parts such as TLP190B only achieve 1%. Probably for cost reasons, as you mention. Condensing LED illumination is hard. Large area photodiodes are expensive, too. \$\endgroup\$
    – tobalt
    Apr 22 at 6:40
  • \$\begingroup\$ @tobalt The TLP190B is inefficient in this case because it's not designed as a power transmission device or a boost converter - its function is isolation, which it does just fine. A device designed to move any significant amount of power would need significant design changes and, yes, would cost much more. \$\endgroup\$
    – J...
    Apr 22 at 13:20
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    \$\begingroup\$ On a similar track, there are probably circumstances where you could connect your DC source to a resistor and then harvest the thermal energy with a series of thermoelectric generators. But I'm guessing that would be even less efficient than the PV approach. \$\endgroup\$ Apr 26 at 19:41
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Other people have answered the headline question, so I'll quickly address what I think you're trying to do (get very low noise DC voltage for a sensitive application at low current. You can get (at least prior to component shortage) DCDC converters optimized for lower current at extremely low noise. For example, TI makes charge pumps designed for ~ 100mA load and matching LDOs that can attenuate the (small) charge pump noise another 40-50dB. Other vendors even integrate the LDO with the converter to reduce package size. I have used similar solutions in things like single photon sensitive detectors and they can generate noise so low it will be hard to measure, at least for low current loads like you need.

I can't recommend any specific product, but generating a new voltage with extremely low noise comes up a lot in portable audio, cell phones, medical devices, etc so there are good stock solutions for this. Take a look at some manufacturer catalogs and see if you can find something that fits your specs.

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Everything that boosts voltage needs switching, whether a boost converter or charge pump.

However after boosting the voltage you can just remove ripple down to any level you require, but you need to know a specification for the ripple.

For example ripple can be filtered by putting enough capacitance to limit voltage ripple, or with RC or LC filters or linear regulators.

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There is a way that does not use switching in the traditional sense, but again it is not very efficient.

NOTE: The context for this response is the TS original post, which clearly shows a high-frequency flyback boost converter as the alternative to a linear regulator.

Use the boosted-opamp circuit you posted to grow a sine wave oscillator. Nothing fancy like a Wein bridge, keep it simple with a phase-shift topology. Neither the frequency nor the distortion are important. Run this signal through an audio transformer, then full wave rectify, filter, and regulate.

If whatever signal you are processing is being sampled with an A/D converter, another trick is to disable the regulator during a conversion cycle, with the circuits running on current from a hold-up capacitor. This works well for keeping switching regulator noise out of measurements.

  • AND -

If you go with the two-batteries-and-linear-regulator approach, you will get better performance with a linear regulator chip designed for that purpose. There are dozens of low-dropout parts for this. From your favorite analog chip manufacturer, search for LDO devices.

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    \$\begingroup\$ This is "switching" in a sine wave instead of a square wave - it's still ripple \$\endgroup\$
    – user253751
    Apr 21 at 12:18
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    \$\begingroup\$ then full wave rectify -> how do you achieve this without switches ? \$\endgroup\$
    – tobalt
    Apr 21 at 12:19
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    \$\begingroup\$ Diodes are switches too \$\endgroup\$ Apr 21 at 12:25
  • \$\begingroup\$ @AnalogKid about the LDOs, they convert excess voltage to heat (since they are actually self-regulated-resistors). It would not help me with my wearable project, but good to know! \$\endgroup\$ Apr 21 at 13:44
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    \$\begingroup\$ "how do you achieve this without switches". The old fashioned way, with diodes. Note that I said "not ... in the traditional sense." There is a huge difference in the harmonic energy content between a high-frequency boost converter and the zero-crossing transitions of a bridge-rectified sine wave. \$\endgroup\$
    – AnalogKid
    Apr 21 at 15:59
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Here is a terrible, unhelpful answer, but I don't see why it wouldn't work: Run a DC electric motor off of the battery, mechanically connected to a generator that outputs AC. Route that AC output through the transformer of your choice, then through a rectifying bridge and then a linear regulator!

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    \$\begingroup\$ Still uses switches! The motor commutator is exactly a switch, and the diodes in the rectifiers are also switches, turning on or off depending on which way current is trying to flow. Still, good for thinking out of the box! \$\endgroup\$ Apr 22 at 5:01
  • \$\begingroup\$ @Reversed Engineer I thought about that. But I figured the DC motor was electrically isolated, on the other side of the device. And the negligible diode "switching" would be easily cleaned up by the regulator and some capacitors. :) Very steam punk! \$\endgroup\$
    – kackle123
    Apr 22 at 19:10
  • \$\begingroup\$ I'm voting you up one to sincerely apologise on behalf for the rude Stack Police Officer who downvoted you without explaining why. No ticket on the windscreen, no summons to appear in court. Just that rude downvote. Please own up! \$\endgroup\$ Apr 25 at 5:07
  • \$\begingroup\$ @Reversed Engineer Haha, thank you, but you didn't have to. The buttons DO say to vote whether the answer is "useful", and this one only is useful when trapped on an island full of random parts! 'Gilligan's answer, if you will (or would that be the professor's)... Edit: Oops, I see "windscreen"; I don't know whether you know of Gilligan. \$\endgroup\$
    – kackle123
    Apr 26 at 13:47

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