Boost voltage without switching

Question

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:

^{simulate this circuit – Schematic created using CircuitLab}

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

2. 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.

^{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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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!