Boosting 0.3V to 2V or higher?


by Artlav
Tags: boosting
Artlav
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#1
May6-12, 06:36 AM
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Hello.

I want to get about 2-5 volt out of a 0.3V DC source, at ľA levels, or get power out of it in any other way that would allow to charge a capacitor up to the said 5V.

Any leads on how can this be done?
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vk6kro
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#2
May6-12, 08:39 AM
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Normally, this would be very difficult because even the best Schottky diodes still require about 100 mV to rectify an AC waveform and most oscillators need a volt or two to get started.

However, I came across this website:
http://www.dicks-website.eu/fetosc/enindex.htm

The author shows FET oscillators oscillating with extremely low supply voltages. Down to 5 mV.

If you built one of these oscillators, you may be able to use a half wave voltage doubler to develop a much larger DC voltage.

I haven't tried it, but it may be worth a shot.
Bobbywhy
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#3
May6-12, 07:02 PM
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Try these:

The MAX1044 and ICL7660 are monolithic, CMOS switched-capacitor voltage converters that invert, double, divide, or multiply a positive input voltage.

http://www.maxim-ic.com/datasheet/index.mvp/id/1017

vk6kro
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#4
May6-12, 07:57 PM
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Boosting 0.3V to 2V or higher?


They need 1.5 volts minimum, though. This was the problem here, because there is only 0.3 volts available.

That web site for ultra low voltage oscillators looks promising, although the oscillators need a transformer that may be hard to find.

Solar powered garden lights use a single solar cell (0.6 volts) to charge up a NiCd cell (1.2 volts), which then supplies an oscillator to light up a white LED (3.5 volts). So there may be a ready made solution there if they will work on 0.3 volts.
Bobbywhy
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#5
May6-12, 08:51 PM
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vk6kro, Thanks for catching that error. I failed to check the 7660's data sheet for minimum inputs.
NascentOxygen
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#6
May6-12, 09:34 PM
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Quote Quote by Artlav View Post
Hello.

I want to get about 2-5 volt out of a 0.3V DC source, at ľA levels, or get power out of it in any other way that would allow to charge a capacitor up to the said 5V.

Any leads on how can this be done?
Hi Artlav! Just out of interest, what is your 0.3V source here?
Artlav
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#7
May7-12, 12:41 PM
P: 138
Quote Quote by vk6kro View Post
The author shows FET oscillators oscillating with extremely low supply voltages.
Nice. Been trying to replicate the first one.
Looks like a 180:8 winding-round-a-toroid transformer is not the correct choice, as it didn't work.

Does this kind of oscillator have a name?
Something to look it up by?

Quote Quote by NascentOxygen View Post
Hi Artlav! Just out of interest, what is your 0.3V source here?
A coin sized solar panel. Gives 0.45V at 2mA in broad sunlight and 0.3V at 100μA when overcast.
I was thinking if it can be of any use on it's own, like to power a weather recorder of some sort.
sophiecentaur
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#8
May7-12, 01:50 PM
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Having such a small area means that the actual Power out is very limited (of course) boosting the volts will not help you there. If you want to power a useful circuit then this circuit could only be expected to operate for a very small proportion of the time. Digital watches sometimes claim to be 'solar powered' but I wonder how much time they need to be in bright sunlight for them to work in fact.
Bobbywhy
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#9
May8-12, 02:54 AM
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Quote Quote by Artlav View Post
A coin sized solar panel. Gives 0.45V at 2mA in broad sunlight and 0.3V at 100μA when overcast.
I was thinking if it can be of any use on it's own, like to power a weather recorder of some sort.
Artlav, if you are expecting to get some usable power from this coin-sized solar cell (not panel), then I suggest you go to the wiki page and study the following eccerpted information, and don't expect more than a few microwatts!

"Open-circuit voltage and short-circuit current

When the cell is operated at open circuit, I = 0 and the voltage across the output terminals is defined as the open-circuit voltage. Assuming the shunt resistance is high enough to neglect the final term of the characteristic equation, the open-circuit voltage VOC is:
(formula)

Similarly, when the cell is operated at short circuit, V = 0 and the current I through the terminals is defined as the short-circuit current. It can be shown that for a high-quality solar cell (low RS and I0, and high RSH) the short-circuit current ISC is:
(formula)

It should be noted that it is not possible to extract any power from the device when operating at either open circuit or short circuit conditions."

http://en.wikipedia.org/wiki/Theory_of_solar_cells
Artlav
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#10
May11-12, 11:49 AM
P: 138
Maybe it is possible to make a charge pump-like voltage multiplier that would extract more power than it consumes, provided that there is already an existing reserve?

I don't need a cold start capability.

Simplified (maybe excessively so, but that works manually) what it would take is this:


The source's connection moves back and forth along the smaller capacitors, and the big capacitor is charged as the result.

Problem is, i can't find any way to make a relay that would take less than 2mA each, and that thing would need 16.

Is there any practical way to do this kind of switching, or practical way of doing some other kind of charge pump within an order of 10μA power budget?
Bobbywhy
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#11
May11-12, 05:31 PM
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These switches might allow you to design the proposed charge pump. Unfortunately, it appears that the lowest operating voltage is 1.65 Volts. So, you would need a voltage multiplier to power your voltage multiplier!

Selecting the Right CMOS Analog Switch
http://www.maxim-ic.com/app-notes/index.mvp/id/638

Ask The Application Engineer—34
Wideband CMOS Switches
http://www.analog.com/library/analog...nd_switch.html

http://dkc1.digikey.com/us/en/tod/AD...s_noaudio.html
sophiecentaur
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#12
May11-12, 06:03 PM
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Rather than getting carried away with interesting sidelines, why not take an engineering approach? Ask yourself what you want to drive with this minute power source (i.e how many W, mW or uW you need) and then calculate (using 1kW/msq as the solar constant, for the absolute maximum power you could ever hope to gather) what size of collecting area you would need. The actual Volts that you might get are irrelevant if there is basically not enough Power available. That is the bottom line - as in all things to do with Energy and Power supply problems.

Using a voltage multiplier is only a solution in circumstances where you have a large amount of surplus power available but it just happens to arrive at an embarrassingly low voltage. You can then afford to chuck away most of that power in an inefficient system that will supply you with your required value of higher voltage. In the case of this minute supply voltage, your problem would be all the greater.
Meizirkki
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#13
May13-12, 07:30 AM
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If you can get your hands on some germanium transistors with sufficiently low forward voltage, a super simple single transistor "joule thief" circuit can be used. There are very good instructions in wikipedia:


In the place of the LED, you can use a capacitor and a zener diode to stabilize the output to 5V:


http://en.wikipedia.org/wiki/Joule_thief

EDIT:
A carefully constructed joule thief on youtube:
http://www.youtube.com/watch?v=8T9HQkDnIuU
It will keep the LED (Uf 3.3V) blinking at 18Hz until the input voltage drops below transistors forward voltage. With a germanium transistor such joule thief can run from 0.3V and the output pulses should be sufficiently high to achieve well over 5V with the few uA input current.
sophiecentaur
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#14
May13-12, 11:27 AM
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A shunt regulator (zener) is a bit on the wasteful side for an application where the available power is so marginal.
A series regulator would make more sense, as would a battery - to tide you over poor lighting conditions.
Artlav
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#15
May13-12, 01:23 PM
P: 138
Ok, some simplifications later i got a proof of concept on the charge pump idea:


The flip-flop consumes about 6μA, the switch side consumes less than i can measure.
From 1.5mA at 0.45V this gets about 250μA at 1V of raw output, and is quite capable of maintaining itself once the big capacitor is charged over 2.4V.
The voltage gain is close to clear 8x.

I haven't tried to optimize the design yet - frequency (~70Hz) or capacitor sizes might be way off optimal.
Also, the power leaks out if the source voltage drops - might need a schottky diode between the last stage and the big capacitor.

So, how can this be improved?
Anyone knows how to make a sub-μA square wave oscillator for ~100Hz?

Quote Quote by sophiecentaur View Post
Rather than getting carried away with interesting sidelines, why not take an engineering approach?
Engineering approach would be to buy a battery. :)
For the cost of tapping this "free energy" i can get a box of watch batteries which would each supply a small sensor board for years.

So, i'm mostly interested in how much i can get out of the cell.
It seems to be capable of 1mW total under best conditions, and 100μW at worst i consider.
Meizirkki
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#16
May15-12, 08:32 AM
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Quote Quote by sophiecentaur View Post
A shunt regulator (zener) is a bit on the wasteful side for an application where the available power is so marginal.
A series regulator would make more sense, as would a battery - to tide you over poor lighting conditions.
A series regulator would only add unnecessary complexity to the circuit. The output voltage (=> switching frequency) can be adjusted, quite accurately to constant load, from the base resistor of the switching transistor. The 5V zener would only be there to make sure maximum voltage is never exceeded.
Mike_In_Plano
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#17
May20-12, 10:15 PM
P: 557
People who construct "joule thiefs" are dealing with this issue by taking advantage of a clever trick associated with synchronouse boost converters -

Since a synchronous boost uses MOSFETs, they don't have a junction drop to creat a lower limit.

Also, synchronous boos converters typically use the output voltage to supply drive to the MOSFET gates. So, after the device is started at a sufficeintly high voltage, it will continue to operate at much lower voltages.

Of course, the issue then is getting enough voltage to start...
NascentOxygen
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#18
May21-12, 11:34 AM
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Quote Quote by Mike_In_Plano View Post
Of course, the issue then is getting enough voltage to start...
In dry weather, could you get enough starting voltage just by brushing it on the cat?


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