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Boosting 0.3V to 2V or higher?

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Ok, some simplifications and part cross-referencing later i got a proof of concept for the charge pump:
boost_cp.png


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) and up to 3.3V of raw output, and is quite capable of maintaining itself once the big capacitor is charged over 2.4V.

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 - needs a diode between the last stage and the big capacitor.

Any obvious improvements?
At the least i'm looking for some kind of a sub-μA oscillator at ~100Hz to drive it.
 

I would prefer a CMOS gate oscillator RC oscillator. ADG736 power supply connections are missing in your schematic.
 

I would prefer a CMOS gate oscillator RC oscillator.
Details please.
If you mean a 555 timer or 4000/74 series chips, then i'm still at loss on what to believe about power consumption.
That ADG736 was chosen simply because it was the only one left after cross-referencing all the 2xSPDT chips from AD and TI that had the right supply voltages with what was available in local stores.

ADG736 power supply connections are missing in your schematic.
I tried to avoid extra clutter.
They connect to the common ground and the 3V line, i.e. supplied from the same 10mF capacitor their action charges.
 

Getting back to that idea, it mostly worked.
I'm getting about 10µA out of it in broad sunlight, with up to 9V of voltage.
That gives efficiency of about 1%?
Well, that's a start. Efficiency sounds higher than 1% though. If you get 10uA @ 5V out for 1mA @ 0.5V in, that's 10%.

The transformer i got is 1:70, 10Ω primary, 7KΩ secondary.
:???: 10Ω primary, 7KΩ secondary gives a voltage ratio of about 26.5 to 1

The problem is, it does not self-start and collapses when the secondary gives less than about 5V - it needs to draw just under 1mA from the source to start oscillating.
[snip]
Anything meaningful i can do to improve it?
  • A capacitor across the power source to keep the impedance down at high frequencies, otherwise the photocell's internal impedance will damp the oscillation.
  • Maybe try different JFETs. I suspect something like a BF245A would be better at self-starting.
  • I think the transformer step-up ratio is critical. You could try a small mains transformer e.g. 240V to 3V or 6V, or perhaps wind your own transformer on a ferrite ring.

What did you use to simulate it?
I use the free version of SIMetrix SIMPLIS.
 
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:???: 10Ω primary, 7KΩ secondary gives a voltage ratio of about 26.5 to 1
I always thought it was the proportion of the number of turns that define the transformer's ratio. How does winding resistance come into play?
Or does the Ohms there mean something else?
I've seen the resistances mentioned and the ratio=sqrt(primaryΩ/secondaryΩ) equation here - http://www.dicks-website.eu/fetosc/enindex.htm , but couldn't find where it came from.

I think the transformer step-up ratio is critical. You could try a small mains transformer e.g. 240V to 3V or 6V, or perhaps wind your own transformer on a ferrite ring.
It is a mains 220V to 3V transformer.
Winding my own on a ring didn't yield any results before since i did at most 180:4 windings before the wire tangle too much.

I use the free version of SIMetrix SIMPLIS.
Thanks, that looks interesting.
 

Perhaps a different oscillator circuit would work better. This one uses two separate non-coupled air-cored coils.

Each coil can be made as follows:
Diameter = 10mm
Length = 10mm
N = 20 turns

A total of about 2 feet of 0.5mm diameter enameled copper wire would be needed to make both. The idea was to get away from the uncertainty and losses associated with transformers.

L2, C1 and C3 set the voltage gain to 100 and frequency to about 10MHz. C1 and C3 can be increased ten-fold to reduce the frequency to about 3MHz.

 

WHAT's ALL THIS STUFF ABOUT BOOSTING HIGH IMPEDANCE low level sources to 5V?

When a source is a few hundred millivolts at micro-amp source levels, it has an equivalent circuit of an ideal voltage gen. with a series resistance R. COnsidering the original question was µA's of current, let's be generous with 100 µA

R= V/I = 0.3 / 100 µA =3kΩ

- Now which of the solutions will not suppress this source with the load impedance?
- What is the time constant of charging an ideal Cap at say 100µF from 0.3V?

Pipe dream?

THe Jewel Thief design is about the only way to step up a peanut power level to drive 5V.
Joule_thief.png

File:Joule_thief.png


Example of a joule thief circuit driving an LED. The coil consists of a standard ferrite toroid core with two windings of 20 turns each using 0.15 mm (0.006 inch) diameter wire (38 swg) (34-35 AWG). The circuit can utilize an input voltage down to about 0.35 V and can run for weeks using a 1.5 V LR6/AA. The battery voltage is usually 1.5 V. The resistor is ~1 kΩ, 1/4 W. The transistor could be a BC547B, 2SC2500, BC337, PN2222, 2N4401 or other NPN. Vceo = 30 V, P = 0.625 W. A white light-emitting diode with Vf = 3.2 V might be used." ..wiki

You can search for GaAs transistors or Ge but may be happy to get it working down to 300mV with common small signal transistors.

have fun squirting out the last electrons from those dead batteries. White LEDS are around 3V. same as Blue.
 

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Probably, but an interesting one.
It would certainly be easier to just use a few extra photocells.

Or a single lithium coin cell... but that takes the fun out of it I guess.
 

Some stats on the charge pump.

Pumping commences at 0.37V and 630µA, when it heaves over approximately 0.1V at the first doubler cap.
Once going, pumping is sustainable down to about 20µA at 0.28V, keeping 1.1V at the output, sub-µA current.
Reducing capacitor sizes reduces performance.
For a single stage pumping works for as low a current as it can go, but adding the second stage stops it all together, which suggest some design issues with stage addition. Any ideas?

Sustains itself (consumption of 11µA at 2.6V) with 1200µA at 0.39V, 6% efficiency?
Charges 100µF cap zero to 3V in about 1.5 seconds (200µA?) with 1800µA at 0.433V. 72% efficiency?

I'm not really sure how to estimate efficiency - i usually measure how long it takes to charge a capacitor from A to B and go from here.
Is that a valid method?
Then why is there so much discrepancy?

Or a single lithium coin cell... but that takes the fun out of it I guess.
By now it's obvious that this "free energy" sucks - for the cost of such a boost circuit i can get a pack of watch batteries, each of which could supply that sensor board for about a year.

So now i'm basically just interested what kind of power can be drawn from that cell, and at what cost.
 

I always thought it was the proportion of the number of turns that define the transformer's ratio. How does winding resistance come into play?
Yes, it's the ratio of the number of turns on the primary and secondary. Sorry for the confusion, I didn't realize you were talking about winding resistance.

Or does the Ohms there mean something else?
I've seen the resistances mentioned and the ratio=sqrt(primaryΩ/secondaryΩ) equation here - http://www.dicks-website.eu/fetosc/enindex.htm , but couldn't find where it came from.
That's what I thought you meant. The impedance ratio is often quoted, especially for audio transformers, but it's not to do with winding resistance.

Consider an output transformer for a valve amplifier, and lets say it has a turns ratio of 20 to 1. When the output from the secondary winding is 1A @ 8V into an 8 Ohm loudspeaker, the input to the primary winding from the output valve must be 50mA @ 160V (ignoring losses), so the valve thinks it is driving a load of 3.2K. This would be advertised as a "3.2 K to 8 Ohms" transformer.

There's some other examples here: http://www.esr.co.uk/electronics/transformer-audio.htm. I see they got the specs for the LT700 wrong, but the rest look OK.

Sustains itself (consumption of 11µA at 2.6V) with 1200µA at 0.39V, 6% efficiency?
Charges 100µF cap zero to 3V in about 1.5 seconds (200µA?) with 1800µA at 0.433V. 72% efficiency?

I'm not really sure how to estimate efficiency - i usually measure how long it takes to charge a capacitor from A to B and go from here.
Is that a valid method?
Then why is there so much discrepancy?
I would calculate efficiency as the useful output power delivered to an external load divided by the total power drawn from the photocell.

If I understand your first example correctly, the charge pump was just sustaining itself, not delivering any output. In that case, the efficiency is zero. It is similar to a car's fuel efficiency, measured in miles per gallon, being equal to zero when the car is parked with the engine running.

In your second example:
Input power = 1800µA * 0.433V = 780uW
Average output power = 200µA * (average voltage = 1.5V) = 300uW
So average efficiency = 300 / 780 = 38.5%
I'm not sure how you got 72% - maybe I'm missing something.

Earlier you said: "From 1.5mA at 0.45V this gets about 250μA (at 1V)". In that case:
Input power = 1.5mA 8 0.45V = 675uW
Output power = 0.25mA * 1V = 250uW
Efficiency = 250 / 675 = 37%

I'm surprised the charge pump idea works so well! I expected it to have difficulty raising it's own supply voltage.
 
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Concerning godfreyl's oscillator #2:
The result is only slightly more than zero.
There is nothing on the load if there is a load, and slightly over 1V of essentially static charge if the load is removed.
The current is not quite enough to feed the voltmeter.
I don't see any kind of activity anywhere with the scope, no idea what is not working.

Concerning charge pump:
Putting stage 2 on the opposite side of the clock makes the thing start pumping from as low as 80µA, and produce some current at 2V with 200µA input.
Putting a schottky diode between the last pump capacitor and the supply line prevents the big supply capacitor from draining back if the source current drops.

From 2mA at 0.45V it sustains itself at 2.7V, and charges a 1F supercap with 60µA across 1V of voltage difference.
Trying to draw more current drops the voltage below the oscillator's threshold.

With the setup being self-sufficient, the supercap being at 1.8V, and the clock provided by a microcontroller on the same power bus, it provides 120µA of current towards charging the supercap with the source at 0.45V and 2mA.
So it's efficiency of about 24%?

All in all, it's marginally practical to power things in the 10-100µA range @3V in broad sunlight from a coin cell if you have the means to shut off the oscillator for the night, or provide the clock for free from the thing being powered.
But if the power runs out (a few cloudy days?), it essentially dies.
Just like a living thing.

I'm surprised the charge pump idea works so well! I expected it to have difficulty raising it's own supply voltage.
These are raw numbers - it charges that capacitor in a second while being powered externally, so it's 37% for conversion.
But if it is self-sustaining and a current is drawn, then it can only provide about 60µA on the imperfect oscillator or 120µA with a MC-driven clock - the higher the voltage, the less is the pumping efficiency, and it's top voltage of 3.3V is very close to the lower operational one of 2.5V or 1.8V.
You can't really get it all from there - 6% to 24% as far as i understand it.

The impedance ratio is often quoted, especially for audio transformers, but it's not to do with winding resistance.
Now that makes sense, thanks.
 

It's easy to make a switch that turns off the charging circuitry when the photocell voltage is too low to be useful but, as you say, the whole thing will die after a few cloudy days.
 

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