Charge pump and single pulse creator, how to?

[neazoi]

Here's my Mk3 version.

Using the same UV LED (Vf = 3.3V) it gives a bright flash at 0.9V supply with a 'just works reliably' supply voltage of 0.65V. I tried it up to 1.5V but not higher, it should be OK up to about 2V before the LED starts to draw it's own current.

I salvaged the ferrite core from a broken CFL, it seems to be used quite commonly in many different types but I do not have it's specifications or part number. You can get a slight increase in brightness by adding a 100nF capacitor from the junction of the 100R resistor and transformer to ground but it probably isn't worth fitting and it could risk the circuit self-biasing, in other words not shutting down after it has been triggered. It was OK for me but make sure it stops drawing current after a few seconds if you add it to your build. Increasing the 1M timing resistor will make it take longer to re-trigger but don't drop it below about 220K as it pass enough current to turn the first transistor on whenever the switch is closed.

Brian.

That is a different concept from the previous circuit isn't it?
As far as I understand it, the BC212 is turned ON, only when the vibraswitch is closed but only if the 10uF is mostly charged.
This means that a few vibrations are needed before the led can flash, assuming an initially discharged capacitor.
Have I understood that right? I think I am missing something here.

The other thing I am affraid of, is the joule thief efficiency. Would it draw more power than we want to save, because of it's bad efficiency compared to the capacitor charge pump?

 

[betwixt]

You figured it the wrong way around! The first transistor is PNP and the flash occurs when it passes base current IN to the charging capacitor. The resistor across it is to discharge it again so it is ready for the next time the switch closes.

Efficiency isn't really important here but the battery life is. The circuit draws negligible current when not being used (< 1uA) and a higher current peaking at about 30mA as the LED flashes. The average current over a long period is very low. In reality, it actually is quite efficient, I did not calculate it but I would guess while the LED is lit, it would by > 50%. The power needed to light the LED is going to be about the same which ever way you do it but this method avoids the need for active timing circuits which inevitably draw current all the time and therefore reduce overall circuit efficiency. The flash is quite bright and lasts about 0.5 second which should be enough to 'charge' your phosphorescent indicator.

Brian.

 

[neazoi]

You figured it the wrong way around! The first transistor is PNP and the flash occurs when it passes base current IN to the charging capacitor. The resistor across it is to discharge it again so it is ready for the next time the switch closes.
Brian.

Thank you Brian!
So when the electrolytic is charged (up to a level) then the led cannot light up. When it is discharged, the next vibration will charge it but also trigger the joule thief for as long as it takes for the cap to charge?
Am I getting it right now?

Also can I use pn2222 and pn2907?

 

[betwixt]

Yes, you got it!

I used transistor to hand but the types are not critical, the highest voltage oin the circuit is only 3.3V. As before, you will get more output if the VCEsat of the second transistor is as low as possible because it is effectively unavailable as usable voltage. It should be OK with the ones you mentioned but if you have others, it would be worth experimenting with the second (NPN) transistor to see which works best.

Brian.

 

[neazoi]

Yes, you got it!

I used transistor to hand but the types are not critical, the highest voltage oin the circuit is only 3.3V. As before, you will get more output if the VCEsat of the second transistor is as low as possible because it is effectively unavailable as usable voltage. It should be OK with the ones you mentioned but if you have others, it would be worth experimenting with the second (NPN) transistor to see which works best.

Brian.

Can I also experiment with the transformer turns ratio, to drain the battery even more?
Once I have build a joule thief that could work with as low as 86mV (although not so bright).
Again if you go to **broken link removed** look for "Experiments for Energy Harvesting and Efficiency". Near the middle of this page you will find the "US Patent 4,734,658".
I am sorry I cannot give a direct link but the page runs on a dynamic IP.

 

[betwixt]

Of course you can experiment. I make no claims to the design of the 'Joule Thief', I think the name might belong to someone but the electronics is just a basic blocking oscillator. I wound 20 turns on the secondary (collector side) and 10 on the primary (base side) as 'best guesses' at what might work optimally but I doubt you would see a great improvement by changing them.

I note the schematics have appeared on your web page already :-D

Brian.

 

[neazoi]

Of course you can experiment. I make no claims to the design of the 'Joule Thief', I think the name might belong to someone but the electronics is just a basic blocking oscillator. I wound 20 turns on the secondary (collector side) and 10 on the primary (base side) as 'best guesses' at what might work optimally but I doubt you would see a great improvement by changing them.

I note the schematics have appeared on your web page already :-D

Brian.

Yes, I usually keep good designs that work there! I hope you do not mind
My intention is to build this "tritium-alternative" hybrid illuminator for key chains.
I am thinking about it a few months but your help on the electronics is great!
I have some thoughts about the mechanical and rest of construction but this is out of the purpose of this thread. But if you want we can discuss it further.
I am really curious for how long will the small AG1 (silver oxide) battery last before it needs replacement. Who knows, it might take years to figure it out

 

[neazoi]

Here's my Mk3 version.

Using the same UV LED (Vf = 3.3V) it gives a bright flash at 0.9V supply with a 'just works reliably' supply voltage of 0.65V. I tried it up to 1.5V but not higher, it should be OK up to about 2V before the LED starts to draw it's own current.

I salvaged the ferrite core from a broken CFL, it seems to be used quite commonly in many different types but I do not have it's specifications or part number. You can get a slight increase in brightness by adding a 100nF capacitor from the junction of the 100R resistor and transformer to ground but it probably isn't worth fitting and it could risk the circuit self-biasing, in other words not shutting down after it has been triggered. It was OK for me but make sure it stops drawing current after a few seconds if you add it to your build. Increasing the 1M timing resistor will make it take longer to re-trigger but don't drop it below about 220K as it pass enough current to turn the first transistor on whenever the switch is closed.

Brian.

Brian, I built the circuit but it does not work ok. The circuit is never deactivated. I tried to replace the electrolytic with a 10nF and still nothing.
It seems that the cap takes too much time to charge. Whick resistor should I decrease? the 1M at the top side?
What could the problem be?

- - - Updated - - -

Ok I found it, I replaced the top 1M with 10K and the 10uF with 0.47uF and the bottom 1M with 10M. This provided more reasonable times, but was that ok that I altered the top 1M, in terms of power drawn by the battery?

 

[Audioguru]

PWM is used as a light dimmer. When the full current pulses are wide then an LED looks bright but when the duration of the pulses drop less than 30ms the LED begins to look dim.
You need a pretty big battery to charge a capacitor in a reasonable time to blink an LED at 100mA(!). The voltage stepup circuit multiplies the capacitor charging current by the amount stepped up. A battery almost dead at 0.9V stepped up to 3.3V is a stepup of 3.67 times. Try getting 367mA from a little battery that is almost dead or use it to charge the capacitor for a few hours.

 

[neazoi]

You mean that with the alteration of the 1M top resistor to 10K I draw much current from the battery right?

 

[Audioguru]

You mean that with the alteration of the 1M top resistor to 10K I draw much current from the battery right?

Reducing the top 1M resistor to 10k reduces the duration of the transistors turning on and charging the transformer. Therefore less power is taken from the battery.
But I do not think a "dead" battery can supply much current so the LED will be dim from the short pulse duration and dim from the low current.

 

[neazoi]

I have an idea. To minimize power consumption, I think it is better to use a smaller value of charging capacitor and a very large value of discharging resistor. The top charging resistor defines the capacitor charging time, so for a smaller cap a larger resistor can be used and since the cap is smaller now, less current is needed to charge it in the same desired time.
The drawback is that the bottom discharging resistor must be very large (100Mohm?) to ensure the small capacitor will not be discharged instantly.

How does my thought sound to you all?

 

[Audioguru]

When the switch is turned on, the capacitor is already discharged and its charging current immediately turns on the top transistor. If the capacitor has a small value then the transistors turn on the LED for a very short duration that might not be visible or might appear to be dim. As the capacitor continues to charge, its charging current becomes low enough for the transistors and LED to turn off then the upper resistor finishes the charging.

 

[neazoi]

When the switch is turned on, the capacitor is already discharged and its charging current immediately turns on the top transistor.

It is this current that I refer to. If using a 10nF capacitor instead of a 10uF, then it will take shorter time to charge.
To keep the time the same as before a larger value of top resistor (1M) can be used. But at the same time of duration, the current drawn by the battery to charge this capacitor will be much lower.
In other words more current will be drawn to charge a 10uF capacitor through a 1K resistor, than charging a 10nF capacitor through a 1M resistor, assuming the same charging time for both. (valies are just for comparison

If the capacitor has a small value then the transistors turn on the LED for a very short duration that might not be visible or might appear to be dim.

I have tested if with 10nF and 1M top resistor. The flash is quite bright and can be seen with the eye easily, so I assume the diration is not too small. there are 2-4 pulses before the first transistor is turned off (pulse numbers depend on the input voltage of 0.7-1.5v)

As the capacitor continues to charge, its charging current becomes low enough for the transistors and LED to turn off then the upper resistor finishes the charging.

Yes. The only problem I found with the values used (1nF, 1M top resistor) is that the small capacitor needs an impractically large value of discharging (low side 1M) resistor, so as to prevent if from discharging too quickly. 10M in place, is not enough, it would probably fir in the 100M range.
I have tried it without the discharging resistor and just short circuit the capacitor leads to manually discharge it.
Any ideas here for a large value of resistor without using too many 10M in series, are appreciated.

I insisτ in using a small cap and a large CHARGING resistor to keep the battery current drawn small. Is this thought correct?

 

[Audioguru]

Before I should have said that when the switch is turned on then the discharged capacitor immediately turns on the transistors that charge the inductance of the transformer, not turn on the LED. The LED turns on only when the transistors turn off and it is the flyback voltage from the inductance of the transformer that turns on the LED.

The circuit wastes battery power when the capacitor value is too high because when the inductance of the transformer is charged, the transformer is saturated and draws a very high current from the battery. The circuit is missing feedback from the transformer to the base of the first transistor to turn it off when the transformer becomes saturated that happens in a Joule Thief circuit.

 

[betwixt]

Sorry - I've been away for a couple of days with no internet connection.

Audioguru is 'half' right. The LED does light when the second transistor turns off, it is the stored energy in the inductor that light the LED, not the current through the transistor. However, that isn't how it works! The second transistor and the transformer make a blocking oscillator, on each cycle of the oscillation the transistor turns off and the LED flashes, I didn't measure the frequency but I would guess it's a few KHz. It depends on the core properties as to how fast it runs.

If you short the collector and emitter of the first transistor it will run continuously and the LED will stay lit all the time, that's how the original Joule Thief was intended to be run. The RC network around the switch is to crudely pulse the bias to the second transistor on and off again. The value of the 'top' 1M resistor isn't too important, it's purpose is only to ensure the transistors own leakage and potential pick-up from the oscillator doesn't keep it conducting. It makes very little difference to the timing because almost all the current flowing into the capacitor is through the transistor's B-E junction. The 1K resistor is to protect the transistor from excessive base current if the capacitor is fully discharged, without it, there could be 0V on the base and >0.6V on the emitter which could momentarily exceed maximum allowed base current.

The real timing components are the capacitor and resistor in parallel. The capacitor value determines how long it takes to charge and therefore how long the oscillator is allowed to run (LED lights up). Once charged the first transistor has no bias current so it stops conducting and the LED goes out. If the capacitor stayed charged, which it would until it's internal leakage let it decay, the circuit would never re-trigger so the resistor across it is to deliberately leak the charge away, in other words set the time before the circuit will trigger again.

Current is only drawn to charge the capacitor and to operate the oscillator when triggered, after a few seconds without being triggered the current consumption drops to virtually zero.

The concept in your original IC design was to boost the voltage and store it so it was ready to discharge to the LED. The idea is OK but in the real World it wouldn't work as intended because even if you didn't trigger the circuit, it would have to keep restarting to keep the charge topped up. That means the average current consumption would be higher on the basis of low current but for a long time.

You can try a modification to my circuit, add a capacitor (~10uf) from the collector of the first transistor to 0V, that would extend the duration of the pulse but it also risks the oscillator 'free running' in a self biasing fashion as the impedance in the circuit from base to 0V through the transformer secondary becomes much lower.

Brian.

 

[neazoi]

The value of the 'top' 1M resistor isn't too important, it's purpose is only to ensure the transistors own leakage and potential pick-up from the oscillator doesn't keep it conducting. It makes very little difference to the timing because almost all the current flowing into the capacitor is through the transistor's B-E junction.

Brian.

Brian, the problem I have is that the circuit as is (with the top 1M in place) is unable to charge the 10uF/16v capacitor, so the joule thief always operate with the vibrations. I used 2n2907 for the PNP.

When I changed the value of the top 1M resistor to 1K, then I was able to notice the cap charging, so I assumed current flows through it as well and so it plays a role as the charging element of the capacitor.

Apart from the above thing, I wonder if using a smaller capacitor than 10uF, wouldn't it save battery current? ((capacitor charging current)

 

[betwixt]

The charging current should pass through the emitter-base junction of the transistor, not the resistor. If it didn't, the transistor would not turn on and activate the oscillator. Any current flowing through the resistor will reduce the amount flowing through the E-B junction, that's why the value is so high.

I wouldn't worry about the capacitor charging current, it charges at the same time the oscillator is running so the few uA extra it draws will be insignificant. At most, a fully discharged capacitor would draw (supply - Vbe)/series resistor or about (0.9 - 0.6)/1000 = 300uA from a 0.9V supply and about 900uA from a 1.5V supply. The charging current drops to zero after only a few mS anyway.

Brian.

 

[neazoi]

The charging current should pass through the emitter-base junction of the transistor, not the resistor. If it didn't, the transistor would not turn on and activate the oscillator. Any current flowing through the resistor will reduce the amount flowing through the E-B junction, that's why the value is so high.

I wouldn't worry about the capacitor charging current, it charges at the same time the oscillator is running so the few uA extra it draws will be insignificant. At most, a fully discharged capacitor would draw (supply - Vbe)/series resistor or about (0.9 - 0.6)/1000 = 300uA from a 0.9V supply and about 900uA from a 1.5V supply. The charging current drops to zero after only a few mS anyway.

Brian.

I see. Ok I will leave the top 1M untouched.

Have you got any guess why the circuit, as you drawn it, does not switch off after a while? It seems to me the capacitor never charges, or it is discharged too fast for me to notice.
Any things to try out?

 

[betwixt]

It should flash the LED for no more than about 1 second then it should go off completely and the circuit then draws no current. I suspect what is happening is the voltage in the transformer secondary (base side) is too high and it is enough for the transistor to maintain oscillation by itself when the bias is removed. Try removing turns from the secondary winding and see if that fixes it. Maybe the difference in the core I used is responsible for the extra feedback you are seeing. If all is well, if you short the emitter and collector of the PNP transistor, the LED should stay lit and it should go out as soon as you remove the short.

Brian.