My first Joule Theif - Questions Galore :)

[PeterDove]

Hi All,

Just made my first working Joule theif today following instructions in

**broken link removed**

I measured the current in the circuit coming from the battery to be 40mA, I then added another led in series... and it again showed about 40mA.

Can anyone let me know how this whole thing works, I read about it on the site and it explained a lot but I have listed the questions I have below - maybe someone knows,

1) How does one bump up the amount of mA available - ie how do I bump up the current in this design?
2) as The current seems the same after I put 2 LEDs in series does that mean that the coil will keep powering X number of LEDs in series? What would be the max, anyway to know mathematically?

A lot of these questions probably stem from the fact I dont really know about coils/transformers.

Cheers

Peter Dove

 

[PeterDove]

My first clue

My first clue in attempting to understand this circuit comes from how I have managed to get a control on the current in the circuit. The bigger resister I used the less current is released. Now if I could just understand the mechanics of how the coil is working in conjunction with the resister and NPN! Anyone?

Peter

 

[Hero999]

The current consumption much when the load is varied, it's mainly determined by the value of the inductor, the saturation flux, power supply voltage and to a small extent the transistor gain and switching speed.

Adding extra LEDs will reduce the current consumption rather than increasing it.

For more current use slightly thicker wire and less turns on the inductor, you'll probably need to use a larger core too.

 

[PeterDove]

The current consumption much when the load is varied, it's mainly determined by the value of the inductor, the saturation flux, power supply voltage and to a small extent the transistor gain and switching speed.

Adding extra LEDs will reduce the current consumption rather than increasing it.

For more current use slightly thicker wire and less turns on the inductor, you'll probably need to use a larger core too.

After reading a document explaining some it, it said that as the current is drawn through the resitor and turns on the base of the transistor. This then causes a current to flow through the transistor via the other winding and effectively turns off the transistor, causing voltage to build in the coil until it is enough to turn on the LED. This is what I dont understand - what/why does voltage build? Does anyone know of a good internet resource which can explain this?

Cheers

Peter

 

[audioguru]

When current in an inductor is stopped, the collapsing magnetic field cuts across the turns in the coil and generates a high voltage. That is why we connect a reversed diode across a relay coil or other kind of inductor to arrest the high voltage spike. In this stepup circuit the LED or LEDs limit how high the voltage goes.

Two or more LEDs in series use the same current. But the battery for the stepup circuit must supply twice the power so its current is doubled.

 

[PeterDove]

When current in an inductor is stopped, the collapsing magnetic field cuts across the turns in the coil and generates a high voltage. That is why we connect a reversed diode across a relay coil or other kind of inductor to arrest the high voltage spike. In this stepup circuit the LED or LEDs limit how high the voltage goes.

Two or more LEDs in series use the same current. But the battery for the stepup circuit must supply twice the power so its current is doubled.

Thanks - one more question - why does it only generate enough voltage to light the LED, what I mean is - why doesnt the LED burn out due to excessivly high voltage? Or is the current fixed by the value of the resistor going into the base of the transistor? errr - I know - I do sound slightly confused - but I can see a little light already - har har little light..LED... urg sorry

Cheers

Peter

 

[audioguru]

An LED is a diode. Diodes limit the voltage across them. It is current that LEDs use and too much will burn them out.

A few things in the circuit limit the LED's current. The current in the inductor rises in a ramp. Its operating frequency shuts off the current when it has risen to a certain amount. The resistance of the inductor limits its current, also the resistance of the transistor limits how much current goes into the inductor.

 

[PeterDove]

An LED is a diode. Diodes limit the voltage across them. It is current that LEDs use and too much will burn them out.

A few things in the circuit limit the LED's current. The current in the inductor rises in a ramp. Its operating frequency shuts off the current when it has risen to a certain amount. The resistance of the inductor limits its current, also the resistance of the transistor limits how much current goes into the inductor.

You know I said 'just one more question'

a) In the diagram in the link I gave, why does reducing the 1K resistor increase the current to the LEDs? Does turning on more current to the base of the BC548 allow more current to flow on the other winding tthrough the collector and emmiter, or is it simply that the current size on the resistor winding is then mirrored on the other side?

b) Is there a mathematical model behind this that I can use to create variations. ie how can I work out the inductance etc etc of the toriod 20 wind transformer. What info do I need to know. What would allow me to get more current etc.

It would be nice if I could create a system where I knew the values of the components and could closely match the current with what I wanted.

Cheers

Peter

 

[audioguru]

a) In the diagram in the link I gave, why does reducing the 1K resistor increase the current to the LEDs? Does turning on more current to the base of the BC548 allow more current to flow on the other winding through the collector and emitter?

Yes. The collector current of a transistor is a multiple of its base current.

b) Is there a mathematical model behind this that I can use to create variations. ie how can I work out the inductance etc etc of the toriod 20 wind transformer. What info do I need to know. What would allow me to get more current etc.

There are the frequency, core size and material, wire size and number of turns, current gain of the transistor and probably many more things that would affect how much output current.

 

[PeterDove]

Yes. The collector current of a transistor is a multiple of its base current.

Great, I was looking at the datasheet for it, I assume this info in the DC Current Gain section? am I reading it right, it seems to be saying that for a 2ma current on base you would induce 200ma in collector/emitter ( though they have a large range stating a min and max )

There are the frequency, core size and material, wire size and number of turns, current gain of the transistor and probably many more things that would affect how much output current.

Just as a 'for example' what would I expect to happen if I increased/decreased the number of turns?

Thanks once again - I'm afraid I cant seem to help myself with the questions

One thing I can promise is that I will write all this up into a good tutorial when I understand it

Peter

 

[Hero999]

Increasing the number of turns will reduce the frequency and current but will increase the inductive kickback you you'll be able to connect more LEDs in series. Reducing the number of turns will increase the frequency and current but reduce the inductive kick back so you won't be able to connect as many LEDs in series.

 

[audioguru]

Hi Peter,
The transistor can be purchased with its current gain selected into 3 smaller ranges by adding a suffix letter to its part number. So a BC548C has a high current gain.

Adding turns would increase the inductance which would increase the magnetic field in the inductor so the current would increase if the transistor and battery can supply the increase. But the extra turns increases the resistance that reduces the current.

Add more turns of heavier wire for low resistance on a bigger core so the core doesn't saturate. Drive it with a darlington (two transistors connected together in one package with a very high current gain) power transistor and use a high capacity battery. Then it will have more current for a bigger LED.

Every semiconductor manufacturer make ICs for flashlights and cell phones that are much more efficient than this extremely simple circuit. Some of them even regulate the current so the LED remains bright as the battery voltage runs down.

 

[Hero999]

A darlington won't work here because the battery voltage is too low to turn it on.

 

[audioguru]

A darlington won't work here because the battery voltage is too low to turn it on.

Oh yeah, it is only 1.5V when the battery is new and 1V or less when it is finished.

Zetec make high gain, high current transistors that saturate well.

 

[PeterDove]

Oh yeah, it is only 1.5V when the battery is new and 1V or less when it is finished.

Zetec make high gain, high current transistors that saturate well.

Thanks! Does anyone have any idea of the efficiency of these type of circuits?

I am measuring that mine gives me a kick of 14V. So if I were to just take that PLUS the current... and attempted to work out how long my 1.2V 2000mAH rechargable battery would last.. would the calc go like

( 14v / 1.2V ) * ( 100ma = TotalCurrentDrawn ) and this will give me the real current outage - 1160mAH appox used - so a 2000mAH battery would last like 1.72hrs - but I imagine that the transistor, coil and resistors steal some of that.

Peter

 

[Hero999]

Easy, measure the average current and voltage across the LED and power supply, then work out the power in vs power out.

 

[audioguru]

Efficiency is the amount of power doing the job divided by the total power from the battery. This very simple circuit creates heat in the transformer, the transistor and the resistor. The heat is wasted power that is not used for the job of lighting the LED.

The power that is doing the job of lighting the LED is the average voltage across the LED times the average current through the LED times the duty-cycle of the pulses.

It is difficult to measure how much power is going to the LED and is coming from the battery because they pulse.

I would guess the efficiency is only about 60%. Nearly half the battery's power is wasted.

 

[Hero999]

Come on audioguru, it's more efficient than your FM transmitter.

I haven't built this circuit before, but I've achieved about 80% efficiency with a similar step-up switching regulator.

 

[audioguru]

This circuit is the simplest and smallest voltage stepup one that I have seen, with only 3 parts. Other circuits on the web use a few or more than a few more parts and are much more efficient than this one.

 

[PeterDove]

This circuit is the simplest and smallest voltage stepup one that I have seen, with only 3 parts. Other circuits on the web use a few or more than a few more parts and are much more efficient than this one.

I have seen one which uses an inductor and two transistors ( plus a few others ) - sadly I didnt see an explanation of how it worked.

Peter