Joule Thief 101

[Magluvin]

The second circuit below will drain more of the remaining energy from the battery--can you work out why?. Can you see how in the original circuit(your fav circuit)that the battery is being drained even during the Off time of the transistor,while the second circuit only draws power from the battery during the on time.

Brad

Its funny. How many times have we talked about this. I had seen it before I built it. It is kinda easy to see that the battery is in series with the led when across the transistor, and the direction of that current through the led is depleting the battery during transistor off time.

Did TK ever show his findings of the difference of input with the led across the coil then the transistor? Remember the right way and the wrong way?

Im just not sure why they cant see it.

Hey Brad. Can you repost the circuit with the led on the base side coil. 

Mags

[tinman]

Its funny. How many times have we talked about this. I had seen it before I built it. It is kinda easy to see that the battery is in series with the led when across the transistor, and the direction of that current through the led is depleting the battery during transistor off time.

Did TK ever show his findings of the difference of input with the led across the coil then the transistor? Remember the right way and the wrong way?

Im just not sure why they cant see it.

Hey Brad. Can you repost the circuit with the led on the base side coil. 

Mags

Ah yes.
The first one below is the circuit i posted some time back. It is just the SS/SSG circuit,without the death spike driving the LED.
The second is the new beaut DSJT (deathspike joule thief)circuit--works a treat
But the third(new circuit) is the king of JTs. We call this one the Triple DSJT 
In the Triple DSJT,all most all the energy is used driving LED's. The two LEDs across the VR clamp the voltage across the VR,so as the VR may only have a voltage drop across it that is equal to the clamping voltage of the LED's. As the battery voltage drop's,the base resistance can be decreased. This will insure that the LED on L1 remains lit as bright as can be at low battery voltages. It also removes the wasted energy dissipated by the VR,and uses most of that energy to drive the other two LED's.


Brad


Brad

[MileHigh]

MH
Some of what you say is true--if we build your standard JT,where it has a fixed base resistor value.
But with a higher turn ratio on L2 to that of L1,and a variable base resistance,then the (your) JT circuit is just as efficient,as i have already done what you say above.

The whole point was to be able to maintain maximum light output,as the battery voltage drop's,and the !death spike! version dose that very well. The other advantage is that by lowering the base resistance,the circuit will run normally at a much lower voltage that your standard fixed resistor circuit. In fact,your standard JT circuit is quite inefficient in it self,and the second version where the LED is across the L1 coil only,is far more efficient,as it excludes the battery in the current loop during the off time of the transistor.

The second circuit below will drain more of the remaining energy from the battery--can you work out why?. Can you see how in the original circuit(your fav circuit)that the battery is being drained even during the Off time of the transistor,while the second circuit only draws power from the battery during the on time.

Brad

I don't think the standard Joule Thief was ever intended to keep the light output of the LED steady as the battery voltage dropped over time.  And I will say it again because I believe it is important; the output impedance of the battery increases over time also.  I have never seen a single person on the forums try to measure the output impedance of a given battery to understand what they were working with.  If you play with batteries all the time and want to extract the maximum energy from them, how could you NOT want to do this measurement, but that's going off on a bit of a tangent.

I believe the Joule Thief was just a fun little circuit that demoed how to extract more than the "normal" amount of energy from a battery.  It's nothing more than a timing circuit to energize an inductor and discharge the inductor though an LED.

Now, if you want to keep the LED illumination level manually adjustable to compensate for the dropping battery voltage that makes perfect sense.  However, clearly if you add extra turns to L2 to allow the transistor switching to still run at low battery voltages you run into the problem of too high an EMF from L2 causing a death spike and shorting out the transformer by punching through the transistor.  In any design situation there are trade-offs and compromises that have to be made.  Then you have the base resistor connected to L2.  From what I have seen so far, having a variable base resistor is a very poor way of adjusting the LED brightness.  It is not a "brightness control" by a long shot.  Is there any other way to adjust the LED brightness in a standard Joule Thief?  I am not sure you can, but nothing is stopping anybody from experimenting with all of the parameters.  Don't lock yourself into a box and just assume that varying the base resistor is the only way to do it because in fact it looks like a crappy way to do it.

MileHigh

[MileHigh]

Brad:

In fact,your standard JT circuit is quite inefficient in it self,and the second version where the LED is across the L1 coil only,is far more efficient,as it excludes the battery in the current loop during the off time of the transistor.

The second circuit below will drain more of the remaining energy from the battery--can you work out why?. Can you see how in the original circuit(your fav circuit)that the battery is being drained even during the Off time of the transistor,while the second circuit only draws power from the battery during the on time.

No, you are wrong.  The standard Joule Thief circuit is more efficient than the second circuit.  Also, the standard Joule Thief circuit will do a better job at draining the battery compared to the second circuit.  It's not a huge difference in both cases but that's not the point.  So you and Magluvin are foiled again because you did not think it through.  So now the two of you now have an opportunity to think it through and find the error in your ways.

MileHigh

[tinman]

Well the tripple DSJT is a winner

At just 200mV supply voltage,the transistor(a 2n3055) is still switching on cleanly and fully--no spastic operation seen here

LED 2 is the only one still lit brightly 
P/in 2.12mA @ 200mV
Scope shot across base/emitter junction.
The standard JT using the very same inductor,stop's working at 340mV
Thank god for the death spike



Brad