Joule Thief 101

[Magluvin]

Brad:

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

" The standard Joule Thief circuit is more efficient than the second circuit."

You made the claim. So now you should explain and or show why. Someone else makes a claim and you pounce to push them to explain their claims. So follow your own rules.


" Also, the standard Joule Thief circuit will do a better job at draining the battery compared to the second circuit."

Again. Explain your claim.


"So now the two of you now have an opportunity to think it through and find the error in your ways."


No. You made 2 'claims'. You need to prove them. Otherwise nobody here needs to explain themselves to you. Simple fairness. Otherwise you are just playing make believe, silly talk, falsehoods, fairy tales. Right?   Prove your statements. Otherwise its just wishy wash.

Mags

[tinman]

Brad:

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

You have lost your marbles MH,and it once again comes down to not being able to understand how the two circuits work.

The first circuit with scope shot,clearly shows more current flowing from the battery. With your circuit,the battery is in series with L1, and the LED. So with your circuit,the battery is having power drawn from it for an entire cycle. The scopes channel A is across L1,and reads 2.32v max(top of the spike during off time. Then the battery voltage is also added to that,so as there is enough voltage to drive the LED--all while energy is being drained from the battery--it gets no rest phase between cycles.

In the second circuit-along with scope shot,we can clearly see that the current draw is less,and the battery is not supplying any power to the circuit while the transistor is switched off. Only L1 is supplying the power needed to drive the LED during the OFF period.

The second circuit will drain more of the remaining energy from the battery that your JT circuit will,as the battery gets a rest phase during each cycle.
Your circuit looses efficiency as more components are included during the off time of each cycle.
There will be small losses through -,the battery-- will dissipate more heat due to current always flowing through it--more losses.
Your circuit !during the off time!,includes L1,the LED,and the battery.
The second circuit includes !During the off time! L1 and the LED.

So try again guru
Which circuit is more efficient?.

Brad

[sm0ky2]

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... MileHigh

the Joule Thief was never intended to "light an LED"....
This simple minded assumption is what keeps many people
from understanding what it is that they have built.

The JT is intended to charge the ferrite, the field of which, then collapses.
What you do with that afterwards, is your own business.
If it is "light an LED", great.
Some people have more in mind to do with it.
But the device still functions, even when you throw the LED in the trash can.

the LED has nothing to do with the oscillator.
it serves no more purpose than the little green light
on the front of your computer that tells you it's "on".

[tinman]

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

Well oddly enough,i would have to agree with everything you stated above.
I guess the aim here,is to make the most efficient circuit we can,where losses are reduced as much as possible . Some of which would be to drain as much energy as we can from the !otherwise! dead battery,and obtain maximum light output while doing so.

Funny thing is,so far,i have not been able to make a JT that out performs the circuit you find in a $2.00 garden light . The only problem with them,is they stop working around 660mV,but they will drive a white 10mm LED at 1.2 volts very brightly for only 3mA of current. The circuit for the LED christmas lights will light 100 LED's at 1.2 volt's for only a 7mA current draw.

Seems we have a ways to go yet,before we can come close to !off the shelf! JTs
In fact,i might throw one on the scope,and have a close look at what is happening with one of them.

Brad

[MileHigh]

So you guys are given the opportunity to pause, reflect, and think it through and both of you balk.  You don't even try.  And one wonders why knowledge flows like a glacier in warm weather around here.

Let's assume that both circuits pulse an identical amount of energy into the LED, and let's explore the losses associated with each circuit in simple terms.  Let's make a reasonable assumption that less energy is lost in the battery itself to energize L1 as compared to the resistive losses in L1 when it energizes or discharges.

For the first circuit, there are looses in the battery to energize L1 of B_LOSS.  Then there are losses in L1 as it's being energized of L1_LOSS.  Then there are losses in L1 as it discharges into the LED of L1_LOSS.  When the discharge into the LED occurs, there are losses again in the battery.  Let's say those losses are 0.3xB_LOSS.

Total losses first circuit are:  1.3xB_LOSS + 2xL1_LOSS.

For the second circuit, more energy has to be put into L1 to light the LED because you don't have the battery "helping" the discharge cycle.  Therefore there is a longer energizing cycle for L1. So lets say the battery losses are B_LOSS + DELTA_B and the L1 losses are L1_LOSS + DELTA_L1

For the second circuit, there are looses in the battery to energize L1 of B_LOSS + DELTA_B.  Then there are losses in L1 as it's being energized of L1_LOSS + DELTA_L1.  Then there are losses in L1 as it discharges into the LED of L1_LOSS + DELTA_L1.

Total losses second circuit are:  B_LOSS + DELTA_B + 2xL1_LOSS + 2xDELTA_L1.

We are assuming that B_LOSS is less than L1_LOSS.  For sure DELTA_B is very small.

To summarize:

Total losses first circuit:  1.3xB_LOSS + 2xL1_LOSS.
Total losses second circuit:  B_LOSS + DELTA_B + 2xL1_LOSS + 2xDELTA_L1.

Let's remove B_LOSS and 2xL1_LOSS from each equation:

Differential losses first circuit:  0.3xB_LOSS
Differential losses second circuit:  DELTA_B + 2xDELTA_L1.

For the fist circuit 0.3xB_LOSS is the energy lost in the battery during the "helper phase" as the battery and L1 light the LED together.

For the second circuit, DELTA_B is the small amount of extra losses incurred in the battery to put extra energy into L1 and 2xDELTA_L1 is that extra battery energy that is lost twice in the resistance of the L1 coil.

My sense is that the differential losses in the second circuit are higher because the assumption is that the battery internal impedance losses are less than the L1 coil resistive losses.

Then the other thing that gives the first circuit an efficiency advantage is that the discharge cycle to light the LED will have a "flatter top."  A flatter top means more of the discharge to light the LED will be in the sweet spot range for the LED.

The first circuit will be able to discharge more of the battery energy because the battery EMF added to the discharging coil gives you more overall discharge EMF "push" than just the coil alone.  So the assumption is that the first circuit will be still running with a dim LED when the second circuit craps out and dies.

MileHigh