Joule Thief 101

[MileHigh]

Some basic colour commentary about my previous posting:

I am making an assumption that the identical amount of energy is discharged in to the LED in both cases.  This is a thought experiment so nothing is stopping us from making slight timing changes in the operation of the circuit, and these slight timing changes can be realized in real life.

Since the energy that is discharged into the LED for the first circuit is a combination of battery energy and L1 energy, then is goes without saying that for the second circuit you have to put more energy into L1 and energize it for a longer time to get the same amount of energy in the discharge pulse.

And that is were the second circuit loses efficiency.  You have to cycle more energy twice through L1 for the second circuit and that means you incur more resistive losses.

[tinman]

Some basic colour commentary about my previous posting:

I am making an assumption that the identical amount of energy is discharged in to the LED in both cases.  This is a thought experiment so nothing is stopping us from making slight timing changes in the operation of the circuit, and these slight timing changes can be realized in real life.

Since the energy that is discharged into the LED for the first circuit is a combination of battery energy and L1 energy, then is goes without saying that for the second circuit you have to put more energy into L1 and energize it for a longer time to get the same amount of energy in the discharge pulse.

And that is were the second circuit loses efficiency.  You have to cycle more energy twice through L1 for the second circuit and that means you incur more resistive losses.

Only that is not the case,as can clearly be seen in the two scope shot's.
As the battery it self has an internal resistance that increases as the battery voltage drop's,the resistive losses will increase in the first circuit,as the battery voltage drop's--this you know MH.
The second circuits resistive losses become less as the battery voltage drop's,as the battery is omitted from the circuit when the transistor switches off,and thus the resistive losses of the battery are also omitted from the circuit during the off period.
This makes the second circuit more efficient,as both circuits see the resistive losses from L1 during the on time,but only the first circuit incur the resistive losses of the battery during the off period of the transistor. L1 may have a resistance value of .1 ohm if your lucky--i would suspect even less than that. So the resistive losses through L1 are very low,but the resistive losses of the battery can be very high. The internal resistance of a depleted 1.5 volt battery would be 4 to 6 ohm's--or even greater. This we can test without to much trouble.

Like i said MH--it pays to think a little before making claim's that you cannot even back up--where as i can through experiments.

You are also about to learn about resonant systems being far more efficient than a non resonant system.

In about 1 1/2 hours,i will be posting a video,showing you that energy can be drawn from a resonant system,that decreases the required input power of that system.
You will also see that when the system is out of resonance,less power can be drawn from it,but the system draws more power.
This system was inspired by Mag's system he is working with now,where as i have replaced the electric motor with a coil,and pulse that coil at the resonant frequency of the mechanical side of the system,and draw energy from that system,and have the P/in go down when doing so.

I will post the video in the mechanical resonant systems thread,but will post it here as well for you to comment on.


Brad

[MileHigh]

Brad:

Only that is not the case,as can clearly be seen in the two scope shot's.

How do you even know the two scope shots are comparable?  The truth is you don't.  The two scope shots are not a valid comparison because you are not doing a controlled experiment.  That's just you in your typical mode of blindly moving forward.  You have to think more about what you are saying and what you are doing.  The most logical controlled experiment would be to have the two setups put identical amounts of energy into the LED per pulse, and that has to be measured with a DSO.  Then make your other measurements.

As the battery it self has an internal resistance that increases as the battery voltage drop's,the resistive losses will increase in the first circuit,as the battery voltage drop's--this you know MH.

I know that but I don't know what the numbers are and I am presuming that you don't either.

The second circuits resistive losses become less as the battery voltage drop's,as the battery is omitted from the circuit when the transistor switches off,and thus the resistive losses of the battery are also omitted from the circuit during the off period.

Without a controlled experiment that statement is meaningless.

L1 may have a resistance value of .1 ohm if your lucky--i would suspect even less than that.

I am operating under the assumption that the resistance is much higher than that.  Such a low resistance would put lots of stress on the transistor and/or the battery.

Like i said MH--it pays to think a little before making claim's that you cannot even back up--where as i can through experime

It pays to think before you just hook up the two circuits to your scope and declare victory.  No controlled experiment putting the two circuits on an equal playing field, then you can't say anything.  Garbage in - garbage out.

My gut feel is telling me that the first circuit loses less energy to heat per LED pulse.  I am not going to make the precise measurements and neither are you.  You blindly took some scope captures and thought that it was a valid test, in typical Brad fashion.

MileHigh

[MileHigh]

Since you are back Brad, time to close the loop on this:

The red dots show the current flowing back into the battery.

Show us your smarts.

[tinman]

 author=MileHigh link=topic=8341.msg478541#msg478541 date=1459004901]




MileHigh

I know that but I don't know what the numbers are and I am presuming that you don't either.

Oh,so now we put forward valid arguments based on assumptions.

Without a controlled experiment that statement is meaningless.

Says he who dose no experiments at all.

I am operating under the assumption that the resistance is much higher than that.  Such a low resistance would put lots of stress on the transistor and/or the battery.

Like i said,you dont understand how the JT circuit work's,and yet here you are trying to tell us all about it.
I see you are making more assumption's--fantastic. Well here is something to think about.
How many turns of wire would you say the standard JT has around that small toroid core?.
Let's say L1 has 20 turns,and the circumference of the cross section of the core is 40mm-- a thick core i know,but as an example we'll use that. The smaller cross section would mean less wire for the 20 turns. Anyway,20 x 40mm = 800mm + lets say another 100mm for the angled winding.
I use .6mm copper wire in most of my JT's,so lets go with that. Thicker wire of course has a lower resistance per meter. so .6mm copper wire has a resistance value of .059 ohm's per meter,and we just used 900mm for a very thick core. 90cm of .6mm copper wire has a resistance value of .053 ohm's.

http://chemandy.com/calculators/round-wire-resistance-calculator.htm

So now lets look at the scope shot below on my JT,where i used the .6mm wire. Now we would be lucky if i used 1/2 the amount of wire i calculated above,but we will stick with the 900mm,with a resistance of .053 ohm's. We can see the voltage across the coil is close to 1 volt at the maximum current. As we are using a 1 ohm CVR,we can calculate the peak current at about 24mA.
So how is it MH,that the L1 resistance value is less than .053 ohm's,and we have 1volt across it,but the peak current is only 24mA?,when we should have about 18.6 amps.
So you see how silly your statement is?.
You simply do not understand how a JT work's.

It pays to think before you just hook up the two circuits to your scope and declare victory.  No controlled experiment putting the two circuits on an equal playing field, then you can't say anything.  Garbage in - garbage out.

What is garbage MH,is you basing your guru claims on assumption's--that are all wrong.
You have posted this your self--i am going on assumption's you said. Now,what kind of blasting would you give the rest of us if we based claims on assumption's  .

Oh-by the way,the internal resistance of an eveready extra heavy duty battery at 1.1 volt's is 5.7 ohm's--just did the test for you,so as you know.

My gut feel is telling me that the first circuit loses less energy to heat per LED pulse.

If the first circuit was using the battery i stated above,then it will include the resistive losses of the battery during the off time,where that resistance is 5.7 ohm's. This value increases as the voltage gets lower in the battery,so the first circuit will become less efficient as it drains the battery.

The second circuit excludes this loss.

I am not going to make the precise measurements and neither are you.

I already have,and unlike you ,i can look at a circuit,and see all the losses involved,where as you cannot--like the batteries internal resistance.

Now,can you work out why we only have 24mA peak current,with 1volt across a coil that has less than .053 ohms of resistance?
Why dose the transistor switch of before we get anywhere near peak current?
Answer that,and then you may start to understand as to how the JT work's,and know why only a very low current flows into a coil with a very low resistance value.
Then you wont have to make silly assumptions that the coils resistance must be much higher than it actually is,because it will put great stress on the transistor and battery.

You blindly took some scope captures and thought that it was a valid test, in typical Brad fashion.

I base all my findings on experiment's,where you have admitted to basing your claims on assumption's.

How the hell have you made it this far MH,when you post garbage like you have above.
I cannot believe you have been arguing with myself and Mag's (who do experiment,and base there findings around those experiments),and making  claims (with some sort of authority) ,based around assumptions--that are way out to lunch .


Brad