A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?

[tinman]

You bring up lots of valid points wrt "efficiency" of a JT type flashing/pulsing light. There are many many variables to be considered, both in the circuit itself (like number of turns, core material, impedance of power source, etc) and in the behaviour of the LED, the lightmeter, etc. As far as I can tell this lightmeter is doing a pretty good integrating of the 9-13 kHz flashing of the LED so I mostly believe in its readings. I don't have "threshold" operation levels yet, that is I don't know the low voltage limit of the two circuits. It seems to be the case that the supply impedance affects the efficiency (my definition, Lux/watt input power) but I'll have to do more testing to see how much. (Power supply vs battery have different impedances.) Then there is the matter of the turns ratio of the inductor. Then again, the different waveforms that the oscilloscope sees on the input current measured across the 0.1 ohm CSR in the two cases may be affecting the calculations of the average input power, although I've made the computation using DMM measurements and they agree with the scope measures to within 10 percent or so.

So nobody should take my measurements as "solving" the issue or supporting either side of the argument _at this point_. At best, I am trying to establish a reasonable testing protocol that can be used to track down the effects of the various variables in the experiment.

@ TK

Please know that these two circuits are not the set parameters to this !friendly! competition.
As far as im concerned,both of those circuits are quite poor efficiency wise.

Also,could you post a schematic of your power measurement points,as circuit 1 is much harder to calculate,as the battery is also in series with the inductor and LED,and so some of that light output also includes energy from the battery,and so that must also be taken into account when calculating P/in. Circuit 2 is much easier to calculate total P/in,as it excludes the battery during the off part of the cycle(transistor open). So when you are calculating P/in,in circuit 1,you must also include the energy that the battery supplies to the LED during the off time of the transistor,as it is in series with the other current source--that being the inductor,and so the LED is driven by both the stored energy in the inductor,plus some from the battery as well.

This gets even harder to calculate,as during the off time of the transistor,the battery is also trying to recover a little as well,which can also created the elusion that the battery is actually being charged during the off time of the transistor.


Brad

Thanks

[MileHigh]

Great test TK,and of course the results are as expected.
As  i pointed out to MH,the second circuit eliminates the losses associated with charging the battery also,when a battery is used in place of your PSU. The battery will produce more waste heat,when being discharged and charged continuously,and also there are the internal resistive losses that grow as the battery voltage drops.

Brad

Nope, what you say above doesn't make any sense at all.  First of all, you had a serious serious case of scrambled brains when you wrote that.  It's a double whammy.  I will unscramble it first, and then explain to you how it is still "scrambled" and doesn't make any sense at all.

For starters, I am going to put on my secret decoder ring:

<<<
As I pointed out to MH, the second circuit eliminates the internal resistive losses associated with discharging the battery during the transistor OFF cycle when the LED is being lit.  In the first circuit the battery will produce more internal resistive losses when being discharged continuously.  For both circuits the internal resistive losses increase as the battery voltage drops.
>>>

Brad, your text is so messed up that I am not even sure if I decoded it properly.

Moving on, you are not seeing the forest for the trees.  It doesn't really make a difference with respect to the internal losses in the battery for either circuit, your logic is flawed.  In BOTH cases, when power is being drawn from the battery, there are internal losses in the battery.  The fact that the current draw for the second circuit toggles ON and OFF makes no difference, there are STILL internal losses in the battery when the transistor is ON.

Look at TK's numbers.  The first circuit draws roughly twice the amount of current and produces roughly twice the amount of light.  If you could change the timing in the first circuit and slow it down so it draws the same amount of current as the second circuit, then chances are the LED would be about the same brightness.

So, supposing that you do that, then what do you have in terms of internals losses in the battery for each case?  Well, the current draw is the same so presumably the internal losses in the battery are approximately the same.

I am not going to speculate on further subtleties, that would have to be investigated for real on the bench.  The point that I am making is that your logic is flawed and doesn't make sense.  You have to think these things through and not just jump on what you first think is true.

MileHigh

[tinman]

 author=MileHigh link=topic=16225.msg488141#msg488141 date=1468100075]

Well, I can easily see Brad having a braingasm from TK's posting so I will get a few words in edgewise.

Oh MH--please.
No need to be so childish. I know circuit 2 is more efficient through bench tests,and also for reasons i have already given to you regarding I/R battery losses.

One of the classic weaknesses on the forums is to use the term "efficiency" without even defining what it means.  Brad is someone that does this all the time.

When it comes to the JT,then it would be maximum light out as determined by visual reference-per watt in. It would also be the ability to produce actual maximum light out per watt in,and last but not least,be able to drain as much as the remaining battery energy as possible.

My line of thinking was as follows:  In circuit #1 when the LED is lit it is based on a discharge of the LED in series with the battery.  So the EMF from the battery is a "helper" to keep the LED lit.

That is correct,and i suspect that TK did not take this into account when making his calculations on P/in in circuit 1,as the battery is also delivering energy to the LED during the off time of the transistor,as it is in series with the inductor and LED. When this is added to the P/in calculations,you will see circuit 1's efficiency drops right off.

So, it suggests to me that circuit #1 may be able to extract more energy from a nearly dead battery because it looks like it will run at lower battery voltages than circuit #2.  That will likely translate into a longer run time from the same battery.

That would be easy to test.

How did I define "efficiency" for the "most efficient Joule Thief?"

The answer was the most efficient Joule Thief by my definition would be the one that has the longest run time and extracts the most possible energy from the nearly-dead battery.

I stated this to Brad multiple times but it never sank in.

LOL--now there is a big load of crap MH.
Was it not me that suggested using the J/FET ,in order to make a JT that would extract most of the remaining energy from a battery,only to have you say that that made no sense at all.
Was it not me that posted videos showing how low i could get the voltage,and yet still have the JT circuit running-->remember your death spike saga ?. Would not a JT circuit capable of running on the lowest voltage,not extract the most remaining energy from a battery?.

In fact,your statement above is absolute rubbish,and those that were a part of that JT thread will know your talking crap. When most of us tried to explain to you the relevance of having a variable resistor on the base,so as it could be adjusted as the battery voltage dropped,you just got your back up,and said that the JT circuit has a fixed 1k ohm resistor to the base,as it is designed to run on 1.5 volts.

So stop your crap talk and lies MH,as people can see that what you are saying is absolute rubbish.

Quote post 254: 
And the pen is mightier than the bench - it is.  Suck on that and stop sounding like a miserable sourpus.

Quote post 286:

The pen is mightier than the bench Brad and my words have real meaning.

Quote post 304:
That all sounds fine and dandy except for the fact that if I wanted to do it I would need to work on a bench to iterate on a design.  Since I have no bench and no desire to do it, it's not going to happen.  I can't just put something on paper without testing it and iterating on it myself.

Now we can go back into the holding pattern waiting for Brad's expected tsunami braingasm.

Well some one had one MH,and im pretty sure it was not me

Seems your pen is no longer so mighty MH.


Brad

[tinman]

Nope, what you say above doesn't make any sense at all.  First of all, you had a serious serious case of scrambled brains when you wrote that.  It's a double whammy.  I will unscramble it first, and then explain to you how it is still "scrambled" and doesn't make any sense at all.

For starters, I am going to put on my secret decoder ring:

<<<
As I pointed out to MH, the second circuit eliminates the losses associated with discharging the battery during the transistor OFF cycle when the LED is being lit.  In the first circuit the battery will produce more waste heat when being discharged continuously. For both circuits there are the internal resistive losses that grow as the battery voltage drops.
>>>

Brad, your text is so messed up that I am not even sure if I decoded it properly.

Moving on, you are not seeing the forest for the trees.  It doesn't really make a difference with respect to the internal losses in the battery for either circuit, your logic is flawed.  In BOTH cases, when power is being drawn from the battery, there are internal losses in the battery.  The fact that the current draw for the second circuit toggles ON and OFF makes no difference, there are STILL internal losses in the battery when the transistor is ON.

Look at TK's numbers.  The first circuit draws roughly twice the amount of current and produces roughly twice the amount of light.  If you could change the timing in the first circuit and slow it down so it draws the same amount of current as the second circuit, then chances are the LED would be about the same brightness.

So, supposing that you do that, then what do you have in terms of internals losses in the battery for each case?  Well, the current draw is the same so presumably the internal losses in the battery are approximately the same.

I am not going to speculate on further subtleties, that would have to be investigates for real on the bench.  The point that I am making is that your logic is flawed and doesn't make sense.  You have to think these things through and not just jump on what you first think is true.

MileHigh

MH

You really are lost,and it is becoming more apparent each day as to how little you understand the simple JT circuit.
The battery is also a resistor,and as the voltage drops,the internal resistance grows,and so do the I/R losses associated with that battery.

In the second circuit,these losses are only had when the transistor is on.
In the first circuit(your circuit) these battery I/R losses are not only included when the transistor is on,but they are also included when the transistor is off.

During the off time of the transistor in circuit 2,the current loop excludes the battery,and thus the losses associated with the batteries internal resistance.
In the first circuit(your circuit) the battery becomes part of the current loop during the off time of the transistor,and so also includes the I/R losses associated with the battery.

Why you find this so hard to understand-i guess we will never know.\


Brad

[MileHigh]

Brad:

Like I just did an unscramble and decode of what you said in posting #324, your logic is flawed and makes no sense.

I seriously doubt that TK made a mistake in his power measurements.  It doesn't make any sense.  All that he had to do was scope the voltage and current from the power supply for both circuits.

Was it not me that suggested using the J/FET ,in order to make a JT that would extract most of the remaining energy from a battery,only to have you say that that made no sense at all.

Like I said, I think the chances are nearly zero that that will ever happen.

In fact,your statement above is absolute rubbish,and those that were a part of that JT thread will know your talking crap. When most of us tried to explain to you the relevance of having a variable resistor on the base,so as it could be adjusted as the battery voltage dropped,you just got your back up,and said that the JT circuit has a fixed 1k ohm resistor to the base,as it is designed to run on 1.5 volts.

I was not talking crap at all.  I tried desperately to explain to you what the true reason for having a base resistor was but you would have nothing to do with that, it was pure willful ignorance on your part.

So stop your crap talk and lies MH,as people can see that what you are saying is absolute rubbish.

You are just talking stupid gratuitous foolish idiocy, you need to blow off some steam and thar she blows!

Seems your pen is no longer so mighty MH.

My pen is absolutely fine.  You need to work on thinking straight and being able to string five sentences together that make sense.  Those are some of your biggest problems.

MileHigh