Author Topic: Joule Thief 101 (Read 944419 times)

tinman · « **Reply #1170 on:** March 24, 2016, 07:39:10 PM »

author=picowatt link=topic=8341.msg478352#msg478352 date=1458842767]

Quote

I thought you two were "discussing" whether or not decreasing the base resistance increased the collector (and L1) current or just produced additional waste heat in the base resistor. Perhaps I misunderstood. As I said, I have no dog in this hunt.

My claim is that an increase in current flow through L2 will increase the magnetic field strength in the core. As we now know that the current flow dose indeed increase as we reduce the base resistance in L2,and we know that the LED puts out more light when we reduce the base resistance,then how can that not mean that the magnetic field strength has increased in the core?.
The only way to get more light output from the LED is an increase in magnetic field strength in the inductor,and this can only happen when we increase the current flowing through L2and the base of the transistor. So how MH can say that increasing the current flow through L2 dose not increase the magnetic field strength in the inductor is beyond me

This would of course depend on whether Q1 is fully saturated at both the lower and higher base resistance.
Measuring Vce and Ic at the two base resistances should help clarify is Q1 is fully saturated (turned on) at both base resistance values or is turned on harder with one more than the other.

Oh-dont go there with MH. There is no such thing as switching on the transistor harder in the JT circuit.

It is late here now,so i will get to those test as soon as i can PW.

Brad

MileHigh · « **Reply #1171 on:** March 24, 2016, 08:04:00 PM »

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You are going to drag him down with you,and the outcome will be he leaves because of your lying ways.

LOL

Look at post #1210, PW was courteous enough to find the reference for you.

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Oh-dont go there with MH. There is no such thing as switching on the transistor harder in the JT circuit.

Put your brain in gear and read PW's quoted text again. Plus I made a fairly detailed posting about the same subject.

Magluvin · « **Reply #1172 on:** March 25, 2016, 02:00:40 AM »

Guy shows output of microwave transformers set up with and without a capacitor. With capacitors, resonance is just above 50hz.

https://www.youtube.com/watch?v=XlG4NyQf3T0

Mags

SeaMonkey · « **Reply #1173 on:** March 25, 2016, 04:48:38 AM »

Quote from: Miles Higher

Put your brain in gear and read PW's quoted text again. Plus I made a fairly detailed posting about the same subject.

Quote from: TinMan
There is no such thing as switching on the transistor harder in the JT circuit.

Whenever a Power Transistor such as the 2N3055 is used in a
switching circuit there is always the possibility of driving the
transistor into (a) non-saturation, (b) soft-saturation or
(c) hard-saturation.

Hard saturation is preferred since it will result in best circuit
efficiency, providing it is accomplished without excessive (just
enough) base drive. The drawback of Hard-Saturation is
charge-storage within the base region which slows turn-off
unless some means of applying momentary reverse-bias across
the base-emitter junction is incorporated to speed up turn-off.

The base drive winding of the Joule Thief transformer will normally
accomplish that to some degree. Placing a speed-up capacitor
across the base-drive-resistor will assure that the reverse bias
impulse is sufficiently strong to make a difference.

By the way, most transistors will operate reasonably well in
their inverse mode where it is "upside-down" in the circuit.
The Emitter is connected as the Collector and the Collector
is connected as the Emitter. Current gain is not good in
this configuration but that it does work is sometimes made
use of in exotic circuits.

picowatt · « **Reply #1174 on:** March 25, 2016, 05:25:22 AM »

Quote from: SeaMonkey on March 25, 2016, 04:48:38 AM

SeaMonkey · « **Reply #1175 on:** March 25, 2016, 05:30:47 AM »

Please accept my apology for the mis-quote. Actually it was
TinMan who made the statement. Sarcastically, I'm quite sure.
The quote attribution in the offending message has been
corrected.

picowatt · « **Reply #1176 on:** March 25, 2016, 05:33:10 AM »

Quote from: SeaMonkey on March 25, 2016, 05:30:47 AM

Please accept my apology for the mis-quote. Actually it was
TinMan who made the statement. Sarcastically, I'm quite sure.
The quote attribution in the offending message has been
corrected.

No problem,
Thanks,
PW

tinman · « **Reply #1177 on:** March 25, 2016, 05:38:01 AM »

@PW

Below is a pic of the brand of 2n3055 i use,as it is what our local electronics store stocks.

I have carried out the V/EB breakdown voltage test,and for this particular brand of transistor ,it is 15.3 volt's before the voltage is clamped.
The test was carried out with a 10k ohm resistor in series,and gave a very clean and clear result.
I tried 4 out of the 10 transistors i bought ,along with the transistor being used in the circuit ATM,and all clamp the voltage at 15.3 volt's.

So now we have the answer as to why my reverse voltage spikes are around the 16 volt mark.

@ MH
Credit where credit is due,you hit the nail on the head with that one.

Now,this reverse current that is now flowing through the base resistor and L2--what is it doing?,where is it going?
Once that is worked out,then what happens when the base resistors resistance is reduced ?-where is this current flowing to?

We are looking at only the off period of the transistor at this point in time.

If we look at the schematic below,i have drawn in two options for L2s current path.
The red dots show the current flowing back into the battery.
The blue dot's show the current from L2 flowing into L1.
It is my belief that the later is correct,and that the current flowing through L2 flows into,and is additive to the current flowing through L1,and the LED.
This gives rise to an interesting situation,where we now have the current flowing through L1 and L2,where the flow of current in L1 is always in the same direction,but the current flowing through L2 during the off period is now flowing in the opposite direction through windings that are wound opposite to that of L1 from the common tap of the two windings(bat + in)

You have said-along with PW,that during the on time,the current flowing through L2 would subtract field strength from L1. But what about during the off period,where now the current flow in L1 remains in the same direction,but the current flow through L2 is now in the opposite direction. Would not the magnetic field being built by the current flow in L2 now be additive to the magnetic field that is now collapsing in L1. Could this be the reason for the brighter LED when the base resistance is lowered,in effect raising the current flowing through L2. It makes sense tome,as when we lower the base resistance,and increase the current flowing through L2 during the off period of the transistor,the LED on time(conducting time) increases.

I can test this by way of placing a CVR between B+ and the common connection of L1 and L2,and another CVR between L2 and B+ common. This will tell us if the current flowing from L2 is flowing into L1,or back into B+ during the off time.

Brad

SeaMonkey · « **Reply #1178 on:** March 25, 2016, 05:52:19 AM »

Quote from: Magluvin

Guy shows output of microwave transformers set up with and without a capacitor. With capacitors, resonance is just above 50hz.

https://www.youtube.com/watch?v=XlG4NyQf3T0

Mags

Excellent video Mags. That technique of placing a capacitor
in series with a high voltage winding was commonly utilized
in the Lamp Driver circuit where a Mercury Vapor Lamp was
used as a source of UltraViolet Emissions. The capacitor
effectively doubles the voltage peaks applied to the load
which, in the case of the Mercury Vapor Lamp, resulted in
reliable starting and improved operation. In the case of
the Arc in the video, the length of the arc is lengthened
considerably.

The arc is a pretty strong source of both UltraViolet radiation
and Nitrogen Oxides/Ozone. Sustained arcs such as that
in the video were once used in Norway to make Nitric Acid
and Nitrates for commerce.

tinman · « **Reply #1179 on:** March 25, 2016, 07:01:25 AM »

@ Mag's

Throughout my testing today so far,i have found that circuit 2 below is indeed more efficient than circuit 1. There is more light output for less P/in.

Testing another circuit ATM.
Will post findings soon.

Brad

sm0ky2 · « **Reply #1180 on:** March 25, 2016, 09:07:57 AM »

Do you see why?

Johan_1955 · « **Reply #1181 on:** March 25, 2016, 09:36:13 AM »

Quote from: sm0ky2 on March 25, 2016, 09:07:57 AM

Do you see why?

Yes, and even better with the NPN Collector to the +, and Emitter to the coil, so the NPN above the Flipped Coil, and a small C over CE junctions.

MileHigh · « **Reply #1182 on:** March 25, 2016, 11:37:13 AM »

Brad:

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Credit where credit is due,you hit the nail on the head with that one.

And look at what you first said:

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There is no death spike-you need to understand the circuit, and why there is a spike across L2 when the transistor becomes open. It all has to do with the number of turns on each coil-the more turns, the higher that L2 spike will be-there is nothing out of the ordinary with that spike.

Then you said:

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It is not a massive anomaly,and is present in every JT--as seen in Mag's scope shot's.
You simply cannot work out why it is of a higher value in my setup.

Quote

It's all coming back now to bite you on the ass--and you are the one showing your own errors lol.

It's a never-ending slog with you. It's like driving down a road with a continuous series of speed bumps, and even worse, "don't back up" road spikes pointed in your direction. I am not going to be a bobbing duckie for you in a straitjacket, but I am also not going to do this kind of debate forever. Once this is done I am jumping off the train and you can go back to doing your experiments and leading yourself down a garden path with your bobbing duckies all nodding in agreement. Or I can hope that you become more open minded and get over this obsession with always being right no matter what. That's for you to decide because I won't be around.

MileHigh

MileHigh · « **Reply #1183 on:** March 25, 2016, 11:38:51 AM »

Getting back to the matter at hand...

Your "red dot" current flow speculation is incomplete for starters, and wrong. If you are going to show current flow you have to show a complete loop of current. The blue dots are correct like you speculated, with the exception of the blue dots that lead to the LED and then stop and go nowhere. Like I said before the Joule Thief transformer is temporarily shorting itself out and burning off energy and stealing that energy away from the LED. This whole time your Joule Thief has been crippled.

Quote

But what about during the off period,where now the current flow in L1 remains in the same direction,but the current flow through L2 is now in the opposite direction. Would not the magnetic field being built by the current flow in L2 now be additive to the magnetic field that is now collapsing in L1. Could this be the reason for the brighter LED when the base resistance is lowered,in effect raising the current flowing through L2.

When you get the reverse current flow that is an energy burn where energy is burnt off in the N-P collector-base junction and also in the base resistor. So no go for a mechanism for making the LED brighter.

As I previously stated when you lower the base resistance, it's really a secondary effect that makes the LED brighter. If we assume that for both a 1k base resistance and a 500-ohm base resistance that the transistor is "fully ON" (a.k.a. "hard ON") then the rise in current through L1 when the transistor switches on is identical in both cases. However there is a secondary effect that makes the time period when the transistor remains on slightly longer, and that allows for a smidgen of increased current to flow through the L1 which results in an LED which is a smidgen brighter.

Now, we can't forget that your Joule Thief is not functioning properly. So it would be worthwhile to repeat the variable base resistor tests after the problem has been fixed and see what the results are. I am going to suspect we will see something similar with the LED slightly increasing in brightness for the same reasons. And of course when the problem is fixed presumably the LED will be brighter overall.

MileHigh

tinman · « **Reply #1184 on:** March 25, 2016, 12:04:56 PM »

Quote from: MileHigh link=topic=8341.msg478425#msg478425 date=1458902233

MileHigh

[quote

And look at what you first said:

MH
It is not a death spike at all.
It is in fact present in all JT circuit's--it's just that no one has taken any notice of it as of yet.
I have now tested 3 different JT circuit's,and it is present in all of them.
This !so called! death spike actually helps in raise the brightness of the LED.

Quote

It's a never-ending slog with you. It's like driving down a road with a continuous speed bumps, and even worse, "don't back up" road spikes pointed in your direction. I am not going to be a bobbing duckie for you in a straitjacket, but I am also not going to do this kind of debate forever. Once this is done I am jumping off the train and you can go back to doing your experiments and leading yourself down a garden path with your bobbing duckies all nodding in agreement.

You may do what you please MH,but as you only made this !death spike! discovery (that you didnt know existed until i showed some test results),it is clear that you do not really know all that is going on within the circuit.

Quote

Your "red dot" current flow speculation is incomplete for starters, and wrong. If you are going to show current flow you have to show a complete loop of current. The blue dots are correct like you speculated, with the exception of the blue dots that lead to the LED and then stop and go nowhere.

Im not going down this babying road with you MH. If you cannot work out where the current flow continues from where the marked dot's are,then you clearly do not belong here.

Quote

Like I said before the Joule Thief transformer is temporarily shorting itself out and burning off energy and stealing that energy away from the LED. This whole time your Joule Thief has been crippled.

In fact it is not.
The biggest waste of energy is actually in your 1k ohm resistor.
My testing today has confirmed this,and it also confirms my statements about being able to reduce the base resistance as the supply voltage drop's,that enables us to keep the LEDs light output at a constant(or near to)brightness.
Turns out that what i said is correct,in that the current flow in L2 dose actually add to the the current flow in L1 during the off period,and this is what is causing the LED to go brighter when the base resistance drop's.
So you see MH,the 1k ohm base resistor is actually dissipating power that could be otherwise used to drive the LED.

Quote

When you get the reverse current flow that is an energy burn where energy is burnt off in the N-P collector-base junction and also in the base resistor. So no go for a mechanism for making the LED brighter.

A contradiction on your behalf MH.
You clearly state that energy is being burned off in the base resistor--i say the same,and once that base resistance is removed,then that energy is added to the energy of L1,and sent to the LED.

Quote

As I previously stated when you lower the base resistance, it's really a secondary effect that makes the LED brighter. If we assume that for both a 1k base resistance and a 5000-ohm base resistance that the "fully ON" (a.k.a. "hard ON") then the rise in current through L1 when the transistor switches on is identical. However there is a secondary effect that makes the time period when the transistor remains on slightly longer, and that allows for a smidgen of increased current to flow through the L1 which results in an LED which is a smidgen brighter.

Partly correct,but only while the battery is still reasonably healthy. As the battery voltage reduces to around .8 volts(in my test setup),the secondary effect is from the current flow in L2.
As battery voltage drop's,the frequency rises,but as we reduce the base resistance,that frequency can be reduced,and the light output maintained--even though the P/in is now less.

N

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ow, we can't forget that your Joule Thief is not functioning properly.

My JT is functioning correctly. It functions the very same way the other two i constructed-with only very minor differences due to things like number of turns of each coil,and the size of the toroid core.
All three have the very same operational characteristics,and near identical scope traces.

Quote

So it would be worthwhile to repeat the variable base resistor tests after the problem has been fixed and see what the results are. I am going to suspect we will see something similar with the LED slightly increasing in brightness for the same reasons. And of course when the problem is fixed presumably the LED will be brighter overall.

There is no problem with the way the circuit is operating,nor the other ones i built today.
In fact,i will be presenting a JT circuit that is most efficient based around this !now found! reverse current flow.

Brad

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Author Topic: Joule Thief 101 (Read 944419 times)

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