Joule Thief 101

[SeaMonkey]

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]

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]

@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]

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]

@ 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