Joule Thief

[xee2]

@ jadaro2600

This doesn't solve the diode problem,

The voltage across the diode when it is forward biased (turned on) is only about 0.7 volts. The only problem possible when forward biased is too many amps. But almost all diodes will handle 1 amp pulses. The problem cones when the diodes are reverse biased. The reverse breakdown voltage can be increased by putting diodes in series. But putting diodes in series will also increase the forward voltage required to turn them on. The 1N4007 will block reverse voltages up to 1000 volts. To handle larger reverse voltages just put several diodes in series. Two 1N4007 in series will handle 2000 volts of reverse bias and have forward drop of about 1.4 volts (it will take 1.4 volts to get current to flow through the two diodes in series).

I do not think diodes are the problem. Transistors can not handle large collector - emitter voltages. This is specified as the Vceo voltage. Exceeding this voltage can destroy the transistor. The Joule thief generates high voltage pulses at the collector, this is most likely source of problems.

[innovation_station]

finished winding the MK HV!

turned out ok this is wound on the same cores as all other mk coils i have made ...  same amount of jt primary turns   5 turns ..

i have not ran it yet but i will mount it today .. and compleat it and well lets light some CFL'S.......     or charge some self timing caps .... 

you see i have this awsome hv cascade built ....  just not BRAVE .....TO FIRE IT UP ......   this is where the DANGER RESIDES!!!! HVDC CAPS ....

when you discharge through a gap ....   and can varry the distance of the gap ... 

does this effect conserved power !?!?!?!?   lol

ist   

now this leads to a cap spark gap young effect ELECTRIC motor idea im working on in my mind ...   hv but ....  well    few can handle simple things ..  so why go too wild with it yet

[weri812]

Hello All

nice work going on.

@jadaro2600
edited wer

50 WV DC means that is the maxuming working voltage that you can use with that device without burning it up. you all ways want to use less voltage or large WV [ working volts] to not burn up your device. This really applies to capacitors real nasty when they explode. In some cases you have to look at max amps this also can  be a mess crash and burn.

God Bless All
wer

[TheNOP]

I will show you an example of one of my favorite sites. Barry's coil gun java applet. In this example the volts are 100v and on the graph the amps are shown as the wave height.
In ac the electron charge goes first one way then another way. This is why I made the connection to the back emf caused by the collapse that produces the spike.

oops.
spikes, not spark.
sorry about that.

bemf is caused by the colapse of a mag field.
and if i interpret Lenz's law correctly, it oppose the colapse.
so it try to keep the current going in same direction, rising the volts trying to do that.
velocity of the collape is what i think give the volts boost.
i might be wrong tho.

yes, what you call spike are known as peek
E=IR = I(R*R)
the I and E are true rms values unless peek is stated.
Ep is not the true power.
it is a maximum power value reached for a short period of time, not a value that is constant.

you are right when thinking it is probably a peek current that light the jt leds
even if the rms approximation values we measure at the output of our diode bridge is less then what the manufacturer state for their leds.
the frequency of these peek current is what give us the impression that the leds are always on and is also playing a role in the brightness of the leds.

[TheNOP]

Can someone explain what the notation 50WVDC max means?

50 Watts Volts DC perhap ?
usually there are no words abbreviated to a single character in a part number, only units type.
but it really depends in what context this was taken from.

@any / all
Measuring the current must be done by inserting the ammeter into the circuit.  ..just put it inline upfront.   I don't want to insult though - I had to be told this - I was getting odd readings.  Sometimes inserting the ammeter can cause the circuit to behave differently.

I = V / R ohms law http://en.wikipedia.org/wiki/Ohm's_law
with this you can get the amps used in a circuit.
put a 1 ~ 10 ohms resistor in place of where you would normally hook your ampmeter.
measure the real value of your resistance first.
read the voltage drop at the resistor then calculate.