Joule Thief

[jadaro2600]

I am confused.

There are other things about this drawing that are different, but you have now put the base resistor where most of us have been putting it all along.

I see there are other differences like the cap and diode parallel to the led which is still in the expensive bjt position. and that cap at the base, well, [edit.] I guess it was gadget that did something like that to his fuji that ran the cfl for 12 hours.

I think way back jesus posted his circuit of his charging jt. I think Bill has already posted it on the circuits pages early, like page 2, maybe.

But really Jadaro, get rid of that led. It just takes power from the battery. A secondary of almost any number of turns - like even one- will light the led for virtually nothing. In fact usually, the amps use goes down when you add the led on the secondary. It certainly is less to run it on the secondary than that bjt place.

Try it, You'll like it! 

jeanna

(I have been installing software today. It is a welcome relief to look at a jtc!)

I actually modified thw wrong circuit diagram and still posted it, it was late.  I will create a new drawing - my problem is all my circuit diagrams are 'saved-as' from an original...I need to do some serious diagram work before I can repost my idea.  It's too late for me to withdraw the design at this point.

Jadaro:

I would not worry too much about the transistor being overdriven.  Gadgetmall said he sometimes uses 3 or 4 transistors wired together for protection.  I just saw a youtube vid of a guy that had a simple Bedini motor and, God only knows why, it wasn't turning all that fast, he had something like 50 of the transistors all mounted on a board and wired together.  I am guessing in parallel so I guess that would not raise the operating level that the transistors function, as far as start voltage.  If you are interested, let me know and I will try to hunt up that vid.  Maybe Al can chime in here with a thought or two.

Bill

He probably had that many in series to drop the voltage across the multiple components - which means jack when a high voltage pulse slams into a PN junction wall ...it would still have the first junction to deal with.  But I've been advised on the characteristics of components, they're capable of handling a bit more than what's reported, in well-manufactured pieces, that is.

-weather is bad here, will post more later;

[MrMag]

i think all coil will have 120 10 amp each.

You said your self 1 for 1 ratio so there equal, no gain here. but that is with 1 primary and one secondary , but the laws say the current is the same in all the coil , so 10 amps is 10amps in each coil , it doesn't say to devide the output per secondary, you do !

Can you prove that ? That is the question.
There is nothing that show the the amps is divided by number of secondary .

Mark

In a transformer, power in = power out. They are rated in volt amps. 120 volt in at 10 amps is 1200 volt/amps. The output no matter how may secondaries can only supply you with this amount be it 12 volts at 100 amps or 1200 volts at 1 amp.
If what you were saying is true, everyone would have a transformer in their house.

Sorry, but this is basic transformer theory. How would you like me to prove it? I would be more than happy to prove it to you but since this is basic theory, I think it more important for you to prove to me otherwise.
I really hope you can. I will then run out and buy a transformer for my house.

Tim

[Mk1]

@MrMag

Basic,lol . its all i can say.

If the transformer ration is 1 to 1 , why would you get less? 1 for 1 on every secondary coil , what magic law get the current down per/coil , where is that in that basic law that also say all coils are the same .

Think about it before answering.

Mark

[MrMag]

@MrMag

Basic,lol . its all i can say.

If the transformer ration is 1 to 1 , why would you get less? 1 for 1 on every secondary coil , what magic law get the current down per/coil , where is that in that basic law that also say all coils are the same .

Think about it before answering.

Mark

First off, 1:1 ratio is in regards to the number of windings which reflect the voltage in vs. voltage out. I think you need to do some further research. I am sorry if it seems that I am picking on you, i am not. I jst want to set the record straight to those who may not know.

Quote from Electrical Power System and Transmission Network.

As you know, the amount of power used by the load of an electrical circuit is equal to the current in the load times the voltage across the load, or P = EI. If, for example, the load in an electrical circuit requires an input of 2 amperes at 10 volts (20 watts) and the source is capable of delivering only 1 ampere at 20 volts, the circuit could not normally be used with this particular source. However, if a transformer is connected between the source and the load, the voltage can be decreased (stepped down) to 10 volts and the current increased (stepped up) to 2 amperes. Notice in the above case that the power remains the same. That is, 20 volts times 1 ampere equals the same power as 10 volts times 2 amperes.

quote from the link you posted

Current and Voltage Ratios.
In a well-designed induction coil the energy in the secondary, i.e., the induced current, is for all practical purposes equal to that of the primary current, yet the values of the voltage and the amperage of the induced current may vary widely from the values of the voltage and the amperage of the primary current. With simple periodic currents, such as the commercial alternating lighting currents, the ratio between the voltage in the primary and that in the secondary will be equal to the ratio of the number of turns in the primary to the number of turns in the secondary. Since the energy in the two circuits will be practically the same, it follows that the ratio between the current in the primary and that in the secondary will be equal to the ratio of the number of turns in the secondary to the number of turns in the primary. In telephony, where the currents are not simple periodic currents, and where the variations in current strength take place at different rates, such a law as that just stated does not hold for all cases; but it may be stated in general that the induced currents will be of higher voltage and smaller current strength than those of the primary in all coils where the secondary winding has a greater number of turns than the primary, and vice versâ.

Note that the use of the word current is not used to mean amperage but total power.

I hope this will satisfy you, or at least all others. Does anybody else disagree with me ?

[Mk1]

@MrMag

I am not feeling picked on , don't worry about that , but i am afraid you are missing my point.

But thank for trying.

Mark

All laws are with 2 coil design and not many people saw the need to put more coil since , they can easily understand it.
But its all there.