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Overunity Machines Forum



Joule Thief behavior question.

Started by Legalizeshemp420, October 24, 2013, 03:04:05 AM

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0 Members and 2 Guests are viewing this topic.

crowclaw

Quote from: ddredar on November 12, 2013, 03:34:31 AM


Also, can someone tell me if there is a way to measure the amp output of the circuit?  My meter just reads 0.0.  I'm assuming its because it cant keep up with the oscillation speed of the circuit.
Hi, ddredar and welcome.

Xee2 has answered your question well, if you wish to experiment further with these or similar circuits, I would strongly suggest you invest in an oscilloscope. You should be able to pick up a decent one reasonably cheaply which will give you the opportunity to analyse the waveforms and measure the signal amplitudes etc. Digital meters don't perform well when inputing these type of signals.
Regards Crow


TinselKoala

That's right.

It appears that there is still some confusion about the LED ratings. If your package says "3.3 v, 28 mA" that means that the LED will produce its rated light output at a DC current draw of 28 mA. The LED is never a "direct short", in fact it is a non-linear load to the power supply. When it is supplied with a voltage below its "forward voltage" then it looks like an open circuit and no current flows. At the "fwd voltage" a current begins to flow, and at voltages slightly above the fwd voltage the LED begins to look like a short but with that fwd voltage drop included. To calculate the current through the LED you do an Ohm's Law calculation. I = V/R, for example. Around a circuit, the whole supply voltage is "dropped" (Kirchoff's circuit rules). But the LED drops the voltage some already, so your series resistance drops the rest.
(Supply voltage - LED fwd voltage)/series resistor = current in the circuit.
So if you have a LED with a 3.3 volt fwd voltage, and a 12 volt supply, and you want 28 mA in the LED (to get the rated brightness) you do R = V/I :
(12 -3.3)/0.028 = about 310 Ohms. 330 Ohms is a standard value for a resistor, so use that and accept a slightly lower current in the circuit.
The above is for the DC case.

Pulsed LEDs are the same... but different. The LED can withstand a lot more voltage if the pulse is kept very short. For example I have made some LED strobes to freeze motion of my MHOP pulse motor, and they are getting a full 12 volts pulse with no series resistance, but this is very short, so the LED doesn't fail. It also doesn't get very bright!

So the JT works like this: it makes high voltage pulses when not loaded. The capacitor-diode arrangement will charge the cap to the maximum output voltage of the JT, given enough time. When you hook up the LED, though, the voltage will only go as high as the fwd voltage of the LED, because then the LED looks like a "short" (clipping the voltage rise of the JT), and the light comes on. It might be very bright, but only for a short time, then the JT shuts off until it makes another pulse. The eye "integrates" the bright flashes into what looks like a constant light at some brightness less than the maximum. The eye is a very non-linear "brightness" measurement and what you perceive is a complex function of the true pulse brightness, the duration of the pulse and the time between pulses.

To measure the current during these very short pulses, the proper way is to use an oscilloscope to monitor the voltage drop across an inline resistor in series with the LED. A one-ohm resistor will show a voltage drop across it that can be directly converted to the current by Ohm's law. I = V/R, where the V is the voltage drop read on the scope, and the R is 1 ohm... so the volts indicated = amps. A 0.1 ohm resistor will disturb the circuit less, but then you need to be able to divide by 0.1 instead of 1 !
But this current only flows for a short time out of the total process. Say it flows for 1/3 of the whole cycle as shown on the scope. The _average_ current for the whole cycle  then is 1/3 of the value you have calculated based on the voltage drop across the "current sense resistor" CSR, sometimes called CVR (current viewing resistor, the term I prefer).

Nowadays, even cheap DMMs can be surprisingly accurate at "average" current and voltage measurements even in pulsed circuits. There are videos and documents available that show this to be true. Poynt99 has published a very good demonstration/test of this, but I can't find the link at the moment.

ddredar

Thank you all for the information.  Even though I don't completely understand this circuit, I do have a much better idea of what is going on.  Would any of you have a recommendation for an oscilloscope?  I will probably order it online due to the horribly over priced local electronics store.  Also are there any documents or threads you would recommend for reading?

Legalizeshemp420

Quote from: ddredar on November 13, 2013, 01:38:43 AM
Thank you all for the information.  Even though I don't completely understand this circuit, I do have a much better idea of what is going on.  Would any of you have a recommendation for an oscilloscope?  I will probably order it online due to the horribly over priced local electronics store.  Also are there any documents or threads you would recommend for reading?
Analog or digital?  I personally would recommend a used analog 60mhz, or above, for 50-100 dollars.  Get a 100mhz analog scope for 75-100 and that is all you would need for most JT work.

ddredar

Quote from: Legalizeshemp420 on November 13, 2013, 02:46:07 AM
Analog or digital?  I personally would recommend a used analog 60mhz, or above, for 50-100 dollars.  Get a 100mhz analog scope for 75-100 and that is all you would need for most JT work.


Oh my, please tell me where you are finding those prices.  When I look online, most of them are in the $600 to $5000 range.  I found a few in the $250 area, but not too many and they are not good for 60mhz or higher.