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  • boostedbb
    replied
    diodes

    I've been following this thread for months. I'm finally getting ready to try to build a circuit for my truck. I can't seem to decide which or how many diodes I am going to need. I see the nte517's that gmeast is using, but after pricing them at nearly $7.00 a piece x 32 I was hoping there would be a version that would be as effective at a substantially less cost. How many and which "cheap" diodes would work as well as the four parallel nte517's?

    Thanks for the help.

    Warren

    Leave a comment:


  • insane4evr
    replied
    In the plasma spark circuits, the idea of connecting 1N5408 diodes in a string is to divide equally the total reverse voltage among them. So, if the voltage is 25,000 and the strings are 25 diodes, each diode theoretically will be exposed to 1,000 volts.

    However, if each diode has different reverse voltage leakage currents, then the 25,000 volts will not be shared equally. The diode with the lowest leakage current will be exposed to the highest voltage being shared.

    ON semiconductor rates their diode at reverse voltage of 1000 volts at 25 deg C with leakage of 10 uA max. and at 150 deg C with leakage of 100 uA max. No values given for typical or minimum. So, in a mix of diodes, each one can have different leakage currents from each other.

    Example only: 10,000 volts and 10 1N5408 string, first diode has leakage of 1 uA, next 2 uA, etc. For analysis only, at reverse voltage condition, let us replace each diode with a calculated resistance value.

    diode D1: R1 = 1000/.000001 = 1000 megohms
    diode D2: R2 = 1000/.000002 = 500 megohms
    diode D3: R3 = 1000/.000003 = 333 megohms
    diode D4: R4 = 1000/.000004 = 250 megohms
    diode D5: R5 = 1000/.000005 = 200 megohms
    diode D6: R6 = 1000/.000006 = 167 megohms
    diode D7: R7 = 1000/.000007 = 143 megohms
    diode D8: R8 = 1000/.000008 = 125 megohms
    diode D9: R9 = 1000/.000009 = 111 megohms
    diode D10: R10 = 1000/.0000010 = 100 megohms

    Sum of R2 thru R10 = 1929 megohms. Add R1 = 2929 megohms. Reverse current for string is 10000/2929000000 =3.4 uA. Voltage across R1 or D1 = I x R = .0000034 x 1000000000 = 3400 volts. This is over 3 times its rated reverse voltage and will fail usually in shorted condition. Then the next diode could fail like a domino effect. End of example.

    Now if each diode has the same value resistor across it lower than 100 megohms but still high enough not to affect performance like 22 megohms or 10 megohms. then the effect of differences in leakage currents will be minimized making the 1N5408s survive. Just my analysis, I don't know when I can test.
    Last edited by insane4evr; 12-07-2008, 04:36 PM.

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  • magnetO
    replied
    @gmeast

    Originally posted by gmeast View Post
    Hi everyone,

    I'm not getting the point of this thread, but I will say that in my present circuit I am discharging 350 VDC in a time similar to the trace I posted earlier. I use 4 NTE 517's in parallel. Almost no heat. Also look at the VexUs circuit and note the direction of the diode. It is like the Tero arrangement but only one diode and an air gap ... counter to others.

    Greg

    Hi Greg,

    thanks for your contris.

    Yes, the standard ignition in my car does require the "inverse" direction of the diodes (diode string) too.
    And due to the resistance wires and dissi air gap, the second diode(s) is/are not necessary too.

    Interesting, you only need 1 x 15kV in series (4 parallel) diode to block HV-spark.

    magnetO,
    who immediately needs a scope!
    Last edited by magnetO; 12-07-2008, 02:35 PM.

    Leave a comment:


  • magnetO
    replied
    @Kinetix

    Originally posted by Kinetix View Post
    Second, I don't think your diodes died cause of the high current. It is more likely it is too high voltage to blame. I used a string of 1N4007 (max 30A) and they had no problem with the heat (as you know, Q=IxIxt, and that means that the generated heat depends firstly upon the current). The high voltage killed them. There were only 25 diodes in my string (that could block only 25kV), and my ignition coil was capable to output as much as 35kV... When I accelerated too fast (recquiring more kV from the ignition coil), they died.
    There is another issue here. The diode string has a certain resistance in itself. Their reistance is not zero. There are maybe a few tens of Ohms, and that lowers the current running through them when capacitor discharges.
    For example, if your capacitor is charged at 400V and the diode string resistance is only 10 Ohms, the current couldn't reach more than 40Amps.

    Hope these thoughts will help.
    All the best.

    Hi Kinetix,

    will do a little edit on the old post regarding spark duration. All you are saying seems right.

    Started the latest postings, because I have some problems with 2 of my 4 diode strings.

    Now, I know how to read scope screens , and the fastness of capacitive discharge. And because a lot of diodes (in my case 4 x 25 1N5408) do have more resistance than a single one or two (4 x (1..2) NTE517 e.g.), and are less expensive, I tried the 25 diode strings

    Both strings on cylinder 1 and cylinder 2 already run for over 500km, but never had a working string for more than 100km on cylinder 3 or cylinder 4. So, looking for solution.

    As you already have described, it seems the diodes die because of not being capable to block high voltage anymore. Dying always begins when already having driven some km and going "full throttle", then missfire occurs. When going back to "partly throttle" missfire disappears in the first time, but getting worse and worse.

    Will buy some NTE517 and put it on cyl.3 and cyl.4. I think will add two 517 inline with resitor (few ohms, ~50Watt).


    Does the demanded Voltage really go that high up during "full throttle" in your car, 35kV?
    Does in contrary Greg's bug stay below 15kV?
    Interesting!

    magnetO

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  • gmeast
    replied
    isolation

    Originally posted by Gibs View Post
    Hi Greg,

    The transformer itself provides enough isolation already, I don't understand why it should be isolate again.


    Gibs
    Hi Gibs,

    Yes, you're right. On the circuit posted I mistook the doubler on the output as part of the core circuit. It's naturally isolated ... perfect for our application. I've ordered the complimentary transistors to build the circuit ... can't wait.

    Greg

    Leave a comment:


  • Gibs
    replied
    Isolation

    Originally posted by gmeast View Post
    Hi Mike,

    I'm not sure right now, but the secondary is hooked to ground via a diode in Phil's inverter. That leaves us without isolation. What's your take on that?

    Greg
    Hi Greg,

    The transformer itself provides enough isolation already, I don't understand why it should be isolate again.


    Gibs

    Leave a comment:


  • ashtweth
    replied
    Hi All. decided to do some more runs , we just switched over to the super deluxe booster (Wouters -3LPM at 30amps) and have had these plugs using Rev'z circuit just over 4 weeks.
    Using the circuit with the hydroxy the first circuit was using Rev's normal configuration(50uF), we found a tiny bit of where. The second one, we use put the cap value down to 20uF.
    Little more wear as you can see from the original in the PDF, these will be in there shortly, we still enjoyed more power and extra 5% at least on top of the normal hydroxy booster
    ImageShack - Image Hosting :: rimg0309tw6.jpg
    ImageShack - Image Hosting :: rimg0310mo4.jpg
    Gonna keep these plugs and give them a hiding on the GEET ready next weekend
    Last edited by ashtweth; 12-07-2008, 02:43 AM.

    Leave a comment:


  • gmeast
    replied
    Nte517


    Hi everyone,

    I'm not getting the point of this thread, but I will say that in my present circuit I am discharging 350 VDC in a time similar to the trace I posted earlier. I use 4 NTE 517's in parallel. Almost no heat. Also look at the VexUs circuit and note the direction of the diode. It is like the Tero arrangement but only one diode and an air gap ... counter to others.

    I have never lost any NTE517's either in lab development or vehicle testing. The diodes are not seeing the current that is at the spark gap. The diode sees voltage and not current in these circuits. The spark gap self-modulates the power thereby "producing" the current "within" the spark gap. The conversion of power (volts to amps) happens "BY" the spark ... not the other way around.

    It's very hard to explain.

    Greg

    Leave a comment:


  • Ansis Freimanis
    replied
    I agree with Kinetix

    Originally posted by Kinetix View Post
    Friend, I think there is a little misunderstanding here...
    First, I don't think the electrons run faster in a plasma type discharge than in a regular one. They have almost the same speed allways.
    It would be better to speak about the plasma being "short duration" than being "fast". But there is a "fast" phenomenon taking place here: it's the discharge of the capacitor through a low resistance circuit. That doesn't mean that the electrons are faster this case...
    And it's true that a regular spark (made by a transistorized ignition) is "longer" than a capacitor discharge one, and that means that it lasts longer. And I think it's also true that even between capacitor discharges there is a difference made by the capacitor's value. 1uF discharges faster than 100uF on the same resistor, I think.
    You can hear bigger bang and louder sound in plasma discharge cause it's about more energy per time unit. That's all. That's why bigger caps make bigger bangs when discharges on the same circuit, being charged at the same voltage.
    Second, I don't think your diodes died cause of the high current. It is more likely it is too high voltage to blame. I used a string of 1N4007 (max 30A) and they had no problem with the heat. The high voltage killed them. There were only 25 diodes in my string (that could block only 25kV), and my ignition coil was capable to output as much as 35kV... When I accelerated too fast (recquiring more kV from the ignition coil), they died.
    There is another issue here. The diode string has a certain resistance in itself. Their reistance is not zero. There are maybe a few tens of Ohms, and that lowers the current running through them when capacitor discharges.
    For example, if your capacitor is charged at 400V and the diode string resistance is only 10 Ohms, the current couldn't reach more than 40Amps.

    Hope these thoughts will help.
    All the best.
    It’s correct. I think there is something what we do not understand completely. It is summary of HV AC pulse+ LV DC pulse. Maybe this big resistance of HV diode strings make a something like "resistance = heat" situation. And I think these HV diodes is made from material which is not good for that operation.
    P.S. MagnetO, make corrections!

    Leave a comment:


  • Kinetix
    replied
    Friend, I think there is a little misunderstanding here...
    First, I don't think the electrons run faster in a plasma type discharge than in a regular one. They have almost the same speed allways.
    It would be better to speak about the plasma being "short duration" than being "fast". But there is a "fast" phenomenon taking place here: it's the discharge of the capacitor through a low resistance circuit. That doesn't mean that the electrons are faster this case...
    And it's true that a regular spark (made by a transistorized ignition) is "longer" than a capacitor discharge one, and that means that it lasts longer. And I think it's also true that even between capacitor discharges there is a difference made by the capacitor's value. 1uF discharges faster than 100uF on the same resistor, I think.
    You can hear bigger bang and louder sound in plasma discharge cause it's about more energy per time unit. That's all. That's why bigger caps make bigger bangs when discharges on the same circuit, being charged at the same voltage.
    Second, I don't think your diodes died cause of the high current. It is more likely it is too high voltage to blame. I used a string of 1N4007 (max 30A) and they had no problem with the heat (as you know, Q=IxIxt, and that means that the generated heat depends firstly upon the current). The high voltage killed them. There were only 25 diodes in my string (that could block only 25kV), and my ignition coil was capable to output as much as 35kV... When I accelerated too fast (recquiring more kV from the ignition coil), they died.
    There is another issue here. The diode string has a certain resistance in itself. Their reistance is not zero. There are maybe a few tens of Ohms, and that lowers the current running through them when capacitor discharges.
    For example, if your capacitor is charged at 400V and the diode string resistance is only 10 Ohms, the current couldn't reach more than 40Amps.

    Hope these thoughts will help.
    All the best.
    Last edited by Kinetix; 12-06-2008, 08:58 PM.

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  • magnetO
    replied
    a 200 VDC drop in a few nanoseconds

    Hi Greg,

    thanks for the screen shot. You also say, 1 div = 1 square, so you are three. Seems to be so.

    That means:
    Spark duration is fast (~100us)
    Plasma spark duration or capacitve discharge duration goes much faster on. (depends on capacitance, and resistance).

    In my case there is about 1.6Ws per discharge.
    VDC is 440V;
    I = W/t *1/U, when asuming t == 11us, getting I = 330A (its a peak I)
    this is asuming only 10 times normal spark, but Lee's PLASMA scope shot already showed, it's maybe faster. And you also say its only few nanoSeconds, not thousand, or even some us.

    So, this peak i is higher, probably much higher than Peak Forward Surge Current of diodes. No matter if 1N5408 (200A) or NTE 517 (200A). (Average Forward current is not the matter, max. 180mA @ 6000rpm.)

    As I have written above, I "like" destroying diodes, and dont know why. Having a "chain" of 25 x 1N5408 on cylinder 1 and cylinder 2. Both were never an issue.
    But the 2 other one's....

    Will look for NTE517, and mount it on, no far away from cylinder 3 and 4.


    magnetO
    Attached Files
    Last edited by magnetO; 12-07-2008, 02:09 PM.

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  • Ansis Freimanis
    replied
    to Greg

    Yes the discharge is very fast- it is like shoot with fire arms- sound barrel is broken, and we can hear "bang".
    P.S. In oscilloscope we can’t see that, but we see absolutely precise- like a cut with super sharp knife- discharge- absolute "zero" in nanoseconds, from 174 Volts...
    Cheers!

    Leave a comment:


  • Ansis Freimanis
    replied
    to MagnetO

    I will say- intensity of discharge. In moment when you start hear loud sound - than your spark energy is more than speed of sound- 343 m/s.
    P.S. I am not engineer of sound barrier, but it is true- You hear it with your ears.
    Be happy!

    Leave a comment:


  • gmeast
    replied
    EXAMPLE: VexUs Charge / Discharge Trace

    Originally posted by insane4evr View Post
    Volts/div or Time/div:

    On the scope screen (graticule) markings, use the squares, not the tiny division lines on the cross-hair. These tiny division lines are fractions of your volts/div or time/div.

    It is like your tape measure. It has numbers every centimeter which have larger lines but it also has smaller lines every millimeter but no numbers. Not sure if this clarifies or confuses.
    Hi all,

    Here is an example of one of the charge / discharge curves from my 4-cylinder VW engine with the VexUs plasma spark. You can glean a HUGE amount of information about "what's goin' on" from a simple trace like the one below:

    The horizontal and vertical scales are:
    50 VDC for each major division / square on the vertical scale
    and
    .0025 seconds (2.5 ms = 2.5 thousandths of a second) per division / square on the horizontal scale.

    It shows a rather classic charge of the 'dump' capacitor (curvy part) and then the plasma discharge when the spark plug fires (nearly vertical). The plasma lasts until the voltage across the gap has diminished to about 10 - 20 VDC ... that's amazing to me. It only lasts a few nanoseconds (very steep slope)-a 200 VDC drop in a few nanoseconds !

    It shows a pulse period of (3 x 0.0025) = 0.0075 and that is a frequency of:
    1 / 0.0075 = 133 Hz.

    There are 4-cylinders so:
    120 / (4 x 0.0075) = 4000 RPM
    (the "120" is all of the constants and conversions reduced to a single term)

    ... just some real-world application sort of stuff. Peace,

    Greg


    Last edited by gmeast; 12-06-2008, 05:17 PM.

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  • magnetO
    replied
    110us and use squares. okay

    Originally posted by insane4evr View Post
    Volts/div or Time/div:

    Hi Ansis Freimanis,
    hi insane4ever,

    so you are saying both the same: answer is a).

    (Not the charging time for cap or frequency of re-charging is in the moment important to me. But spark duration, and Lee said plasma spark is 10 times or even more faster, but how fast exactly.)
    http://www.energeticforum.com/renewa...html#post26418

    Thank you
    magnetO
    Attached Files

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