Tesla coil + Blocking-oscillator as driver ("Slayer device") - testing Pin, Pout

[PhysicsProf]

Quick background -pls see this vid from earlier today:
http://www.youtube.com/watch?v=1hCLOkODTQs

Running the Tesla coil at about 660 KHz output, 30.7V DC @15mA input --  with a one-transistor blocking oscillator to drive the primary coil.  Then I light up LED lamps by placing them along SIDE the secondary coil!   See what you think -- lots of fun is what I think!

http://www.youtube.com/watch?v=YEzdUq8_iFs&feature=youtu.be

Question --
1.  Pinput -- is approx. I*Vin, since close to DC, but there is ripple on the 30V bias... Looking at it,  I think this ripple would provide a fairly small correction to I*Vin.

2.  Poutput is more difficult to measure, approx 660 KHz, and I would appreciate suggestions.  I have some ideas... later.

ps -- waiting for build or re-build of "friction heater" by Gabriel...

[PhysicsProf]

Schematic attached.  I think this is fairly standard for Slayer exciters, but if you have any ideas how to further REDUCE the idling power draw, pls let me know. 
The primary is approx 29 turns of ~ 12 AWG copper. The secondary is approx 1280 turns of 32 AWG.

[PhysicsProf]

ps -- I know Slider has done a lot of work in this area and related wireless transmission and perhaps others of this team also.... and I look forward to discussion -- and hope to contribute to measuring Pin and Pout.

[Itsu]

Hi Steve,

nice build, and amazingly low input.
I build some Kacher coils, but they pulled several amps and heated up the transistor considerably.

Would love to build this too, but am working on some other stuff right now, maybe later.
Any info on the used transistor/diodes?

I think the coil/osc is radiating a steady output and whether this radiated output is being "wasted" (or better "used up") in the radiated waves or being used to light up some led lamps is indifferent to the input.

I do agree that the low input (0.5W) also puzzles me, but don't get fooled by the brightness of fluro's and led lamps when feeding them with HF.

Thanks,  regards Itsu

[PhysicsProf]

Hi Steve,

nice build, and amazingly low input.
I build some Kacher coils, but they pulled several amps and heated up the transistor considerably.

Would love to build this too, but am working on some other stuff right now, maybe later.
Any info on the used transistor/diodes?

I think the coil/osc is radiating a steady output and whether this radiated output is being "wasted" (or better "used up") in the radiated waves or being used to light up some led lamps is indifferent to the input.

I do agree that the low input (0.5W) also puzzles me, but don't get fooled by the brightness of fluro's and led lamps when feeding them with HF.

Thanks,  regards Itsu

I have further reduced the input to 3 mA @30V, so about 0.09W, and it still lights the LED lamps with no change in input power...
I did this by increasing the 68Kohm R1 to 330 Kohm (using a pot.).

Here's the vid:  http://www.youtube.com/watch?v=pTOMWZsir1w&feature=youtu.be

Interesting stuff, and I hope you'll jump in Itsu and @all are welcomed!

MJL21194 was used, and two 1N4001 diodes (turns given above).  I will be playing with the turns on the primary also...

[ion]

May I suggest a simple method to test the output power? Testing with LED light output is too arbitrary for my taste. I like precision measurement.

A more exacting method would be to use a 1/4 Watt resistor in this manner:

Attach one of the thermocouples from your Fluke dual thermometer with tape to the resistor and insulate the combination from ambient temp with some styrofoam, and tape it all together well.

Now you can calibrate your thermal power meter using a known input from your power supply. Temperature rise above ambient versus power input will be recorded at several power inputs. Don't exceed 1/4 Watt input.

Now having "characterized or calibrated" your thermal power meter you can proceed to connect the resistor to your DUT and get a reasonable value of power output indication by noting the temperature rise above ambient and comparing it to your previously constructed table of data points.

The resistor will accurately integrate any input, pulse or abnormal waveform AC or DC into a heating value. You can scale this up with a larger resistor if necessary, but 1/4 watt should be ok for such flea power systems.

If your input causes the Fluke to have erratic readings, use a simple thermometer or isolate the Fluke electrically with a ground foil.

If you enjoy measurement as much as I do, you will appreciate the simplicity of this arrangement. It beats trying to guess from the brightness level and non-linearity of an LED.

[PhysicsProf]

ION -- yes, I've been thinking of doing this also.  What R-value would you suggest (ohms)?

Main question -- Not quite sure how to draw power from the field around the secondary.  I could simply connect one end of R to the HV "top" wire and the other to ground; that might work best.  But wirelessly -- I could try some kind of antenna to harvest in the near field, connect that to one end of R and the other to ground... 

[ion]

ION -- yes, I've been thinking of doing this also.  What R-value would you suggest (ohms)?

Main question -- Not quite sure how to draw power from the field around the secondary.  I could simply connect one end of R to the HV "top" wire and the other to ground; that might work best.  But wirelessly -- I could try some kind of antenna to harvest in the near field, connect that to one end of R and the other to ground... 

For best power transfer the "R" value should be impedance matched to the coil. While this may be difficult to ascertain you can get a rough idea from the voltage output of your coil e.g. if your open circuit coil puts out around 100 volts, a 40 k resistor would be in the range of 1/4 watt and would probably be a reasonable load. E^2 / R.

The idea is to get in the range and try a few values until you think you have a good match to your system.

A tuned antenna system would also work but even in the near field you won't be harvesting all of the radiated energy so it would be a qualitative measurement only. Best to measure directly across the secondary.

[Itsu]

Steve,

i put something together this evening, but need to know some more, like what direction your coils are, both CW, CCW or different?

Also how is the L1 coil hooked up to the collector, bottom side or the top side?

It makes a hugh difference with me.

As your transistor is an audio transistor, i wonder if you not better should use one with a higher frequency rating.

Anyway fun stuff, and comparable with the kacher circuit when hooked up L1 a certain way (streamer on top L2 coil, overheated transistor etc.)

Regards Itsu

[PhysicsProf]

Answers in CAPITALS, NOT SHOUTING.

Steve,

i put something together this evening, but need to know some more, like what direction your coils are, both CW, CCW or different?
SAME, CCW (MEANING THE WIRE TRACES A CCW PATH ON BOTH)

Also how is the L1 coil hooked up to the collector, bottom side or the top side?
BOTTOM CONNECTED TO + RAIL, TOP TO COLLECTOR

It makes a hugh difference with me.

As your transistor is an audio transistor, i wonder if you not better should use one with a higher frequency rating.


Anyway fun stuff, and comparable with the kacher circuit when hooked up L1 a certain way (streamer on top L2 coil, overheated transistor etc.)

I THOUGHT I WOULD HAVE AN OVERHEATED TRANSISTOR, SO CHOSE THE BIG ONE WITH OPTION TO HOOK TO ALUMINUM CHANNEL -- BUT THERE WAS NO PROBLEM AT ALL; SO YES, I MAY TRY OTHER XSISTORS.

Regards Itsu

STEVE

[TinMan]

Hi Steve
Looks like a simple and interesting setup.I like how the current draw didnt rise when lighting the LED's.
So i think i might give it a shot tomorrow.I will try and scale it up a bit though-if i have enough wire lying around.

[Chet K]

Steve
Have you got a figure on your coil impedance ?

The Fluke method sounds very good for out put !!

ION's suggestion here
Quote,
May I suggest a simple method to test the output power? Testing with LED light output is too arbitrary for my taste. I like precision measurement.

A more exacting method would be to use a 1/4 Watt resistor in this manner:

Attach one of the thermocouples from your Fluke dual thermometer with tape to the resistor and insulate the combination from ambient temp with some styrofoam, and tape it all together well.

Now you can calibrate your thermal power meter using a known input from your power supply. Temperature rise above ambient versus power input will be recorded at several power inputs. Don't exceed 1/4 Watt input.

Now having "characterized or calibrated" your thermal power meter you can proceed to connect the resistor to your DUT and get a reasonable value of power output indication by noting the temperature rise above ambient and comparing it to your previously constructed table of data points.

The resistor will accurately integrate any input, pulse or abnormal waveform AC or DC into a heating value. You can scale this up with a larger resistor if necessary, but 1/4 watt should be ok for such flea power systems.

If your input causes the Fluke to have erratic readings, use a simple thermometer or isolate the Fluke electrically with a ground foil.

If you enjoy measurement as much as I do, you will appreciate the simplicity of this arrangement. It beats trying to guess from the brightness level and non-linearity of an LED.
-----------------------------------------------

Thx
Chet
PS
I spoke briefly with Brad about a similar method his group uses during pulse motor efficiency tests.
they utilize a one ohm resistor on the battery output to the  pulse motor,after that I am not quite sure how
he utilizes the other one ohm resistor he mentioned?

However understanding the method to match and scale the data so it is appropriate for each application would be a huge gift to this community,as well as help advance and focus true "anomalies'.

[Itsu]

Ok, made a replication, using:

2x 1n4001 diodes
BU508AF transistor
56K Ohm base resistor
prim. coil 22 turns awg 26 ccw
sec. coil  900 turns awg 13 ccw
30V/30mA input (getting a streamer on top of L2 coil when cranking up the voltage to 40V)

Frequency = 615Khz, but when putting my hand closer it seems to flip over the double that,
like it flips from 1/2 wavelength resonance to 1/4 wavelength resonance.

Looking into the resistive measurement ION proposed......

Video here:  http://www.youtube.com/watch?v=KFV6ZOgE9Iw&feature=youtu.be


Regards Itsu

[Itsu]

Followed up on IONs proposal for a calibrated 1/2 W resistor.
Used a 330K Ohm one as i need to measure high voltages from this BOT thing.

Still the main problem is that it impacts the output way to much (perhaps use a 1MOhm resistor?).

Also the need for better input measurements is there as the 30V / 0.01mA my PS shows is way to inaccurate

Video here: http://www.youtube.com/watch?v=kzna4gG4kvY&feature=youtu.be

Regards Itsu

[ion]

Followed up on IONs proposal for a calibrated 1/2 W resistor.
Used a 330K Ohm one as i need to measure high voltages from this BOT thing.

Still the main problem is that it impacts the output way to much (perhaps use a 1MOhm resistor?).

Also the need for better input measurements is there as the 30V / 0.01mA my PS shows is way to inaccurate

Video here: http://www.youtube.com/watch?v=kzna4gG4kvY&feature=youtu.be

Regards Itsu

Hi Itsu

Be aware that your Tesla coil will put out high voltages at  microampere or sub microampere current levels. As such it will load down with even high value resistors. Very little heating will be produced.

There is also a danger of arcing through the resistor to your temperature meter and burning out the input amplifier. I do not recommend using the resistor power measurement technique for such high voltage devices. It is nearly impossible to insert a ground shield and still get reliable temperature values.

We will have to devise a better method, but I surmise your T coil output will be on the same order of magnitude as the input power to your driving circuit minus losses thereof.

If you must measure the output power, I would do it this way: Create a string of resistors of high value e.g 100k to one meghohm. The total string must be able to support the terminal voltage of your T coil. At the bottom of the string and going to ground you will place a 1000 ohm resistor. Now you will read the current flowing through the 1000 ohm by measuring the voltage across it. You can also compute the ratio of resistors above the 1000 ohms to determine the top terminal voltage.

Knowing the loop current and terminal voltage you can now compute power output.

Your meter may have difficulty reading the HF AC signal across the 1000 ohm so you may have to use your scope. I'll try to post a schematic for you.

[Itsu]

I do not recommend using the resistor power measurement technique for such high voltage devices.

Thanks for the warning ION, i thought that the purpose of your post #5 above was to propose a method to measure Steve his output of his Tesla coil which probably puts out more HV (1280 turns vs mine 900 turns) then mine.

Anyway, i will await your diagram and see if i can use that method, but don't feel pressed about it, i will see it coming  :-)


Regards Itsu

[PhysicsProf]

Thanks for your ideas, Ion, and for your replication, Itsu!!

See attached photos:  I've played with this until I get a bright LED bulb for ~60mW input to the entire circuit, 2mA @ 30.0V. 

There is the problem of HF high-voltage that Ion points out, for using the resistor-heat method.  Then there is also the problem of determining where to hook up the resistor -- note from the second photo, I get best results (brightest) when I hook up a lead wire AND a coil/inductor/antenna touching the +on the base of the LED bulb.  There is "no connection" to the other bulb-terminal.

Also, the input power does NOT vary (measurably) when I pull the LED bulb business out of the field -- 60mW seems to be approximately what is used by the circuit; but I need a more accurate ammeter to tell for sure.

Suggestions welcomed...

PS -- I can read the light output in LUX with a light meter; but the readings seem to vary -- I think the light meter is affected by the HF field...

[ion]

Thanks for the warning ION, i thought that the purpose of your post #5 above was to propose a method to measure Steve his output of his Tesla coil which probably puts out more HV (1280 turns vs mine 900 turns) then mine.

Anyway, i will await your diagram and see if i can use that method, but don't feel pressed about it, i will see it coming  :-)
Regards Itsu

Hi Itsu

Although my post was a reply, it was more of a general purpose procedure for low power measurement. It can be used with such a device if you are aware of the dangers.

A more exacting method for HV measurement and power computation is in the schematic I supplied in my reply to you. It may have been missed as I uploaded it after you read the post.

To sum it up, you create a HV divider resistor string. This will have a ratio of applied voltage (top terminal) to measured voltage (1000 Ohm resistor). Voltage measured across the 1000 ohm converts directly into milliamps.

When we know the milliamps flowing in the series string, it is easy to compute terminal voltage (by the divider ratio) and power distributed in each resistor by Ohms law.

e.g. if our divider string ratio is 100 to 1 ( a total 99, 000 ohms total  in the top resistors and 1000 in the R1) we measure the voltage across the 1000 Ohm R1 and multiply by 100 to get our top terminal voltage. Then  Power Output = E^2/R, where R string = 100,000 ohms total.

To efficiently transfer power to a load, the load should be impedance matched to the source. Depending on how any particular load extracts power from the source depends on this factor. This can be approximated empirically by adjusting your top string of resistors and recomputing power in the total string. You will then determine if you are trending towards or away from a perfect match.

[PhysicsProf]

  Looking for a good "True RMS" meter.  Found 2 on sale at Amazon; comments?

[ion]

 Looking for a good "True RMS" meter.  Found 2 on sale at Amazon; comments?

For the type of circuits you are testing (pulse circuits) it will be important to know the frequency bandwidth (error vs Freq) and crest factor capability. A very high input impedance would also be a plus.

The Fluke is only good to about 1 kHz Trms AC volts and amps , couldn't find crest factor in the brief.

[PhysicsProf]

The frequency is about 650KHz -- way above the capability of these meters it seems...

So how do we measure the output energy?  without using light output, which I guess seems questionable, although the number of photons has got to tell us something ??

See attached photos:  I've played with this until I get a bright LED bulb for ~60mW input to the entire circuit, 2mA @ 30.0V. 

There is the problem of HF high-voltage that Ion points out, for using the resistor-heat method.  Then there is also the problem of determining where to hook up the resistor -- note from the second photo, I get best results (brightest) when I hook up a lead wire AND a coil/inductor/antenna touching the +on the base of the LED bulb.  There is "no connection" to the other bulb-terminal.

Also, the input power does NOT vary (measurably) when I pull the LED bulb business out of the field -- 60mW seems to be approximately what is used by the circuit; but I need a more accurate ammeter to tell for sure.

Suggestions welcomed...

PS -- I can read the light output in LUX with a light meter; but the readings seem to vary -- I think the light meter is affected by the HF field...

[ion]

The frequency is about 650KHz -- way above the capability of these meters it seems...

So how do we measure the output energy?  without using light output, which I guess seems questionable, although the number of photons has got to tell us something ??

You could use the light output provided it is spectrally matched to your photometer, and if you create a calibration curve for it.

Take a look at the divider circuit I posted for Itsu. It can be optimized to impedance match to the coil to measure power.

Also note that the thermal resistor method priorly proposed will integrate into heat just about any frequency until you approach the inductive component of the resistor, at which point it will be absorbing energy and returning some reactive energy to the source. This is way up in frequency though, depending on the type of resistor.

You could also put a number of taps on the secondary of your coil and select the one that lights the lamp the brightest. This is impedance matching.

Itsu:
After reviewing your video it is easy to see that the top terminal of your coil is very high impedance causing the signal collapse when loaded with your 330k resistor.

Perhaps your divider string will need to be in the megohms total resistance in order to provide a better match and maintain oscillator amplitude. You can expect a progressive collapse as you load the coil more and more heavily. This is normal. Trick is to find the sweet spot (best impedance match) for max power transfer. A tedious job.

The high impedance secondary top terminal of a Tesla coil provides a nice capacitive coupling to the surrounding air when a CFL is touched to it. Very difficult to measure in this mode though. You could put a shunt resistor on the low side and try to measure voltage across that, but without differential probes, the scope ground will load the circuit skewing it.

[Itsu]

ION, thanks for the diagram and explanation, i have to go looking for some 1MOhm resistors  :-)

Meanwhile i made some scope measurements on the input, measuring the input current via a CSR of 0.1 Ohm in the return lead and the voltage with the other probe.

The current is highly ac (20mA) in the 1.2Mhz range, so i doubt a DMM Will be able to measure it correctly.

I also put my HV probe on the open output lead of the L2 coil, and measure 4160V p2p at 495Khz compared to ground.

My HV probe specs says it has 100MOhm impedance.

Here the video: http://www.youtube.com/watch?v=tQCvk21aapg&feature=youtu.be

Regards itsu

[ion]

ION, thanks for the diagram and explanation, i have to go looking for some 1MOhm resistors  :-)

Meanwhile i made some scope measurements on the input, measuring the input current via a CSR of 0.1 Ohm in the return lead and the voltage with the other probe.

The current is highly ac (20mA) in the 1.2Mhz range, so i doubt a DMM Will be able to measure it correctly.

I also put my HV probe on the open output lead of the L2 coil, and measure 4160V p2p at 495Khz compared to ground.

My HV probe specs says it has 100MOhm impedance.

Here the video: http://www.youtube.com/watch?v=tQCvk21aapg&feature=youtu.be

Regards itsu

OK ITSU , this tells us a lot.

First we find the RMS voltage from the Peak to peak by dividing the 4160 V by 2.8 which equals 1485.7 Volts RMS

Now we square the 1458.7 and divide by 100 megohm load of your probe (P = E^2 / R). This comes out to approximately 0.2207 Watts dissipated in your probe acting as a load.

Now we look at the input which you say is 20 milliamps at 30 volts or 0.6 Watts.

Your COP is Pout / Pin or 0.2207 / 0.6 or COP = 0.3678 Efficiency is 36.78 %

Check my math for errors, I did this quickly.

It is very difficult to read the voltage across the 0.1 ohm resistor properly. This voltage would be exaggerated by slight inductance in the resistor itself, clip lead inductance and the ground lead of your scope, which also has inductance. It requires very careful layout and build with a ground plane and removing the ground clip lead entirely and using the ground very close to the tip of the probe with a tiny piece of wire. It is not easy to get accurate readings at these frequencies.

You will probably need a better way to read the input current accurately as you are fluctuating down in the least significant digit of your power supply ampere readout, but these values are as expected for an air core transformer operating at high frequency. The rest of the losses are in the transistor switching etc.

P.S. I didn't know you had a high voltage probe so no need to build the HV divider.

Edit : Should be 0.0212 Watts, cheap calculator truncation error.

[Itsu]

Check my math for errors, I did this quickly.

Hi ION,

when i do the calculations (P = E^2 / R) i end up a factor 10 lower:
1458.7^2 = 2127805.69 divided by 100000000 = 0.0212W

This makes the COP very low:
Pout / Pin or 0.0212 / 0.6 or COP = 0.0354 Efficiency is 3.5 %

On the other hand i do not trust the input (0.02) on my PS amp indicator, as i made some wrong calculations on the current presented by the scope.
We see 220mV p2p over 0.1 Ohm which means (i=U/R) =220/0.1 = 2200mA = 2.2A (not 22mA)

But as you said, this csr method with 1.4Mhz signals need better laid out techniques, so i will redo that with my probe tip directly across the csr (which should be almost induction free) like you proposed in the "DSRD pulse generator" thread (post #16)


Regards Itsu   (editted the cop to read 3.5%)