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Author Topic: Janost's "Self-runner" Device: Replicating and Testing  (Read 180696 times)
Group: Guest
Draw is .032 A x 4.5 V = 0.144 watts
Output is .0086 A x 250 V = 2.15 watts
Output / Input = 2.15 / 0.144 = 14.93 more out than in - but I'm just comparing watts not Joules so maybe that means nothing.  I was asking.  I don't know all this in depth but I thought it sounded good and encouraging.  Almost too good so I was ASKING.  I do not know why you would take offense for that. 

I shorted the output to a multimeter,
And I measured the input.

But I did not claim anything.
   
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No worries janost.  Everything is cool   O0
   
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Never mind.
My orignal circut is still running after the first stop.

In my circuit the magic voltage is 2,81v

I dont know why.
   
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@Janost

About diode choice, it is not because your diode choice works that it means it is the best one to use. Very often out of 10 diodes, only one will give double the output then the others. I use what I call a diode carousel that enables you to switch from one model to the other very quickly to find the best output or effect you are looking for.

The other point again is frequency. You could be 1500Hz away from greatest result and never know it if you cannot hunt for the best frequency to run your device. This again could double your output.

Once you find the best frequency, you should then again hunt for the best diode just to make sure the one you have found is still the best.

About the 2.81 volts. Seems like at higher battery volts the system works but generates an unrecoverable loss, until it self stabilizes to a neutral condition.

From what I know of batteries, you can only recharge them up to 20% of the amp rating. Any more and it is wasted since the battery cannot store more in that time interval. Chances are if you put 2 sets of batteries in parallel, this will increase your amp rating and should increase the battery recharge ability to absorb more juice and may give you more power to work with at your neutral condition and may light your led brighter and now instead of 2.81 volts as the neutral condition, it could be higher. But of course it is all up for trials.

Keep on man.

wattsup



---------------------------
   
Group: Guest
I shorted the output to a multimeter,
And I measured the input.

But I did not claim anything.


If you short the output with a multimeter and measure 0.086A, then the O/P voltage cannot be 250V. To get a better idea use a range of load resistors and plot a curve to see maximum output voltage across the load resistor. you will see from this that the output power will be much lower than the input. Also, if you think that your earth lead is drawing up electrons, then fit a very low value resistor in series with the lead and scope across it to see if you have a voltage / current waveform that can be measured. Also, try to use good quality analogue meters, not DVM's when measuring HV/HF circuits.

Hoppy

   
Group: Guest
@Janost

About diode choice, it is not because your diode choice works that it means it is the best one to use. Very often out of 10 diodes, only one will give double the output then the others. I use what I call a diode carousel that enables you to switch from one model to the other very quickly to find the best output or effect you are looking for.

The other point again is frequency. You could be 1500Hz away from greatest result and never know it if you cannot hunt for the best frequency to run your device. This again could double your output.

Once you find the best frequency, you should then again hunt for the best diode just to make sure the one you have found is still the best.

About the 2.81 volts. Seems like at higher battery volts the system works but generates an unrecoverable loss, until it self stabilizes to a neutral condition.

From what I know of batteries, you can only recharge them up to 20% of the amp rating. Any more and it is wasted since the battery cannot store more in that time interval. Chances are if you put 2 sets of batteries in parallel, this will increase your amp rating and should increase the battery recharge ability to absorb more juice and may give you more power to work with at your neutral condition and may light your led brighter and now instead of 2.81 volts as the neutral condition, it could be higher. But of course it is all up for trials.

Keep on man.

wattsup



Why didnt I think of that?  :)

I'll try 2 or more batterys parallell.
   

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@ Hoppy,

Quote
Also, if you think that your earth lead is drawing up electrons, then fit a very low value resistor in series with the lead and scope across it to see if you have a voltage / current waveform that can be measured

I did that in my replications using a 0.1 Ohm resistor, and indeed there was a waveform across the resistor indicating current draw from ground.
It was mA's, but still.

Waiting for my 220V GDT's to arrive, but when available i will repeat this measurement and record it on video.

Regards Itsu
   
Group: Guest
If you short the output with a multimeter and measure 0.086A, then the O/P voltage cannot be 250V. To get a better idea use a range of load resistors and plot a curve to see maximum output voltage across the load resistor. you will see from this that the output power will be much lower than the input. Also, if you think that your earth lead is drawing up electrons, then fit a very low value resistor in series with the lead and scope across it to see if you have a voltage / current waveform that can be measured. Also, try to use good quality analogue meters, not DVM's when measuring HV/HF circuits.

Hoppy


If you compress a spring and let it go the exact force used to compress it comes out when it expands, right?
If you compress it and place it on the ground the same force comes out but it also jumps up in the air.

Why doesnt it only expand and stay on the ground?

Because the ground is a mass that pushes back against the spring.

The same goes for the ground-rod.

If you take a capacitor and connect one end to ground and the other end to a HV supply it will charge.
Same with a LED it will light even though the HV supply has no connection to ground other than through the LED.
 
   
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If you short the output with a multimeter and measure 0.086A, then the O/P voltage cannot be 250V.

If I put a 15w 230V bulb on the output it glows dimmly.

These type of vibrator inverters was used in the 50ties to make vacuumtube radios run on batteries.

   
Group: Guest
...
Same with a LED it will light even though the HV supply has no connection to ground other than through the LED.

There is always at some degree, a capacitive coupling between any conductor and the ground. For instance the winding conductor of a HV supply constitutes the "plate" of a capacitor and the other plate is the ground or any near conductor. Even if this capacity is low, on the order of pF, its impedance is not negligible at high frequencies or when signals have sharp transitions. This explains why a current can pass through a "single wire": an apparently open circuit is in fact never open but looped, there is always a capacitive connection to the ground or to a return path. As you seem to be skilled in electronics, I'm sure you understand what I mean and should agree, isn't it? So this explains why it is difficult to modelize your setup or predict its behavior: we would need to know the characteristics of the capacitive coupling of the "ground rod" with all other components around it, and this is not trivial (and the problem is even worse than it seems, because a measurement disturbs the circuit, a simple voltmeter or an oscilloscope probe adding parasitic capacities!)  :(.

   
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There is always at some degree, a capacitive coupling between any conductor and the ground. For instance the winding conductor of a HV supply constitutes the "plate" of a capacitor and the other plate is the ground or any near conductor. Even if this capacity is low, on the order of pF, its impedance is not negligible at high frequencies or when signals have sharp transitions. This explains why a current can pass through a "single wire": an apparently open circuit is in fact never open but looped, there is always a capacitive connection to the ground or to a return path. As you seem to be skilled in electronics, I'm sure you understand what I mean and should agree, isn't it? So this explains why it is difficult to modelize your setup or predict its behavior: we would need to know the characteristics of the capacitive coupling of the "ground rod" with all other components around it, and this is not trivial (and the problem is even worse than it seems, because a measurement disturbs the circuit, a simple voltmeter or an oscilloscope probe adding parasitic capacities!)  :(.



I know what you mean by capacitive coupling.
But the effect is larger than what could be explained with that.

And the input current is not affected when you load the output and should be if it was a capacitive coupling.

Care to explain my spring therory?
Why doesnt it just expand and stay on the ground?
What makes it jump into the air if the force is same as free expanding?
« Last Edit: 2012-08-03, 12:21:01 by janost »
   
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If I put a 15w 230V bulb on the output it glows dimmly.

These type of vibrator inverters was used in the 50ties to make vacuumtube radios run on batteries.



Your 15W bulb is not a short circuit. Put your DVM across the bulb (set on the same current range you used to measure the 0.0086 Amps), then with another DVM measure the voltage across the bulb. Also, measure the voltage across the bulb without the short circuit across it. It will not be anything close to 250V in either case.

Hoppy
   
Group: Guest
@ Hoppy,

I did that in my replications using a 0.1 Ohm resistor, and indeed there was a waveform across the resistor indicating current draw from ground.
It was mA's, but still.

Waiting for my 220V GDT's to arrive, but when available i will repeat this measurement and record it on video.

Regards Itsu

Itsu,

Is the current source from the ground or from the battery powering the circuit?

Hoppy

   
Group: Guest
...
Care to explain my spring therory?
Why doesnt it just expand and stay on the ground?
What makes it jump into the air if the force is same as free expanding?

I understand your spring theory but it modelizes only roughly the behavior of charges. I agree that you can see the ground rod as a reservoir of charges that would constitute a steady state of potential, or a voltage reference, on which charges could lean "for jumping", as a spring. This is approximately what is happening.
But not exactely. The charges in a conductor are very mobile. They don't constitute a block unlike the earth which is like a steady block allowing the spring to jump. Any change of the topology of charges near a ground conductor, for instance when the voltage of a HV transformer rises to a peek, influences the charges of that ground conductor. These charges move toward a new pattern minimizing the electric field and their potential energy (principle of least action). This change of the charge pattern of the ground conductor influences step by step (but at near the speed of light) the other charges of the environment including these of the circuit that were cause of the first change. This is equivalent to a current looping in a circuit and physically explains the electronics viewpoint of the circuit looped by the capacities that always exist between near conductors.

   

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Hoppy,

Quote
Is the current source from the ground or from the battery powering the circuit?

this current source is from ground (ground rod).

Regards Itsu
   
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Hoppy,

this current source is from ground (ground rod).

Regards Itsu

Itsu,

How do you know / be sure that the current is not being sourced from the battery and being capacitively coupled to ground through the earth rod?

Hoppy
   
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It's turtles all the way down
Due to the extreme non-linearity of a tungsten filament, lamps do not make good loads for circuits that may be very sensitive to load impedance.

In certain lamps, a cold filament may be in the order of milliohms, which can be a near short on the DUT. After it heats up a bit the resistance moves into the ohmic range and the energy can then have a more reasonable load to work into.

Oftentimes the DUT cannot provide enough surge amperage to overcome the cold resistance of the incandescent lamp.

For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms).

Twelve volt automotive lamps, such as brake llamps have lower initial resistance.

A variable resistor as the load can be a better help in impedance matching the DUT, and makes calculation of power easier.

In some cases though, an incandescent bulb will try to automatically find the impedance sweet spot, providing the DUT can provide the current to overcome the cold resistance and if the lamp is suitably sized to the DUT output.

In the final analysis, an incandescent lamp can provide "rough eyeball" indication of power output without DSO's true RMS multimeters etc. so are not to be ignored.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

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Hoppy,

Quote
Itsu,

How do you know / be sure that the current is not being sourced from the battery and being capacitive coupled to ground through the earth rod?


I don't, i only see a waveform across a 0.1 Ohm resistor in the ground path.
Normally this waveform across a 0.1 Ohm resistor comes from the current going through this resistor, but it could be inductively or capacitive coupled onto this ground wire/resistor, who knows.

It was just an observation i did after you mentioned to do this as an extra measurement.

When i have my GDT's i will repeat this and show the results.   

Regards Itsu   
   
Group: Guest
Wattsup, that's a brilliant idea about a Diode Carrousel.

Thanks for sharing.
   
Group: Guest
I'm going to run my ticker like this with a Lead-Acid battery and 2 ground-rods.
Pumping the 350V pulses through ground pulse-charging the battery.

Perhaps even using a GDT between the bridge-rectifier and the battery.

The 3 diodes are only used as a voltagedrop to the relay-vibrator.

   
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Janost,  Please tell us what kind of software you create your circuit schematics with.
Is there a place where we can download that schematic drawing software?

THANK YOU for sharing your great circuits.
   

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tExB=qr

If you take a capacitor and connect one end to ground and the other end to a HV supply it will charge.
 

Try it.
   
Group: Guest
Janost,  Please tell us what kind of software you create your circuit schematics with.
Is there a place where we can download that schematic drawing software?

THANK YOU for sharing your great circuits.


It is here

http://www.expresspcb.com/expresspcbhtm/Free_schematic_software.htm

And it is free for personal use.
   
Group: Guest
It is here

http://www.expresspcb.com/expresspcbhtm/Free_schematic_software.htm

And it is free for personal use.


Good software for free! PCB service is expensive for small quantity boards but reasonable on large runs.

Hoppy
   
Group: Guest
Good software for free! PCB service is expensive for small quantity boards but reasonable on large runs.

Hoppy

The really nifty thing about it is that if you want to make a PCB from your schematic it will do so automagically  O0
   
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