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Author Topic: Bedini SSG measurements  (Read 9081 times)
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This thread is about measuring the input and output of the Bedini SSG circuit under a series of controlled tests.
   
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Hi Fausto,

This is a great idea for a thread.  I would be happy to offer some suggestions.

MileHigh
   
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I would offer that a more general thread be started entitled: A Guide To Input / Output Power Measurements, of which the SSG could be a sub-category with variations on the theme specific to the SSG.

I would hate to see all MileHigh's good information get buried in the SSG thread and be unread once the SSG is exposed for what it is.


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

Let's start with measuring the power consumption of the motor.

If you use a digital multimeter on current measurement, and put it between the source battery and the Bedini motor, it should be quite accurate with one important limitation.   The limitation is that the pulse frequency of the motor must be above about 20 Hz.  The digital multimeter is very good at making an average current consumption measurement, as long as the pulsing current waveform is above about 20 Hz.  If the frequency of the pulsing falls below about 20 Hz, then you can expect that the display will become unstable.

So you can verify this by simply looking at the digital display.  If the digital display is stable and not jumping all over the place then you can be quite confident that the average current consumption is accurate.  If the display is not stable then you have to ignore the reading from the digital multimeter.

We all know that analog meters do a pretty good job of measuring the current consumption also, because they simply average the current because the analog needle movement itself acts as a low-pass smoothing filter.  However, analog meters themselves have their limitations.  If the current waveform has very sharp transients in it, I would be suspicious of the reading of the analog meter.  If you see the needle moving up and down, then it would be similar to the case of the digital multimeter and you must be careful.  The smart thing to do would be to act conservatively and ignore the readout because the needle is not stable.

In both cases you are really more interested in the power consumption of the motor.  If the battery has voltage fluctuations because of the pulsing load of the Bedini motor then you must put a very large capacitor in parallel with your source battery to smooth out the voltage fluctuations.  You have to verify this with your scope.  That way, when you measure the average current consumption of the motor, and you know that the battery voltage has been stabilized with a big capacitor, then and only then can you get an accurate power consumption calculation.

All of the above is standard practice, with the caveat that you must have stable displays on either type of meter.  It is a huge mistake to ignore this and just pick one of the changing numbers on your digital display or to mentally average a fluctuating needle display in your head.  In addition, for very irregular waveforms I think that an analog meter can give an inaccurate display, even if it is stable.

How do we overcome these limitations?  My suggestion is to uses simple techniques that make use of standard components which I will discuss in the next posting.

MileHigh
   
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So let's discuss a very simple technique to get a very accurate power consumption measurement for your Bedini motor.  There are many ways of doing this but let's start with one simple example.

Here is what you need:

1) A large electrolytic capacitor.  Perhaps something like 30,000 uF at 25 volts.  That might be the size of a 300 ml Coke can or larger.
2) A 1% precision resistor, let's say 100 ohms @ 1/2 watt.  Alternatively you can take a regular 100 ohm resistor and measure the resistance of it with your multimeter.
3) A bench power supply that has a variable output voltage from 0 to 25 volts.

With these components you are going to simulate a 12-volt power supply with the large capacitor.

The circuit is very simple:  You connect the output of the power supply to the 100-ohm resistor, and that is connected to the positive of your big capacitor.  The positive of the big capacitor is also connected to the power input of the Bedini motor.

You set your power supply to 12.6 volts and you charge up the big capacitor to 12.6 volts.  Then you connect your Bedini motor and get it running.

When you do this you are monitoring the voltage across the big capacitor.   You notice the voltage across the capacitor starts to drop.  So you compensate by increasing the voltage from your variable power supply.

Let's assume that after a few minutes, the Bedini motor is running normally at it's stabilized speed.  You have also stabilized the voltage in the big capacitor back up to 12.6 volts to simulate a normal battery.   To do this let's assume that the voltage from the power supply has to be set to 13.84 volts.

You have to check the voltage across the capacitor with your oscilloscope to see if it is relatively stable.  If the ripple voltage is less than +/- 0.1 volts, then let's say that the voltage is stable enough.  If there is too much voltage ripple, then simply add a second big capacitor in parallel to the first capacitor.

So how do you calculate the power consumption of the motor?  It is very easy, you know the voltage supplied to the motor, it is 12.6 volts.  And you know that the average current that you are feeding into the capacitor from the power supply is identical to the average pulsing current that is going into the Bedini motor.  The current in must be equal to the current out if the capacitor voltage is stable.

So the current is (13.84 - 12.6)/100  = 12.4 milliamperes.

Therefore the power consumption of the Bedini motor is (0.0124 x 12.6) = 0.156 watts.

Note that you don't actually measure 13.84 volts and 12.6 volts and do the subtraction.  You simply put your multimeter across the 100-ohm resistor and measure 1.24 volts.  You are taking advantage of the fact that the multimeter is more accurate when measuring lower voltages.

Let's check the power dissipation in the resistor = (0.0124 x 0.0124 x 100 ) = 0.015 watts.  So you are fine using a 1/2 watt resistor in this example and you are sure that it will not burn up.

Here is the big advantage with this setup:  It will be able to measure the power consumption of a Bedini motor or any other device no matter how crazy the current waveform is.  As long as your big capacitor is big enough to smooth out the voltage then it will work.  It will allow you to make very accurate power consumption measurements without worrying about the limitations of digital and analog meters.

The disadvantage is that you have to work with a variable voltage power supply and you have to wait a few minutes to let everything stabilize before you make your final measurements.  If you are doing research this should not be a problem.

MileHigh
   
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Ion:

Thanks for your suggestion but I prefer to leave the thread title as is.  It is better 'marketing' to have the measurements closely connected to Bedini motors.  It will draw more attention that way.

MileHigh
   
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Fausto:

Now I will discuss a method for measuring the power output to the charging battery for a Bedini motor.

The way a Bedini motor charges the charging battery is via a discharging inductor.  This is discussed in detail in the Bedini 10-coil Alternative Discussion thread on this forum so I will only mention the summary points here.  Anyone that is truly interested in understanding how their Bedini motor works should read that thread and try to understand all of the concepts discussed there.

A discharging inductor outputs a fixed amount of energy for each current pulse.  If the battery is sulfated and has a very high input impedance then the discharging inductor will output a sequence of short current pulses at high voltage for a low average current.  If the battery is new then the discharging inductor will output a sequence of longer current pulses at a lower voltage for a high average current.

Therefore, depending on the condition and resting voltage of the charging battery, the average current output by the drive coil of the Bedini motor can be high or low.  Therefore it does not make sense to measure the average current output into the charging battery.

I know that this statement will not seem to make sense to almost all of the Bedini experimenters, but it is a fact.  The sequence of discharge pulses of the drive coil of the Bedini motor will output a fixed amount of average power, but the average current in the current pulses can be variable, and the associated voltage spikes due to the current pulses can be variable.

Another fact in the above statement that may be surprising to new readers of this thread is that the discharging drive coil in a Bedini motor outputs pulses of current, and not pulses of voltage.  For more information please read the Bedini 10-coil Alternative Discussion thread on this forum.

If you have been working with your Bedini motor and have always thought that the drive coil outputs voltage spikes after the transistor switches off, it's time to take a fresh look at this issue.  The drive coil actually outputs current spikes, and you see voltage spikes on your scope as a result of that.

So we know that whether the charging battery is old and sulfated and has a high input impedance, or if it is new and has a low input impedance, the discharging drive coil of the Bedini motor will output a fixed amount of power into either type of charging battery.  Therefore the logical thing to measure is the average power output by the drive coil.

When the drive coil discharges, the current flows through the diode and then into the charging battery.  Therefore some of the output power is also dissipated in the diode as heat power.  When we make our measurement of the average output power of the Bedini motor's drive coil, we need to account for the average power dissipated in the diode also.  To do this measurement we are going to swap out the charging battery with a large capacitor in parallel with a resistor.  If we keep the voltage on the large capacitor approximately the same as the charging battery voltage, then the power dissipated in the diode will be approximately the same for the case when the charging battery is in the circuit.

So to measure the power output from the Bedini motor you do the following:  Swap out the charging battery for a large capacitor in parallel with a variable resistor.  Run the Bedini motor and watch the voltage across the large capacitor start to increase and then stabilize at a certain voltage.  Then adjust the variable resistor so that the capacitor voltage stabilizes at the normal voltage of the charging battery.  Measure the voltage across the large capacitor with your multimeter and make a note of it.  Then shut down the Beini motor and disconnect the variable resistor making sure that you do not change the setting.  Take your multimeter and measure the resistance of the variable resistor.  The average output power of the Bedini motor is then the capacitor voltage squared divided by the measured value of the resistance.  This measurement will be quite accurate and does not include the power that is dissipated in the diode.

It may seem strange to remove the charging battery from the Bedini motor to make a measurement of how much average power is going into the charging battery but it is valid.  The reason it is valid is because you know that a discharging inductor will discharge all of its energy independent of the load.  So the load can be a charging battery or a capacitor in parallel with a resistor, it does not matter.

So far in this thread you have been given some techniques to make precise measurements of the average power consumption of a Bedini motor and the average power output into the charging battery.   If you play with your base resistor and let the motor RPMs stabilize, for each setting of the base resistor you will get different readings for the average power consumption of the Bedini motor and the average power output into the charging battery.  In reality, you can make this measurement quite simple.  All that you have to do to get the maximum power output to the charging battery is play with the base resistor and look at the resulting voltage across the capacitor in parallel with the resistor that is in place of the charging battery.  When you find the base resistance setting that gives you the highest voltage across the capacitor, that is the base resistor setting that gives you the maximum average charging power.   More sophisticated experimenters would want to watch both the motor's average power consumption and the average charging power to find a "sweet spot" that factors in both of these variables.

MileHigh
   
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Yesterday I purchased more items to build a new 4 coil Bedini motor for this thread and today I purchased on EBay 2 large 15000uf 100v caps to do the filtering the MileHigh proposed for the measurements.

The coils have been ordered straight from Bedini's site so that I have the correct coil with 8 windings for 8 transistor per coil. So this is not a regular SSG but not a 10 "coiler" either. It will be middle ground.

I am still wondering about the load test if we should use 2 or groups of 3 or 4 batteries on front-end and back-end so rotation becomes possible, but we can leave that for after i have the machine build and the first proposed measurements done.

This is much fun and thank you MileHigh for your input and desire to teach.

Fausto.
   
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Hi Fausto,

That sounds great and I wish you good luck with your experiments.

A few comments about big capacitors used in the charging battery position.  Even though the spikes can be high enough in voltage to fire a neon when there is no load, some experimenters might feel the need to get capacitors that can sustain a very high voltage.  This is not truly necessary because when you put a fully discharged capacitor in place of a charging battery, the spikes from the coil start off at zero volts and start charging the capacitor from zero volts.  In other words the "voltage spikes" completely disappear.  I assume that some experimenters may not be aware of this. The reason for this of course is that they are not voltage spikes, they are current spikes.

So a 25-volt or 35-volt electrolytic capacitor is perfectly good for running these types of experiments also.  The advantage is that you can get a larger capacitance in the same sized capacitor can if the voltage rating is lower.

Whatever type of capacitor you use, there is a real danger for overcharging them to too high a voltage when they are charged by the coil's inductive current spikes.  You really have to be careful and watch the voltage.  Even if you have a resistor in parallel with the big capacitor, until you are familiar with your setup, you really should check the capacitor voltage frequently and ensure that you don't damage the capacitors.  I assume that if you over-voltage a big electrolytic can capacitor and it blows, it is very nasty and could be dangerous.

Finally, I sent you some emails about your circuit on OU for you to contemplate.

MileHigh
   
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Quote
I assume that if you over-voltage a big electrolytic can capacitor and it blows, it is very nasty and could be dangerous.

That is true if a low source impedance and high currents are available for the cap such as line operation off a bridge rectifier. The SSG is a rather high source impedance device, as such the big electrolytics will get very leaky and only charge up to and maybe a little beyond their rated voltage.

You can check this out by simulating the SSG with a resistor characteristic of the SSG DCR, and hook it to the power supply and cap. Start running the voltage up and at some point close to the voltage rating of the cap you should start to see the current increase. Don't stay there too long...not good for the cap but will give you a good idea how bad these caps are slightly above rated voltage.,

The area above the normal rating is the surge rating, typically 110% of DC rating,  a very leaky area that will cause heating effects and danger of blowing if sustained.

The SSG is a very poor motor / generator as such it's coupling from the primary battery through rotation and a very leaky magnetic structure should pose no problems for the cap with a suitable load resistor.


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

I am not quite sure what you mean by simulating the SSG's DC resistance and stuff like that.

Permit me to be more clear with the dangerous scenario that I was implying:

You have a 20,000 uF 35-volt electrolytic cap, and you put it in the charging battery position and there is no bleeder resistor across the capacitor, and the initial capacitor voltage is zero volts.

You then start the SSG and the drive coil starts pulsing current into the big cap.  So the voltage starts to climb on the cap in steps, and the higher the voltage gets on the cap, the smaller the voltage steps get.

You know the reverse breakdown voltage on the diode is going to be high, let's say 100 volts.  So if you leave the Bedini motor running like this unattended, then the cap will keep on charging and the voltage in theory can get as high as the diode breakdown voltage.

I am assuming that the 35-volt electrolytic cap will start to croak in the 40-ish volt range.  I just don't know how they will fail, it is something that I have never looked into.  All that I know is that they are a potential powder-keg, with lots of electrical energy stored up in them.

Plengo is getting 15,000 uF caps rated for 100 volts.  Let's check the energy at 110 volts, = 1/2 * 0.015 * 100 * 100 = 90.75 Joules.  It's not a huge amount of energy but it is easily enough to make a 1-watt low-valued resistor explode if it was placed across the terminals.  Just an ordinary piece of wire would probably explode if it was placed across the terminals.  So I am just wondering what would happen if the capacitor is sitting there charged to 109.8 volts, and then one more inductive spike from the coil comes along through the diode and puts the capacitor over the threshold.

So we are talking about some sort of self-explosion here, where the electrolytic dielectric fails and it has an internal short or something?  I don't know because I have never been there.  I can even imagine the cap spraying hot oil everywhere and going boom!  I just don't know.

MileHigh
   
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Yes,  MH, I haven't been exactly clear. Lets say the capacitor in question has a soft knee above it's rated voltage where it actually starts to dissipate energy due to the increase in leakage current. In some ways the electrolytic capacitor can be thus compared to an MOV (metal oxide varistor) as you exceed it's rated voltage.

As long as the power input to the capacitor is below that which causes the temperature rating to be in the safe area (80 C or 105 C), the clamping action of the leakage current will probably safely dissipate the heating, and the voltage will not continue to rise.

My statement is based on operation with a SSG of normal build where there is just not that much current available to cause a rapid overheating and explosion.

My statement does not hold if you have a lot of current available to rapidly charge the capacitor past it's rated voltage.

Most of the video's on youtube that show the fireworks are seriously pumped with lots of current, such as this one:

http://www.youtube.com/watch?v=JCPXckfT-6g&feature=player_embedded

Yes there is considerable energy stored in a large electrolytic, but usually when they short within a circuit that has limited applied current, they generally do not explode.


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Ion:

Awesome stuff thank you.  I was punch-drunk stooopid and I totally forgot about YouTooobe.

Here are two more clips for fun and fear:

http://www.youtube.com/watch?v=_WheLp0RdLQ&feature=related

http://www.youtube.com/watch?v=-W1LVnFrDZQ&feature=related

Fausto and anyone else, especially beginners:

When you connect a capacitor where the charging battery normally goes, it's almost like a slow version of when the neon lights up.  In theory any Bedini motor output can charge any capacitor to several hundred volts.  This means that if you leave the Bedini motor unattended connected to a capacitor, there is a risk of the capacitor exploding from over-voltage.

If you look at the video clips you will see small electrolytic capacitors explode violently.

This means that a coke-can sized electrolytic capacitor could cause a very dangerous explosion if you leave it connected to the Bedini motor unattended.  Watch the capacitor voltage with your multimeter and do not exceed the voltage rating!!!

Play safe!

MileHigh
« Last Edit: 2010-09-17, 03:36:35 by MileHigh »
   
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Perhaps the test I ran this morning will shed a little light on electrolytic capacitors and how they work.

The object of the test was a 4700uF 16 Volt Nippon capacitor rated 85 C

I hooked it up to my HP6181C current source which has three settings of 2.5 mA, 25 mA and 250 mA with a vernier for each range and has up to 115 volts source capability in the max setting. I monitored current through the cap, voltage across it with various Fluke meters and case temp with a Fluke 52

I stepped through various current settings and recorded current, voltage across the cap and case temperature.

Note that voltage setting was maxed at 115 volts.

As you can see from the data, the internal leakage rises dramatically with overvoltage and attempts to self limit when fed from a current source. The overvoltage creates temperature rise which contributes to but is not the only source of increased leakage current. All these factors are interrelated.

When fed from a current source you would expect the capacitor to continually rise to the source voltage (115 volts in this case) and explode. At the current settings shown, this is not the case because of the increase in leakage current.

Notice from the data the capacitor in some ways acts like a zener diode neatly limiting it's voltage to 25 V despite large current change.

When this capacitor did let go (on the last test), it was not an explosion and fire, but a snap like a cap gun. They are designed to fail safely in their normal failure modes. Tops of smaller caps have triangular weak spots designed to release pressure safely. Larger caps have vents or seals that let go.

What you see in the videos on youtube with fire, smoke and explosion are gross overloads of the capacitors.

 Still best to play it safe and monitor capacitor voltage, try not to exceed their ratings.

Now can anyone tell me how many Joules per pulse a typical Bedini puts out and what is the typical pulse repetition rate?
« Last Edit: 2010-09-17, 23:19:59 by ION »


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Ion:

Thanks for your explanations and your test data.  It's very much appreciated.

It's also VERY cool to see that you did a test using a current source type of power supply for your testing.  The concept of a "current source" with an infinite output impedance is a difficult one for beginners to grasp.  It's much easier for people to understand what a voltage source power supply is with a zero output impedance.  I have been hammering away on this concept in the 10-coiler thread because it is central to how a Bedini motor works.  So it is really great that you showed that you can use one on your test bench,  i.e.; they really exist in real life.

Quote
Now can anyone tell me how many Joules per pulse a typical Bedini puts out and what is the typical pulse repetition rate?

What a GREAT question!!!  Any keeners out there with their Bedini rigs that can answer that one?  It's time to move beyond measuring your RPMs!

MileHigh
« Last Edit: 2010-09-18, 03:14:46 by MileHigh »
   
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Hi every1,
First off thanks for the invite MileHigh, nice forum ;)
About the input power consumption, i have a power supply 0 to 30vdc.
I have a bunch of 25.000uf 15vdc caps and a few 1Farad 20vdc audio power caps aswell.
Been using for power input measurements 2 25.000uf caps in parallel with the power supply and a analog amp meter, make that 2 amp meter.
Only thing i dont have is the 1% precision resistor 100 ohms @ 1/2 watt.
Will get one and put it in place.

At this moment my power input connection are like this... :P
Power supply is at 22 volts going to the 2 25.000uf caps in parallel than through a Voltage Regulator 18V and than through a Daftman Hall/Mosfet and than through 2 amp meters and last to my run coils. The caps are rated 15vdc but can and have handle countless times volts around 30vdc without any damage. i`ll be carefull ;)
I have 2 amp meters(1 3amp & 1 100Ma) becuase when i start up my device it draws more so i switch to the 3 amp meter and afther a few minutes or so when it stabilize i switch to the 100Ma meter to see more accurate the amp draw.
The Daftman circuit has been modify, no neon bulp. It dont collect B-emf or use the B-emf.
The B-emf if any goes to a small cap and than back to the run coils so i hope... ;D

One question about that resistor, so it goes on the positive wire coming from my power supply and going to the positive on the cap. Is this right ???
In other words, can i like connect it directly on the positive of the cap and the ohter side of the resistor goes to the positive wire?? Or do i need to like cut the positive wire in 2 pieces and sandwich that resistor in between??
Oke, so when i`m done with this just meassure the voltage across the resistor and i`ll have the input Milliamps ??
I will be comparing the voltage reading from across that resistor and the Milliamp on my analog meter to see if they are kinda the same.

At this moment even if i let my device running for a long time the 100Ma meter still moves back and forth a little bit...not really stabilizing ???

A question about measuring a Gen coil or more Gen coils in series, do i need a large cap and a 1% 0.5watt resistor??
And how do i connect that resistor??series, parallel adn where... ???
Oke, thanks in advance for the help on this and sorry for all noobie questions.

Peace!
VZ2DAY


So let's discuss a very simple technique to get a very accurate power consumption measurement for your Bedini motor.  There are many ways of doing this but let's start with one simple example.

Here is what you need:

1) A large electrolytic capacitor.  Perhaps something like 30,000 uF at 25 volts.  That might be the size of a 300 ml Coke can or larger.
2) A 1% precision resistor, let's say 100 ohms @ 1/2 watt.  Alternatively you can take a regular 100 ohm resistor and measure the resistance of it with your multimeter.
3) A bench power supply that has a variable output voltage from 0 to 25 volts.

With these components you are going to simulate a 12-volt power supply with the large capacitor.

The circuit is very simple:  You connect the output of the power supply to the 100-ohm resistor, and that is connected to the positive of your big capacitor.  The positive of the big capacitor is also connected to the power input of the Bedini motor.

You set your power supply to 12.6 volts and you charge up the big capacitor to 12.6 volts.  Then you connect your Bedini motor and get it running.

When you do this you are monitoring the voltage across the big capacitor.   You notice the voltage across the capacitor starts to drop.  So you compensate by increasing the voltage from your variable power supply.

Let's assume that after a few minutes, the Bedini motor is running normally at it's stabilized speed.  You have also stabilized the voltage in the big capacitor back up to 12.6 volts to simulate a normal battery.   To do this let's assume that the voltage from the power supply has to be set to 13.84 volts.

You have to check the voltage across the capacitor with your oscilloscope to see if it is relatively stable.  If the ripple voltage is less than +/- 0.1 volts, then let's say that the voltage is stable enough.  If there is too much voltage ripple, then simply add a second big capacitor in parallel to the first capacitor.

So how do you calculate the power consumption of the motor?  It is very easy, you know the voltage supplied to the motor, it is 12.6 volts.  And you know that the average current that you are feeding into the capacitor from the power supply is identical to the average pulsing current that is going into the Bedini motor.  The current in must be equal to the current out if the capacitor voltage is stable.

So the current is (13.84 - 12.6)/100  = 12.4 milliamperes.

Therefore the power consumption of the Bedini motor is (0.0124 x 12.6) = 0.156 watts.

Note that you don't actually measure 13.84 volts and 12.6 volts and do the subtraction.  You simply put your multimeter across the 100-ohm resistor and measure 1.24 volts.  You are taking advantage of the fact that the multimeter is more accurate when measuring lower voltages.

Let's check the power dissipation in the resistor = (0.0124 x 0.0124 x 100 ) = 0.015 watts.  So you are fine using a 1/2 watt resistor in this example and you are sure that it will not burn up.

Here is the big advantage with this setup:  It will be able to measure the power consumption of a Bedini motor or any other device no matter how crazy the current waveform is.  As long as your big capacitor is big enough to smooth out the voltage then it will work.  It will allow you to make very accurate power consumption measurements without worrying about the limitations of digital and analog meters.

The disadvantage is that you have to work with a variable voltage power supply and you have to wait a few minutes to let everything stabilize before you make your final measurements.  If you are doing research this should not be a problem.

MileHigh
   
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Hi BiDaDiKuNuKu,

Welcome to the forum and you are very welcome!

Quote
Power supply is at 22 volts going to the 2 25.000uf caps in parallel than through a Voltage Regulator 18V and than through a Daftman Hall/Mosfet and than through 2 amp meters and last to my run coils. The caps are rated 15vdc but can and have handle countless times volts around 30vdc without any damage. i`ll be carefull Wink
I have 2 amp meters(1 3amp & 1 100Ma) becuase when i start up my device it draws more so i switch to the 3 amp meter and afther a few minutes or so when it stabilize i switch to the 100Ma meter to see more accurate the amp draw.

Can you link to one of your clips that shows this?  Or perhaps you have a link to a picture that shows this setup on your web site or even draw a schematic of your power measurement circuit?  I am not sure what the "Daftman Hall/Mosfet" circuit is also, do you have a link for that or perhaps there is one on the TEEP forum?  Using two amp meters also sounds strange to me.  So I can't comment on that setup right now.

If your 15,000 uF caps are electrolytic caps please read the last few postings by Ion.  He is stating that when you over-voltage electrolytic caps then they can self-discharge.  If you charge one up to 30 volts and leave it alone for one hour and then measure the voltage, is it still 30 volts?

Quote
The Daftman circuit has been modify, no neon bulp. It dont collect B-emf or use the B-emf.
The B-emf if any goes to a small cap and than back to the run coils so i hope...

I saw one of your clips where you collect B-emf into a capacitor and if I recall correctly you measure about 30 volts.  I am not sure what size capacitor you are using to collect the B-emf.  Then you demonstrate putting what looks like a 3-volt white LED across the capacitor.  That would indicate that by adding the LED you reduce the capacitor voltage to 3 volts, and all of the B-emf power then is used to light the LED.  Again, it's hard to comment on exactly what you are doing.

If you are not collecting the B-emf and you have no capacitor then in theory the B-emf energy is being dissipated in your main drive transistor(s) and that is something that you don't want to do.

Quote
One question about that resistor, so it goes on the positive wire coming from my power supply and going to the positive on the cap. Is this right?
In other words, can i like connect it directly on the positive of the cap and the ohter side of the resistor goes to the positive wire?? Or do i need to like cut the positive wire in 2 pieces and sandwich that resistor in between??

Yes the resistor goes between the positive output terminal of your power supply and the positive terminal of your capacitor.  Certainly you can add a wire to make the physical connection.

Perhaps something like this:  <power supply positive terminal> <wire> <resistor> <capacitor positive terminal> <wire> <motor positive input terminal>

Note that your motor (the one in your clips) is quite big and may consume a lot of power.  In your case it may be necessary to use a resistance of higher wattage.  For example you could put two 1-watt 200-ohm resistors in parallel to make a 100-ohm 2-watt resistor.  They don't have to be 1% precision resistors, all that you have to do is measure the resistance with your best multimeter and write it down.

Quote
Oke, so when i`m done with this just meassure the voltage across the resistor and i`ll have the input Milliamps ??

Yes, as long as you make sure everything is stable like it is explained in the procedure.  Good luck!

MileHigh
   
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Now to discuss measuring the power output from your generator coils:

Quote
A question about measuring a Gen coil or more Gen coils in series, do i need a large cap and a 1% 0.5watt resistor??
And how do i connect that resistor??series, parallel adn where...

For the generator coils, you don't need to use a cap at all, you just need a load resistor.  It's a bit complicated because the amount of power you measure from the generator coils depends on the RPMs of the motor and the value of your load resistor.

Here is a suggestion for a simple test to start learning about the process:  Get two 100-ohm 1-watt (or 1/2 watt) resistors.  They don't necessarily have to be 1% resistors.  In all cases you can always measure the resistances of the resistors with your best multimeter and write that information down.

You will need a "True RMS" multimeter.  You may already have one right now.

The resistor goes across the two terminals of your generator coil setup.

You will run three tests with three different load resistor configurations:

1.  With the two resistors in series for a load resistance of 200 ohms.
2.  With one resistor only for a load resistance of 100 ohms.
3.  With the two resistors in parallel for a load resistance of 50 ohms.

For each test you put the load resistance across the two output terminals of your generator coil setup.  Then you run the motor and with your True RMS multimeter you measure the voltage across the load resistance.  The True RMS multimeter must be set on AC voltage measurement.   We are going to hope and pray that the motor will be running fast enough to give you a stable voltage readout on your multimeter.

For each test the power output from your generator coils is = (voltage-squared)/(load resistance).

You should get different power measurements for the three different load resistor configurations.  If it all works out that would be a good start!

MileHigh
   

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It's not as complicated as it may seem...
A handy way to measure input power. A diagram might be helpful. I drew this up for gotoluc.

.99


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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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Hi Poynt,

Can you describe how to use that circuit?  It's not too clear to me, I don't see and input, not even sure what you mean by "input power."

Thanks,

MileHigh
   
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Thanks for the input "MileHigh and poynt99"...

@ poynt99...a diagram was just what i needed all though i dont really understand it yet... ???

@ MileHigh...TheDaftman circuit can be found on the teep forum.
I`ve modify it by removing the neon and resistor, in place of the resistor theres an 1k pot now. All though i dont really use that pot...but its there just in case.
I get some B-emf from my run coils and that is beying directed to the cap in the system.
Its a cap with some diodes so the mosfet doesnt get any B-emf and the energy in the cap is being use when the run coil is activated.
It does work, the cap does collect some energy and feed it back to the coil...:)

About my Big setup using alot of energy, not really.

@ first it ran with a 12vdc motor...12vdc @ around 3 amps...36 watts.
Second time it ran on standard bedini, bedini darlington pair, half bipolar and full bipolar (see the to vids of the FLYMAG device of mine)
It ran at 30vdc max @ around 0.23amps...about 6 to 7 watts.

But now i`ve rebuild the whole setup "man that was some work" still aint done yet...anyway
Its running now with that modify circuit with alot of good upgrades...
So 18vdc @ 0.06 to 0.200 depending on what ratio i use and how fast i wanted to go.
But i am aiming to have it running below 100 milliamps...
So lowest energy use is 18vdc @ 60milliamps...1.08 watts...:)
Highest is 18vdc @ 0.2...3.6 watts...:)
About the 2 amp meters, thats prettey simple i jsut have them both in parallel along the same wire but both have a on/off switch so i can switch from 1 to the opther depending on the amp draw... ;D

Big setups doesnt necesery means big amp draw...i`ve seen some tiny bedini`s using like 2 or 3 amp 2 12vdc.
Aint no way one gonna get even close to get that from the output...;-)
Anyway...hope i make some sence here above...

Btw, i will be making a vid on alll this above when i am done with my upgraded FLYMAG setup...:)
And now that i know how to meassure the output(not really, i`ll have to read it a few times for it to sink in) hopefully i can meassure that aswell with the input.

My page is : http://www.youtube.com/user/BiDaDiKuNuKu

Peace!
VZ2DAY
   
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Hi BiDaDiKuNuKu ,

Ok I looked at the Daftman circuit layout and I am pretty sure that I understand.  You say that you are feeding the energy in the cap back to the run coil.  I remember seeing that in your clip.  There is no rush but perhaps one day you can add that to the standard Daftman circuit layout drawing and take a picture and upload it here or put it on your web site.

I understand that pictures are really helpful, even for my simple power measurement circuit for measuring the power consumption for the Bedini motor.  Just for completeness, I will say that the ground for the large capacitor is connected to the power supply ground, and that is also the source battery ground relative to the Bedini circuit.

Quote
@ first it ran with a 12vdc motor...12vdc @ around 3 amps...36 watts.
Second time it ran on standard bedini, bedini darlington pair, half bipolar and full bipolar (see the to vids of the FLYMAG device of mine)
It ran at 30vdc max @ around 0.23amps...about 6 to 7 watts.

Yes, for the 12 vdc motor @ 36 watts, that's a lot of power.  It would be best to measure the power consumption for the standard Bedini configuration and other lower power configurations.  With 6 or 7 watts of average power consumption, you should be fine with a 1/2 watt 100-ohm resistor.  If the 100-ohm resistor burns up you can try a higher power resistor or try a 50-ohm resistor.  The actual value of the resistor is not critical, I just picked 100 ohms as a typical value.

Quote
About the 2 amp meters, thats prettey simple i jsut have them both in parallel along the same wire but both have a on/off switch so i can switch from 1 to the opther depending on the amp draw..

I get it now!

Quote
Big setups doesnt necesery means big amp draw...i`ve seen some tiny bedini`s using like 2 or 3 amp 2 12vdc.
Aint no way one gonna get even close to get that from the output...;-)
Anyway...hope i make some sence here above...

You are making sense for sure.  Just to elaborate, we can never forget that a Bedini motor is a pulse circuit, and consumes current in irregular pulses.  Therefore what we are trying to do is measure the equivalent average DC current consumption.  Then we can calculate the power consumption.

Quote
Btw, i will be making a vid on alll this above when i am done with my upgraded FLYMAG setup...Smiley
And now that i know how to meassure the output(not really, i`ll have to read it a few times for it to sink in) hopefully i can meassure that aswell with the input.

Looking forward to seeing your video.

A few more comments about measuring the output from your generator coils.  Most experimenters want to connect a full-wave bridge rectifier (FWBR) to the generator coil output and then add a capacitor the the output of the FWBR and then measure the rectified DC output voltage and also put a load resistor on the DC output and measure like that.  It sounds sensible to most people to do it that way.  In fact, doing this FWBR + capacitor setup unnecessarily complicates things and actually makes it more difficult to measure the power output.  I will not discuss all of the technical details to explain why.

In my posting I said just add a load resistance to your generator coil output and measure the voltage with a "True RMS" multimeter.   We can call this the "bare bones" approach.  This approach is simpler and should give better results.  It is more like what we see in the real world.  For example, if you are making toast with a toaster, it's the identical setup.  A toaster is simply a device that connects AC mains voltage to a load resistor.  No capacitors, no FWBRs, no-nothing, just a bare-bones heating element connected to the AC mains power.  "Keep it Simple" is a good way to go here.

MileHigh
« Last Edit: 2010-09-19, 17:12:12 by MileHigh »
   

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It's not as complicated as it may seem...
Hi Poynt,

Can you describe how to use that circuit?  It's not too clear to me, I don't see and input, not even sure what you mean by "input power."

Thanks,

MileHigh

MH & Hi BiDaDiKuNuKu,

It's the same circuit I recommended the RA group use for measuring the power supplied by the battery. I have made this circuit as well for my tests, you may remember. It's quite simple, and V1 is your source battery or power supply:

1) Pi = Vi x Ii
2) measure "Vbat" on the diagram. This is Vi
3) measure with a digital voltage meter the DC voltage across the current sense resistor R3.
4) Ii (input current) is VR3/10 (or 1 Ohm if you choose a lower value resistor)
5) Pi(average) = Vi x Ii

Now you can calculate the average DC power being supplied by the source.

.99



---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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Thanks Poynt,

Yes it is essentially the same thing that I described in my posting.  You may note the "in your own fishbowl" phenomenon at play here.  If I may suggest, it would be great if you could annotate your diagram, it would be a great help.

Your description makes sense, but my mind was still dealing with multiple names for the same variables and even V1 and Vi make things confusing because those are not the same voltages.  Also your diagram has "1.5" under "V1" making it look like a standard 1.5-volt battery, when in fact that's where you connect the variable power supply.

Nonetheless, your circuit is essentially the same thing as I outline in my posting.  Relative to your diagram I do not use C1 in the setup that I describe, I just have the variable power supply output.  I am also suggesting that C2 be a much larger capacitor, like 15,000 uF or greater, considering that the Bedini motor may suck quite a bit of current when the transistor switches on.

Thanks again!

MileHigh
   
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@ MileHigh...
About a pic, will make one when i fixied my camera. The bat holder broke of... >:(
Or better yet, i`ll take a drawing of the hall/mosfet from the Daftman and use photoshop to modify it... ;)

I havent done it with a standard bedini, what i was doing with the bedini was simply removing the noen and ad a HV cap to collect the B-emf.
With that i lid up a Led...

btw, whats a true RMS multimeter... ??? got a few multi meters here but if they are true RMS
About that gen coils meassurements, i`ll just hook up a resistor and meassure it/them... ;)

About the pulsed devices and their amp draw in pulses, that means that a device is pulsing (example 0.5amp) for the period of time u adjust it by...interesteing ;D

So like to really meassure it properly, one should take in account the times pulsed for each revolution and also the time of each pulse duration.
And also the lenght of the cirkel being made by the magnetes that pulses the unit.
Example, if u got like 3 magnetes to pulse on a cirkel of 5cm diameter (5x3.14=15.7 cm) and the 3 magnetes are like (example 1/5 of that toatal(15.7cm)..
Does this means that u only pulsed for 1/5 of each revolution... so if 1 revolution is 100 % so 1/5 is like 20% pulsed...
How can one convert that 20% amp draw into the "genuine" amp draw.
in other words, if we balanced out these numbers for input and output we just might crack the BIG OU thingy...LOL... :P
Now this is really getting beyound my understanding...will get it though ;D

I see u like the "KISS" approche..."keep it simple stupid"...haha...i do to ;D
   
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