1 ohm resistor test applicable to larger systems too?

[Woody]

Hello all,

In the docs we all used to get when we joined the old yahoo group, there was "the 1 ohm resistor test" which we would use to determine if our vanilla SSG was putting out radiant, or standard positive current. It involved simply putting a 1 ohm resistor in place of the charging battery. If the resistor stayed cool, the energy produced was predominantly radiant. If it was hot, you were putting out mostly positive current. I'm curious if the 1 ohm resistor test applies to the larger systems as well. Would J.B.'s ten coiler, or 6 coiler be able to pass the 1 ohm resistor test, or do they output enough positive current mixed in with the radiant to heat up the resistor? Just curious, as my mini 6 coiler does heat up the 1 ohm resistor, even though it is the first unit I've had that was able to charge a single battery to around 14.75v on a single charge of an identical battery and still have some charge to spare. My bike wheel never did heat up the 1 ohm resistor at all, but was never even close to being able to do a 1 for 1 charge. Any thoughts? Anyone? Bueller?

Thanks,

-Woody

[sephiroth]

I've wondered about this test as well... All the 1 ohm test shows is whether 1 watt of measurable power is being output and I would have thought this would be largely dependent on the size of the system.

[John_Koorn]

The purpose of the 1 ohm resistor test was to test if the machine was tuned. If the voltage drop over the resistor was under 1V then the test passed. This showed that there was little current on the output and it is the radiant energy that is charging the battery. On the most basic bifilar SSG the output current in generally only around 50mA, so a 1W resistor would not heat up.

It can also be used for the bigger systems but you just need to use a higher wattage 1 ohm resistor as the output current is proportionally higher.

John K.

[James Milner]

John,
The bifilar SSG is a #23 and #26 I presume?
That being the case the current draw is likely to be bigger for those running #18 gauge wire.
What does your superpole run at? My 5 filar #18 superpole tested with a 1.5ohm 10w resistor dropped by 0.75 v. Which suggests 500mA. Or 125mA for each power strand which seems a little too high based on this test.
I'd appreciate your advice.
Thanks
James

[sephiroth]

The purpose of the 1 ohm resistor test was to test if the machine was tuned. If the voltage drop over the resistor was under 1V then the test passed. This showed that there was little current on the output and it is the radiant energy that is charging the battery. On the most basic bifilar SSG the output current in generally only around 50mA, so a 1W resistor would not heat up.

It can also be used for the bigger systems but you just need to use a higher wattage 1 ohm resistor as the output current is proportionally higher.

John K.

Hi John,

But if there is a higher current then obviously there will be higher voltage across the resistor. Measuring less than one volt across a one ohm resistor just means that there is less than 1 amp of current running through it. And although there may be only 50ma output from a very basic and small motor with a 12v battery as a load (we're talking less than 3 watts input here), from what I understand the 1 ohm test is supposed to be performed without the charging battery attached. Without the charging battery in place, the amount of current flowing through the one ohm resistor will be far higher because the the duty cycle of the field collapse will be much higher without the 12v impedance of the charging battery.

So if you need to use a higher than 1 watt rating resistor, it is obvious that you will be measuring over 1v across that resistor and so it would fail the test. Any moderate to high power SSG device will.

[John_Koorn]

John,
The bifilar SSG is a #23 and #26 I presume?
That being the case the current draw is likely to be bigger for those running #18 gauge wire.
What does your superpole run at? My 5 filar #18 superpole tested with a 1.5ohm 10w resistor dropped by 0.75 v. Which suggests 500mA. Or 125mA for each power strand which seems a little too high based on this test.
I'd appreciate your advice.
Thanks
James

James, that is right. The current draw and output current will be higher so you will need to adjust the voltage drop accordingly to pass the test. The question is how can that small amount of current charge your battery?

my super pole draws 5A @12v or 8A @24v.

John K.

[John_Koorn]

Hi John,

But if there is a higher current then obviously there will be higher voltage across the resistor. Measuring less than one volt across a one ohm resistor just means that there is less than 1 amp of current running through it. And although there may be only 50ma output from a very basic and small motor with a 12v battery as a load (we're talking less than 3 watts input here), from what I understand the 1 ohm test is supposed to be performed without the charging battery attached. Without the charging battery in place, the amount of current flowing through the one ohm resistor will be far higher because the the duty cycle of the field collapse will be much higher without the 12v impedance of the charging battery.

So if you need to use a higher than 1 watt rating resistor, it is obvious that you will be measuring over 1v across that resistor and so it would fail the test. Any moderate to high power SSG device will.

Yes Sephiroth there will be higher than 1v across the resistor with bigger machines. But you are incorrect about the duty cycle changing, the duty cycle is dependant on the speed of the rotor. The speed of the rotor however is determined by the impedance of the load, as the 1 ohm resistor is significantly higher than a 12v battery the rotor speed will be much lower.

Don't get too hung up about this test, it is there just to show there is something else charging your battery besides hot current.

John K.

[sephiroth]

Sorry, John. No disrespect but I'm sure you are incorrect. Although the internal resistance of a lead acid battery is many times smaller than a 1 ohm resistor, to the energiser the 12v battery is a higher impedance load. We know this because of the 1 ohm resistor test. If the one ohm resistor was a higher impedance to the energiser then the voltage across the resistor will be over 12v. Look at a neon. An infinitly higher impedance than both a lead acid battery and a 1 ohm resistor. If the neon alone is the load on the back end then the rotor will accelerate.

And the duty cycle of the output pulse does increase as the impedance of the load is lowered. Take a look at your scopes. Substitute the 12v charging battery with a 6v charging battery. The duration of the output pulse will double what it was with the 12v charging battery.

Here are some scope shots to illustrate.

This is the scope shot of a tuned and built to spec Bedini SSG with 12v input and around 13v-14v charging battery (it is pretty much fully charged at the moment)



Here is the scope shot for when I substitute the charging battery with a one ohm resistor. I did not adjust the pot. The ssg is set the same as it was in the previous shot.



Now, it's not very clear what is happening in that shot, so this time I put the scope probes across the 1 ohm resistor so we can get a better look.



As you can see the duty cycle for the 1 ohm resistor test is far greater than it was when it was charging a 12v battery. It is possibly longer than what I have labelled in the shot, though it overflows into the peak induced voltage from the rotor magnets passing the stator.

[John_Koorn]

James, that is right. The current draw and output current will be higher so you will need to adjust the voltage drop accordingly to pass the test. The question is how can that small amount of current charge your battery?

my super pole draws 5A @12v or 8A @24v.

John K.

Sorry James, my bad. Those figures were for my 10 coiler. I don't have the 5-filar coil on my superpole bike wheel anymore, but from memory it drew about 0.5A.

John K.

[John_Koorn]

Sorry, John. No disrespect but I'm sure you are incorrect. Although the internal resistance of a lead acid battery is many times smaller than a 1 ohm resistor, to the energiser the 12v battery is a higher impedance load. We know this because of the 1 ohm resistor test. If the one ohm resistor was a higher impedance to the energiser then the voltage across the resistor will be over 12v. Look at a neon. An infinitly higher impedance than both a lead acid battery and a 1 ohm resistor. If the neon alone is the load on the back end then the rotor will accelerate.

And the duty cycle of the output pulse does increase as the impedance of the load is lowered. Take a look at your scopes. Substitute the 12v charging battery with a 6v charging battery. The duration of the output pulse will double what it was with the 12v charging battery.

Here are some scope shots to illustrate.

This is the scope shot of a tuned and built to spec Bedini SSG with 12v input and around 13v-14v charging battery (it is pretty much fully charged at the moment)



Here is the scope shot for when I substitute the charging battery with a one ohm resistor. I did not adjust the pot. The ssg is set the same as it was in the previous shot.



Now, it's not very clear what is happening in that shot, so this time I put the scope probes across the 1 ohm resistor so we can get a better look.



As you can see the duty cycle for the 1 ohm resistor test is far greater than it was when it was charging a 12v battery. It is possibly longer than what I have labelled in the shot, though it overflows into the peak induced voltage from the rotor magnets passing the stator.

Hey Sephiroth,

No disrespect taken, it's nice to have a conversation with someone who knows what they are talking about. Thanks for posting your scope shots, they are a real help in seeing what is going on. That's probably one of the best scope shots of a tuned SSG I've seen. (Everyone else, that's what you are aiming for)

I guess we are somewhat disagreeing in what we define as the duty cycle. The way I define it is the the on time of the transistor (the input pulse in your scopes). Apart from that I agree with what you are saying. It is interesting to see (assuming you didn't change the time scale on the scope shots) that the duty cycle seems lower when using the 1 ohm resistor as the load. In all my tests I have seen that the rotor slows down, but I don't recall ever scoping it at the same time.

The way I understand it is that the slower the rotor is, the higher the duty cycle because the passing magnet is spending more time over the stator. Therefore the the transistor is on for longer and hence the average draw current is higher.

Like I said in my previous post the 1 ohm resistor test is simply to show that there is not enough conventional current to be able to charge the battery, which is great as a tool for beginners to understand what is actually charging the battery. I'm sure that you understand that it is the difference in potential between the the collapsing magnetic field and the charge battery that is causing the ions to move backwards which the battery sees as a potential time charge.

I'm sure you'll agree that this flies in the face of convention which states that you needs gobs of current to charge a battery, which is simply untrue and well proven by the Bedini technology. All you need to do is create a difference of potential between the source (stator) and the load (battery). Of course the energizer sees the battery as a very high impedance because compared to the relative impedance of the spike the battery's impedance is massive. In fact the spike we see on the scope (if it is fast enough) is the amount of energy that cannot be absorbed by the battery because the potential difference is too much for the battery's chemistry to take.

To prove this you only need to use a load with a lower impedance, such as a capacitor, to see that the spikes are lower when charging a capacitor than compared to a flooded lead acid battery. On the other end of the scale if you place a sulfated battery with a higher impedance as the load the spikes are a lot higher because the battery's chemistry cannot absorb the potential difference.

Simply put, the battery and the HV spike caused by the collapsing magnetic field of the coil wants to balance each other out. So if the spike is 300v and the battery is 12v both the spike and the battery want to balance at ~156v. Obviously the battery's chemistry is incapable of 156v so it charges as high as it can to the best of its ability. As the spike is of significantly lower impedance (or the energizer sees the battery as a higher impedance) compared to the battery, most of that spike is absorbed by the battery - the better the condition of the battery the more it will absorb the spike.

John K.

[sephiroth]

Hey John,

I did change the time scale on the scope since the rpm dropped from around 3000 to around 1000 and so it wouldn't have show up very well without adjusting that. The duration of the input pulse is approximately the same (only slightly shorter) but the duty cycle is a lot lower, as you said due to the lower rpm. It was just the duty cycle of the output pulse I was refering to which is far greater when the impedance of the load is lowered.

i pretty much agree with your interpretation of the spike I posted my understanding of it a few days ago and it is more or less the same. http://www.energyscienceforum.com/be....html#post3724

There was something else I was going to mention about the input pulse duration but it is taking this thread off topic so I'll start a new one