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Author Topic: twofer  (Read 5477 times)

Offline webby1

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Re: twofer
« Reply #15 on: November 07, 2016, 09:24:13 PM »
I put the caps back in and the free wheel part came back,, well it is sort of free wheel.

If I spin it really fast ( a relative speed thing when spinning it by hand) I can charge one cap up to say 9V and the other will go DOWN to 2V but the higher the high voltage the easier it is to turn,, then when I stop spinning and the caps are far apart in voltage they will balance out and for a while when they are doing that the motor is being spun by the caps and not me.

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Re: twofer
« Reply #15 on: November 07, 2016, 09:24:13 PM »

Offline webby1

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Re: twofer
« Reply #16 on: November 07, 2016, 09:40:38 PM »
Here is a pic of my testbed,, messy

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Offline webby1

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Re: twofer
« Reply #17 on: November 07, 2016, 10:24:53 PM »
I reversed one of the coil sets and have the motor connected between the positive plate of one C2 and the negative of the other,, PSU turned off.

I can spin the motor one way and get a fair amount of resistance  and charge,, or at least try to charge the C2 that has its negative plate hooked to the motor negative,, and the other positive,, I can not get much of a charge on them,, spin the motor in the other direction and the higher the voltage the easier it is to turn,, but both C2's charge up positive.

When I ran the system from the PSU with the motor hooked up like this one C2 would be the drive voltage to the motor and the other one would rise up a little bit,, from 20 to 22 while the other was at 16,, but with the coils in this orientation both caps are almost the same value and below the input,, 13.7V in and 11.2V driving and 11.6V on the other ,, disconnected the motor,, the cap that was driving is now 12.4 and the other has dropped to 10.7.

I am guessing that this is to do with the turns ratio which was not showing up as a difference with the coils setup the other way.

Offline webby1

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Re: twofer
« Reply #18 on: November 07, 2016, 11:55:29 PM »
I tried to compare by hand spinning the motor up with just the cap connected to it and using the system with the PSU off,, and it feels like about the same thing,, except for that one way easier than the other part.

I again tried using the motor on one side without the added system to see if it still charged the cap on the other side and then if discharging the cap on the other side would still spin the motor,, yes that all still happens,, so I am guessing that there is a high inductance to this toroid transformer allowing for a very LOW frequency to be passed through,, less than 1hz.

That still does not explain the sudden increase in force needed to spin the motor,,


Offline webby1

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Re: twofer
« Reply #19 on: November 09, 2016, 07:43:01 AM »
T4 and S2 and S3,, well the switches are so I can pulse only 1 coil directly,, if I want to :)

T4 is so that when I do manually pulse the single coil I only pass about 2.14A,, it will peg my PSU

Q5 and Q4 are run by my little astable multivibrator,, 47uf and apr. 7500ohm resistors for the timing parts.

When I pulse the single coil just right,, and it is just right,, then I can drive both caps up to over twice input voltage,, I am using 10V in right now and I had them over 30V,, well with 2.14A pumping in there I would guess that those caps could reach a fairly high value.

What is interesting is both the rate of charge and charge voltage when I discharge one of the caps or place a resistor across one of the caps,, as well as when I use a 560 ohm resistor across both caps,, this one is funny and it might be due to me blowing the smoke out of one of my transistors and so I replaced them with some darlingtons,, and they are not the same,, any way, one cap will go up high and one will go down low,, so I will get one at 9.3V and the other at about 6.4V. (that is with 10V from source) ((idle draw is 17mA loaded is 49mA and right now the caps are 6.2V and 7.5V, 560 ohm each, I have the resistance in there as well with T4))

When a cap is shorted the other side will climb over source voltage and it will climb much faster,, so it will charge up the cap in less than 1\2 the time when the other side coil is shorted and no transistor feed,, it still works shorting the cap but that raises the input and I am playing nice with the npn's :)

All 3 caps are 14000uf 30V ( I have 2 in the pic for the source,, but it is only one on my testbed,, it was easier to draw 2 than have all the connections to do 1)

1n1007 for all the diodes,, surprisingly I have not blown one of those yet,, and yet I have smoked 3 560 ohm resistors.

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Re: twofer
« Reply #19 on: November 09, 2016, 07:43:01 AM »
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Offline shylo

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Re: twofer
« Reply #20 on: November 09, 2016, 08:57:56 AM »
Hi Webby, I tried to modify your schematic from page 1 ,but gave up. I was trying to add diodes in the drawing to explain a way of collecting more out of a system.
In the latest I see you got 4 in the right spot(but wrong polarity), put pairs at each coil lead, put big caps between all the pairs.
artv

You have room for 12 more diodes, but they should be placed opposite each others' polarity.

Offline webby1

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Re: twofer
« Reply #21 on: November 09, 2016, 01:53:33 PM »
Not sure what you mean?

They are how I have the testbed,, the white stripe going to negative for flow to happen.

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Re: twofer
« Reply #21 on: November 09, 2016, 01:53:33 PM »
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Offline webby1

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Re: twofer
« Reply #22 on: November 09, 2016, 07:24:45 PM »
Dumb question time,

When C4 and C5 are charged up to say 7V and with my PSU being B2 and that set to 10V does that mean that I am pumping 3V @ whatever current,, in this case it is 44mA,, into my load?

I am holding the caps at 7V but the resistor I am using between the coil legs is 560 ohms, one on each side, and it is showing 6.1V @ 11.5mA per side DC,, if so then what about the losses in the circuit stuff,, that takes 17mA with no coils @ 10V,,


Offline webby1

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Re: twofer
« Reply #23 on: November 09, 2016, 09:25:34 PM »
Maybe it is just my view that is messed up.

In a sense I have a DC converter using batteries,, so if the source is 10V and the caps are 7V batteries then my load out is 3V @ 44mA,, that load is taken as 2X 6V @ 11mA,, so that is 66*2 for 132 and my input is 3*44 for 132,, they are the same.

Offline webby1

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Re: twofer
« Reply #24 on: November 09, 2016, 10:45:28 PM »
I swapped out 2 14000uf C2 (C4 and C5) with 270uf caps,, same load resistors as before, both caps read 7V and the input current is moving between 44mA and 45mA @ 10V, but it is staying mostly at 45mA.

Dropped the input down to 6.9V,, it does not like going down to low,, so input 6.9V @ 23mA caps at 4V load voltage is 3.3V @ 6mA
20V @ 118mA 16.5V on caps 15.8V across load @ 28.5mA
30V @ 195mA 26.1V on caps 25.5V across load @ 47mA
Back to
10V @ 44mA  7V on caps 6.3V across load @ 11.5mA

10V is the most stable,, the other voltages had the input current varying as much as 10mA.

I have 2 other caps in the circuit that I failed to mention ,, one a 14000uf cap and the other a 10000uf cap, these are across the base legs of the coil pairs,, so one is connected to T3 pin 1 and T2 pin 2 and the other one is across T3 pin 3 and T2 pin 4,, doing that caused my input draw to decrease to 45mA,, the output caps are at 7V and they are holding 6.3V across the load of 560 ohms.

If I only use one of those added caps the input goes up,, it swings between 75mA and 41mA but hanging out around 50 to 65 mostly,, and the output drops and the output shows a 1.5V AC current,, if I do not use any the input goes up and the output goes down,, 5.8V and 6.1V,, depends on which one I disconnect first.

To me then this is then working just like a DC transformer and or a resistive voltage divider with the twist that the voltage is not the difference between the batteries and the current is not that either BUT the output matches the input.

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Re: twofer
« Reply #24 on: November 09, 2016, 10:45:28 PM »
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Offline webby1

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Re: twofer
« Reply #25 on: November 09, 2016, 11:27:17 PM »
Now I am wondering what it would look like if I could get rid of the transistor costs,, would loosing that 14mA have me only passing 3V at 31mA?

Offline webby1

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Re: twofer
« Reply #26 on: November 09, 2016, 11:44:43 PM »
I think this is how I have it currently.


Offline webby1

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Re: twofer
« Reply #27 on: November 10, 2016, 12:22:09 AM »
My first new twist is to get rid of T4.

Input went down and output went up,, 10V @ 41 to 42 mA in and 6.3 to 6.4V @ 11.6mA  I would remind myself at this time that the meter for current could be off,, but at these values it has been the same as 2 other meters,, within about 0.05mA,, I still need to double check the input current since I am using the PSU meter for that value.,, and the resistors are supposed to be 560 ohm,, but they are usually under that so I need to check that as well.

Offline webby1

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Re: twofer
« Reply #28 on: November 10, 2016, 07:31:24 AM »
I thought I would run a few tests on the PSU meter reading and it appears to be 3mA low,, I used a current meter as well as a voltage drop to compare the PSU meter.

Re-running a few tests with that consideration shows that my input to output are close,, as well as the 11mA draw for the transistors,, I think.

10V @ 33.7mA in C2 at 5.92V  load voltage across 545 ohms shows 5.25V @ 10mA or 52.5*2 for 105,, 10-5.92 for 4.08 @ 22.7 for 93 if I can loose the transistor costs,, not sure if I can look at it that way.  Very close for rough stuff :)

Now I know to add 3mA to the meter reading at values above 20mA,, it goes up to 4mA down under 10mA showing on the PSU.

Time to run more setups :)

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Offline webby1

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Re: twofer
« Reply #29 on: November 10, 2016, 03:14:45 PM »
My simple thought on this setup was that the load would see the difference in voltage between the C2 and source,, so if C2 was charged up to source voltage my load would see nothing, but if C2 was lower, there was a resistance allowing current to flow, then the load would be passing that current and voltage,, the larger the load the larger the difference until C2 was down to 0V then maximum current would be flowing through the load and the source would only be showing the same voltage but with an increase in current,, the load would be seeing an increase in both voltage and current.

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Re: twofer
« Reply #29 on: November 10, 2016, 03:14:45 PM »

 

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