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Author Topic: The Non-Sense Pulse Motor.  (Read 83460 times)

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I am not sure what 12v on the collector has to do with it,as when the transistor closes,current will flow through the circuit via the LED(now up the right way),and the capacitor.

But anyway,how about this circuit?


Hmmm,   like Gyula mentioned,  D1 will prevent any current to flow.



Anyway, your new circuit with the load resistor (1K in my case), i got the below screenshot.
 
Yellow voltage across the battery/C1 (12.5V)
purple current through csr1  =  718uA rms  (0.718mA)  / 633uA average
Blue voltage current through csr2 = 8.3mA rms   /  2.12mA average

The input current DMM shows 0.57mA = 570uA (average).


So also with this circuit we measure more current through csr2 then csr1.

Itsu
   
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But the average current through CVR1 still should be the same as the average current through CVR2,as a capacitor is not an energy source.

When the voltage is a constant,then the same amount of current has to flow into the cap (CVR1)as what flows out of the cap (CVR2).

Here are the test results from having the scope math give us the average power,instead of the RMS power.

First scope shot is P/in from the power supply--CH2 across CVR1.
Second scope shot is CH2 acoss CVR2.


Brad

Brad,

Your Pin and Pout measurements in these scope shots are correct and would indicate an apparent COP = 1.81/1.09 = 1.66.   This does not include any power in the load from the field collapse.

Regards,
Pm
   

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I removed the csr's which enabled me to move the FG ground lead to the battery minus lead.
The current probe now shows a different signal for the input (ex csr1 position).
The ex csr2 position shows still a similar signal as before.

Still the ex csr2 signal (current) is higer then ex csr1 signal.

Green is ex csr1 signal
white  is ex csr2 signal

Will restore the full circuit and repeat this csr-less measurement.

Itsu   
   

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Original circuit restored without the csr's

leds are much brighter now without csr's, currents also stronger.

Green signal again ex csr1 current
White signal ex csr2 current.

input current DMM shows 50mA (average), input voltage still 12.5V.


So in any case i have tested (with or without csr's, with or without coil, etc.) i find that the csr2 position
always shows more current compared to csr1 position.

I know current alone is not power, but i cannot explain this.




Itsu
   
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  wow O0   Thank you for trying such things, Brad!  clever, I must say.

A very straightforward circuit now, with the inductor removed... surprised you still get more current through CVR2.
  At what frequency does the circuit operate (resonate) with this change?
   

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Original circuit restored without the csr's

leds are much brighter now without csr's, currents also stronger.

Green signal again ex csr1 current
White signal ex csr2 current.

input current DMM shows 50mA (average), input voltage still 12.5V.


So in any case i have tested (with or without csr's, with or without coil, etc.) i find that the csr2 position
always shows more current compared to csr1 position.

I know current alone is not power, but i cannot explain this.




Itsu

Hi Itsu

With the 5% duty cycle you only need to use one CSR to see power from the battery to the cap and from the cap to the coil. Why? well a cap does not work the same as a battery as we know, a cap will give up whatever we put across it much quicker than a battery.

So lets turn % into seconds :) The cap is being charged for 95 seconds to replace the power taken out of the cap in 5 seconds. So if we measure on the CSR1 1mA = 95mA in 95 seconds, the CSR2 would measure  19mA, 19 X 5=95mA to keep the balance. Naturally this is all without a load, so real tests will vary a bit. This is why CSR2 is a lot more.

As I have said, it is all a question of TIME FRAME (@ 50% duty things will be a lot different, but you would have to increase the frequency or blow your transistor).

Regards

Mike 8)


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Mike,

i get what you are saying, and it sound logical, however, the currents shown are averaged over a long timeframe,
so like Brad said, there comes a time that the high average current through csr2 can not anymore being topped
of by the lower averaged current through csr1.



Anyway, to be complete, my setup has:

C1 = C2 = 330uF / 450V electrolytical
L1 = 2.7mH / 2.6 Ohm ferrite cored coil
LED = 4x 3W power leds in series
D1 = UF4007
Frequency = 86Hz (not in resonance).
Source = 12V battery
FG = 3.5V DC square wave @ 5% duty cycle


Itsu
 
   

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Mike,

i get what you are saying, and it sound logical, however, the currents shown are averaged over a long timeframe,
so like Brad said, there comes a time that the high average current through csr2 can not anymore being topped
of by the lower averaged current through csr1.



Anyway, to be complete, my setup has:

C1 = C2 = 330uF / 450V electrolytical
L1 = 2.7mH / 2.6 Ohm ferrite cored coil
LED = 4x 3W power leds in series
D1 = UF4007
Frequency = 86Hz (not in resonance).
Source = 12V battery
FG = 3.5V DC square wave @ 5% duty cycle


Itsu

Thanks for the reply Itsu.

What if you replaced C1 with the same size battery as the source, both batteries being exactly equal, what would you see?

The number of samples does not make any difference, the duty cycle is what makes the difference, CSR1 is on all the time "apart from when C1 is full", CSR2 is only on 5% of the time (but dumping a lot of current in that 5%, it is a cap).

Regards

Mike 8)


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"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.
   

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It's not as complicated as it may seem...
Original circuit restored without the csr's

leds are much brighter now without csr's, currents also stronger.

Green signal again ex csr1 current
White signal ex csr2 current.

input current DMM shows 50mA (average), input voltage still 12.5V.


So in any case i have tested (with or without csr's, with or without coil, etc.) i find that the csr2 position
always shows more current compared to csr1 position.

I know current alone is not power, but i cannot explain this.

Itsu

I would say your DMM current measurement is probably the most reliable and accurate at this point. If you display 50 current pulses rather than 9, you might find the scope mean computation of 37mA approaches 50mA. So your Pin is indicating about 625mW. What is your Pout (and which component(s) is your load?).

I would only be worried that CSR1 and CSR2 aren't showing the same current if C1 wasn't present.
   

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Poynt,

indeed, when i increase the number of pulses in the screen, then the mean current approaches 50mA.

I normally do that (more pulses), but that is not the issue here, i know that the csr1 together with
the voltage across the battery is the real input to this circuit (as you mention about 625mW).


The issue is that the current through csr2 is constantly higher (factor 2 by 4) then through csr1.
If we then calculate the power using this 4x higher current together with the same 12.5V battery voltage
we get a much higher power there.

The question is why do we see such a higher current (power?) at csr2 compared to csr1?

Pout is not measured / calculated (yet).

Itsu
   

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I build something in LTspice XVII,  is it useable?

Itsu
   

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Thanks for the reply Itsu.

What if you replaced C1 with the same size battery as the source, both batteries being exactly equal, what would you see?

The number of samples does not make any difference, the duty cycle is what makes the difference, CSR1 is on all the time "apart from when C1 is full", CSR2 is only on 5% of the time (but dumping a lot of current in that 5%, it is a cap).

Regards

Mike 8)

Mike
Regardless of that,the average current over every full cycle should be the same.

If you look at my circuit,and where i have the ground leads for the scope,you will see that i have CH1 across the cap-not the power supply. We are measuring power delivered to the cap from the supply when ch2 is across CVR1,and then measuring power being delivered to the circuit by the cap when CH2 is across CVR2.

How can the average current through CVR2 be greater than the average current through CVR1,when the voltage is a constant 12 volts--in my case,24 volts.

Average current X peak voltage--we do not average twice to calculate power.


Brad


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It's not as complicated as it may seem...
Thank you Itsu for confirming the scope lines up with the DMM, that is a good thing!

In regards to the CSR currents, and as I already alluded, why is it surprising to anyone that the currents aren't equal?

What is the current in the bottom leg of C1?
   

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It's not as complicated as it may seem...
I build something in LTspice XVII,  is it useable?

Itsu

Yes. What have you found with it so far?
   
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Hello Tinman, good investigation work so far. 

I guess the best way to mesure OU is bypass whats happning in the circuit. Insted. Just mesure input and output.

I have a simple request to mesure the input power supply, and the CVR1. This is the total input.

And the total output. Load. In this case its the LED. So mesure across the LED, and place the CVR2 in line with the LED.

If there is OU this will sum it up. Input vs output.

Please see attached drawing.

If the circuit was making energy. This will sum it up quite quickly.

Blessings! ~Russ

 :o 8) :P O0
   

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Then of course we must not overlook Transistor Base Drive Current.
This drive current will appear in the Emitter Circuit of the Transistor.

If the Transistor is being driven into Hard Saturation then the Base
Current could be quite substantial.


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Thank you Itsu for confirming the scope lines up with the DMM, that is a good thing!

In regards to the CSR currents, and as I already alluded, why is it surprising to anyone that the currents aren't equal?

What is the current in the bottom leg of C1?

Why would the average current through each be different ?,when the voltage is a constant.

So, there is nothing out of the ordinary when the source is delivering 12v @ 36mA to the cap,and the cap is delivering 12v @ 78mA to the circuit,where there is no energy from the circuit being returned to C1  ???


Brad


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Hello Tinman, good investigation work so far. 

I guess the best way to mesure OU is bypass whats happning in the circuit. Insted. Just mesure input and output.

I have a simple request to mesure the input power supply, and the CVR1. This is the total input.

And the total output. Load. In this case its the LED. So mesure across the LED, and place the CVR2 in line with the LED.

If there is OU this will sum it up. Input vs output.

Please see attached drawing.

If the circuit was making energy. This will sum it up quite quickly.

Blessings! ~Russ

 :o 8) :P O0

Hey Russ,great to see you here.

Were not to interested in the electrical output ATM.
What wete interested in is the current through the coil,and the voltage across the coil.
This is because the coil is the drive coil of a pulse motor with a difference--read whole thread for explanation.

So,the higher the current through the coil,the stronger the magnetic field.

If we use the CVR1 value for average current of 36mA,the coil will build a magnetic field of a certain strength.
If we use CVR2s average current of 78mA,then the field built by the coil will be much greater.

So that is what we are looking at here,as well as why the 2 CVRs show different values in !average! current values.


Brad


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Then of course we must not overlook Transistor Base Drive Current.
This drive current will appear in the Emitter Circuit of the Transistor.

If the Transistor is being driven into Hard Saturation then the Base
Current could be quite substantial.

Covered this in my first video.
VP from the SG is 3v.
Base resistor is 100 0hms.
The tip3055 starts to conduct with a base voltage of about .7v.
So max voltage across the 100 ohm base resistor will be about 2-2.3v. Max current is there for 23mA at a 5%duty cycle.
Average current is therefore 1.15mA from the SG


Brad


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Ok, i understand your motivation.

However is there any chance you can take the measurement? Post the screen shot here ?

Im trying to wrap my head around things here and the more data you can post the better we can analyze the system and understand it.

If you don't have time so be it. But it sure would be a good thing to measure.

Thanks. ~Russ
   

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Ok, i understand your motivation.

However is there any chance you can take the measurement? Post the screen shot here ?

Im trying to wrap my head around things here and the more data you can post the better we can analyze the system and understand it.

If you don't have time so be it. But it sure would be a good thing to measure.

Thanks. ~Russ

Sorry Russ,i cannot test your circuit,as my scope and SG share a common ground,so CVR2 in your schematic is a !no can do!.

I must also add that the LED is not total power out as described in your schematic.
Total power out is power dissipated by L1 + LED

Power dissipated by L1 is VxI(CVR2)-LED.
We are of course disregarding the small amount of power dissipated by CVR2,transistor,and D1.


Brad


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Thank you Itsu for confirming the scope lines up with the DMM, that is a good thing!

In regards to the CSR currents, and as I already alluded, why is it surprising to anyone that the currents aren't equal?

What is the current in the bottom leg of C1?

Poynt,

its not surprising to me they are not equal, but its surprising to me the csr2 current is constanly higher.


The current in the bottom leg of C1 is similar as through csr2, see this point in my earlier video:
https://youtu.be/xTDG_xs8FNU?t=280



Concerning the LTspice sim, i just put it together, i was/am not able to have it run, probably some
run parameters are not specified correctly.

Not sure about the FG parameters either (3V square, 86Hz, 5% duty cycle, DC).

Will try lateron today to get it run.

Itsu
   

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Below is the black box scenario,and is what needs explaining.

CVRs 1 & 2 average current as calculated by the scope.


Brad


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Poynt,

its not surprising to me they are not equal, but its surprising to me the csr2 current is constanly higher.


The current in the bottom leg of C1 is similar as through csr2, see this point in my earlier video:
https://youtu.be/xTDG_xs8FNU?t=280


Itsu

Exactly as it should be O0

CSR2 will have a constant current at this 5% duty, it is the CSR2 which is giving the real current from both sources (battery+C1 in parallel).

The scope can't measure the CSR1 current to the C1 because the battery is 100% of the time across C1, CSR1 sees a current when C1 drops in voltage compared to the battery only.

As I have said T (time) is not taken into consideration when two sources (battery+ cap) do not work the same on discharge and the battery is 100% all the time connected to the cap.

Regards

Mike 8)

PS. The very interesting thing for me is what is happening on the other side of the transistor, the coil and magnet, magnetic compression of a magnetic field, one field to compress and two fields springing back :D




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"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.
   

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Brad

Look at the top of your CSR1 waveform in relation to CSR2, You can see the wave where the battery is putting power back into the C1, over a longer period of time.

You might think that current flows in two directions at the same time :D but we know it does not.

Regards

Mike 8)


---------------------------
"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.
   
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