Effects of Recirculating BEMF to Coil

[rave154]

another oddity

another curious thing...

as ther whole thing is running, if i touch the heat-sink part of the mosfet( no actual heat sink attached but just the metal part on the back of the mosfet, you know what i mean ).....if i touch this..the bedini speeds up slightly....stop touching it.bedini slows down a lil...

[rave154]

i have moved the position of the "input" bulb ( see atatchment) to be before entrance to the coil... bulb barely glows maybe 5%-10% max, output bulb is about 60% bright.

these two bulbs are effectively next to each other in series , the only thing between them is the coil and recycling diode...does this prove that the recycling is adding "energy" into the "output" bulb?

of course, if i switch OFF recycling, the output bulb goes out completely and "input" bulb goes a lil dimmer but still just about glowing dim.

with recycling switched ON..
---------------------------

voltage across output bulb 5.3VDC / 3.5VAC

voltage across input bulb 0.35VDC / 0.53VAC

standing voltage on cap 98VDC

with recycling switched OFF..
----------------------------

voltage across output bulb 0 VDC / 0 AC

voltage across input bulb 0.36VDC / 0.50VAC

standing voltage on cap 94VDC

[gotoluc]

Hi David,

I wrote you a post a few hours ago and I don't see it posted

Anyways, I was asking you to remove the input bulb as it is a waste of energy since you can use the Bedini output cap voltage to find the best tuning efficiency. With no input bulb on my circuit tune the duty cycle and frequency till the output bulb (on my circuit) is at maximum but keep the Bedini cap at the highest voltage as possible without loosing light output. Just make sure the Bedini cap is rated to the voltage you can bring it up to.

Let me know if this helps boost the bulb output. Also ad a capacitor at the output bulb so you are measuring real DC. It will also help in fine tuning to be able to have the DC voltage at the bulb.

Thanks for sharing.

Luc

[rave154]

GOTO, i saw your post ( i think ),

before i change the circuit from how it is in my last schematic, here are some current readings using analog meter...

input bulb 7mA

output bulb 56mA

ok, with the input bulb STILL in the same place as my last schematic, i have added a 100uF,25V cap in parallel with the output bulb, reading gives 6.05VDC ( zero VAC )

[poynt99]

Easier Input Power Measurement

Luc,

Here's a simple circuit suggested by aethertech and similar concept by gyula to ease input power measurements.

I had never tried this before so I PSpiced it up to check against the more direct method.

I'm happy to report that it seems to work quite well. All you need is a couple fairly large electrolytic capacitors and a sufficient wattage 10 Ohm resistor (or close). Connect as shown in the diagram here. If you use a wire-wound 10 Ohm it will filter even a little more.

Take two cheap DC voltage meters (both set on "DC Voltage"), one across R3 to measure the voltage across/current through it (I=V/R), and one across the Vbat point on the diagram to measure output voltage V.

Input power to the circuit is now Pi = V x I

Hope that helps.

.99

[rave154]

ok, now i have removed the input bulb, still have cap across the output bulb, voltage is 6.32VDC ( zero VAC ), bedini Cap voltage is standing at 101 VDC,

the 6 VDC across the bulb seems correct as the does appear to be about 50% bright ( being a 12V bulb of course )

amps through the input side 6.5 mA with no input bulb,

amps through output bulb, 56mA

[poynt99]

A Similarity?

Guys,

this might help in understanding pulsed coils, specifically how we're using them here to step down voltage and step up current. This is formally called a Buck Converter. Read down to about Figure 3 or so:

Buck converter - Wikipedia, the free encyclopedia

and one that is also very good, perhaps better:

Buck-Converter Design Demystified Page of

and the best of all (looks very similar to your circuit Luc), see page 3. The duty cycle determines whether the circuit operates as a buck (step down) or boost (step up) converter. For duty cycle less than 50% (what we are using here), it is a buck (step down) converter. Looks like an excellent article!:
http://www.electronics.dit.ie/staff/...Converters.pdf

.99

[gotoluc]

Luc,

Here's a simple circuit suggested by aethertech and similar concept by gyula to ease input power measurements.

I had never tried this before so I PSpiced it up to check against the more direct method.

I'm happy to report that it seems to work quite well. All you need is a couple fairly large electrolytic capacitors and a sufficient wattage 10 Ohm resistor (or close). Connect as shown in the diagram here. If you use a wire-wound 10 Ohm it will filter even a little more.

Take two cheap DC voltage meters (both set on "DC Voltage"), one across R3 to measure the voltage across/current through it (I=V/R), and one across the Vbat point on the diagram to measure output voltage V.

Input power to the circuit is now Pi = V x I

Hope that helps.

.99

Thanks .99 for looking into a more simple approach for me to use

The highest identical value caps I have that can handle hi voltage are 3900uf. Do you think it would still work?

So I measure voltage across R3 and divide it by the value of R3 which is 10 and measure the voltage across C2 and then what ... I'm not kidding! this is all I can understand from those letters, symbols and your instructions. I never learned this stuff so how would I know to use them.

One formula I learned and understand is: V x V / R = P

If you say I=V to me that means that amps is equal to voltage then divide it by the resistance value. It's like your writing an equation backwards.

Sorry for not understanding what may seem so simple to you.

Is this circuit basically keeping the voltage to a flat DC even though the circuit is being pulsed?

Thanks for your help.

Luc

[gotoluc]

ok, now i have removed the input bulb, still have cap across the output bulb, voltage is 6.32VDC ( zero VAC ), bedini Cap voltage is standing at 101 VDC,

the 6 VDC across the bulb seems correct as the does appear to be about 50% bright ( being a 12V bulb of course )

amps through the input side 6.5 mA with no input bulb,

amps through output bulb, 56mA

Okay David,

thanks for doing this test.

I had asked you to do a test with the bulb attached to the Bedini cap only and you said you measured 6.5 vdc. Now with my circuit attached and without the input bulb wasting energy you can get the output voltage up to 6.32 vdc right!

What is happening here is a conversion is taking place by using my circuit, however we are no creating more energy since the output bulb would need to go above 6.5 vdc which is the voltage measured with the same bulb at the source (Bedini circuit output)

Until you can go over the voltage using the same load connected at the source there is nothing out of the ordinary we can claim that is going on in the circuit.

Are you understanding why I asked you to do the tests? it is to establish the power available from your Bedini energizer source so we can compared to power we reached once you connect my circuit.

Now that you know what you need to pass you can see if there are any other changes or adjustments you can do to beat the 6.5 vdc. Remember to also recheck the Bedini circuit output with the same bulb in case something changes occur.

Hope this helps you and thanks for taking the time to test and share

Luc

[rave154]

can you guys explain something to me,

when both circuits are running, ie the bedini CAP powering the coil / flyback circuit....also the pwm circuit running to turn on & off the coil....the bulb glowing 50%........if i disconnect the negative lead of the pwm circuit from the negative of the 9V battery that is powering it.....the bulb flashes BRIGHT.....almost photo-flash brilliance......ive done this over & over..manually flicking the negative lead across the negative terminal of the battery to produce this flash...i was watching the amp draw of the bedini circuit, it doesnt change as the flashes occur.

the charge on the bedini cap, its harder to see...sometimes the voltage doesnt appear to change much at alkl.other times it drops maybe 5-10V,

as well as that......if i remove the coil completely...and simply have the bulb in its place, ie using the filament of the bulb as the "coil"..... the bulb flashes when i CONNECT the power to the PWM circuit...and doesnt flash when i DIS-connect.....unlike when using the coil AND the bulb.....where it flashes upon DIS-connection and does not flash upon connection.




thoughts?

[petersone]

Hi Luc
Thanks for answering,the only pos. differance is I'm only using 1 batt.but only monitoring the voltage with a dmm,which won't tell the whole story.I'll try some more tests,thanks again.
peter

[poynt99]

Thanks .99 for looking into a more simple approach for me to use

The highest identical value caps I have that can handle hi voltage are 3900uf. Do you think it would still work?

Yes, these should work well too.

So I measure voltage across R3 and divide it by the value of R3 which is 10 and measure the voltage across C2 and then what ... I'm not kidding! this is all I can understand from those letters, symbols and your instructions. I never learned this stuff so how would I know to use them.

Ohm's law is the minimum all need to learn Luc. Without it, it's like walking in the dark without a light. Use this as your starting point if it helps:

V = I x R (easy to remember "VIR")
Which is: Voltage is equal to current times resistance.

By simple manipulation, you can solve for any of the 3 parameters. For example: I = V/R (current is equal to voltage divided by resistance).

So for R3, you have measured the voltage across it, and you know the resistance, so you can calculate the current I from the above equation.

You measure the voltage across the output capacitor C2, and that gives you another voltage to use in the power equation which I keep writing in all my posts (hint hint ):

P = V x I (Power is equal to voltage times current)

So from the above, we have calculated the current I in R3, and we measured the voltage across C2 which now provides us the two values to plug in to the power equation.

In the diagram I posted, let's say the measured R3 voltage was 100mV and the measured voltage at C2 was 1.45V.

First the current: I = 100mV/10 Ohms = 10mA
Second we already measured VC2 as 1.45V
Third: P(in) = 1.45V x 10mA = 14.5mW of power.

One formula I learned and understand is: V x V / R = P

Yes, this the P = V x I I have been chanting all along. V/R is I (as I mentioned above).

If you say I=V to me that means that amps is equal to voltage then divide it by the resistance value. It's like your writing an equation backwards.

I hope I didn't say that, that is not correct

Sorry for not understanding what may seem so simple to you.

No apologies necessary

Is this circuit basically keeping the voltage to a flat DC even though the circuit is being pulsed?

Thanks for your help.

Luc

It's actually keeping the current through R3 "smooth" (or flat) so it is easy to read with any meter. The voltage at C2 should already be fairly steady, so again easy to read with a meter. So now you have two easy to read values which when crunched through your calculator will give you an accurate value of power taken from the source battery or source supply.

.99

[rave154]

Hi all,

with regards to the circuit presented in post #137.....

im still puzzled WHY when i disconnect the negative lead of the PWM circuit ( powered by a 9v battery ).....the "output" bulb flashes brilliant..

i have dabbed the lead on & off by hand..producing the flashes.......but i want to "transistorize" / "PWM-erize" the effect.....in other words i want a pwm / transsitor that continually connects & disconnects that negative lead from the negative of the 9v battery powering the PWM circuit....

any ideas how to do this ?

David.D

[gyula]

Hi David,

I think your output lamp can flash brilliantly whenever the MOSFET switches off and the flyback pulse thus created makes it brilliant.
Does this make sense ?

I assume your present pwm circuit should do the on / off switching of the MOSFET?

rgds, Gyula

[rave154]

gyula, yes my pwm circuit switches the mosfet on & off....

what i cant understand is..why when i shut down the whole pwm circuit ( disconnect the negative lead from the battery )........does the bulb flash bright?......shouldnt it just go off?

im confused.......but......i really want to try putting a controlable breaker inbetween the negative lead and the negative terminal of the battery...so i can produce the flash over and over..and control frequency etc

help......please

David. D