More Watts Out Than In? P2

[stivep1]

what you need to do is to connect light bulb at the end and measure luminescence .

Than connect the same light bulb at the input and measure luminescence as well and compare.

Than

take DC circuit and check luminescence for two instances based on previously calculated power.

while checking luminescence put light bulb in the tube to insure the same distance to the light strength meter.

I do not like approach of rob790 . IGNORANT

Get a lab. I did it.And I never stated I'm the best.

[SuperDisbelief]

Are there complex formulaes? I guess it is easy to calculate things wrong. Did you put a load on the putput (something that really uses the power)? That might be the missing piece.

[kapierenundkopieren]

Get a scope with RMS reading... that circuit is similar to a joule thief and i bet there are more spikes which your scope doesen't see.

[Mushketer888]

big thanks for vid!

[fgleich]

why not increase the output number of turns to get a greater output voltage, then add a filter capacitor to the output to make a DC supply, and then use that as the supply for the entire circuit ? ditch the battery and use the output generated, if that worked, you would know it's the real thing.

[blaargenheimer]

Have you tried turning up the juice a bit to see if you can get enough back through to run the 555 timer?

[klexn]

I agree with KokomoJ0.. Measure DC in then convert it to a clean DC out and measure your power differences. Don't use a resistor on your input or you will need to add it to your power consumption calculations. it's much easier to measure DC power in and out any day of the week.

[Drevtoobe]

This clip is way way off the mark. I hope that you have been going up the learning curve in the two years since it was made.

[openmindmbj]

Also, consider compensating for the values due to the waveform. (ex. If it were a sinusoidal waveform, rms expressed by amplitude x sqrt of 2, would be sufficiently accurate.) Or better try computing power per pulse.

You could use the oscilloscope to read values of V & I, & the displacement. Thank you for sharing. I hope this helps.

[openmindmbj]

Camster,

You have complete resources for your experiment. I would suggest you remove the resistor in sec A.Take input voltage, V1 and current, I1 readings including the displacement angle between V1 & I1. Take output voltage,V2across the resistor in sec B & current, I2 in series, include displacement angle. Then express your computations in vectors. Very basic yet very effective.

[Jobob542]

That's very interesting I built a circuit just like that except it was a step up instead of a step down triodal transformer and it had a huge out put compared to the in put, it was 12volts input at .560A when the out put leads were shorted, and on the output was 50 or 60 volts probably more at .230A.Which was 6.72watts and the input was 11.5 watts {at 50 volts} which is almost TWICE the in put. But I find it hard to believe these readings, and I dont have an oscilloscope to test it:(

[narivasant]

The first resistor's watts is not the power going into the coil and on over to the other resistor. It is only the power lost through the first resistor, with all the rest of the power going on to the 2nd resistor.

Sorry but, no big deal.

You have to measure the power going into the coil and over to the other resistor.

Dissipately some power with the first resistor does not do that.

[bill99959]

??? WHAT????? P=EI is what is being sunk or lost in the resistor as heat, you are losing some on the input then losing even more at the output...

[KokomoJ0]

meter error. you need to convert both the input and output to a pure dc then measure the dc. meters dont know how to measure the threashy wave you are putting into it. get a tranmission line ic and configure it to a unity gain buffer then calibrate your bridges for both and and measure the dc value in and compare to the out. Its the only way to get accurate measurements.

[mcpotplant]

use a better bridge rec and not use so m,much

[rob790]

Camster, you NEED to read this document. You are not correctly calculating the power.

edn (dot) com (slash) contents (slash) images (slash) 46857.pdf

What you need is a thermal TRUE RMS meter my friend. Go to eBay and get a used one for less than $100.

[camster6]

So what we need to do is get rid of the bridge and calculate rms on the coil output to determine true current. There is also a formula to calculate the current of the periodic wave form.

Perhaps a revisit to this experiment is needed. Great comments thanks.

[rob790]

You need to use the same resistor, place a thermal true RMS meter across it and only THEN use P = Vrms^2/R to measure the power. You don't need to get rid of the bridge, you just need to correctly measure the energy, taking into account the true RMS value across R.

[rob790]

You CAN NOT use P = IV in this situation! That equation is strictly for DC cases. You obviously have a very non standard AC type of current and voltage going on with your resistor. Power is better measured with the proper equipment where P = V^2/R where V^2 is averaged over the time of measurement. In other words, you need specialized circuits to properly measure the power. Sorry, but your calculations are worthless.

[camster6]

The input is pulsed DC, the output is that pulsed DC and BEMF bridged to DC. The first resistor is on the Battery the second on the bridge. The resistors only see DC as the electron only flows one way through both resistors. Perhaps I did not explain that part well enough, sorry for any confusion.

[PszemoI]

Tell me, please, what exactly do you mean when you say "the resistors only see DC"?

Thanks.

[IceBerggeo]

You still have the varying voltage across the resistor, and current through the resistor. Yes, it might only SEE a POSITIVE voltage, but it is certainly NOT DC. Also, were you able to factor in the voltage drop across the bridge rectifier? That should also have been a part of your calculations, especially if you're only dealing with 3 volts across the resistor and dropping across two diodes.

[IceBerggeo]

I'm going to disagree with you as you CAN use P = I*V ONLY if you are using the rms measurements from the meters. In which case, they all need to be able to measure true rms or, you use the DC values given and calculate the rms from the length and duty cycle of the waveform.

[wcernuska]

Camster,

The 555 Timer outputs 5Vp-p and the final output is 3Vp-p. Is it OK to calc Watts with DC Volts and Amps? Don't you need to consider volts accross coil as well as resistor and bridge as well as load resistor? I want to learn more please. Very interesting stuff. I watched many times.