Re-Inventing The Wheel-Part1-Clemente_Figuera-THE INFINITE ENERGY MACHINE

[gyulasun]

Hi NRamaswami,

Thanks for your answers, and regarding my answers, here are some of them, I will try to consider the rest of your further questions in your Reply #2188 tomorrow.

With the electromagnet coil of 4 times as long wire for the unfiltered DC output of the diode bridge (you use the expression "pulsed DC") you increase the DC resistance of the coil also 4 times. This insures that the fuse or the circuit breaker would not blow out as you defined the condition for a stable electromagnet.

Without the diode bridge, the electromagnet coil receives normal 50 Hz AC voltage, the coil's inductive impedance is already high enough not to blow out the fuse, so you do not need to use 4 times as long wire for the coil like for the unfiltered DC output of the full wave bridge.

Notice that the output of the diode bridge feeds the coil with a DC voltage too because the positive sine wave peaks have an average DC value which amounts to 63.7% of the peak AC value. This means that in case of a 220V AC input, the positive peak value of the sine waves at the output of the diode bridge will be about 310V (neglecting diode losses) and the average DC will be 0.637*310V=197.5V  this DC voltage biases the electromagnet's core of course with a certain polarity and heat dissipation takes place in the DC resistance of the coil which you increased 4 times for this diode bridge fed case.

See info on the average DC output of a full wave rectifier for a resistive load:
http://www.electronics-tutorials.ws/diode/diode_6.html 

I attached the drawing taken from the link to show the average DC value with respect to the peak sine wave amplitudes at the output of a full wave rectifier (i.e. diode bridge).  In the link, the effect of using a puffer capacitor at the diode bridge output is also shown.

Gyula

[RandyFL]

5V-transformerless-powersupply.png


If you have a switch, plus a 2 amp circuit breaker and a .500 fuse in line...would that be safe?

[gyulasun]

Hi NRamaswami,

First some additions, these are needed because the info below may seem to be in conflict with my previous attached figure which showed the resulting output voltage waveform and the level of the DC voltage it averages to at the output of a full wave rectifier. While this can be true, we have to consider we have an L inductance across the output as a load.
When you connect a coil to the output of a full wave rectifier, in fact this is not exactly a resistive load because of the inductance part of the coil is present in series with the wire DC resistance.
I mention this because yesterday I considered the coil as if its resistive value only which increased when you increased the wire length 4 times to work for you as a 'stable' electromagnet, but of course the longer wire involved an inductance increase for the coil too.
I referred to the output voltage of the diode bridge as an unfiltered DC voltage and normally a puffer capacitor is used to store voltage (charge) between the sine wave peaks, to minimize the voltage amplitude fluctuation.
When you connect a coil across the output of the diode bridge instead of a puffer capacitor (so no capacitor is present but the coil), it should be the current peaks which average out, the coil functions as a current choke (regardless of its assigned electromagnet role). The output voltage of the diode bridge still remains fluctuating between a zero and a positive peak amplitude, while the coil fights against any current change as usual for a coil, this is why it is the output current which can average out and the higher the inductance of the coil, the less fluctuation the current will have, hence the output current can become a cleaner and cleaner DC current with less and less ripple value.

You asked what a pulsed DC was. Because we know you used it to refer to the output voltage coming from a full wave diode bridge rectifier, we know what you meant but if this reference is not stressed or included, the term pulsed DC may be misleading, it means for instance a DC voltage periodically interrupted by a switch, i.e. a series of pulses. (This is why I asked you about a switch if you recall but of course you did not use any switch.)

A 'pulsating DC' could be a much better term instead of the 'pulsed DC' to use, conventional electrical engineering normally uses the term unfiltered DC for naming the voltage or current at the output of a rectifier (both for a half wave or full wave one).
And depending on what kind of load you hook up to the output of a full wave rectifier, now you know when using a coil it is the output DC current which averages, or when using a puffer capacitor it is the output DC voltage which averages, both the output current and the voltage will have less and less fluctuation i.e. less ripple, when the coil's L inductance or the puffer capacitor's C capacitance is increased, respectively.
When you use a purely resistive load across the diode bridge output, both the current in the resistor or the voltage across the resistor remains fluctuating between a zero and a peak value (i.e. as per the series of the positive sine wave peaks show for the voltage in the previous attached figure),  none of them can average out to a clean DC quantity with a minimum ripple (of course the heat dissipated in the resistor averages to a certain value in this case too).

To be continued tomorrow.

Gyula

[NRamaswami]

Hi TK

I'm very grateful and am obliged for the answers and clarifications provided. I agree with your explanations No. 1 and 4. I am not able to understand the clarification No. 3.

I most respectfully beg to disagree with you on the clarification No. 2

Here is your clarification No. 2

Second, the "220VAC" that the wall plug provides is 220 V RMS. This means that the _peak_ voltages are quite a bit higher: about + and - 310 V. When this is run through an unflitered Full Wave diode bridge, the DC Peaks as shown in the waveform below will be at +310V, minus a little bit for the fed voltage drop of the bridge. An averaging meter will then knock off some more from this value -- 270 volts DC might well be an indicated "average" from this kind of waveform. Once this DC output from the bridge is filtered by capacitors, the voltage will be steady and near the _peak_ value of the AC input... not the RMS value. So it is not surprising that a bridge rectifier can put out DC voltage measurements that are much higher than the "nameplate" voltage input (which will usually be an RMS value.) I think this fully explains the Variac results reported by NRamaswami: nothing unusual happening, just some misunderstandings about FWB action, RMS vs. Peak values, and the averaging functions of meters.

I apologize to disagree with you here but effectively what you are saying here is this..

a. There is no loss in the Variac and it is a 100% efficient device. Not true.

b. There is some loss in diode bridge which you acknowledge and it is around 20-22 watts as measured by us. (Does it include AC to DC conversion losses? Here in India it is normally rated at 15% of the input as a thumb rule.. I'm afraid that You do not provide for this)

c. There is no loss in the coil due to either resistance or inductance or eddy currents or heat dissipation..All these losses are there

d. The losses due to the 5 x 200 watts lamps in parallel are not considered.

e. And after overcoming all the above the FWB (full wave diode bridge rectifier) can on its own increase the output voltage which is DC to 270 volts. In DC V=IR So if V increases I also should increse as the resistance if fixed. If both of them increase wattage should increase. So if I agree with you a simple FWB can act as a device that can save not only a lot of energy but can boost the energy output as well when connected directly to mains.

f. Please connect a FWB alone and let us remove the variac and the coil and see if you can get 220 volts to be increased to 270 volts..It should happen if what you say is correct. It would not.

Most times theoretical explanation is correct. Some times practical observations differ from theory. With due respect I would request you to check if you are able to practically observe if connecting the FWB to the mains at 220 volts automatically boosts the output voltage in the meter to 270 volts. It would not. There is another phenomena that is involved here in my very humble opinion and observation. But I do not know if it correct also and it is only a guess. I would post my guess after Gyula has completed.

I again remain very obliged and am grateful for your explanations. My knowledge is not sufficient to understand what you have stated in No. 3 of your explanation and I apologize for my lack of knowledge at this time.

[TinselKoala]

Hi TK

I'm very grateful and am obliged for the answers and clarifications provided. I agree with your explanations No. 1 and 4. I am not able to understand the clarification No. 3.

I most respectfully beg to disagree with you on the clarification No. 2

Here is your clarification No. 2

Second, the "220VAC" that the wall plug provides is 220 V RMS. This means that the _peak_ voltages are quite a bit higher: about + and - 310 V. When this is run through an unflitered Full Wave diode bridge, the DC Peaks as shown in the waveform below will be at +310V, minus a little bit for the fed voltage drop of the bridge. An averaging meter will then knock off some more from this value -- 270 volts DC might well be an indicated "average" from this kind of waveform. Once this DC output from the bridge is filtered by capacitors, the voltage will be steady and near the _peak_ value of the AC input... not the RMS value. So it is not surprising that a bridge rectifier can put out DC voltage measurements that are much higher than the "nameplate" voltage input (which will usually be an RMS value.) I think this fully explains the Variac results reported by NRamaswami: nothing unusual happening, just some misunderstandings about FWB action, RMS vs. Peak values, and the averaging functions of meters.

I apologize to disagree with you here but effectively what you are saying here is this..

a. There is no loss in the Variac and it is a 100% efficient device. Not true.

b. There is some loss in diode bridge which you acknowledge and it is around 20-22 watts as measured by us. (Does it include AC to DC conversion losses? Here in India it is normally rated at 15% of the input as a thumb rule.. I'm afraid that You do not provide for this)

c. There is no loss in the coil due to either resistance or inductance or eddy currents or heat dissipation..All these losses are there

d. The losses due to the 5 x 200 watts lamps in parallel are not considered.

e. And after overcoming all the above the FWB (full wave diode bridge rectifier) can on its own increase the output voltage which is DC to 270 volts. In DC V=IR So if V increases I also should increse as the resistance if fixed. If both of them increase wattage should increase. So if I agree with you a simple FWB can act as a device that can save not only a lot of energy but can boost the energy output as well when connected directly to mains.

You are apparently agreeing with something that I never said. A FWB has losses, as you have noted above, and the more current you draw through it the more heat (power) it will dissipate. Voltage is not energy, current is not energy, power (especially peak power) is not energy. Every stage in your system has losses: the Variac, the FWB, any coils connected, the interconnecting wiring, and of course the power dissipated in the load. I am trying to explain the _voltage readings_ you have reported. I have not dealt with any losses caused by load connections or wastage in the bridge, except to point out that there will be a slight drop in _voltage_ below the peak value due to the forward voltage drop of the diodes in the bridge. The _energy_ available at the load will be less than the _energy_ delivered by the mains to the system, because of the losses in the bridge, Variac, wiring and other components. The peak _voltage_ can be much greater than that delivered by the mains. Instantaneous output current, and instantaneous output power levels, can also be higher than the average, or RMS values at the input. This does not mean that energy is increased.

f. Please connect a FWB alone and let us remove the variac and the coil and see if you can get 220 volts to be increased to 270 volts..It should happen if what you say is correct. It would not.

I suggest that YOU do this test yourself, and report your results. You may be surprised.

Most times theoretical explanation is correct. Some times practical observations differ from theory. With due respect I would request you to check if you are able to practically observe if connecting the FWB to the mains at 220 volts automatically boosts the output voltage in the meter to 270 volts. It would not. There is another phenomena that is involved here in my very humble opinion and observation. But I do not know if it correct also and it is only a guess. I would post my guess after Gyula has completed.

I again remain very obliged and am grateful for your explanations. My knowledge is not sufficient to understand what you have stated in No. 3 of your explanation and I apologize for my lack of knowledge at this time.

I suggest you try the following test yourself, and report your results here completely.

Simply connect the AC input to the FWB to your mains supply, not using a Variac. Then read the DC output voltage of the FWB (no load) using your meter. Does it read "220 VDC" or something higher?

To measure the voltages and power balance properly in circuits such as you are working with, you really need an oscilloscope and the skill to use it correctly. Meters are a poor substitute and have often led people astray in these matters.

As far as the #3 goes, the issue of DC and inductances... as I have tried to explain, the leading and trailing edges of a DC pulse will indeed interact with inductances in the normal way. As the current through the inductor increases at the leading edge of a DC pulse, the magnetic field builds in the inductor and this takes a certain amount of time (more inductance, more time, and vice versa) and as the current decreases at the trailing edge of a DC pulse the magnetic field collapses and tries to keep the current going, and this also takes a certain amount of time. At the trailing edge of the DC pulse, the energy that was stored in the magnetic field of the inductor can "ring" by passing back and forth between the inductance (magnetic field) and the inherent capacitance of the circuit (electric field) and will produce a characteristic "ringdown" waveform. The frequency of this ringdown waveform can then be used to calculate the inductance of the inductor.  Perhaps the following video will help to clear up this idea:
http://www.youtube.com/watch?v=Qx3B89379eQ