[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
http://www.circuitstoday.com/half-wave-rectifiers *(ii)**Disadvantages:1.* The output current in the load contains,* in addition to dc component*, *ac components of basic frequency equal to that of the input voltage frequency*. Ripple factor is high and an elaborate filtering is, therefore, required to give steady dc output. *(iii)*2.The power output and, therefore, rectification efficiency is quite low. This is due to the fact that power is delivered only half the time. *(iv)*3.Transformer utilization factor is low. *(v)*4.*DC saturation of transformer core resulting in magnetizing current and hysteresis losses and generation of harmonics.* transformer are not perfect and saturation solve the DC component problem. 2013/5/27 David Roberson dlrober...@aol.com The concept mentioned below by Duncan is not correct. The DC current that flows into the resistor from the wall socket finds a short circuit to ground in the power transformer center tap in most cases.
Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
Half wave rectifiers are not the way to go. They have been all but abandoned in the electronic world because of the issues you have found. Full wave bridges eliminate the DC component from the mix and should be used. This does not suggest that accurate power measurements can not be obtained from the AC waveforms. This can be done and is the reason for much of the discussion taking place. Dave -Original Message- From: Alain Sepeda alain.sep...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Mon, May 27, 2013 2:19 pm Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments http://www.circuitstoday.com/half-wave-rectifiers (ii)Disadvantages:1. The output current in the load contains, in addition to dc component, ac components of basic frequency equal to that of the input voltage frequency. Ripple factor is high and an elaborate filtering is, therefore, required to give steady dc output. (iii)2.The power output and, therefore, rectification efficiency is quite low. This is due to the fact that power is delivered only half the time. (iv)3.Transformer utilization factor is low. (v)4.DC saturation of transformer core resulting in magnetizing current and hysteresis losses and generation of harmonics. transformer are not perfect and saturation solve the DC component problem. 2013/5/27 David Roberson dlrober...@aol.com The concept mentioned below by Duncan is not correct. The DC current that flows into the resistor from the wall socket finds a short circuit to ground in the power transformer center tap in most cases.
[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
On Mon, May 27, 2013 at 12:33 PM, Andrew andrew...@att.net wrote: ** No separate DC power source is necessary if Duncan's diode fudge is used. Yes, but the point is that you'd need to intentionally tamper with the mains to pull it off, i.e., it implies fraud, if I'm understanding the diode fudge, in light of the DC short to ground in the transformer. Put another way, is there any way to draw hidden current from the mains that does not entail tampering between the transformer and the mains? Earlier Duncan seemed to think that there was a way, but this particular approach does not appear to be an example. But perhaps I'm misunderstanding an important detail? Eric
Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
You are correct Eric. For some reason the skeptics amoung us do not want to understand this issue. I suspect that it is some form of game they are playing. Dave -Original Message- From: Eric Walker eric.wal...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Mon, May 27, 2013 3:38 pm Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments On Mon, May 27, 2013 at 12:33 PM, Andrew andrew...@att.net wrote: No separate DC power source is necessary if Duncan's diode fudge is used. Yes, but the point is that you'd need to intentionally tamper with the mains to pull it off, i.e., it implies fraud, if I'm understanding the diode fudge, in light of the DC short to ground in the transformer. Put another way, is there any way to draw hidden current from the mains that does not entail tampering between the transformer and the mains? Earlier Duncan seemed to think that there was a way, but this particular approach does not appear to be an example. But perhaps I'm misunderstanding an important detail? Eric
