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15th August 2010, 03:17 PM  #21 
diyAudio Member

Power factors when dealing with nonsinusoidal voltage waveforms are a bit more complex to understand. A rectified AC signal when fed to a reservoir cap will have a current waveform which doesn't look anywhere near a sinewave so its hard to estimate the power factor which is normally just calculated from the phase difference between the current and voltage waveforms.
The general principles are the same though  the larger the cap is, the lower the ripple voltage and therefore the shorter the time the rectifiers will be conducting. During this short time, the charging current into the caps must be high enough to recharge the caps for the next mains cycle (10mS with fullwave). Take a simple example  if the rectifiers are conducting for 20% of the time (2mS), the current pulses through the rectifiers will need to be 5X bigger than the DC output load current. Fitting a bigger cap will decrease the 2mS time and therefore increase the size of the current pulses. With me so far?
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'The total potential here must be nothing less than astronomical.' 'Nothing less. The number 10 raised almost literally to the power of infinity.' 
15th August 2010, 03:20 PM  #22 
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Join Date: Jul 2004
Location: Scottish Borders

I have no problem with that.
As I said psud makes complete sense to me.
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regards Andrew T. 
15th August 2010, 03:26 PM  #23 
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Join Date: Sep 2002
Location: Lakewood, Ohio

A rectified AC signal when fed to a reservoir cap will have a current waveform which doesn't look anywhere near a sinewave
OK The general principles are the same though  the larger the cap is, the lower the ripple voltage and therefore the shorter the time the rectifiers will be conducting. During this short time, the charging current into the caps must be high enough to recharge the caps for the next mains cycle (10mS with fullwave). OK Take a simple example  if the rectifiers are conducting for 20% of the time (2mS), the current pulses through the rectifiers will need to be 5X bigger than the DC output load current. Fitting a bigger cap will decrease the 2mS time and therefore increase the size of the current pulses. OK Now go on about the power factor part. And why should we even care about power factor.
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Kevin 
15th August 2010, 03:36 PM  #24 
diyAudio Member

The power factor part is just another way of expressing that you've got much higher currents in your secondary circuit than you would have if there was only a resistive load, not a rectifiercapacitor load.
Why you should care about power factor is for the reasons already explained  your transformer's secondary current is specified into a resistive load, not a capacitive one. Hence the derating Andrew has already mentioned. <edit> Having a poor power factor in this case means your mains waveform is going to get distorted (3rd/5th/etc harmonic distortion)  flattened tops. This might have implications for other appliances you have connected.
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'The total potential here must be nothing less than astronomical.' 'Nothing less. The number 10 raised almost literally to the power of infinity.' Last edited by abraxalito; 15th August 2010 at 03:40 PM. 
15th August 2010, 03:46 PM  #25 
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Join Date: Sep 2002
Location: Lakewood, Ohio

When I think of "Power Factor" I think of phase (timing) differences between the Voltage and Current waveforms. I'm not sure that "3rd/5th/etc harmonic distortion" will cause great phase differences.
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Kevin 
15th August 2010, 03:48 PM  #26  
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Quote:
Quote:
Quote:
Abrax, Schem has gone very quiet. Maybe he's still asleep. Let him justify what he alleged.
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regards Andrew T. Last edited by AndrewT; 15th August 2010 at 03:51 PM. 

15th August 2010, 03:54 PM  #27  
diyAudio Member

Quote:
Quote:
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'The total potential here must be nothing less than astronomical.' 'Nothing less. The number 10 raised almost literally to the power of infinity.' 

15th August 2010, 04:01 PM  #28 
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Join Date: Sep 2002
Location: Lakewood, Ohio

Andrew, I think that I see what the problem is!
There are other definitions for "Power Factor". We think of it only as phase shift but another definition is of a resistance to impedance ratio. I have to a football (soccer) game for 6 year olds now.
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Kevin 
15th August 2010, 05:43 PM  #29  
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Join Date: Nov 2003
Location: Brighton UK

Quote:
Optimising power factor is part of optimising conversion efficiency. But optimising conversion efficiency is not optimising power factor, unless that is the only issue being considered, e.g. for a motor. Here they are saying a "too big cap" reduces conversion efficiency. (AFAICT) rgds, /sreten. Last edited by sreten; 15th August 2010 at 05:45 PM. 

15th August 2010, 10:43 PM  #30 
diyAudio Member

There are two kinds of power factor, PF and DPF (displacement power factor). Displacement refers to the fundamental only. DPF can be corrected with PFC inductors/capacitors; in general, PF cannot.
For linear loads, PF == DPF and you get simple sinusoidal power factor, phase angle, etc., all familiar from Power Systems. You can use L and C for power factor correction. A rectifier is nonlinear. It produces strong harmonics in the current draw. The fundamental is nearly in phase, so DPF is very close to 1.00, even though PF can be less than 0.5. Distorted current waveforms cannot, in general, be corrected to any useful amount by adding capacitors or inductors to the line. The problem arises from harmonics, which must be filtered using a lowpass filter. Such a filter is obviously going to be hard to construct at line frequency and impedance. Better is to simply avoid harmonics altogether. Using a minimum of filter capacitance helps. PFC circuits are designed for ideal behavior, while allowing any arbitrary amount of filter capacitance (as long as stability is not impaired). Notice that a PFC circuit always has sinusoidal, doublelinefrequency ripple at the output. The DC voltage can be stabilized (generally something like 400V with good regulation), but the ripple is a necessary part of the circuit's nature. The general definition of power factor is this: S = Vrms * Irms S is Apparent Power, measured in VA. In general, S is NOT watts (work, heat, etc.). Real power can be measured with respect to time, in which case, it varies up and down with demand. Reactive components cause negative power, as they deliver stored energy back to the supply every cycle. The average is real power, in watts. Instantaneous power: p(t) = v(t) * i(t) Real power: P = p(t) averaged over any number of whole cycles PF = P / S. If you have 3A RMS current draw at 120V, with a real power consumption of 200W, the apparent power is 120*3 = 360VA and the power factor is 200/360 = 0.56. Example: If current is drawn in spikes, p(t) is also in spikes. A spike of 1A at a peak of ~340V (i.e., 240VAC RMS) is 340W peak (instantaneous) power. If that spike is 2ms long, triangular shape (a typical approximation for a capinput filter), and the half cycle is 10ms (50Hz line frequency, full wave rectified), the average is 34W. The RMS current is 0.258A (Irms = Ipk * sqrt(t_on / (3*T)) for a triangular waveform of period T with dead time T  t_on), so S = 62VA. Then, PF = 34/62 = 0.549, a very reasonable value. Tim
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