Power Conversion Electronics

Tim Williams wrote:

Eeyore wrote:

It's mentioned in a paragraph or two.

A search couldn't find it.

Second paragraph from link,

"In circuits having only sinusoidal currents and voltages, the power
factor effect arises only from the difference in phase between the
current and voltage. This is narrowly known as "displacement power
factor". The concept can be generalized to a total, distortion, or
true power factor where the apparent power includes all harmonic
components."

You said initially "displacement factor" without the power hence the
absence of hits.

It would be good if phase only related PFC ( motors etc ) could be
better distinguished from harmonics. The average bod gets confused about
the whys and wherefores of the two.

Graham

Franc Zabkar wrote:
On Sun, 26 Apr 2009 22:48:56 -0700, UCLAN put
finger to keyboard and composed:

There is a big 400vdc capacitor (or two 200vdc caps in series) just after
input rectifiers on the AC input. Note C5 and C6 on:

http://www.pavouk.org/hw/en_atxps.html

They charge to the peak value of the input
AC voltage, or 1.414 times the RMS value. Since the cap(s) draw their maximum
current when at lowest charge (zero cross-over point), and draw their least
amount of current when charged to their highest point, the current waveform
*leads* the voltage waveform by 90 degrees. [Maximum current is at the same
time as minimum voltage; minimum current is at the same time as maximum voltage.]

As much as I dislike the man, he's right.

Look at Fig 6 on page 2 of the application note you linked to
elsewhere in this thread.

It does show that the current leads - look at how it's not positioned
symmetrically about the 90 degree point, for example. It's not a 90
degree lead, of course, but it's still a lead.

Mind you, I think the graph is wrong. When it's flowing, the current
into a capactor should be proportional to the rate of change of voltage.
So you should see a rapid rise once the input voltage goes above that
already on the capacitor, followed by a progressive reduction towards
zero (plus the load current) at the voltage peak. I suppose you'd get
something closer to what's shown if there were an inductor in series
with the supply.

Sylvia.

On Wed, 29 Apr 2009 12:28:36 +1000, Sylvia Else
put finger to keyboard and composed:

Franc Zabkar wrote:
On Sun, 26 Apr 2009 22:48:56 -0700, UCLAN put
finger to keyboard and composed:

There is a big 400vdc capacitor (or two 200vdc caps in series) just after
input rectifiers on the AC input. Note C5 and C6 on:

http://www.pavouk.org/hw/en_atxps.html

They charge to the peak value of the input
AC voltage, or 1.414 times the RMS value. Since the cap(s) draw their maximum
current when at lowest charge (zero cross-over point), and draw their least
amount of current when charged to their highest point, the current waveform
*leads* the voltage waveform by 90 degrees. [Maximum current is at the same
time as minimum voltage; minimum current is at the same time as maximum voltage.]

As much as I dislike the man, he's right.

Look at Fig 6 on page 2 of the application note you linked to
elsewhere in this thread.

It does show that the current leads - look at how it's not positioned
symmetrically about the 90 degree point, for example. It's not a 90
degree lead, of course, but it's still a lead.

A bigger capacitor would result in less ripple, which means less lead.
If you were relying on intuition alone, you would expect that the PF
would move closer to unity. In fact the PF actually becomes worse.

See the calculations in Phil's example.

- Franc Zabkar
--
Please remove one 'i' from my address when replying by email.

Franc Zabkar wrote:
On Wed, 29 Apr 2009 12:28:36 +1000, Sylvia Else
put finger to keyboard and composed:

Franc Zabkar wrote:
On Sun, 26 Apr 2009 22:48:56 -0700, UCLAN put
finger to keyboard and composed:

There is a big 400vdc capacitor (or two 200vdc caps in series) just after
input rectifiers on the AC input. Note C5 and C6 on:

http://www.pavouk.org/hw/en_atxps.html

They charge to the peak value of the input
AC voltage, or 1.414 times the RMS value. Since the cap(s) draw their maximum
current when at lowest charge (zero cross-over point), and draw their least
amount of current when charged to their highest point, the current waveform
*leads* the voltage waveform by 90 degrees. [Maximum current is at the same
time as minimum voltage; minimum current is at the same time as maximum voltage.]
As much as I dislike the man, he's right.

Look at Fig 6 on page 2 of the application note you linked to
elsewhere in this thread.
It does show that the current leads - look at how it's not positioned
symmetrically about the 90 degree point, for example. It's not a 90
degree lead, of course, but it's still a lead.

A bigger capacitor would result in less ripple, which means less lead.
If you were relying on intuition alone, you would expect that the PF
would move closer to unity.

In fact the PF actually becomes worse.

With a larger capacitor, you get less lead, but higher instantaneous
currents. My intuition would be that the net result is far from obvious.

However, the higher current for shorter periods means higher harmonic
currents, which don't contribute to the power, so it wouldn't surprise
me that the PF would go down.

Sylvia.