# PFC circuit at end of a long resistive line

M

#### mook johnson

Jan 1, 1970
0
Gents,

Have any of you run a PFC boost circuit from a 50 or 60Hz AC line with a
loop resistance on the order of 100 ohms?

We are currently looking for the most efficient method of delivering
1Kwatt through a lossy cable.

Method one is the simple step down transformer followed by a rectifier
bridge. The concern here is the poor PF of the rectifier bridge had
large capacitor bank.

The other option being considered is a PFC after the transformer. This
should bring the PF closer to unity but the concern is of there will be
a stability issue with the resistive input power.

Anyone have experience or care to speculate about the doability of a PFC
running from a resistive source in the 100 ohm range?

D

Jan 1, 1970
0
Yes.
Convert to high voltage DC.
Run DC through cable.
Step down at load; convert to AC if necessary.

VLV

"Step down" is typically an AC term. DC (power wise) is usually
rectified AC, so the starting AC feed would dictate the voltage. The
load end would need to go into a switch mode of sorts for whatever final
voltages it would need to produce for the load device(s).

M

#### mook johnson

Jan 1, 1970
0
How much voltage is applied at the powered end of the line?

If 120 volts, the most power you can get at the far end is 36 watts. To get a KW
at the end, you need at least 632 volts in, and even then you'd have a kilowatt
of line loss.

So think kilovolts. DC maybe.

This is a HV application. We have up to 750VRMS on the supply end with
4KW available. The cable that long can handle the 1KW of dissipation.

On the other end we are stepping down 5:1 to generate somewhere about
100V needed for the load (100V @ 10A)

U

#### Uwe Hercksen

Jan 1, 1970
0
John said:
OK, sounds like a 60 Hz stepdown transformer could drive a PFC-corrected
switching power supply. You should test it, but I suspect the usual PFC supply
won't mind the fairly high source impedance.

The transformer will see the full line voltage, 750, when unloaded. It would be
prudent to use a transformer of about 3:1 ratio or so, so the switcher never
sees more than 250 RMS if it's lightly loaded.

Hello,

if the cable ist not very long (shorter than a quarter wavelength on the
cable), it may be better to use 400 Hz instead of 60 Hz. The capacitors
after the rectifier may be much smaller then.
What about using 3 phase transmission and a 12 pulse rectifier bridge?
Capacitors may be again much smaller.

Bye

B

#### Bill Sloman

Jan 1, 1970
0
Gents,

Have any of you run a PFC boost circuit from a 50 or 60Hz AC line with a
loop resistance on the order of 100 ohms?

We are currently looking for the most efficient method of delivering

1Kwatt through a lossy cable.

Method one is the simple step down transformer followed by a rectifier
bridge. The concern here is the poor PF of the rectifier bridge had
large capacitor bank.

The other option being considered is a PFC after the transformer. This
should bring the PF closer to unity but the concern is of there will be
a stability issue with the resistive input power.

Anyone have experience or care to speculate about the doability of a PFC
running from a resistive source in the 100 ohm range?

I'm always happy to speculate. One approach would be to put a bridge rectifier on the driven end of the cable, and a DC-to-DC converter at far end.

A 100% efficient DC-to-DC converter would be sucking out about 1.73A rms, dropping 173V RMS in the cable, leaving 577V rms at the far end. If you are going turn this into a steady 10A at 100V, while pulling half-sine waves ofcurrent through the cable you are going to have to store enough energy in your reservoir capacitor during the high "quarter" segments of the waveformto fill in the low "quarter" segments of the waveform.

For a 50Hz source that's - crudely - about half of 5msec worth of 1kW or 10joules.

You couldn't charge your your capacitor to higher than 816V at the peak of the sinewave (at the far end of the cable) nor let it sag to lower than 100V before you started charging it up again which makes your minimum capacitance about 7.6uF. Farnell Australia stocks some 1uF 1kV capacitors which might do the job for about $A5 each in small quantities. It sounds practical, but I've got no idea what the DC-to-DC converter wouldlook like - a web search should show up something without too much effort,but I'm still jet-lagged after getting back to Sydney on Sunday night, andit's quite likely that I've already screwed up somewhere in the above narrative, so putting in any more effort at this point wouldn't be a good idea. M #### mook johnson Jan 1, 1970 0 I'm always happy to speculate. One approach would be to put a bridge rectifier on the driven end of the cable, and a DC-to-DC converter at far end. A 100% efficient DC-to-DC converter would be sucking out about 1.73A rms, dropping 173V RMS in the cable, leaving 577V rms at the far end. If you are going turn this into a steady 10A at 100V, while pulling half-sine waves of current through the cable you are going to have to store enough energy in your reservoir capacitor during the high "quarter" segments of the waveform to fill in the low "quarter" segments of the waveform. For a 50Hz source that's - crudely - about half of 5msec worth of 1kW or 10 joules. You couldn't charge your your capacitor to higher than 816V at the peak of the sinewave (at the far end of the cable) nor let it sag to lower than 100V before you started charging it up again which makes your minimum capacitance about 7.6uF. Farnell Australia stocks some 1uF 1kV capacitors which might do the job for about$A5 each in small quantities.

It sounds practical, but I've got no idea what the DC-to-DC converter would look like - a web search should show up something without too much effort, but I'm still jet-lagged after getting back to Sydney on Sunday night, and it's quite likely that I've already screwed up somewhere in the above narrative, so putting in any more effort at this point wouldn't be a good idea.

Unfortunately DC cannot flow in this cable. AC only. Sounds wired I
know but there actually is a valid reason due to the application.

It did stir up the old tinker. I never considered sending rectified AC
over the line with the filter caps on the far end. The resistance and
inductance of the line would help smooth/average the DC output using the

Unfortunately this is an AC only application.

B

#### Bill Sloman

Jan 1, 1970
0

Unfortunately DC cannot flow in this cable. AC only. Sounds weird I
know but there actually is a valid reason due to the application.

Stray magnetic fields? Ground loops?

So stick the bridge rectifier at the far end of the cable, next to the DC-to-DC converter and the load.

<snip>

F

Jan 1, 1970
0
M

#### mook johnson

Jan 1, 1970
0
Stray magnetic fields? Ground loops?

So stick the bridge rectifier at the far end of the cable, next to the DC-to-DC converter and the load.

<snip>

You guessed it. magnetic fields. An AC field will eventually sum to
zero. A DC field will of course not. This messes with some magnetic
field measurement instrumentation we have nearby.

B

#### Bill Sloman

Jan 1, 1970
0
You guessed it. magnetic fields. An AC field will eventually sum to
zero. A DC field will of course not. This messes with some magnetic
field measurement instrumentation we have nearby.

Twist - or perhaps better - braid the leads in your power line. The external magnetic field is then the weighed sum of a lot of little alternating dipoles, and remarkably small. IIRR it also decreases unexpectedly rapidly as you move away from the cable, as either the fourth or the sixth power of displacement once you get beyond the period of the twist/braid - fourth powerwas in the plane of the dipoles, sixth power at right-angles, or somethingequally cranky.

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