PFC circuit at end of a long resistive line

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.

--

John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    

Precision electronic instrumentation 
Picosecond-resolution Digital Delay and Pulse generators 
Custom timing and laser controllers 
Photonics and fiberoptic TTL data links 
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Multichannel arbitrary waveform generators

I suspect that as long as you can get that much power out of the line at all, you can stabilize the PFC circuit against your 100 ohm line. You may have to design your PFC circuit specifically to be stable in that circumstance, though, and accept that you'll have a pretty tight tradeoff between the size of your input caps and the limits to the dynamic behavior of the supply output.

But if I'm getting my math right, to get 1kW out of a supply with a 100 ohm impedance you'd need to start with over 633V, and at that voltage you'd burn up 1kW in your line. Do you have that much to start with?

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Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

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).

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)

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.

--

John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    

Precision electronic instrumentation 
Picosecond-resolution Digital Delay and Pulse generators 
Custom timing and laser controllers 
Photonics and fiberoptic TTL data links 
VME  analog, thermocouple, LVDT, synchro, tachometer 
Multichannel arbitrary waveform generators

John Larkin schrieb:

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

I'm always happy to speculate. One approach would be to put a bridge rectif ier 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, d ropping 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 100 V before you started charging it up again which makes your minimum capacita nce about 7.6uF. Farnell Australia stocks some 1uF 1kV capacitors which mig ht 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 narr ative, so putting in any more effort at this point wouldn't be a good idea.

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Bill Sloman, Sydney

thanks for sharing your thoughts.

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 parasitic for free. Clever.

Unfortunately this is an AC only application.

a s e

FC

ar

s, 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 wave s 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 wave form to fill in the low "quarter" segments of the waveform.

r 10 > > joules.

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 capa citance about 7.6uF. Farnell Australia stocks some 1uF 1kV capacitors which might do the job for about $A5 each in small quantities.

ould look like - a web search should show up something without too much eff ort, 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 i dea.

Stray magnetic fields? Ground loops?

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

--
Bill Sloman, Sydney

:-)

--
"Design is the reverse of analysis" 
                   (R.D. Middlebrook)

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 externa l magnetic field is then the weighed sum of a lot of little alternating dip oles, 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 di splacement once you get beyond the period of the twist/braid - fourth power was in the plane of the dipoles, sixth power at right-angles, or something equally cranky.

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Bill Sloman, Sydney