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Power Factor using a variac and DC filter caps

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IamSmooth

Mon Jan 25 2010, 09:59PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

When I take power from the MAINS through a variac, rectify it and smooth it will 1500uf of capacitors, how do I tell what my PF is without buying a device?

Can I put a current sense coil in between the variac and capacitor and scope the current and output voltage? If my power factor is off, what can I do to improve it? Could I add an inductor in parallel with my filter capacitors?

radiotech

Mon Jan 25 2010, 11:52PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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A trick to see easily if power factor is low is to add AC capacitors
across the input of your variac and see if this reduces the line amps.
Sense the AC amps at the input of the variac, Overlay that trace
on scope with voltage trace on the other channel and use the timebase markers to see lead (or lag).

An inductor *across* your capacitor bank will make a splendid electromagnet and source of extra heat.

Unless you are metered on a Kvar rate,correcting the PF will save you heating of wires and transformers, the losses being far less than the penalties imposed for low pf by the utility.

Also considered series choke (inductor) input to your capacitor
smoothing? This might reduce the need for large value caps, and, with selection of the gap of the inductor, offer higher peak current.

GeordieBoy

Tue Jan 26 2010, 12:19AM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
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> When I take power from the MAINS through a variac, rectify it and smooth it will 1500uf of capacitors, how do I tell what my PF is without buying a device?

Trust me it will be low!

> Can I put a current sense coil in between the variac and capacitor and scope the current and output voltage?

All you need to do is measure RMS input voltage and current on the AC side, and DC voltage and current on the DC side where the massive smoothing caps are. The DC volts multiplied by DC amps gives you the real power. The AC RMS volts multiplied by AC RMS amps gives you the VA. Divide real power by VA to get PF.

> If my power factor is off, what can I do to improve it? Could I add an inductor in parallel with my filter capacitors?

Not in parallel. You could put an inductor either in series with the AC input or in series between the rectifier and the DC bus capacitors. Both will increase the conduction angle of the rectifier diodes, decrease peak current, and increase power factor. The downsides are that this choke will be bulky and heavy to avoid saturation, and it wrecks the load regulation of the DC bus voltage. The more elegant solution if you require a semi-regulated DC bus voltage from an AC supply is to replace the simple rectifier with a complex switching power supply called an Active PFC boost converter. However...

If this is for an induction heater, the best option by far is just to keep the DC bus capacitance as low as possible. A decent dose of ripple on the DC bus really doesn't matter for heating applications. If you think about it, most AC space heaters and heating elements run straight off the AC mains and the time constant of whatever you're heating will completely smooth out the ripples anyway. It just means that all of the voltages and currents in the power section and work coil etc will be amplitude modulated with some degree of 100Hz or 120Hz ripple.

The DC bus cap in an induction heater only really needs to be beefy enough to support the HF AC current drawn by the inverter. In that respect its current handling ability is more important than its actual capacitance. It's real purpose is to provide a local source for the HF switching currents drawn by the inverter and to keep them out of the mains supply wiring, where they would otherwise cause horrendous RFI.

Minimising the DC bus capacitor also has the added advantages of reducing the stored energy if a fault occurs in the inverter as well as reducing size, weight, cost and inrush current problems.

I hope this helps,

-Richie,

IamSmooth

Tue Jan 26 2010, 01:35AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

Richie, thank you for the help.

I did a quick check at low power: 3A x 120vac input; 76vdc x 4.3A HVDC supply output
PF = 0.9

That doesn't seem too bad?

My doubler uses a rectifier and two 1500uf caps for a net capacitance of 750uf.

What would you consider an appropriate capacitance value for my rectifier? Should I use 100uf or less?

I have heard of a PFC boost converter. Is its benefit in that it does not rely on capacitors, but uses the inverter to produce the required DC power?

radiotech

Tue Jan 26 2010, 03:41AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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Two things I'm not clear on in the previous thread:
Choke input power supplies,(choke between rectifier output and capacitor) don't they furnish a more regulated output over load swings? I've never heard the term conduction angle applied to (non SCR) diodes. I thought they started conduction when the forward bias was exceeded, about 0.6 volt, the angle of that would depend on the peak of the applied (sin) source.

For the system , overall, what is the desired power factor?

Dr. Slack

Tue Jan 26 2010, 07:58AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
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I've never heard the term conduction angle applied to (non SCR) diodes. I thought they started conduction when the forward bias was exceeded, about 0.6 volt, the angle of that would depend on the peak of the applied (sin) source.

They do conduct at 0.6v drop. However, when there's an inductor in the circuit generating a voltage due to the varying current through it, and that voltage is added to the difference between the input sinusoid and the DC cap voltage, then it changes things. Best bet is to put the few components into a simulator, and plot currents and voltages.

Steve Conner

Tue Jan 26 2010, 11:03AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

IamSmooth wrote ...

PF = 0.9
That doesn't seem too bad?

The leakage inductance of your variac will be helping. That varies with the dial setting and is a minimum when the output voltage is the same as the input.

If you have an isolation transformer in there, the leakage inductance of that will help too.

If your line current meter isn't true RMS, then you get the "rectified average" current, and by conservation of charge, that's the same as the DC bus current, so you'll always see PF=0.9, taking into account the meter's form factor calibration.

In my experiments I got PF=0.6 with direct connection to the mains, and PF=0.8 with a series inductor, or with a variac/isolation transformer combo.

Oh and Kill-A-Watt meters are cheap, and read PF.

IamSmooth

Tue Jan 26 2010, 04:22PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Steve McConner wrote ...

If your line current meter isn't true RMS, then you get the "rectified average" current, and by conservation of charge, that's the same as the DC bus current, so you'll always see PF=0.9, taking into account the meter's form factor calibration.

It is not true RMS. How do I get the average current? Is there some math with this?

Steve Conner

Tue Jan 26 2010, 04:25PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

If it's not true RMS, you can't use the measurements from it to calculate power factor. You'll just get the answer 0.9 (or maybe 0.45 for a doubler?) irrespective of what the real PF is.

My above post was an explanation of why this error happens, not an explanation of how to fix it. To fix it, get a true RMS current meter.

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