Re: Power Factor Correction methods
Subject: Re: Power Factor Correction methods
From: Terry Fritz <twftesla-at-uswest-dot-net>
Date: Sat, 31 Jul 1999 12:11:08 -0600
In-Reply-To: <Pine.GSO.3.95-960729.990729144911.7578A-100000-at-aidan.ncl.a c.uk>
Many thanks for the equations here. They work very well. The real power
can be found fairly accurately by finding the "Tank Cap Watts"
Watts = F x 1/2 x Cp x Vfire^2
and then adding the loss in the filter and neon transformer. In my case
the neon does about 30 watts and the filter about 100.
The VA input (in my case) is 1323 watts and the real power is 864 watts.
The neon input is 115 volts at 11.5 amps. Now by adding a 200uF power
factor cap, I can reduce the current to 7.5 amps! Something that is really
neat is with the power factor cap in place, any old voltmeter and any
ammeter will measure the real voltage and real current into the TC. Thus,
no fancy watt meter would be needed.
The power factor caps will allow me to run the coil and the gap off a 15
amp power strip and reduce the voltage droop in the AC line. It will also
reduce the stress on the variac and contactors that tend to arc more with
the inductive loads. It would also allow me to use two neons in the
charging circuit and still be able to run the whole system from a 115 volt
20 amp circuit. That would increase the arc energy by 80%!
I still have not played with gap timing vs power factor like you and John
have. But my gap is not easy to alter (unlike Finn's new design) and the
caps do the trick for me just fine.
At 02:51 PM 7/29/99 +0100, you wrote:
>Hi Terry, and all,
>Power factor is something which I have spent some time investigating
>with regard to TCs, as I am trying to optimise my rotary and charging
>Yes, it is possible to calculate the required capacitor value for best
>power factor correction on a practical system. You don't need to run
>a simulation but I find it is useful to check if I did the maths right !
>Any system with poor power factor is drawing real power AND reactive
>power from the supply. The fundamental frequency component of the supply
>current is actually out of phase with the supply voltage. For best
>power factor the TC should appear as a pure resistive load to the supply.
>However for less than unity power factor the system appears slightly
>capacitive or slightly inductive. This is due to an imbalance between
>the transformer leakage inductance and the tank capacitance in the
>charging circuit. This means that a small amount of "reacitive power"
>merely sloshes back and forth in the supply cables (causing heating etc.)
>To improve the power factor we need to cancel out the reactive power bit,
>so that only the useful real power is left.
>In order to calculate the required Power Factor Correction (PFC)
>capacitance, you need to know two things about your TC:-
>1. The VA drawn by the system form the supply. This one is easy. You
> just measure the supply voltage and current and multiply them.
>2. The Real Power being drawn from the supply. This is a little harder.
> You need to use a WATT METER. (It is possible to get an approximate
> value for the real power by calculating the "Tank Cap Watts" if you
> know the break rate and peak tank cap voltage. This method is less
> accurate however, as it does not take resistive losses into account.)
>When measuring these make sure your system is set up correctly, as any
>change in ballasts, capacitors or rotary phase will change these values!
>Now the maths bit: (Don't panic, only 3 steps.)
>P = Real power, Q = Supply VA, R = Reactive VA, Z = Impedance of PFC,
>V = RMS supply voltage, pf = Power factor, F = Supply frequency,
>C = PFC correction capacitance.
> pf = P / Q, and Q = sqrt ( (P*P) + (R*R) )
>To get unity power factor we need to make the reactive power R = 0, so
>that Q = P and pf = 1.
>STEP 1: Calculate the reactive power R.
>R = sqrt ( (Q*Q) - (P*P) )
>A large reactive power R here indicates that the TC is either highly
>inductive or highly capacitive, and will have poor power factor.
>STEP 2: Now we need calculate the required PFC impedance Z to
> cancel the reactive power R.
>R = V * V / Z (this is from ohms law p = v*v/r)
>rearranges to give:
>Z = V * V / R
>STEP 3: Calculate the correction capacitance C to give the required
> impedance Z accross the line, (just like when choosing a mains
> resonant cap for a neon.)
>C = 1 / 2 / pi / F / Z (Remember C is in Farads)
>The PFC capacitor should be connected as close as possible accross the
>LV supply side of the step-up transformer. You should repeat the two
>measurements of supply VA and real power. If the PFC value is correct
>the real power will be the same as before, but the supply VA should
>have dropped to little more than the real power value.
>I have tried to make this somewhat confusing subject of power factor as
>clear as possible, by showing the steps involved. However I still get
>confused so someone please check my maths !
>I think there are simpler, "rule of thumb" type calculations around,
>however this method will give "perfect" power factor correction. The
>accuracy depends only on how accurately you can measure P and Q.
>If you don't want to go through the maths, try adding progressively more
>and more PFC capacitance until you begin to notice very little decrease
>in current drawn from the supply. This is pretty close. If you add too
>much PFC capacitance the current will start to increase again.
>Also, remember that using a PFC cap only reduces the supply cord current,
>it does not make life any easier for the transformer and ballast parts.
>Comments and corrections welcome,
> - Richie,
> - In Sunny Newcastle.