Test of sub and sup

DISCLAIMER:
The following article is not intended to be anything other than an expression of my own opinion. Take anything you read here with a grain of salt.

First let me refer you to an article that explains what power factor correction is and how it works. The page can be found here. Once you read and understand this, what I do next may make some little bit of sense.

Now lets look at a model of a typical Neon Sign Transformer. A schematic representation is illustrated in figure 1. We will model this as an ideal transformer with an ideal inductor in series with the output.

nst schematic

figure 1

If we choose a 12kV 60mA NST, then the reactance of that inductor can be taken as 200kΩ. This is determined by dividing the open circuit voltage of the transformer by its short circuit current. The input voltage is 120V A/C, so the turns ratio (secondary to primary) is 100 to 1. (This will be important later) Since a NST has many turns of fine wire on the secondary, it will have a significant resistance as well. Let us assume for our model that that resistance amounts to 10kΩ between the high voltage terminals. After all, we are just picking values that make for easy calculations for this model. We do want to keep them somewhat close to reality though.

Let us now list the specifications of our model NST.

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Now we redraw the model. We will also add a jumper to short circuit the secondary.

nst equivelent circuit

figure 2

Now we have something to work with. We will analyze the circuit with the secondary shorted. We will stick to the methods used in the openbookproject.net page referenced above. First we will calculate the load as seen by the primary. Impedance is reflected through a transformer as the square of the turns ratio. The turns ratio (primary to secondary) is 1 to 100 or 1/100. 1/1002=1/10,000 so the primary sees 20 Ω (inductive reactance) + 1 Ω (resistance). Total impedance then is: 


 20Ω at 90 deg. + 1Ω at 0 deg. =
 √(202 + 12) = √(400+1) = √(401) =  20.025Ω
 Angle = Acos(1/20.025)= 87.1376 deg.

The power factor is cos(87.1376 deg.) or 0.04994. Not very good. The power factor correction capacitor required to bring the power factor back in line will have an impedance of 20.025 Ω ∠ -90 deg. We have 60 Hz here, so solving for C gives us: 

 C = 1/2pi(60)(20) = 1/7540 = 132.6µF.

nst model simplified for sourse with pfc

figure 3

So if we add 133µF in parallel with the source, or across the low Voltage input to the NST, we can correct the power factor back to 1.0. This is fine for the condition where the NST is shorted, as when the spark gap is firing. A simplified equivelent circuit diagram is shown in figure 3.

Next let us take a look at a typical application of a NST in Tesla coil service. That would be with a static spark gap and a LTR (larger than resonant) tank cap. Redrawing the model without the PFC capacitor, and with the shorting jumper replaced by a 20ηF tank cap, let us redo our calculations. This configuration is represented in figure 4.

nst equivalent with tank cap

figure 4

We already know the reflected inductive reactance and the resistance. All we have to do is calculate the reflected capacitive reactance, which is 1/10,000 * 133,000Ω or 13.3Ω. Subtract the capacitive reactance from the inductive reactance: 20 - 13.3 = 6.7Ω ∠ 90 deg. Total impedance is then: 


6.7Ω ∠ 90 deg. + 1Ω ∠ 0 deg. =
 √(6.72 + 12) = √(44.89 + 1) =
 √(45.89) = 6.774Ω
 and the angle = Acos(1/6.774) = 81.51 deg.
 and the power factor is 0.1776.

The current can be found using the formula I = E/ZT where I is the current delivered by the source, E is the voltage of the source, and ZT is the total impedance. So I = 120V ∠ 0 deg./6.774Ω ∠ 81.51 deg. = 16.7 A ∠ -81.51 deg. (note that you subtract angles when dividing). So we have a leading power factor here and adding a PFC capacitor will only increase the current drawn from the source.

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At this point I do not think I need to do resonant and smaller than resonant caps. It should be obvious that a resonant size cap will give a unity power factor and horrific input current (in our example 120 Amps!) , and smaller than resonant will result in a lagging power factor. This should be true for NSTs and OBITs, as well as bug zapper transformers and electrostatic precipitator transformers.

Next: A short discussion of different types of transformers and their characteristics

Before I continue, let me just say that I often see folks talk about resonant size cap. for a PT or pole pig. There is no such thing! Well maybe there is, but for our intents and purposes, not. There is no inherent current limiting ability. That is why ballasting is necessary. The tank cap, be it LTR, STR, or resonant, is interacting with the ballast through the transformer, not with the transformer. Reflected impedance again.

I suppose it would be in order to digress a bit here and explain the preceding statement. There are three types of transformers commonly used in Tesla coil power supplies. Perhaps the most common is the self ballasting transformer, usually incorporating "shunts" to increase/control the "leakage inductance". Examples of this type includes Neon sign Transformers (NSTs), oil burner ignition transformers (OBITs), bug zapper transformers, and electrostatic precipitator transformers. The second type of transformer commonly used in Tesla coil power supplies is designed to have good voltage regulation. Examples include polemount distribution transformers (pole pigs), and potential transformers (PTs). The third type of transformer commonly used in Tesla coil power supplies is the microwave oven transformer (MOT) which is technically a member of the first group, but due to the special characteristics of the current limiting inherent in these transformers, require additional ballasting in order to be used as NSTs are used.

In the case of NSTs and OBITS determining the resonant size tank capacitance is fairly straight forward. The impedance of the NST's secondary can be found by taking the open circuit voltage and dividing by the short circuit current. Then find the size of capacitor which has the same impedance at the frequency of interest.

In the case of pole pigs and PTs the process is not quite as simple. Dividing the KVA rating of the transformer by the voltage rating will yield the Full Load Ampere rating of the transformer. Dividing the voltage rating by the FLA rating will _NOT_ give the impedance of the transformer. For that you need short circuit current. The short circuit current can be determined from the nameplate ratings, and some math. The KVA rating and voltage rating can be used to find the FLA rating, and the FLA rating and the % impedance rating can be used to calculate the short circuit current. Once the short circuit current is known, the voltage rating can be divided by that to determine the impedance, and the resonant size capacitor can then be calculated. For more information about percent impedance and calculating short circuit current for a pole pig see these references: http://findarticles.com/p/articles/mi_qa3726/is_200601/ai_n17178925 http://www.brainfiller.com/documents/TransformerandSourceImpedance_000.pdf

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From the findarticals.com reference above:

"A transformer's percent impedance is evaluated by a simple test. The secondary terminals are short-circuited. A low voltage is then applied to the primary terminals, and increased until the current measured in the short-circuited secondary reaches the rated ampere value. The impedance is then the ratio of that primary voltage to the rated voltage; multiplying that by 100 gives the impedance in percent.

In addition to its role in voltage drop calculation, that figure determines the maximum fault current the transformer is capable of delivering to a downstream short-circuit. For example, consider a 1,000 kVA transformer with a 5% impedance rating and a 480 volt secondary. Rated full-load secondary current would be 1,200 amperes (current = kVA/[ 1.73 times 0.48 kV]). The maximum short-circuit current would be 1,200 divided by per unit impedance of 0.05. or 24,000 amperes.

Looking at that another way: If 5% of rated primary voltage results in 100% rated current in the short-circuited secondary, 100% of rated primary voltage would cause 20 times as much secondary current. In this example, that's 20 times 1,200, or 24,000. Of course, that's only a theoretical maximum.

Under fault conditions in a real system, impedance of cables and other circuit components upstream from the transformer primary would drop the primary voltage below 100%."

By Richard L. Nailen, P.E., EA Engineering Editor

Note that this is for 3 phase transformers. Single phase leave out the 1.73 (square root of three) in the calculation. edited 04/02/09

A short example of finding resonant size tank capacitor for a pole pig

Let's take a moment and calculate the resonant size tank capacitor for my 5KVA pole pig. First the nameplate data:

We will do this "backwards", that is secondary is primary and primary is secondary, as that is how we coilers use them. First we need to find the full load amp rating. 5000VA/14400V = 347.2mA Then divide FLA by unit impedance: 0.3472A/0.016 = 21.7A short circuit. Now find transformer impedance: 14400V/21.7A = 663.6Ω Now find the capacitance that will have a reactance of 663.6Ω at 60Hz: Ah, but wait! With a NST the resistance of the secondary is such a tiny part of the total impedance that it is considered insignificant and neglected in calculations more often than not. Will the same hold true for a pole pig? Lets see. I measured the resistance of the high voltage winding at 220Ω, which is a more significant value considering the total impedance is 663.6, than in the case of the 12/60 NST in the first example at the top of the page where the resistance is 10kΩ and the total impedance is 200kΩ. Solving for the inductive reactance: √(663.62 - 2202) = 626Ω So that is the reactance we need to match with the capacitor value. 1/(626*377) = 4.24µF is the resonant size capacitor for my 5kVA pole pig at 60Hz. Input current at 240VAC would be 1302A. A little more than my 150A service can supply.

MOTs are a special case and I will not get into that here, at this time.

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Next: potential and polemount distribution transformers.

balpig.gif

figure 5

Let us say that we have a 14400V 5kVA polemount distribution transformer. For some reason we want it to "act" like a 60mA NST. I just happen to have one of those, and measured between the "horns" with an Ohmmeter, and find the resistance to be 220Ω. The turns ratio is 60 to 1 if we input 240V, and 120 to 1 if we input 120V. Just for grins, let us say we are going to input 120V. So for 60mA short circuit current on the secondary we need 14400/.06 or 240kΩ on the secondary side, or 240kΩ/1202 = 16.67Ω of inductive reactance on the primary side.

The 220Ω secondary resistance looks like .015Ω to the primary. So 120V/16.67Ω = 7.2A ∠ 90 deg. power factor very close to zero, lagging. Power factor correction capacitor would have 16.67Ω reactive, solving for C: C=1/2pi(60)(16.67) = 159µF. Now where do we put the PFC cap? Let us try directly across the input terminals of the pig.

pigpfc1.gif

figure 6

That will give us 16.67Ω ∠ 90 deg. in series with (16.67Ω ∠ -90 deg. || with 0.015Ω ∠ 0 deg.). Do that: 16.67 + (1/(1/16.67)+(1/.015)) = 16.685Ω ∠ 89.999 deg. so the PFC cap does nothing at all here, in the short circuit condition. Now in the case of the tank cap being resonant with the ballast, it will present as the SAME impedance as the PFC cap in parallel with the PFC cap, so will reduce the charging rate of the tank cap. LTR or STR will result in similar effects to varying degrees.

pigpfc2.gif

figure 7

If we put the PFC capacitor across the line before the ballast, then we will have very much the same situation as we saw in the NST example above. The ideal PFC cap. size is a resonant match for the ballast in the short circuit condition.

pigpfc3.gif

figure 8

It would seem that in cap charging mode that the ideal PFC size when the tank cap. is resonant with the ballast is none. If the tank cap is smaller than resonant, then there may well be some compromise size of PFC cap that would provide the most benefit overall.

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Well, that is all for now folks.