Original date 5 December 2000
Revision date 1 February 2001

TRANSFORMER ANALYSIS

The contents of this page are highly technical and may require careful study to follow. If you wish to print it out for that reason I suggest going to my duplicate page which is in black and white and makes much nicer copy. PRINT PAGE

The measurements and analyses herein should only be applied to Boiler Ignition and Neon Sign Transformers having built in current limiting. The objective is to determine what transformer output voltages to expect with Tesla Coil Primary Circuit Capacitors connected.

The approach is to:

a. Measure transformer primary and secondary winding resistances
b. Measure primary (input) current with secondary output) both opened and shorted
c. Measure secondary short circuit current.
d. Measure primary voltage. (Secondary voltage is assumed**)
e. Make calculations using measurements. (Equivalent circuits are included as aids)
f. Substitute results in an equivalent circuit for load analysis. (Primary Circuit Capacitor)
** Measuring secondary voltage should not be attempted as it may be around 8 - 15kV

Figure 1. on the right is an equivalent circuit for transformers. The eddy current and hystereis loss component has been omitted. Rp and Rs are the transformer primary and secondary winding resistances in ohms. Lp and Ls are their inductances in Henries. V1 and ip are the input voltage and current to the transformer primary, n is the transformer secondary to primary windings turns ratio, and k is the coefficient of coupling.

The example used in this exercise is an 8.3kV 20mA Boiler Ignition Transformer.

MEASUREMENTS:

Rp = 16.2 ohms
Rs = 21,200 ohms
Vp = 238 Volts
ip = 66mA (secondary open)
ip = 748mA (output shorted)
is = 18mA (output short circuit current)**
** Great care must be taken when making this measurement.

a. Disconnect the transformer from the AC power input.
b. Connect an AC ammeter, digital or analogue, across the secondary output.
WARNING: Setting the meter to any other function such as Volts, Ohms, etc. will result in its complete distruction!!!
c. Set the ammeter scale to 0-1A range or greater.
d. Apply AC power to transformer and observe ammeter reading.
e. Do NOT touch the meter or transformer.
f. REMOVE AC POWER before changing meter settings.

CALCULATIONS:

Zin1 = 238V/0.069A = 3.45k ohms complex (secondary open)
Zin2 = 238V/0.748A = 318 ohms complex (secodary shorted)

Figure 2. on the right represents the transformer with its output open. It is derived from figure 1 above. Since Zin1 >> Rp, Rp can be ignored. As such, Lp is easily calculated.

XLp = Zin1 = 3450 ohms
Lp = 3450/(2 x PI x 50Hz) =11H

COEFFICIENT OF COUPLING (k)

Figure 4. on the right represents the transformer with its secondary shorted. Again Rp can be ignored because its value is considerably less than the measured input impedance of 315 ohms. The other resistive component in the primary circuit is a function of k, n, and Rs. By estimating n = 8300/238 = 34.5, and k=1, the value of this resistive component is approximately:

21500/34.5^2 = 18 ohms

This effectively shorts out the kLp component leaving only the Lp(1-k^2) component in the circuit. See figure 4A on the right. k is easily calculated as follows:

Effective input inductive reactance = 315 ohms Effective input Lin = 318/(2 x PI x 50Hz) =1.01H
1.01H = 11.0H(1-k^2) Solving for k:

k = SQR(1 - 1.01/11) = 0.953

The final items needed for an equivalent circuit with values are the transformer turns ratio n and and its secondary winding inductance Ls. Using knV1 = 8300V shown in fig. 2, we get 8300V = 0.953 x n x 238 Solving this we get n = 36.6 Knowing that inductance of a coil is proportional to the square of the number of turns, the inductance of the transformer secondary winding Ls is calculated as follows:

Ls=Lp x n^2=11 x 36.6^2=14700H

Figure 5 at the right is the equivalent circuit for the boiler ignition transformer with values inserted. The Zout (transformer load) is represented by Cp, the capacitor used in the Tesla Coil Primary Circuit.

Figure 6 shows how the component, 1475Cp, in the primary circuit was derived.

Resonance occurs at a given frequency in an LC circuit when the inductive reactance (XL) of L equals the capacitive reactance (XC) of C, that is: XL = XC). In our case L = 1340H and f = 50Hz. XL = 2 x 3.14 x 50 x 1340 = 421k ohms. Cp with this reactance: Cp = 1/(2 x 3.14 x 50 x 421000) = 7.6nF.

The reactance of a 7.6nF capacitor will cancel out the reactance of a 1340H inductance at 50Hz in a series circuit. This leaves 8.3kV looking at 21.2k ohms of resistance. I=V/R=0.39A flows through Cp and the transformer secondary. Cp voltage will then be Vc = Ic x Xc = 0.39 x 421000 = 164kV!!!!! I don't know of any capacitors and transformers, except maybe in the National Grid, that come close to standing this kind of voltage. So, connecting this transformer secondary to a 7.6nF capacitor will result in transformer internal arcing and carbonizing, excessive heat due to the I^2R (32kW) and capacitor dielectric break down. In reality, the secondary current won't come close to 0.39 amperes because the transformer iron core will saturate at some lower current. Never-the-less, the transformer will self distruct if no spark gap protection is provided.

Protection against such a thing happening is offered by properly set static spark gaps. They are usually placed across the transformer secondary winding output and across the capacitor (Cp). The gaps should set to arc over at least at about 1 and 1/2 times the peak ac voltage rating of the transformer. Experienced coilers use settings depending on the type of normal operating spark gaps they use such as synchronized rotating spark gaps, single static gaps, or multiple self quenching static gaps. Anyway, the importance of over-voltage protection can not be over-emphasized.

One can see immediately that larger values of Cp can be used with this transformer to advantage. Smaller values will result in less energy, so it would be pointless to use them. If one uses a larger capacitor which has less reactance, it cancels less of the inductive reactance out. This results in the secondary current being determined mainly by this uncancelled inductive reactance. Ideally you want to use as large a capacitor as possible while still maintaining the output voltage.

The expression to the right would be helpful in determining a size for Cp. It was derived using standard ac mathematics. Purists would take exception to my formula, but that's because I omitted the resistive component (Rs) which requires application of complex numbers imaginary numbers (square root of -1). I felt justified jn not worrying about Rs because its value of 21.2k ohms is quite small compared with the reactive values of Cp and Ls which exceed 200k ohms.

Vo in the formula represents the ac voltage seen by the capacitor Cp. C represents Cp. At resonance the denominator is zero. Theoretically, Vo would go to infinity. Increasing the value of Cp above the resonant value decreases the value of Vo. Vo would be negative, but this means Vo at -180 degrees with respect to Vin and is nothing to worry about. Vo reaches 8300V (-180 degrees) when Cp is set at 15.2nF. above 15.2nF the voltage available o Cp would be below 8300V. A 10nF capacitor results in 28kV, probably enough to wreck the transformer and blow Cp. Of course the spark gap is set to fire at 15kV the transformer may hold up. Using the energy formula given on the "PRIMARY COIL" page 1.1joules is calculated. With 100 firings per second there would be 110 Watts of power being used by the Tesla Coil. Pushing the boiler ignition transformer beyond this would be disasterous.

The next update for this page will include information on a Neon Sign Transformer rated 8kV at 35mA. Thank you for reading this far. Comments would be greatly appreciated. There is a "mail me" link at the bottom of the home page.