Part 4 · Chapter 12

Generation of High Alternating Voltages

An AC machine should be tested with AC. But the things we test — cables, bushings, gas-insulated bus — are giant capacitors, and feeding them at hundreds of kilovolts demands enormous charging current. This chapter follows the answer from the humble testing transformer, up through cascaded units that reach a megavolt, to the elegant series-resonant set that tunes the capacitance away and shrinks the supply by a factor of Q.

High-Voltage Engineering Prof. Mithun Mondal Reading time ≈ 45 min
i What you'll learn
  • Why high alternating voltage is the natural test for AC apparatus, and why the test load is almost purely capacitive.
  • The high-voltage testing transformer — how it differs from a power transformer and the charging power \(S=\omega C V^{2}\) it must supply.
  • How cascaded transformers reach megavolts, and why the bottom unit carries the heaviest duty.
  • The series-resonant circuit: tuning a reactor against the load capacitance so the supply need only cover the losses.
  • The quality factor \(Q\) as voltage magnification, \(V_{\text{test}}=Q\,V_{\text{supply}}\), and resonance's built-in breakdown protection.
  • A first look at the Tesla coil for high-frequency high voltage.
Section 12-1

Why High AC?

The grid runs on alternating current, so the apparatus that serves it — transformers, cables, switchgear, insulators — is built for AC and must be proven at AC. Standards demand it: before a piece of equipment enters service it faces a withstand test, held for a minute or more at a voltage well above its rating, to confirm the insulation survives. Generating that test voltage is the task of this chapter.

There is a twist that shapes everything. The test object is rarely a resistor that draws real power; it is overwhelmingly a capacitor. A long cable, a bushing, a length of gas-insulated bus — each is two conductors separated by a dielectric, which is the very definition of a capacitor. When we raise such a load to high AC voltage, almost no power is consumed, but a large charging current flows in and out every cycle. The source's burden is therefore not watts but reactive volt-amperes, and as we will see, that distinction is what makes high-AC generation interesting — and what the resonant circuit later exploits so cleverly.

The load is a capacitor, not a heater. Hold this fact in mind for the whole chapter. It explains why a test set rated for "10 MVA" may consume only kilowatts of real power, and why tuning an inductor against that capacitance can collapse the supply requirement almost to nothing.
Section 12-2

The Testing Transformer

The workhorse for moderate test voltages is the high-voltage testing transformer: a step-up transformer with a low-voltage primary and a heavily insulated high-voltage secondary, usually with one end of the secondary solidly earthed so that only the live terminal needs full insulation. It looks like a power transformer but is bred for a different life. A power transformer carries large continuous current at modest voltage; a test transformer carries tiny current at enormous voltage, for short periods, with its winding insulation carefully graded from the earthed end to the live end.

core LV HV +V (high voltage) C (test object)
A testing transformer with a graded, single-end-earthed HV secondary, driving a capacitive test object

Because the load is a capacitance \(C\) raised to voltage \(V\) at angular frequency \(\omega = 2\pi f\), the current it draws is the capacitive charging current, and the apparent power the transformer must furnish follows at once:

Charging current and apparent power
\[ I = \omega C V\,, \qquad S = V I = \omega C V^{2} \]

That \(V^{2}\) is the sting. Doubling the test voltage quadruples the volt-ampere demand, so a single transformer's rating climbs steeply with voltage. This is manageable up to a point — but to reach the highest voltages, a single unit becomes impossibly large and hard to insulate, and we must turn to cascading.

Section 12-3

Cascaded Transformers

Above roughly a megavolt, building one transformer with secondary insulation rated for the full voltage becomes prohibitive. The cascade solves this by stacking several identical units of modest voltage so that their high-voltage windings add in series. Each transformer sits on an insulating support raised to the potential of the stage below it, so unit number two floats at \(V\), unit three at \(2V\), and the top terminal reaches \(nV\) for \(n\) stages — yet no single transformer is insulated for more than \(V\) between its own windings.

T1 T2 T3 +3V 0 V 2V ~ LV supply
A three-stage cascade — HV windings add in series to 3V, while each unit is insulated for only V; excitation passes up the stack

The cascade hides a subtlety that governs its design. The excitation for the upper units must be passed up through the lower ones, so the lower transformers carry not only their own load but also the throughput power of every stage above them. If each secondary delivers power \(P = VI\), the transformer at the bottom must transmit \(nP\), the next \((n-1)P\), and so on up to the top unit's \(P\). The total installed capacity is therefore far larger than the delivered power:

Loading of the stack
\[ \text{bottom unit} = nP, \qquad \text{total installed} = P\,\frac{n(n+1)}{2} \quad\text{for}\quad P_{\text{out}} = nP \]
The bottom unit does the heavy lifting. In a three-stage set the lowest transformer must be rated for three times the per-stage power even though every unit reaches the same voltage. This uneven duty — not the voltage — is what limits how tall a cascade can practically grow.
Section 12-4

The Capacitive-Load Problem

Cascading conquers the voltage, but it does nothing about the current. Recall \(S = \omega C V^{2}\): for a long high-voltage cable, \(C\) can be hundreds of nanofarads, and at a few hundred kilovolts the charging current runs to tens of amperes and the apparent power to several megavolt-amperes. Supplying that with transformers means a vast, heavy installation whose entire output is reactive — it is shuttled in and out of the cable every cycle and never consumed.

This is wasteful in the extreme, and it invites a question: if the load is a capacitor that only ever stores and returns energy, why supply that energy from the source at all? Why not give the energy somewhere local to slosh into, an inductor, so the source need only top up the small real losses? That single idea — pairing the capacitive load with an inductor at resonance — is the most important development in modern high-AC testing.

🔑
The reactive burden grows as voltage squared
\[ S = \omega C V^{2} \]

Because the test object is a capacitor, the supply must furnish reactive power rising with the square of the test voltage. Cancelling that reactance with a tuned inductor is the key to a small, efficient source.

Section 12-5

The Series-Resonant Circuit

Place a variable reactor (a high-voltage inductor) in series between the supply and the capacitive test object, and tune its inductance \(L\) until its reactance exactly equals that of the load capacitance \(C\) at the supply frequency. At that point the two reactances cancel — the inductor's appetite for current is met precisely by the capacitor's surplus — and the circuit is in series resonance:

Resonance condition
\[ \omega L = \frac{1}{\omega C} \quad\Longrightarrow\quad f_{r} = \frac{1}{2\pi\sqrt{LC}} \]
~ V_supply L (reactor) R (losses) C V_test
Series resonance — the reactor L is tuned against the load C so their reactances cancel, leaving only the small loss resistance R for the supply to drive

With the reactances cancelled, the supply no longer sees a megavolt-ampere capacitor; it sees only the small resistance \(R\) that represents the reactor and dielectric losses. A modest low-voltage supply, providing just enough to make up those losses, drives a large resonant current around the \(L\)–\(C\) loop — and that current develops a very high voltage across the test capacitor. The reactive energy now sloshes back and forth between \(L\) and \(C\), borrowed from the source only once and thereafter recycled each cycle.

Section 12-6

Quality Factor and Voltage Magnification

How high does the test voltage rise above the supply? The answer is the circuit's quality factor \(Q\), the ratio of reactance to resistance, which here doubles as the voltage magnification:

Quality factor and voltage gain
\[ Q = \frac{\omega L}{R} = \frac{1}{\omega C R}\,, \qquad V_{\text{test}} = Q\,V_{\text{supply}} \]
V_test f → f_r Q · V_supply
At resonance the test voltage peaks sharply at Q times the supply — a high-Q circuit magnifies a few kilovolts into hundreds

A reactor with \(Q = 50\) turns a \(5~\mathrm{kV}\) supply into a \(250~\mathrm{kV}\) test voltage, and the supply provides only \(1/Q\) of the apparent power the load would otherwise demand. Four virtues follow, and together they explain why resonant sets have displaced transformer banks for cable and GIS testing:

VirtueWhy it happens
Small supplySource covers only the losses, \(P = S/Q\) — a fraction of the reactive load
Pure sine waveThe tuned \(L\)–\(C\) loop filters supply harmonics, giving a clean test voltage
Self-protectionIf the object breaks down, resonance is lost and the voltage collapses at once — no destructive fault current
Compact & efficientNo bulky high-power transformer; energy is recycled between \(L\) and \(C\)

Practical resonant test systems come in two flavours. Variable-inductance sets keep the frequency at mains value and tune \(L\) to match each load. Variable-frequency sets keep a fixed reactor and instead sweep the supply frequency (typically \(20\)–\(300~\mathrm{Hz}\)) until resonance is found — lighter and now standard for on-site testing of long cables, where a tunable HV inductor would be impractical to carry.

Section 12-7

The Tesla Coil

One more AC source deserves mention, for high frequency rather than power frequency. The Tesla coil is an air-cored, doubly-tuned resonant transformer: a primary circuit with a capacitor and spark gap rings at high frequency and couples loosely into a secondary tuned to the same frequency, which builds up a very high oscillating voltage. Because the coupling is loose and the gap interrupts the drive, the output is a train of damped high-frequency oscillations rather than a steady sine.

~ supply C gap primary air core (loose coupling) secondary (many turns) HF high voltage
The Tesla coil — a spark-gap-driven primary loosely coupled through air to a tuned secondary, building high-frequency high voltage by double resonance

Tesla coils are little used for routine withstand testing today, but they remain valuable for high-frequency insulation studies and as a vivid demonstration of resonant voltage magnification — the same principle as the series-resonant set, pushed to radio frequencies. They close the loop on a theme that runs through this whole chapter: at high voltage, resonance is not a curiosity but the central tool, because it lets a small source command a large voltage.

Section 12-8

Worked Examples

1 Rating a testing transformer

Problem. A testing transformer must energise a bushing of capacitance \(C = 2~\mathrm{nF}\) to \(V = 100~\mathrm{kV}\) at \(f = 50~\mathrm{Hz}\). Find the charging current and the apparent power required.

Solution. With \(\omega = 2\pi(50) = 314~\mathrm{rad/s}\), use \(I = \omega C V\) and \(S = \omega C V^{2}\):

Working
\[ I = 314 \times 2\times10^{-9}\times10^{5} = 62.8~\text{mA}, \qquad S = V I = 10^{5}\times0.0628 = 6.3~\text{kVA} \]

A modest 6.3 kVA unit suffices here — but note that raising the test voltage to 200 kV would quadruple this to 25 kVA, the \(V^{2}\) penalty in action.

2 Loading of a cascade

Problem. A cascade of three transformers, each rated \(100~\mathrm{kV}\) and each secondary delivering \(P = 100~\mathrm{kVA}\), supplies a load. Find the output voltage, the rating the bottom unit must have, and the total installed capacity.

Solution. Voltages add to \(nV\); the bottom unit transmits \(nP\); the total is \(P\,n(n+1)/2\):

Working
\[ V_{\text{out}} = 3\times100 = 300~\text{kV}, \quad \text{bottom} = 3\times100 = 300~\text{kVA}, \quad \text{total} = 100\times\tfrac{3\cdot4}{2} = 600~\text{kVA} \]

The set reaches 300 kV and delivers 300 kVA, but needs 600 kVA of installed transformers — and the bottom unit alone must be rated for the full 300 kVA. Utilisation is only 50%.

3 Tuning a resonant set

Problem. A long cable of capacitance \(C = 0.5~\mu\mathrm{F}\) is to be tested at \(50~\mathrm{Hz}\). Find the reactor inductance for resonance, and, if the circuit \(Q = 50\), the supply voltage needed to reach \(250~\mathrm{kV}\) across the cable.

Solution. Resonance needs \(L = 1/(\omega^{2}C)\); then \(V_{\text{supply}} = V_{\text{test}}/Q\):

Working
\[ L = \frac{1}{(314)^{2}(0.5\times10^{-6})} \approx 20.3~\text{H}, \qquad V_{\text{supply}} = \frac{250}{50} = 5~\text{kV} \]

A \(20.3~\mathrm{H}\) reactor tunes out the cable, and a mere 5 kV supply, magnified fifty-fold, reaches the full 250 kV test voltage.

4 The power saving from resonance

Problem. For the cable of Example 3, compare the apparent (reactive) power a transformer set would supply with the real power the resonant supply actually draws.

Solution. The capacitive load is \(S = \omega C V^{2}\); the resonant supply provides only \(P = S/Q\):

Working
\[ S = 314\times0.5\times10^{-6}\times(2.5\times10^{5})^{2} \approx 9.8~\text{MVA}, \qquad P = \frac{S}{Q} = \frac{9.8}{50} \approx 196~\text{kW} \]

A transformer set would need nearly 10 MVA; the resonant set draws under 200 kW — a fifty-fold reduction, exactly the factor \(Q\). This is why series resonance, not cascading, is the modern way to test large capacitive objects.

Review

Chapter Summary

Why AC

AC apparatus is proven by AC withstand tests; the load is almost purely capacitive, so the source supplies reactive charging power, not watts.

Test transformer

Graded, single-end-earthed HV secondary; must furnish \(I=\omega CV\) and \(S=\omega CV^2\) — a rating that climbs with \(V^2\).

Cascade

HV windings add in series to \(nV\); each unit insulated for only \(V\). The bottom unit transmits \(nP\) — the heaviest duty in the stack.

Resonance

A reactor tuned to \(\omega L = 1/\omega C\) cancels the load reactance; the supply then covers only the losses.

Magnification

\(V_{\text{test}}=Q\,V_{\text{supply}}\) and \(P=S/Q\): a high-\(Q\) set gives huge voltage from a small supply, with a clean sine and self-protection.

Tesla coil

An air-cored doubly-tuned resonant transformer for high-frequency high voltage — resonant magnification at radio frequencies.

Practice

Problems

For each item, first identify what it tests — the charging-power formula, the cascade loading, the resonance condition, the \(Q\) magnification, or the power saving — then apply it. Difficulty rises down the list.

  1. Explain in one or two sentences why a high-voltage test load is treated as a capacitor, and what kind of power the source must mainly supply.
  2. A testing transformer energises \(C = 1~\mathrm{nF}\) to \(150~\mathrm{kV}\) at \(50~\mathrm{Hz}\). Find the charging current and the apparent power.
  3. By what factor does the kVA demand change if the test voltage in Problem 2 is raised from 150 kV to 300 kV? Justify with \(S=\omega CV^2\).
  4. A four-stage cascade uses identical \(125~\mathrm{kV}\) units. Give the output voltage, and state which unit carries the heaviest duty and by what multiple.
  5. For that four-stage cascade with per-stage power \(P = 80~\mathrm{kVA}\), find the total installed transformer capacity and the utilisation factor.
  6. A reactor must resonate with \(C = 0.2~\mu\mathrm{F}\) at \(50~\mathrm{Hz}\). Find the required inductance \(L\).
  7. The resonant circuit of Problem 6 has \(Q = 40\). What supply voltage produces \(160~\mathrm{kV}\) across the test object?
  8. For Problems 6–7, compute the reactive power of the capacitive load and the real power the supply actually delivers, and state the saving factor.
  9. List the four advantages of a series-resonant set over a transformer bank, and explain in one line why breakdown of the test object is self-limiting.
  10. A field crew must test a long installed cable on site. Explain why a variable-frequency resonant set is preferred over a variable-inductance one, referring to the practicality of a tunable HV reactor.
Tip: every numerical problem here flows from two ideas — the capacitive load draws \(S=\omega CV^2\), and resonance divides the supply burden by \(Q\) while multiplying the voltage by \(Q\). Keep capacitance in farads and frequency in hertz, remember \(\omega = 2\pi f\), and the chain from charging current to resonant supply power stays consistent.