Generation of High Direct Voltages
Parts 1 to 3 asked how insulation fails; from here on we build the machines that push it to the edge. The first is the high-voltage DC source — the rectified, smoothed, multiplied descendant of an ordinary transformer that lets us test a cable at a million volts of steady stress. This chapter follows that idea from a single diode up to the Cockcroft–Walton column and the belt-driven Van de Graaff.
- Why high direct voltage is needed for testing and transmission, and how DC stress differs from AC.
- The half-wave rectifier with a smoothing capacitor — output near \(V_{\max}\), and a diode that must hold off \(2V_{\max}\).
- How to define and compute ripple and regulation for a DC source under load.
- The Greinacher voltage doubler and how stacking it builds the Cockcroft–Walton cascade to \(2nV_{\max}\).
- The cascade's working formulas — ripple \(\propto n^2\), drop \(\propto n^3\) — and the resulting optimum number of stages.
- The Van de Graaff electrostatic generator: very high voltage at tiny current by carrying charge on a moving belt.
Why Generate High DC?
For three parts of this book the high voltage was a threat to be survived. Now it becomes a tool to be made — and the first one we need is a source of steady, high direct voltage. There are two reasons it matters. In the field, HVDC transmission carries bulk power across long distances and undersea cables more efficiently than AC, and its converter valves and cables must be designed and tested at DC. In the laboratory, a DC test set is indispensable because direct voltage stresses insulation in its own distinctive way.
That difference is worth dwelling on. Under AC, the field inside a dielectric divides according to permittivity — the capacitive grading of Chapter 2. Under DC, once the transients die away, the field divides instead according to resistivity, and slow-moving space charge can accumulate deep inside a solid, distorting the field in ways an AC test never reveals. Testing a cable or capacitor at DC therefore probes a genuinely different failure régime. With the rare exception of the electrostatic generator, every source in this chapter works the same way at heart: take the alternating output of a high-voltage transformer, rectify it, smooth it, and — if you need more — multiply it.
The Half-Wave Rectifier
The simplest high-DC source is a high-voltage transformer feeding a single rectifier (a high-voltage diode, historically a valve) into a smoothing capacitor, with the test object as the load across that capacitor. On each positive half-cycle the diode conducts and charges the capacitor up toward the transformer's peak secondary voltage \(V_{\max}\). On the negative half-cycle the diode blocks, and the capacitor holds the load up on its own.
At no load the output settles at the full peak, \(V_{\max}\). A crucial design point concerns the diode: when the transformer swings to its negative peak while the capacitor still holds \(+V_{\max}\), the rectifier sees both in series. Its peak inverse voltage rating must therefore cover
In high-voltage sets this is met by stacking many diodes in series, each sharing a fraction of the reverse stress through grading resistors — itself a small exercise in the voltage-division ideas of Part 1.
Ripple and Regulation
The moment we draw load current the output is no longer perfectly flat. Between charging pulses the load drains the capacitor, so the voltage sags and then jumps back up at the next peak — a small sawtooth riding on the DC. The peak-to-peak size of that sawtooth is the ripple, \(\delta V\). For a half-wave set the capacitor discharges for very nearly a full period \(T = 1/f\) before being topped up, so
Two levers reduce ripple, and both are visible in the formula: a larger smoothing capacitor \(C\), or a higher supply frequency \(f\). Raising \(f\) is the more powerful trick at high voltage, where capacitors are bulky and expensive — which is exactly why cascade sets are often driven well above mains frequency. Separately from ripple, drawing current also lowers the average output through the source's internal impedance; that sag is the regulation of the set, and in the multiplier of the next sections it becomes the dominant limitation.
The Voltage Doubler
A single rectifier tops out near \(V_{\max}\), but a transformer can only be insulated for so much. To go higher we multiply. The key building block is the Greinacher voltage doubler: two diodes and two capacitors arranged so that one capacitor is charged on the negative half-cycle and then used to lift the second capacitor on the positive half-cycle. The two charges add in series, and the output reaches twice the transformer peak:
This little two-diode, two-capacitor unit is not just a way to reach \(2V_{\max}\); it is the rung of a ladder. Stack these rungs and each one lifts the output by another \(2V_{\max}\) — which is precisely the Cockcroft–Walton cascade.
The Cockcroft–Walton Cascade
In 1932 Cockcroft and Walton used a multi-stage version of the doubler to reach the hundreds of kilovolts needed to split a lithium nucleus — the first artificial nuclear disintegration, and a Nobel Prize. The circuit they built is still the standard way to generate very high direct voltage. It is two columns of capacitors linked by a zig-zag chain of diodes; the transformer drives the bottom, and on successive half-cycles charge is pumped up one column and across to the other, stage by stage, until the top of the stack floats at
At no load, with no current drawn, the arithmetic is clean: \(n\) stages give \(2n\) times the transformer peak. A modest \(100~\mathrm{kV}\) transformer and a ten-stage column reach two megavolts on paper. The catch — and the real engineering — appears the instant the set must deliver current, which is the subject of the next section.
Ripple, Drop and the Optimum Number of Stages
Loading a cascade is harsher than loading a single rectifier, because every stage must pass the load current down through the whole stack of capacitors. Two penalties result, and they scale differently with the number of stages \(n\). The first is ripple — the residual sawtooth — which, with all capacitors equal to \(C\), supply frequency \(f\) and mean load current \(I\), accumulates as
The second, and usually larger, penalty is the voltage drop \(\Delta V\) — the amount by which the loaded output falls below the ideal \(2nV_{\max}\). It grows even faster, dominated by an \(n^{3}\) term:
Here is the tension that defines cascade design. Adding stages raises the ideal output linearly (\(2nV_{\max}\)) but inflates the drop cubically (\(\propto n^{3}\)). Push \(n\) too far and each new stage gives back less than it adds — the actual output peaks and then falls. Setting the derivative of the net output to zero gives the classic result for the most stages worth building:
Electrostatic Generators
One family of DC sources owes nothing to rectifiers. An electrostatic generator makes high voltage by mechanically carrying charge against the electric field — converting the work of a motor straight into electrical potential energy. The best-known is the Van de Graaff generator: a moving insulating belt picks up charge sprayed onto it by a corona comb at the base, carries it bodily upward, and surrenders it to a second comb inside a large hollow metal terminal, where it spreads to the outer surface and raises the terminal to a very high potential.
The terminal voltage rises until the charge arriving on the belt is exactly balanced by the charge leaking away through the load and through corona. The charging current is simply the charge carried per second by the belt — the surface charge density \(\sigma\) times the belt width \(w\) and speed \(v\):
That current is tiny — microamperes — so the Van de Graaff is the opposite of a rectifier set: it delivers enormous voltage (several megavolts in a pressurised tank) at almost no current, with extraordinarily low ripple and excellent stability. Those qualities make it the source of choice for particle accelerators and precision physics, where a steady, clean potential matters far more than power. The rectifier cascade and the electrostatic generator thus divide the territory between them — one for current, the other for the very highest, cleanest volts.
| Source | Typical voltage | Current | Best for |
|---|---|---|---|
| Half-wave rectifier | up to ~ \(V_{\max}\) | moderate | Simple DC supply, modest voltage |
| Cockcroft–Walton cascade | hundreds of kV to several MV | mA range | HVDC and cable testing |
| Van de Graaff | up to several MV | µA range | Accelerators, precision physics |
Worked Examples
Problem. A half-wave rectifier with a \(1~\mu\mathrm{F}\) smoothing capacitor runs from a \(50~\mathrm{Hz}\) supply whose peak is \(V_{\max} = 100~\mathrm{kV}\), delivering \(I = 5~\mathrm{mA}\). Find the ripple, the mean output, and the diode PIV.
Solution. Use \(\delta V = I/(fC)\), then \(V_{\text{mean}} = V_{\max} - \delta V/2\):
A microfarad is enough to hold the ripple to a tenth of a percent, but the rectifier stack must still block 200 kV in reverse.
Problem. A Cockcroft–Walton set has \(n = 5\) stages, each capacitor \(C = 0.1~\mu\mathrm{F}\), driven at \(f = 50~\mathrm{Hz}\) from \(V_{\max} = 100~\mathrm{kV}\), supplying \(I = 5~\mathrm{mA}\). Find the no-load output and the peak-to-peak ripple.
Solution. No-load output is \(2nV_{\max}\); ripple is \(\dfrac{I}{2fC}n(n+1)\):
A \(100~\mathrm{kV}\) transformer reaches a megavolt at no load — but already carries 15 kV of ripple under this modest load.
Problem. For the same five-stage cascade, find the voltage drop \(\Delta V\) and the mean loaded output.
Solution. Apply \(\Delta V = \dfrac{I}{fC}\left(\tfrac{2}{3}n^{3}+\tfrac{1}{2}n^{2}-\tfrac{1}{6}n\right)\) with \(I/(fC)=1000\):
The loaded output is about \(1000 - 95 = \textbf{905 kV}\) — the megavolt sags by nearly 10% at just 5 mA, which is why regulation, not ripple, usually limits a cascade.
Problem. With \(C = 0.1~\mu\mathrm{F}\), \(f = 50~\mathrm{Hz}\), \(V_{\max} = 100~\mathrm{kV}\) and \(I = 5~\mathrm{mA}\), find the optimum number of stages.
Solution. Apply \(n_{\text{opt}} = \sqrt{V_{\max}fC/I}\):
Ten stages extract the most voltage from this supply; beyond that the \(n^{3}\) drop wins. The five-stage set of Examples 2–3 is well below optimum — it could be pushed higher, or its \(f\) and \(C\) raised to lift \(n_{\text{opt}}\) further still.
Chapter Summary
HVDC transmission and DC testing both need it; under DC the field divides by resistivity and space charge matters, unlike AC.
Transformer + diode + smoothing C gives ≈ \(V_{\max}\); the diode must block \(\text{PIV}=2V_{\max}\).
Half-wave \(\delta V = I/(fC)\); cut it with larger \(C\) or higher \(f\). \(V_{\text{mean}} = V_{\max}-\delta V/2\).
Stacked doublers give \(2nV_{\max}\) no-load; ripple \(\propto n^2\), drop \(\propto n^3\), so \(n_{\text{opt}}=\sqrt{V_{\max}fC/I}\).
Higher frequency or capacitance lifts both stiffness and \(n_{\text{opt}}\) — far more effective than simply adding stages.
The Van de Graaff carries charge on a belt, \(I=\sigma w v\): megavolts at microamps, very low ripple — built for accelerators.
Problems
For each item, first identify what it tests — the PIV rule, the ripple formula, the cascade output, the drop, the optimum-stage result, or the electrostatic principle — then apply it. Difficulty rises down the list.
- State why high direct voltage is needed for testing, naming one way DC stress differs from AC stress in a solid dielectric.
- A half-wave rectifier is fed from a transformer of peak secondary voltage 80 kV. Give the no-load DC output and the minimum PIV rating of the rectifier.
- A half-wave set with \(C = 2~\mu\mathrm{F}\) at \(50~\mathrm{Hz}\) delivers \(10~\mathrm{mA}\). Find the ripple and the mean output if \(V_{\max} = 150~\mathrm{kV}\).
- Explain in one or two sentences how a Greinacher doubler reaches \(2V_{\max}\) from a transformer whose peak is only \(V_{\max}\).
- A six-stage Cockcroft–Walton cascade is driven from a \(120~\mathrm{kV}\) peak transformer. Find the no-load output.
- For that cascade, with \(C = 0.05~\mu\mathrm{F}\), \(f = 50~\mathrm{Hz}\) and \(I = 4~\mathrm{mA}\), compute the peak-to-peak ripple.
- For the same cascade and data, compute the voltage drop \(\Delta V\) and the loaded mean output.
- Using the same \(C\), \(f\), \(V_{\max}\) and \(I\), find the optimum number of stages, and say whether the six-stage design is above or below it.
- A cascade's ripple is too large. Compare the effect of doubling \(C\) versus doubling \(f\), and explain why raising \(f\) is usually preferred at high voltage.
- A Van de Graaff belt of width \(0.3~\mathrm{m}\) moves at \(15~\mathrm{m/s}\) carrying a surface charge density of \(2.5\times10^{-5}~\mathrm{C/m^2}\). Find the charging current, and comment on why such generators suit accelerators rather than cable testing.