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High voltage power supply

This project is a high voltage power supply, which has an adjustable output voltage of 0 to18 kilovolt. Circuit diagram part 1.

In this part we see from left to right:
- A power supply (TR1, D1, C1, C2) delivering about 23 Volt DC.
- An square-wave oscillator build around the TDA2030 power op-amp.
The oscillator has an output voltage of about 20 Vp-p (Volt peak - peak).
The frequency is about 11 kHz.
- A variac (variable transformer).
The 20 Vp-p is connected to a tap at 70% of the transformer winding.
The transformer output is variable from 0 to about 28 Vp-p.
- Three step up transformers, which are normal mains voltage transformers (230 to 15V/ 900 mA).
However they are designed for 50 Hz operation, they also work at 11 kHz.
The 3 high voltage windings are series connected, giving a maximum of 1700 Vp-p at the output.
With switch S1, you can also select for a lower voltage. Circuit diagram part 2.

This part is a 12 stage voltage cascade.
The first stage is rectifying the input voltage to it's peak - peak value, so in this case maximum 1700 volts.
Every next stage is adding 1700 Volts, so with 12 stages we get more then 20 kV, this voltage can however only be reached when the output of the cascade is not loaded.
In this design there is always some load at the output by the meter circuit (M1, R10 ...R14).
The cascade circuit will give some loss of voltage, depending on load current, capacitor values, number of stages, and input frequency.
Also the transformers will give voltage losses under load conditions.
In this case, I can get 18 to 19 kV out of the circuit.

The higher the input frequency, the lower the voltage losses of the cascade under load conditions.
For this reason, I don't operate the cascade at 50 Hertz mains frequency, but at a much higher frequency.

For low losses in the circuit, I made the capacitor values in the first stages higher.
In the last stages, this is less important.

The circuit is loaded by a meter circuit (M1, R10 ...R14) for monitoring the output voltage.
The meter circuit will also discharge the cascade when the power supply is switch off.
R10 to R14 must be rated at minimal 5 kV breakdown voltage each.

Resistor R15 will limit the output current under all conditions to less then 20 mA.
Which should be a save value, both for human, and for the diodes in the cascade.
This (almost) 20 mA can however only be supplied for a very short period, just after the output is shorted to ground.
When the cascade is discharged, the output current will be limited to maximum 0,36 mA.
For safety reasons, R15 is a special high voltage resistor (30kV breakdown voltage).

All diodes and capacitors in the cascade circuit must be capable of handling the peak-peak value of the input voltage (1700 V).
I used components which can handle 2000 V.

The diodes must withstand this voltage in DC conditions.
Some 2 kV diodes can only handle 2 kV for short periods, but are for DC rated at for instance 1200 V, these diodes are not suitable in this circuit.
The used RGP02-20E diode is a very good one to use here click here for the RGP02-20E datasheet.
The RGP02-20E is obsolete now, but I was able to buy some at: http://www.voti.nl/shop/summary.html . A look inside the high voltage power supply.
With the 4 transformers, the variac, and the oscillator circuit on a small heat sink.
However the oscillator will produce very little heat. At more then 8 kV output, the circuit was making a hissing noise, caused by discharge from the wires to the surrounding air.
For this reason, I covered all high voltage wires with sealant, which removed most of the noise.
Now only resistor R15 (the green one at the right) and the output connector are still hissing. A picture of the input voltage of the cascade.
Every cm screen is 200 Volts, so in this case we have 1200 Vp-p. The completed high voltage supply.
The top is covered with a plate of Plexiglas, so the cascade stays visible.
The right side of the apparatus is made of polypropylene, on this side the high voltage output is positioned.
The connector at the high voltage output is a normal banana socket, so there is nothing special "high voltage" about that.

The power consumption of the supply is 6.5 - 7.5 Watts, depending on output voltage and load current.

The maximum output voltages are:
18 kV with switch S1 in position "18 kV"
5.8 kV with switch S1 in position "5 kV"

When I connected CN1 to CN2 (mains earth connection to cascade ground), the output could reach 19 kV.

The short circuit current of the output is:
0.25 mA with switch S1 in position "18 kV"
0.36 mA with switch S1 in position "5 kV"
The output current is in this design mainly limited by the transformers TR2, TR3 and TR4, which can not supply so much current at 11 kHz.
The cascade also gives some current limiting, but not so much as the transformers.
Maybe the output short circuit current could somewhat be increased by reducing the oscillator frequency, but the measured values are okay for me.

Voltage drop in voltage cascade circuits The input signal is in the first stage of the cascade rectified to the peak - peak value of the input signal.
In fact you lose 2 times the voltage drop of the diode, but for high voltage operation this can be neglected.
Every next stage is adding an equal amount of voltage.

So with n stages, you get n.Vp-p at the output.
This is only true when the output is not loaded.
When the output is loaded with a current ( I ), the output voltage will drop with a value ΔU according to the next formula: Where:
ΔU = Voltage drop of output in V.
I = Load current in A.
f = Input frequency in Hz.
C = Capacitor value in F.
n = Number of stages in the cascade.

For this formula, the input voltage must be a sine wave.

 Load current μA Frequency Hz Capacitor values nF Number of stages 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Voltage drop in cascade: Volt
Calculate the voltage drop of a cascade

With this calculator you can easily calculate the voltage drop of a cascade circuit, it uses the formula mentioned above.
Fill in the yellow coloured fields, and click "calculate".

Of course the voltage drop could never be higher then the unloaded output voltage.
If this happens in the calculation, you have entered a too high load current, which could not be reached in practice.