Saturday, January 19, 2013

High Voltage Pulse Generators

Many applications require a repetitive source of high voltage pulses having fast rise and fall times. These applications include drivers for piezo-electric devices, ion tubes, gas tubes, liquid polarizing cells, beam steering applications, the generation of electric fields in aqueous solutions, and time-of-flight mass spectrometry measurements.
A high voltage pulse generator consists, essentially, of a high voltage power supply and a high voltage switch capable of alternately connecting and disconnecting this voltage source to a load. A simplified pulse generator is shown below. Here the high voltage supply charges a reservoir capacitor while the high voltage switch (HVS) is open. When HVS is closed, this reservoir capacitor rapidly supplies a pulse of current to charge the various capacitances (consisting of internal, cable, and load capacitances, as shown) to the full pulse amplitude and then, with the help of the power supply, supplies a steady current as determined by the load resistance to maintain this amplitude until HVS is again opened.
High Voltage Pulse Generators  in a Circuit
Although a high voltage pulse is the desired waveform, it is important to consider the current waveform required to produce this voltage pulse because it is this current waveform that sets important requirements on both the power supply and the high voltage switch. Current through the high voltage switch is a function of the sum of all internal, cable, and load capacitances and the load resistance. All three capacitances and the load resistance also have a significant impact on the rise and fall times of the pulse. A common problem with many users is in failing to consider these effects when specifying pulse operating requirements. Let’s consider a typical current waveform in more detail.
Component of Current Pulse
This current waveform consists of two parts: an initial sharp pulse and then a steady component. The sharp pulse of current, which determines the voltage pulse 10% to 90% rise time, has an average value as given by the following equation, where dV is the change in pulse amplitude during the rise time and dT is the rise time. As a rule of thumb, the peak current during the rise time is 2 to 3 times this average value.
Iaverage-rise = [Cinternal + Ccable + Cload] dV/dT
The surrent required to maintain the pulse amplitude across the load resistor is as follows, where R is the value of the load.
Iflat top = Pulse Amplitude / R
The 90% to 10% pulse fall time is given by the following equation:
Fall Time = 2.2 R [Cinternal + Ccable + Cload]
Consider this typical application. A 3.2 kV pulse generator with a 1 ns rise time and 12 pF of internal stray capacitance is connected to a Pockels Cell having a capacitance of 1 pF and a resistance of 10 M ohms via a 2-foot length of RG 114, 185 ohm coaxial cable. This cable has a capacitance of 6.5pF per foot when not terminated in its characteristic impedance. The values give:
Iaverage-rise = [12 pf +13 pf +1 pf] [3,200 volts / 1 ns] = 83.2 amps

Ipeak-rise ~ 160 amps

Iflat top = [3,200 volts / 10 M ohms] = 320 µa

Fall Time = 2.2[10 M ohms] [12 pf + 13 pf + 1 pf] = 572 µs
Therefore, to drive a 1 pF Pockels Cell, through 2 feet of 185 ohm cable, from 0 to 3.2 kV in 1 ns requires a peak current of approximately 160 A. The fall time is slow because the various capacitors are discharged through the 10 M ohms resistor when the high voltage switch is opened. The fall time may be made faster by using a shunt resistor across the cell terminals.
As an example, consider what happens when the cable is terminated in its characteristic impedance by adding a 185 ohm shunt resistor across the load. This reduces the current required during the pulse rise time dT by effectively eliminating the cable capacitance, increases the current required from the power suppluy to maintain the pulse amplitude, and reduces the pulse fall time.

Iaverage-rise = [12 pf +1 pf] [3,200 volts / 1 ns] = 41.6 amps

Ipeak-rise ~ 80 amps

Iflat top = [3,200 volts / 185 ohms] = 17 amps

Fall Time = 2.2[185 ohms] [12 pf + 1 pf] = 5.3 ns
Another technique that can be used to reduce the pulse fall time and minimize the effects of load and cable capacitance is to add another high voltage switch from the load to ground. When the first HVS is turned off this HVS is turned on, effectively discharging the load capacitance to ground, or “biting the tail” of the pulse. This generally results in a fall time that is less dependent on the load capacitance and resistance. The average pulse current is also decreased since a shunt resistor across the load is no longer required. The addition of the second switch increases the cost of the pulse generator and, in most cases, has a minimal effect on the rise time.
The HV Pulse 5k-2500 is an example of a pulse generator that utilizes a tail biter to control pulse fall time. This pulser is designed to drive the grid of a high power vacuum tube, which presents a large load capacitance. Assume a pulse amplitude 0-5 kV and a load that looks like 250 ohms in parallel with a total of 1450 pF. The design goal is to keep the pulse fall time below 200 ns. With no tail biter, the approximate fall time would be:
Fall Time = 2.2[250 ohms] [100 pf + 1450 pf] = 853 ns
When a tail biter consisting of a current-limiting resistor of 70 ohms in series with a second HVS is introduced across the load, the fall time is reduced to:
Fall Time = 2.2[250 ohms || 70 ohms] [100 pf + 1450 pf] = 187 ns
It is apparent that load capacitance, including cable capacitance, is a very important consideration in the design and application of high voltage pulsers. Depending on operating requirements, the effects of capacitance can be controlled in several ways: the length of the interconnecting cable can be made as short as possible, a shunt resistor can be added across the load, or a tail biter can be introduced into the circuit.

No comments:

Post a Comment