Laser systems can place extreme demands on their components, especially in power delivery circuits. For the processes of energy storage + discharging, disc capacitors are a great choice because of their high capacitance, low dissipation factor, and more. In this article, get an overview of the latest disc capacitor technology from Vishay, used in high-voltage applications up to 50 kV.
Introduction
High voltage disc capacitors are used in a number of applications that require operating voltage capability up to 50 kV, capacitance of over 5000 pF, low inductance, and a dissipation factor (DF) well under 0.5 %. In this white paper, we will focus on their use for energy storage and discharging in laser system applications. In addition, half-wave voltage doublers, which are voltage multiplier circuits consisting of two diodes, two capacitors, and an AC input voltage source will be considered. Found in a host of applications including X-ray systems, high-voltage power supplies, particle accelerators, and ion pumps, the output voltage amplitude of these circuits is twice that of the input voltage amplitude. In addition, we will explore their use in voltage quadruplers which include an additional diode-capacitor stage.
Half Wave Voltage Doubler
The circuit diagram of a half-wave voltage doubler is shown in the figure below. During the positive half cycle, diode D1 is forward biased so current flows through it. This current will flow to the capacitor C1 and charge it to the peak value of input voltage VM. However, current does not flow to the capacitor C2 because the diode D2 is reverse biased. So, the diode D2 blocks the current flowing to the capacitor C2. Therefore, during the positive half cycle, capacitor C1 is charged, whereas capacitor C2 is uncharged.
During the negative half cycle, diode D1 is reverse biased. Therefore, during the negative half cycle, the capacitor C1 will not be charged. However, the charge (Qm) stored in the capacitor C1 is discharged. D2 is forward biased during the negative half cycle, so capacitor C2 charges to a value of 2 VM because the input voltage VM and capacitor C1 voltage VM are added to the capacitor C2. Hence, during the negative half cycle, the capacitor C2 is charged by both input supply voltage VM and the voltage on capacitor C1.
Therefore, capacitor C2 is charged to 2 VM.
If a load is connected to the circuit at the output side, the charge (2 VM) stored in the capacitor C2 is discharged and flows to the output.
During the next positive half cycle, diode D1 is forward biased and diode D2 is reverse biased. So, the capacitor C1 charges to VM, whereas capacitor C2 will not be charged. However, the charge (2 VM) stored in the capacitor C2 will be discharged and flow to the output load. Thus, the half-wave voltage doubler drives a voltage of 2 VM to the output load.
Design Example
Diodes: 2CL74
Capacitors: HVCC103Y6P202KEAX
For VM = 1000 Vpeak input we have: Vo = 2(1000 V) = 2 kV
Adding Additional Stages- Voltage Quadrupler
The voltage quadrupler can be obtained by adding one more diode-capacitor stage to the voltage doubler circuit. Thus, with this configuration, one can add N number of stages to get an output voltage of Vo = VMN, where N is the number of stages added to the initial voltage doubler. The circuit operation is as follows. During the first positive half cycle of the input AC signal, the diode D1 is forward biased, whereas diodes D2, D3, and D4 are reverse biased. Hence, the diode D1 allows current through it. This current will flow to the capacitor C1 and charge it to the peak value of the input voltage VM.
During the first negative half cycle, diode D2 is forward biased and diodes D1, D3, and D4 are reverse biased. Hence, the diode D2 allows current through it. This current will flow to the capacitor C2 and charge it. The capacitor C2 is charged to twice the peak voltage of the input signal (2 VM). This is because the charge (VM) stored in the capacitor C1 is discharged during the negative half cycle. Therefore, the capacitor C1 voltage (VM) and the input voltage (VM) is added to the capacitor C2. Capacitor voltage + input voltage = VM + VM = 2 VM. As a result, the capacitor C2 charges to 2 VM.
During the second positive half cycle, the diode D3 is forward biased and diodes D1, D2, and D4 are reverse biased. Diode D1 is reverse biased because the voltage at the node of C1 and D1 is negative due to the voltage VM across C1, and diode D2 and D4 are reverse biased because of their orientation. As a result, the voltage (2 VM) across capacitor C2 is discharged. This charge will flow to the capacitor C3 and charge it to the same voltage of 2 Vm.
During the second negative half cycle, diodes D2 and D4 are forward biased, whereas diodes D1 and D3 are reverse biased. As a result, the charge (2 VM) stored in the capacitor C3 is discharged. This charge will flow to the capacitor C4 and charge it to the same voltage (2 VM).
The capacitors C2 and C4 are in series and the output voltage is taken across the two series-connected capacitors C2 and C4. The voltage across capacitor C2 is 2 VM and capacitor C4 is 2 VM. So, the total output voltage is equal to the sum of capacitor C2 voltage and capacitor C4 voltage.
2 VM + 2 VM = 4 VM = VO
Design Example
Diodes: 2CL74
Capacitors: HVCC103Y6P202KEAX
For VM = 1000 Vpeak input we have: VO = 2(1000 V) + 2(1000 V) = 4 kV
Excimer Lasers
Laser action in an excimer molecule occurs because it has a bound (associative) excited state, but a repulsive (dissociative) ground state. Noble gases such as xenon and krypton are highly inert and do not usually form chemical compounds. However, when in an excited state (induced by electrical discharge or high-energy electron beams), they can form temporarily bound molecules with themselves (excimer) or with halogens (exciplex) such as fluorine and chlorine. The excited compound can release its excess energy by undergoing spontaneous or stimulated emission, resulting in a strongly repulsive ground state molecule which very quickly (on the order of a picosecond) dissociates back into two unbound atoms. This forms a population inversion.
In the case of initiating the excited state with electrical discharge, the basic circuit topology is shown below where the storage capacitor, typically a storage bank, is used in a flash lamp.
The capacitor bank mentioned can be configured with Vishay’s 715C series class 2 ceramic disc capacitors, when selected with the appropriate capacitance value, voltage rating, and ceramic type.
The most commonly used methods for charging capacitors in pulsed applications are full discharge and partial discharge. Full discharge, as the name implies, allows the capacitor to be discharged to zero for each shot. The power supply is then enabled, the capacitor is charged to the set voltage, and the discharge cycle is repeated. The high-voltage switch is usually an SCR, or a Thyraton for higher-voltage applications.
The partial discharge method takes advantage of semiconductor switches to turn on and off the discharge from the capacitor to the load, allowing the designer to vary the pulse width along with the energy delivered. The specified capacitor is usually large enough so that only a small percentage of the energy is taken from it during each shot. Hence the name “partial discharge.” In both cases, standard formulas can be used to determine the power supply size and calculate the charge time. The simplest way to estimate the amount of energy needed for an application is to use these formulas.
DESIGN EXAMPLE: 2.5 kV PULSE
Power supply: Lumina CCPF-1500-XX
Capacitors: Vishay 715C10KTD80
For a system utilizing a 20-capacitor bank with 8000 pF per capacitor, we have C = 0.16 μF. And, charging this bank to 2.5 kV results in the following energy per pulse:
Energy/pulse = 1/2 CV2 = 0.5(1.6 x 10-5F)(2500 V)2 = 50 j
The capacitors will be charged and then discharged into the system’s flash lamp. The frequency of this charge / discharge process is the rep rate. Thus, the charge rate is calculated as follows:
Charge rate = (energy/pulse)(rep-rate)
When:
C is the capacitor in farads
V is the charge voltage required
Rep-rate in Hz For a 20 Hz system with the above energy per pulse, we have:
Charge rate = (50 j)(20 Hz) = 1000 j/s
This formula does not allow for any dead time (settling time), which is usually required in most systems. So, in most low rep rate applications, selecting a slightly larger supply is the best choice. In this case, the 1500 j/s power supply would be a good model to specify.
In the case of a partial discharge application, the length of time the capacitor is allowed to discharge determines the amount of energy needed to recharge the capacitor to the set voltage. Pulse widths can vary from several hundred microseconds to tens of milliseconds, with a corresponding droop in voltage. In general, calculating the recharge energy can be done using the formula:
ERecharge = 1/2CL (V2 max. - V2 d)
When:
Vmax. is the maximum voltage
Vd is the lowest droop voltage
Now, for the above system, assuming a drop to 1 kV, we have:
ERecharge = 0.5(1.6 x 10-5F)(25002-10002)=42 j