Monday, January 2, 2012

Discharge Circuit Design

My latest work on the project has centered around the assembly and testing of my charging and discharging circuits. Below is a diagram of the circuit I ultimately implemented for charging and discharging my capacitor bank (click for larger resolution).

When the DPST mechanical switch S1 is thrown to the left, the leftmost DPDT relay S2 is actuated. This then turns the step-up converter on and actuates the SPST charging relay S3. The capacitor C towards the bottom right of the diagram is then charged to the converter's output voltage (approx 6 kV) through the 1.2M resistor. Once the capacitor is fully charged (if C = 0.3 uF, RC = .36 s and the capacitor C will be charged to within 1% of its charging voltage in 5RC = 1.8 s), S1 is thrown to the right, actuating the SPST discharging relay S4 and shorting the capacitor's positive plate to ground. The resultant pulse discharge can be used to initiate a water arc explosion.

The diodes in parallel with each relay are there to protect the switch S1 (and the contacts of relay S2) from the momentary voltage spike produced when the relays are switched off. Relays S3 and S4 must be specially made for switching HV signals. I used two 5kV reed relays connected in series for both S3 and S4. The reed relays are designed so that their contacts are contained in an evacuated chamber, thus allowing them to switch high voltage without sparks arcing between the contacts.

The box labeled "DC-AC-DC Step-Up Converter" will likely comprise the below circuit.

The unit after the step-up inverter is called a Greinacher voltage doubler; it takes an AC signal and outputs DC (with ripple) at twice the peak input voltage. The first capacitor is charged on the negative half of the AC  cycle to the AC peak voltage. On the positive half, the output is a superposition of the input AC waveform onto the capacitor's discharge current. The second capacitor reduces voltage ripple.

The circuit's open-circuit output is nominally 2Vp. Under load, the circuit's output voltage drops with load impedance. The circuit's exact voltage characteristics are given by
Where I(load) is the current through the load (A), f is the frequency of the AC input (Hz), and C is the capacitance of the doubling capacitor (F). Assuming a maximum load current of 6kV/1.2MOhm = 5 mA, an input frequency of 60 kHz, and an input capacitance of 270 pF, the circuit's minimum output voltage comes out as 5691.4 V. As the capacitor C charges, I(load) will decay to zero, and the output voltage will level off at 6 kV.

1 comment:

  1. This is some technical stuff and I mean I am no technical guy but whatever you are intending to develop I hope it benefits humanity at large