Fast High voltage switch - Page 1
Forum section: electronics / beginners

author: mortensent
author_rank: Newbie
timestamp: September 30, 2013, 09:43:28 am
content: I'm looking for help fixing/designing a new a fast high voltage switch. I work in a lab that has a multi-ring penning trap and holds on to charged particles for various amounts of time, from 10 micro-seconds to hundreds of seconds. We bias these electrodes using very high voltage amplifiers that take an input of -10 to +10V and spit out either +1.2kV or -1.2kV depending on the supply.The problem is they're painfully slow to fall back to ground, taking several seconds to discharge. Sometimes we want to ground the trap quickly to ensure it's empty before starting another sequence, but since the amplifiers are too slow that's not an option in the long run, 10s or hundreds of nano-seconds would be nice, but even a microsecond or so would be a huge improvement. We have some professional made fast switches that work well for ejecting or catching particles, but they're expensive, and overkill for what we need. The facility has a workshop where they can build a circuit design, but they generally don't have free staff to sit down and discuss and iterate through ideas about making custom circuits. We want a simple circuit that disconnects the power supply and connects and electrode to ground. A previous group member took a design drawn up by a Japanese collaborator many years ago to them and they built it, but it didn't behave correctly, I've just started working with the group and so I'm tasked with finding out why it doesn't work, or finding a better design.The circuit uses a HCPL-3150 (IGBT driver with photo-couplerhttp://www.avagotech.com/docs/AV02-0164EN) to control a IRG4PH40UD (IGBThttp://www.irf.com/product-info/datasheets/data/irg4ph40ud.pdf), to switch on and off voltage.Vcc is hooked up to +18V and Vee +9V, I don't understand exactly what the IGBT driver does, so the need for 2 power supplies is beyond me, the +18 is needed to drive gate of the transistor, can anyone could explain the purpose of a second power supply? I looked up the datasheet, and the device clearly shows Vcc - Vee has to be above 15V to trigger, so I imagine this is why it doesn't work. But perhaps without the Vee supply it would be above threshold, There's more to the circuit (by-pass capacitors etc), but I'm just wondering if anyone can say if this basic concept should work, or suggest something that would.

author: Marco
author_rank: Super Contributor
timestamp: September 30, 2013, 10:21:23 am
content: Maybe Vee was supposed to be -9 Volt? A negative Vge would turn off the IGBT faster.BTW, do you really need the optical isolation? The propagation delay for it is pretty huge.

author: mortensent
author_rank: Newbie
timestamp: September 30, 2013, 10:40:07 am
content: Quote from: Marco on September 30, 2013, 10:21:23 amI saw an example IGBT circuit diagram with a negative bias on Vee, so I thought maybe that is what should have been done, but the diagram they gave me definitely shows it as +9V which is what confused me. Is there a simple explanation of why a negative bias turns off the IGBT faster?As for the optical isolation, it was in the original design, we're triggering with TTL so there has to trigger the transistor, but not sure why it was optically isolated other than a little paranoia about the kV voltages somehow making it back to the NI equipment that would cost a small fortune to replace. Would a MOSFET and some Zener diodes make a suitable (and more importantly, safe) replacement?

author: Andy Watson
author_rank: Super Contributor
timestamp: September 30, 2013, 11:07:53 am
content: Quote from: mortensent on September 30, 2013, 10:40:07 amThe Vee terminal should be returned to the emitter of the IGBT. Vcc needs to be 15-30V above Vee for the circuit to work.With regard to optical isolation; without the isolation you would have to connect the grounds together, this would generate a lot of problems with ground-loops, especially if you're going to be discharging pulses of energy into them ground(s).

author: Marco
author_rank: Super Contributor
timestamp: September 30, 2013, 11:52:04 am
content: Quote from: Andy Watson on September 30, 2013, 11:07:53 amA common mode choke after the gate driver could compensate for a temporarily raised high current ground. There's always gate drive transformers as well.mortensent, the gate emitter junction acts like a capacitor and forms a RC filter with the gate resistor. At turn-off Vge will reach the threshold voltage faster if Vee is negative.

author: Paul Price
author_rank: Super Contributor
timestamp: September 30, 2013, 04:01:14 pm
content: Why not a simple fast SCR or HV MOSFET circuit to discharge the +-1.2KV through a fast bridge rectifier and a fast pulse from a small toroidial pulse transformer to the gate of the SCR or MOSFET?You've haven't posted your circuit.