Special power supply - High-voltage diodes in series

I am working on a homemade non-transferred arc plasma torch, a device that generates and sustains an electric arc, that is blown in particular conditions.

The power supply is composed of two subsystems: a “low”-voltage power supply (up to 1000 V) running at high power and current-controlled (close to 100 kW at maximum power), and a high-voltage power supply (up to 30 kV) running at low power (up to 50 W).

The high-voltage power supply ignites an arc with a quick discharge (a few tens of µs), and the low-voltage power supply has to react as quickly as possible to sustain the arc once it is ignited.

Both power supplies are connected in parallel, and currently, the high-voltage is going towards the low-voltage power supply as its resistance seems to be lower than the one of the plasma system. To avoid this problem, I want to put in diodes to avoid the current going towards the low-voltage power supply.

For that, I need a diode that can sustain >100 A and >25 kV, which is really expensive to do with only one diode, so I want to use several MDQ 150A 1600V diodes in series, 20 diodes for safety, to increase the reverse voltage capacity of the circuit.

While reverse-biased, the blocking voltage of each diode is different as the diodes need to carry the same leakage current. I read online that this problem can be solved by connecting resistances across every diode. Voltage would be shared equally; hence the leakage current would differ.

However, I’m having trouble calculating the value of the resistors.

I used several sources, for example this calculator. I often come across the value of 3 MΩ per resistor, which is doable, but seems a bit high.

Does anyone have an idea on how I should proceed?

Here is a basic schematic of what I’m trying to achieve:

• The datasheet of the diode you linked does not give any leakage current rating, which makes it unsuitable for your application. You need a diode whose manufacturer actually tells you the leakage current. Commented Mar 10, 2023 at 19:20
• Why not an HF start as most welders use? Commented Mar 10, 2023 at 21:32
• Thank you Jonathan, indeed this info is missing. @TimWilliams I'm using a laser HV power supply that works pretty well. The low-voltage power supply works well as well, all alone. It's when I combine both that problems happens. I could move the electrodes closer to generate the arc with the low-voltage one only, but that would require significant design change in the setup !
– Pwp
Commented Mar 10, 2023 at 22:40
• And 4kW of diodes isn't a "significant design change"..? Commented Mar 10, 2023 at 23:34
• It is, I underestimated the complexity of the diode solution. I will still try to figure out a way to make it work without modifying the mechanical setup too much, as months of numerical simulations were involved!
– Pwp
Commented Mar 11, 2023 at 22:18

What you want to do is pretty difficult to achieve.

Let's take the Vishay VS-VSKE320-20PBF as an example, which is a 2kV, 320A diode that (unlike the one you linked) actually gives a rating for its maximum reverse leakage current: 50mA.

If you apply 30kV to a string of such diodes, the diodes may well dissipate 30kV * 50mA = 1500 W due to their leakage current alone, thereby hopelessly overloading your 30kV power supply. In order for this to work, the leakage current of the diodes has to be around 1mA or less instead.

It seems like silicon carbide Schottky diodes might just work, but they will be pretty expensive.

The GD60MPS17H from GeneSiC is a 1.7kV 122A diode with a typical leakage current of 41µA at 175°C and maximum reverse voltage. This means that you could pass around 200µA through the balancing resistors to keep their reverse voltage drops matched, for a total leakage of roughly 0.25mA. With 1.5kV across each diode, you'd need 20 of them in series, and a 7.5 MOhm resistor across each of them. This will still dissipate 7.5 Watts due to its leakage current, but your 30kV power supply will be able to handle it.

Make sure to cool these diodes appropriately as they will each dissipate almost 300 Watts if you actually pass 100A through them (6kW in total). The case temperature of each diode has to stay below 100°C.

The only downside is that that's $1k in diodes alone, and probably just as much in heatsinks. It might be more realistic to put two of these diodes in parallel each, for a total of 40 diodes ($2k), which would lower the total power dissipation to around 4kW at 100A due to the significantly lower forward drop at lower currents. (Make sure to add current sharing resistors if you go that route.)

Another suitable diode would be the STTH3010D, which is a 1kV 30A type that costs about $2 each. You'll have to put 4 of them in parallel, and 40 in series, for a total of 160 diodes. Additionally, you'll need a 10 Milliohm balancing resistor in series with each diode. The power dissipation in each diode will be 40 Watts (25A per diode), and each resistor will dissipate another 6.25W. In total, that's 7.4kW dissipation for all 160 diodes and resistors combined. The cost of the diodes will likely be negligible compared to the cost of the cooling solution for them, though. You might very well save money by using the more expensive silicon carbide diodes. To put things into perspective: To cool those diodes dissipating 7.4kW, you will need something like a car radiator, or maybe even multiple, plus an expensive water-cooling setup with very expensive coldplates to mount the diodes on. Alternatively, you could use a different ignition voltage source, such as a transformer in series with the 1kV 100A supply's output that can temporarily superimpose a large AC ignition voltage onto the 1kV DC. The transformer will likely have to be pretty insane too, though, since it must handle both the 100A current (thick secondary winding), as well as the 30kV ignition voltage (extremely good isolation). Its power dissipation will be much lower than that of the diodes, and it will likely also be cheaper. It will, however, be more difficult to build. • If I apply 30kV to a string of such diodes, how could they dissipate 30kV * 50mA if the total power of the HV supply is 50W ? I might be missing something, I'm definitely not from an electronic background ! Thank you very much for your tips and interesting insights, I will try to find other diodes that fit in my budget (ideally under$200 in total for the diodes).
– Pwp
Commented Mar 10, 2023 at 22:42
• They would dissipate 1500W if the PSU could deliver that amount of power. Since it can't, the PSU will just give up and its output voltage will collapse. In other words: It'll be overloaded. Commented Mar 10, 2023 at 23:02
• I see, you made some very good points, thanks. I still have a vocabulary question, does the Fig. 5 of this datasheet represent the leakage current ? Is "Instantaneous reverse current" a synonym of leakage current ?
– Pwp
Commented Mar 10, 2023 at 23:22
• $200 is absolutely unrealistic for this kind of project. You might get the diodes for$300, but then you need a cooling solution for them that likely costs five figures. And yes, figure 5 in the datasheet you linked does give the diode's leakage current. You need 100 (better 150) of these diodes in parallel to handle your 100A, so you'll have to multiply this current by 100 (or 150). 20µA * 150 = 3mA -> overload (90W). The diode you linked won't work. You'd ideally stay below 0.5mA because the balancing resistors also draw some additional current. Commented Mar 10, 2023 at 23:26
• Also, even if you were able to use the HVP20 (maybe you buy a better 30kV PSU), you will still have a huge problem: You'll need 150 of these diodes in parallel to handle the current, and two sets of diodes in series to handle the voltage. That's 300 diodes in total. Each diode dissipates about 20W, so you'll have to cool 6kW again. That necessitates a very large industrial chiller unit, which is a five-figure deal. You're basically trying to keep a large oven at 50°C by cooling it with heatsinks and fans. Commented Mar 10, 2023 at 23:31