# Operation amplifier input HV protection

I have a setup consisting of 8x AD8428 ultra-low noise opamps on a PCB to measure very small voltages with a noise floor of around some 100uV and a low-pass filter up to 10kHz. The input is connected to a superconductive solenoid with an inductance of up to 10mH. Normally, the solenoid is in a superconductive state, resulting in no voltage drop across it. However, in certain circumstances, when loss of superconductivity appear a pneumatic switch will kick off feeding current of around 2kA therefor a voltage peak of up to 1kV with exponential decay (20-30ms duration) can appear on the coil and the opamp input.

I am seeking a solution to protect the preamplifier against this high voltage peak without increasing the noise level. The current configuration with input low-pass filter and some existing protection is the following:

My plan is to put directly before the choke (next after the LEMO connector):

• 10 - 10 Ω resistor
• 8.0SMDJ15CA TVS diode (300A peak pulse current)

My calculation indicates that, in the worst case of a 1000V peak, the current through the diode would be limited to 50A for that short duration.

I would like to know if this solution provides the necessary and reliable protection or if there is a better way to achieve this?

UPDATE:

Out of curiosity, the circuit will also function as a detection circuit to detect the resistive state (voltage) of a solenoid. Detecting this early as possible, allows for promptly cutting the feeding current, preventing potential damage to the coil. Therefore, it is crucial to be able to measure the smallest voltage just above the noise floor.

Some asked about the shape of the peak, its a simple circuit consisting of magnetically charged solenoid, diode and discharge resistor with value of about 0.3ohm and with thick cables. So its a fast rump up of voltage as pneumatic switch cut the current then an exponential decay with tau=L/Rdisch. For safety reasons the voltage should be limited to 1kV.

Thanks for the answers, from the ideas they provided, I put together the following circuit. Looks like it would be able to handle +-1kV for several 100 ms which is more than enough: This is the dissipation on one of the MOSFET and the output voltage:

However I still see a space for a slight improvement, if its possible:

1. The dissipation is still around 40-60W, because the MOSFET is not fully closed - it will never be, because then there will be no voltage drop on R1 or R4. Now its value is 10ohms. Is there any trick to decrease further this dissipation without increasing this resistance?

2. The voltage to be blocked is quite high, reaching up to 1kV. It might be beneficial to use multiple MOSFETs in series to reduce the voltage drop across each. However, when I added another MOSFET in series, I observed a voltage drop and power dissipation only on one. This was due to a Vgs difference between the first and second MOSFETs, with the first having 4.69V and the second having 4.76V, resulting in a 70mV difference caused by voltage drop on second MOSFET. Is there a simple method to equalize these voltages? The problem is much clearer from the picture:

UPDATE 2: Are my assumption correct, that until the MOSFET is fully open there is no significant noise?

Thanks

• Do you have any screenshots of the peak? Feb 6 at 17:19
• Ideally, you want as much series resistance as possible before too much adds unacceptable noise to your signal. FYI, that diode your adding only clamps the differential and does nothing for the common-mode. Feb 6 at 17:20
• Is the signal level small enough that you could use back-to-back silicon diodes? Feb 6 at 17:45
• "current of around 2kA therefor a voltage peak of up to 1kV" -- this does not follow; should we understand there is a 2 ohm resistor shunting the coil/switch? Feb 6 at 18:26
• FYI, your edits are changing the direction of the question, after answers have been made to the initial version -- they may be better asked as separate questions, with a backlink referencing this post. Feb 9 at 18:43

Disclaimer: I've never done this before, but that never stopped anyone from giving an opinion on the internet, right?

OK, suppose 1kV appears on the left of this schematic:

If the TVS diode does its job, it will limit the differential voltage at the input of your instrumentation amps to 15V. Indeed, you will have about 50A flowing (1000V/20 ohms and neglecting 15V across the TVS) which means each 10 ohm resistor will have about 500V across it, so it will dissipate 25kW during say 20ms, or about 500 Joules. Let's check the energy meter:

.45 ACP bullet muzzle energy: 432 J

So you need a rather serious pulse power rated resistor, most likely many resistors in series/parallel combination so they withstand the high voltage and power.

The TVS limits voltage so it will get much lower energy, although it's still a substantial amount. In this scenario it gets 50A * 15V * 20ms = 15J which is probably too much. Since your amps have a gain of 2000, I guess your input voltage must be tiny, so a pair of antiparallel diodes would probably be a better choice. With lower voltage, they'll get lower energy.

This is only an estimation, it depends on the exponential decay of your voltage pulse, but I guess it's better to play it safe...

Instead of fixed resistors, you could use some kind of current limiting device. The idea is to keep the resistance as low as possible during normal use (for low noise) while limiting current. One example would be PTC resistors, polyswitch, PTC heating wire... or simply a fast fuse.

Note this is only about differential mode. There is no guarantee the common mode will not blow your opamps. If you remove the current from that superconducting coil in milliseconds there's going to be a huge variable magnetic field which is probably going to make everything in the vicinity act like a transformer and cause a common mode spike on your sensor wires. If you use TVS or diodes to dump it into ground or the supply rails, this current still has to go somewhere and may cause problems for the next equipment in the chain. I guess you'd have to measure it.

Here's a simple current limiting circuit. It only works in one direction so you'd need two circuits in anti-series configuration to limit current in both directions, and it needs an isolated power supply (for example from a photovoltaic optocoupler). So it's a bit cumbersome.

V2 is normally around 0V, so almost no current flows, voltage on the 1R resistor is negligible, the BJT is off, and the MOSFET is fully ON, which means it acts as a low value resistor.

When V2 turns on, applying high voltage, a large current spike flows through the MOSFET, which turns on the BJT, and after a very short time (couple tens of nanoseconds) it becomes a 600mA constant current source.

This is a very unusual problem.

You have 50Ax1kV = 50000W for 30msec this would be 1500J. Over 1 second that would be 1500W, that's a lot of power for a little device to handle.

It looks like the 8.0SMDJ can support 8000W so this is almost 6x what the TVS diode can support.

Not only that but the duration that the device can support goes down with time, it probably won't support that much energy.

Source: 8.0SMDJ datasheet

As far as noise goes, you would be adding some noise from the diode and you would need to parallel them (you would probably need too many for your application) and or add additional protection like a Gas Discharge Tube (although I've never used one to protect an low level analog circuit so I don't know how this would affect your noise).