# Issue with zener diode protection of opamps in high voltage feedback path

#### SiCEngineer

Joined May 22, 2019
439
Hi All.

I am sensing a DC voltage that is divided down to 5V from my power supply. It is a negative voltage power supply so I am also converting it to positive using a unity gain inverting opamp. I also have a buffer opamp between the inverting opamp and the divider chain so that I do not have issues with impedance of the divider chain intefering with the opamp stage.

My question is, what is the best way to protect this circuit from overvoltage? My output is 6kV so I need to be careful with my feedback path to ensure my control electronics are isolated and safe. I will show how I have it at the moment, but my worry is that it is not correct because resistor R3 and R2 will divide due to the virtual ground of the opamp, meaning the voltage I want will not be the voltage I get.

Further, do I need two zener diodes, one in each direction, or is just the negative one suffice? I don't ever expect a positive voltage at that level.

Last question, the inverting OPA does not need protection, if the buffer is protected - correct, or incorrect? I would assume the OPA limiting its max output voltage at 5Vsupply would be sufficient for that, but do let me know if I am mistaken.

Best,
SiC

#### Ian0

Joined Aug 7, 2020
8,939
Have you got V1 the right way round? I thought you said that the voltage was negative?
I would suggest that you make R4 = 100MΩ and R5=83kΩ and delete everything else.
Use an op-amp with a common mode voltage range that includes ground and rail to rail outputs (most of the 5V supply op-amps do this). Connect the non inverting input directly to ground. Place two back-to-back schottky diodes between inverting input and ground.

#### SiCEngineer

Joined May 22, 2019
439
Have you got V1 the right way round? I thought you said that the voltage was negative?
I would suggest that you make R4 = 100MΩ and R5=83kΩ and delete everything else.
Use an op-amp with a common mode voltage range that includes ground and rail to rail outputs (most of the 5V supply op-amps do this). Connect the non inverting input directly to ground. Place two back-to-back schottky diodes between inverting input and ground.
Sorry! Yes, the voltage is shown the wrong way round. I just quickly drew this in LTSpice. So this is the circuit you suggest? For an input voltage of -4.971V, the output voltage is +2.481V. Have attached .asc file.

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#### Ian0

Joined Aug 7, 2020
8,939
No - delete R1 and R2. D1 and D2 should be schottky diodes in inverse parallel.

#### SiCEngineer

Joined May 22, 2019
439
No - delete R1 and R2. D1 and D2 should be schottky diodes in inverse parallel.
If I delete R1 and R2 then the full voltage is placed across the Zener. Maybe you mean this? Do I need the negative supply too? The output will only be positive, but can an OPA invert a negative input voltage without a negative supply?

SiC

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#### Ian0

Joined Aug 7, 2020
8,939
SCHOTTKY not zener.

#### SiCEngineer

Joined May 22, 2019
439
SCHOTTKY not zener.
Can you explain why I need a Schottky diode? In the references I've ever read for OPA, they've always suggested to use Zener diodes for input protection?

#### Ian0

Joined Aug 7, 2020
8,939
An op-amp with common-mode range that includes zero will have a maximum negative input of -0.5V, before substrate diodes start to conduct. A schottky will clamp to -0.2V.

#### SiCEngineer

Joined May 22, 2019
439
An op-amp with common-mode range that includes zero will have a maximum negative input of -0.5V, before substrate diodes start to conduct. A schottky will clamp to -0.2V.
If you would be so kind, could you show a schematic of your alternate solution. I am struggling to match it. Obviously the problem is my input voltage will extend well below -0.2V, so what use is clamping it at the input? Does the OPA need a negative supply in this instance or is this a single supply OPA that has the -0.5V max negative input?

Apologies for all the Q's. I just have trouble envisaging the solution without seeing a circuit.

SiC

#### Ian0

Joined Aug 7, 2020
8,939
The inverting input will be at 0V, because the current through the 100M resistor will be matched by the current through the 83k resistor. It's a virtual earth circuit.

#### SiCEngineer

Joined May 22, 2019
439
The inverting input will be at 0V, because the current through the 100M resistor will be matched by the current through the 83k resistor. It's a virtual earth circuit.
Okay, that makes sense. I suppose I am confused because the results do not match what I expect. The DC voltage at the output of the OPA should be 0-5V for this combination of resistors, but it is 0-2.5V.

Is there a reason that the output introduces a gain of 0.5?

#### Ian0

Joined Aug 7, 2020
8,939
Current it both resistors should be 60uA. That would give +5V at the output for -6000V in.
OPA227's common mode input range doesn't include its negative supply. It needs to be at least 1V above the negative supply.
Try something like an MCP6071

#### MrChips

Joined Oct 2, 2009
29,809
Why do you need an inverting amplifier?
I would go with the original R1 + R2 voltage divider and connect directly to the non-inverting buffer op-amp.
Don't use a zener diode for protection. Low voltage zener diodes have a very soft knee voltage.
Connect a Schottky diode from the input of the opamp to the +5V supply.

#### SiCEngineer

Joined May 22, 2019
439
Current it both resistors should be 60uA. That would give +5V at the output for -6000V in.
OPA227's common mode input range doesn't include its negative supply. It needs to be at least 1V above the negative supply.
Try something like an MCP6071
Problem solved. Thank you very much Ian for taking the time to explain this to me.

SiC

#### SiCEngineer

Joined May 22, 2019
439
Why do you need an inverting amplifier?
I would go with the original R1 + R2 voltage divider and connect directly to the non-inverting buffer op-amp.
Don't use a zener diode for protection. Low voltage zener diodes have a very soft knee voltage.
Connect a Schottky diode from the input of the opamp to the +5V supply.
I need to invert the voltage because the input range to my ADC is 0-3V. I have a voltage divider after this, so I make the output 5V so that the fault voltage at 6kV gives me 3V at the ADC input. Thanks for the input regarding the Schottky diodes, I usually have seen Zener diodes used but I will look into Schottky a bit more to understand why they are better.

#### MrChips

Joined Oct 2, 2009
29,809
I did almost exactly what you need to do.

R1 = 100MΩ
R2 = 50kΩ
and use a Schottky diode to +3.3V supply rail.

#### MrChips

Joined Oct 2, 2009
29,809
Oops!
Sorry, I just read that your HV supply outputs a negative voltage.
In that case you will need a bipolar supply and an inverting amplifier.

#### Ian0

Joined Aug 7, 2020
8,939
Oops!
Sorry, I just read that your HV supply outputs a negative voltage.
In that case you will need a bipolar supply and an inverting amplifier.
Single supply op-amp works, provided that the common mode input voltage includes zero.

#### SiCEngineer

Joined May 22, 2019
439
Single supply op-amp works, provided that the common mode input voltage includes zero.
Hi Ian,

Just revisiting this thread as I am choosing the OPA now for my application. I am looking at OPAx365 from Texas Instruments for my inverting amplifiers and other on-board buffers as it seems a good fit. It says the common mode input voltage range is V- (-0.1) minimum, so I believe for a single supply op-amp, that should mean it includes ground.

But can you elaborate on the reason why an op-amp is able to perform this function despite the -5V signal still being well below this common mode voltage range? What about the VOCM including ground makes this a viable solution?

Best regards,
SIC

#### crutschow

Joined Mar 14, 2008
33,326
But can you elaborate on the reason why an op-amp is able to perform this function despite the -5V signal still being well below this common mode voltage range?
As long as the op amp is operating as an inverter in its linear range (not saturated) the op amp's negative feedback maintains the virtual ground at near 0V, thus the amp input terminals never see -5V.