ESD Protection - Differential Amplifier
More about the circuit
The circuit is sensing (low side sensing for many reasons) current used to power up the lights with 28V(part of the schematics not shown here -> relay with LC filter on the output side to limit the inrush current). The nominal current flowing through R6 is around 0.2A but could be up to 1.5A in peaks(peaks last around 100ms).
First opamp "A1A" is used to convert the differential signal seen on R6(around 200mV) to a single-ended signal with a ~300Hz.
R14, C9, R16, C8, and C7 are here for diff and common mode filtering(cut-off frequencies shown in the bottom right corner of the picture).
Using R12 and R22 around "A1B" and "A1C" I will trim the precise voltage values so that open circuit and overcurrent events flowing through R6 are detectable.
The information whether the circuit is open or there is overcurrent will be reported through Q5 to another device(not on this PCB - most likely industrial computer 5V Input).
Question
How would you position the TVS diodes around the "A1A" inputs so that circuit stays the same but become as immune to ESD events as possible?
ESD that we are talking about here is +-15kV with 330R and 150pF model(IEC6100-4-2 Class 4).
EDIT
Per Andy Aka's suggestion, I am adding more details and maybe further explaining my question.
The load is floating while the bus voltage(28V on the picture) could be between 24V and 32V.
I am worried that +-15kV air discharge will kill "A1A" so I want to protect it with clever positioning of TVS Diodes across the opamp inputs.
Why do I think TVS diodes are needed?
If I would use this circuit as-is and expose it to ESD testing I would blow R14 and R16 because they are 125mW rated. Adding TVS in front of it would dump the usual ESD voltage to a level 200-500V where RC filter on the input could do the job easier.
Are TVS diodes needed in order to keep A1 opamp safe during Class 4 ESD strike? If they are not, how to place them?
3 answers
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Although the voltage drop across the current-sensing resistor R6 is floating and it can "move" between the 28 V rails, the A1A op-amp input voltages can (have to) vary within the 5 V power supply.
So, you can protect the op-amp (and the whole circuit) against any overvoltages by connecting a bidirectional TVS diode between each input and ground. The only problem of this obvious solution can be the diode leakages.
Oh, another problem - the circuit was single-supplied... But it will still work against ESD...
Not a diff amp
Something doesn't make sense here. You want A1A to act as a differential amplifier, but that's not how you wired it:
Here is a real diff amp:
Note how the right side of R1 is tied to ground, not the opamp output.
The diff amp above is really a 3-input circuit. The output is the difference between the two inputs at left, multiplied by the gain (A/B in this example), then added to the reference input. In my circuit the reference input is the right side of R1, which is tied to ground in this example. Therefore, the amplified difference output will be relative to ground.
You have the reference input tied to the output. That makes the output indeterminate for ideal parts. All values of output satisfy the constraints when the inputs are equal.
One way to see this is to analyze your circuit with the left ends of R14 and R16 tied to ground. What will the output be? Whatever value you pick for the output, it will be divided by the same amount at each opamp input. No matter what the output voltage is, the opamp inputs will be the same.
TVSs
You don't need any TVSs.
The zap you specified (15 kV on 150 pF with 330 Ω in series) is going to disrupt the signal. Your only aim therefore is to protect the circuit from physical damage.
You already have 470 Ω followed by 100 nF to ground on each input. That should be enough to limit short zaps to the point where the protection circuitry in the opamp can handle the rest.
Your input capacitors are (100 nF)/(150 pF) = 667 times larger than the capacitance holding the static charge. That alone attenuates the 15 kV to 22.5 V. The input capacitors would only rise from 0 to 22.5 V with nothing else in the circuit.
The time constant of the rise would be (330 Ω + 470 Ω)(100 nF) = 80 µs. Or, we can look at this in frequency space as a single pole low pass filter at 2 kHz. Either way, it should be clear that any active ESD protection circuitry in the opamp can easily react in the time necessary to clamp the voltage.
I didn't look up your opamp since you didn't provide a datasheet link, so it's your job to check how its inputs are protected. But unless this device is unusual, there is no problem here to solve.
[In addition to the answers which have already been posted.]
A current sensing circuit like this lends itself to another scheme for ESD protection, which uses only one TVS. Only the line in which you're sensing the current is exposed to ESD. The small value current-sense resistor will keep the differential inputs at approximately the same potential[1]. So, one TVS on the CURRENT_IN (or the CURRENT_OUT net) will protect both of the differential amp inputs. That TVS will also protect other components which are powered with that current.
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Compared to ESD voltages ↩︎
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