Tool

Can your reference amplifier survive its own actuator?

The current you inject is a loop, and it comes home through the electrode you added to protect your front end. Whether the amplifier can absorb that is a multiplication and a comparison, on a napkin, before a board exists.

uA
The actuation current that comes home through the reference electrode.
Mohm
Total limiting on that leg, as designed.
V
Where the reference holds the body, usually near ground on a low supply.
V
How close to its negative rail the output can actually drive. Zero on a single supply.
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At nominal, every component on a good day
Developed across the limiter
0.96 V
0.8 uA x 1.2 Mohm
Output must reach
-0.36 V
set point 0.6 V minus 0.96 V
It can sink at most
0.50 uA
(0.6 V minus 0 V) / 1.2 Mohm
Over by
1.60x
more than the part can physically do
The amplifier has to reach -0.36 V. On this supply its output cannot go below 0 V, so the voltage being asked for is on the other side of a wall it cannot cross. It rails, stops holding the common mode, and your front end goes blind the instant the actuator fires. You need 0.80 uA and it can deliver 0.50 uA, which is 1.60 times over at nominal. That is not margin you can tune out. The architecture is wrong, not the values.
Sensitivity: what if the current is higher

A real safety review refuses to believe in perfect days. The most common way this gets worse is a mis-assembly: two correct limit resistors placed in parallel instead of in series, which passes every part inspection because both resistors are the right value and the bill of materials is correct. Set the multiplier to whatever your own fault analysis says, and watch the margin move.

x
A sensitivity, not a physical model. Other impedances share the path, so use your own number.
Output must reach
-1.32 V
Over by
3.20x
At 2 times the current the amplifier needs -1.32 V and cannot get there. If this fault is credible for your board, the device does not survive it.
If it does not clear, the fix is a decision rather than a part

You cannot buy your way across that wall, because the wall is the architecture. Give the return current a way home that shares no node with your sensing reference: its own electrode, its own path, going to analog ground directly rather than through the amplifier holding your common mode. Keep the current limit on the drive leg, the outgoing side, so the bound on how much can reach the body holds regardless of what the return path does. Both of those are things you specify, not parts you order.

Send me this result, with where I would look next

Your numbers go with it. You get a straight technical reply about this specific node, including the two design decisions that usually resolve it, from the person who would lead the work.

Your inputs are included so the reply can be specific.

The arithmetic, so you can check it

Start from the loop. Every current you inject is a loop. It goes out through the drive electrode, through the body, and it has to come back. Nowhere on the way does it disappear, and a ground symbol on a schematic is not a sink. It is a node, with real impedance between where the current arrives and where it needs to get.

The current is not malicious, it is lazy: it takes the lowest-impedance path home. On a great many of these designs that path runs straight through the driven reference electrode, precisely because that electrode is a deliberately low-impedance connection to the body, which is what made it good at its job.

Then one multiplication. Voltage equals current times impedance. Microamps times megohms gives volts directly, which is why the numbers are so easy to check.

Take 0.8 microamps, a small, responsible current to put near a body. Take a 1.2 megohm series limit, large on purpose because a large resistor is how you keep the current safe. Neither raises an eyebrow, which is exactly why the next step gets missed: together they develop nearly a volt.

Then one comparison. A volt relative to what? Your reference holds the body at a set point, maybe 0.6 volts on a low-supply system, near ground where the front end wants the common mode. So the amplifier's output has to reach the set point minus that drop, which lands around minus 0.35 volts.

The amplifier runs on a single supply. Its output cannot go below its own ground rail. There is no negative rail to reach toward, so the voltage being asked for is on the other side of a wall it physically cannot cross. It rails, abandons the job of holding the common mode, and the common mode wanders straight out of your front end's range. That is the flat line, and it was never coupling.

Run it backwards to see how far off you are. The most it can sink before railing is the available headroom divided by the resistance. With 0.6 volts over 1.2 megohms that is about 0.5 microamps, against the 0.8 you need. Being 1.6 times over what the part can physically do, with every component at nominal on a good day, is not margin you tune your way out of. The architecture is wrong, not the values.

What this does not model

A single shared node, at DC. It does not model the electrode interface impedance, the body's own distributed impedance, the amplifier's output impedance under load, or anything frequency dependent. Those all make the picture worse rather than better, so treat a comfortable result as necessary and not sufficient, and a failing result as reliable. The fault multiplier is a sensitivity you set, not a physical model of any particular mis-assembly.

If it does not clear

You cannot buy your way across that wall, because the wall is the architecture. It is about where the current goes, not how strong your parts are. Give the return current a way home that shares no node with your sensing reference: a dedicated return, its own electrode, its own path that does not run through the amplifier holding your common mode.

Two properties make that fix correct, and both are things you specify rather than parts you order. Keep the current limit on the drive leg, the outgoing side, so the bound on how much can reach the body holds no matter what the return path does. And make the dedicated return go to analog ground directly, not through the reference amplifier, so the current comes home the short honest way and never lands on the node your front end is trying to hold steady.

The full derivation, the fault cases, and the five-step schematic check are in the return-path bug that kills your biosignal front end.

Questions

Why does my biosignal go flat the moment the actuator turns on?

Because the current you inject is a loop and it comes home by the lowest-impedance path, which is usually the driven reference electrode. Ohm's law then develops a voltage on that node, and if your single-supply reference amplifier cannot reach that voltage it rails, stops holding the common mode, and rejection collapses.

Is a dead front end during stimulation a coupling or shielding problem?

Almost never. Shielding and layout address noise coupling through the air, while this is a DC return-path problem in the loop between your actuation and sensing circuits. You can settle it with arithmetic: compute the return current times the impedance it flows through and check whether the amplifier can reach that voltage.

How do I calculate return-path margin?

Multiply the return current by the series limit resistance to get the voltage developed across it. Subtract that from your front-end set point to get the output voltage the amplifier must reach. Compare it against the amplifier's output low limit, which is the ground rail on a single supply. Then divide the available headroom by the resistance to get the maximum current it can actually sink.

Why would a safety resistor cause a front-end failure?

Because a large series resistor, chosen deliberately to keep the injected current safe, turns even a sub-microamp current into a large voltage by Ohm's law. The safety choice and the failure are the same component seen from two angles.