question about galvanic and DC isolators

[loop123]

9V DC can kill. Did you hear the story of this person who tried using multimeter to test resistance of his skin. He tried to thin his skin then the current of the multimeter probe got bigger and stopped his heart and he got killed.

So in electrode works in the skin where it is prep by abrading it with gel, the equipment must have utmost galvanic isolation.

I'd like to inquire about the different kinds of galvanic isolation.

Some use Opto-isolator to isolate the serial or USB wires. But here the chassis is still not protected, is it?

Then you have the ISO122 where the input signal is transmitted digitally across a high-voltage differential capacitive barrier. Is this safer than just isolating the serial or USB using opto-isolator?

https://www.ti.com/lit/ds/symlink/iso122.pdf?ts=1706404349383&ref_url=https%253A%252F%252Fwww.google.com%252F

Can you also use opto-isolator to isolate the electrodes or only the ISO122, and can the ISO122 be used on the serial or USB? Does it use different chip depending on where you want to put the isolation circuit?

Also isn't there a general commercial single channel differential isolation module where you insert the 3 electrode wires (2 differential plus ground/reference/common) and it has outputs where the 3 can be connected to any bioamplifier unit? Is this possible? Perhaps the safest of all as backup to any existing isolator in the main equipment?

[Andy Chee]

Here you go:

[loop123]

Here you go:

I watched it but it didn't answer my main question. My primary question was, how can you isolate the say Vs or -Vs from the left of the isolated region from affecting the input? Remember 9V can kill a person. So imagine the input is connected to an ECG electrode in you chest and there is fault in the amp that conduct the 20V Vs or -Vs to the input. What kind of isolator can prevent it?

[Andy Chee]

I watched it but it didn't answer my main question. My primary question was, how can you isolate the say Vs or -Vs from the left of the isolated region from affecting the input?

I think a better question is, "how and why does 9V kill a person?".  Answering this question will answer your isolation question.

Your story said the 9V current got bigger and bigger until it stopped his heart.  However this is impossible with a 9V battery.  So post a link to the story so the situation can be analysed. 

For example, was the 9V from a faulty mains transformer?  If so, then it's not the 9V that killed him, it's the 240V from the faulty transformer that killed him.

Post a link to the story.

[selcuk]

You may use an optocoupler or digital isolator on digital I/Os or you may use I2C, SPI etc. specific isolators as well. It depends on your design but you will have the same result. If you want to isolate an analog signal, then there are linear optocouplers. But they are harder to use than digital ones. Of course you need to isolate supply and GND. So you need to use a isolated DC-DC converter to supply the sensor circuitry. You may not have a direct chassis connection to the isolated side, but there may be connections over safety capacitors.

To prevent electrode voltage from going over a threshold, you can use limiting diodes at the input. This is a topic related to medical equipment design and the information I provided is only general purpose. There can be some specific solutions as well.

[loop123]

I watched it but it didn't answer my main question. My primary question was, how can you isolate the say Vs or -Vs from the left of the isolated region from affecting the input?

I think a better question is, "how and why does 9V kill a person?".  Answering this question will answer your isolation question.

Your story said the 9V current got bigger and bigger until it stopped his heart.  However this is impossible with a 9V battery.  So post a link to the story so the situation can be analysed. 

For example, was the 9V from a faulty mains transformer?  If so, then it's not the 9V that killed him, it's the 240V from the faulty transformer that killed him.

Post a link to the story.

Here: https://electronics.stackexchange.com/questions/426452/can-a-9-volt-battery-through-your-bloodstream-kill-you

"This question stems from the rumor that some navy tech wanted to test the conductivity of their body so they pushed the meter beneath their skin and got electrocuted. (at least the version I heard, I'm sure there are many variants by now.)

Yes, it can, it only takes 10-20mA to stop a human heart. A 9V battery can provide much more than that. Your skin has sufficient resistance that it can stop current. If the skin is broken the resistance drops significantly. The current must be across the heart.".

So there is no isolator that can prevent an opamp or instrumentation amp  +VS or -Vs from going into the input and causing shock or electrocution if the skin is punctured when the input is put on the body?

What scenario can an amp fail in this manner? Whatever, no isolator could be design to handle it?

[Nominal Animal]

Galvanic and capacitive isolation means no current flows across the isolation barrier.

An electrode is just a conductor making an electrical connection to something nonmetallic.  In medical uses, it is either a pad or paddle (for skin contact, or placed during surgery on top of muscle tissue, often the heart), or a needle or rod.  The conductivity at the outer side of the human skin (i.e., epidermis) is relatively low, but gets much better deeper in, especially in the corium/dermis.

When current flows through muscle tissue, including the heart, the muscle will contract.  If the current is constant, the muscle will stay contracted no matter what you do.  The thing that tends to kill people is direct or alternating current through the heart, stopping the regular beating and stopping blood flow.  Without an external shock provided by e.g. an defibrillator, it may not automatically regain its own rythm, because the heart itself contains specialized cells that generate the electrical pulses for the beat rhythm.  The central nervous system can affect the rate to a degree, but the pulse cycle signals are generated within the heart itself.

Voltage potential without any current flowing is not dangerous, up to a few kilovolts at least.  For example, your body can accumulate such a large static charge that your hair stands out (since each hair having a similar charge will repel each other).  It is just that when you have a large static charge, it is very easy for that charge to discharge to ground or other objects, which means a rather large but short current pulse, which can be painful.  In the case of lightning, it can be deadly.

The resistance or conductivity of human tissues vary, and it also varies between direct and alternating current.  So, it is extremely difficult to say categorically what is safe and what is not, because it all depends on the path of the current, the amount of current, the frequency if alternating current, the (peak) voltage, and so on.  For that same reason, it is also not just voltage or just current that is dangerous; it is the combination and the context.

To make a safe electrode to be used with biological systems, you provide a controlled amount of current at a controlled voltage to the otherwise isolated side.  Note that that means an electrical circuit is formed between the supply and the otherwise isolated device. The voltage is typically a few volts, since there likely is a couple of opamps for amplifying and buffering the measurements, and very little current; typically under ten milliamps, most of that powering the signal isolators.  MOPP standard basically says that anything under 10mA is considered safe, as that is the amount of current MOPP allows to leak through to ground (say, via a human body; forming a second circuit through the ground).

A battery will typically happily deliver an amp of current, which definitely isn't safe anymore if the outermost skin layers (epidermis) is pierced; i.e. the current is injected directly to muscles, or to the corium/dermis layer (containing blood vessels, nerves, et cetera, and being rather conductive too).  If you use two needle-like electrodes and a minimal leakage controlled supply, you can typically limit the current path to between the needles.  Similarly, passing current through your hand (from one finger to another in the same hand) is unlikely to be fatal, whereas passing the same current from a finger in one hand to a finger in another hand can be, because in the latter case the current passes through or close to the heart, and the heart is particularly sensitive.

In electrical muscle stimulation, the electrodes placed on the skin providing impulses similar to the nervous system causing muscles to contract, uses pairs of electrodes forming a circuit, placed on the skin on the opposite sides of the muscle being affected.  They are designed to have very specific voltages and current, and minimal leakage: the current that goes out from one electrode must come back in from the other, so only the "zone" around the pairs of electrodes are affected.

[Andy Chee]

I watched it but it didn't answer my main question. My primary question was, how can you isolate the say Vs or -Vs from the left of the isolated region from affecting the input?

I think a better question is, "how and why does 9V kill a person?".  Answering this question will answer your isolation question.

Your story said the 9V current got bigger and bigger until it stopped his heart.  However this is impossible with a 9V battery.  So post a link to the story so the situation can be analysed. 

For example, was the 9V from a faulty mains transformer?  If so, then it's not the 9V that killed him, it's the 240V from the faulty transformer that killed him.

Post a link to the story.

Here: https://electronics.stackexchange.com/questions/426452/can-a-9-volt-battery-through-your-bloodstream-kill-you

"This question stems from the rumor that some navy tech wanted to test the conductivity of their body so they pushed the meter beneath their skin and got electrocuted. (at least the version I heard, I'm sure there are many variants by now.)

Yes, it can, it only takes 10-20mA to stop a human heart. A 9V battery can provide much more than that. Your skin has sufficient resistance that it can stop current. If the skin is broken the resistance drops significantly. The current must be across the heart.".

So there is no isolator that can prevent an opamp or instrumentation amp  +VS or -Vs from going into the input and causing shock or electrocution if the skin is punctured when the input is put on the body?

What scenario can an amp fail in this manner? Whatever, no isolator could be design to handle it?

Firstly, 10-20mA AC current stops a human heart, NOT DC.  For DC current you need about 300mA.

Secondly, a resistor can be permanently placed in series with the +VS.  Using your 9V example, a 470 ohm resistor would limit maximum current to 19mA, no matter the blood or flesh resistance.

[loop123]

Galvanic and capacitive isolation means no current flows across the isolation barrier.

An electrode is just a conductor making an electrical connection to something nonmetallic.  In medical uses, it is either a pad or paddle (for skin contact, or placed during surgery on top of muscle tissue, often the heart), or a needle or rod.  The conductivity at the outer side of the human skin (i.e., epidermis) is relatively low, but gets much better deeper in, especially in the corium/dermis.

When current flows through muscle tissue, including the heart, the muscle will contract.  If the current is constant, the muscle will stay contracted no matter what you do.  The thing that tends to kill people is direct or alternating current through the heart, stopping the regular beating and stopping blood flow.  Without an external shock provided by e.g. an defibrillator, it may not automatically regain its own rythm, because the heart itself contains specialized cells that generate the electrical pulses for the beat rhythm.  The central nervous system can affect the rate to a degree, but the pulse cycle signals are generated within the heart itself.

Voltage potential without any current flowing is not dangerous, up to a few kilovolts at least.  For example, your body can accumulate such a large static charge that your hair stands out (since each hair having a similar charge will repel each other).  It is just that when you have a large static charge, it is very easy for that charge to discharge to ground or other objects, which means a rather large but short current pulse, which can be painful.  In the case of lightning, it can be deadly.

The resistance or conductivity of human tissues vary, and it also varies between direct and alternating current.  So, it is extremely difficult to say categorically what is safe and what is not, because it all depends on the path of the current, the amount of current, the frequency if alternating current, the (peak) voltage, and so on.  For that same reason, it is also not just voltage or just current that is dangerous; it is the combination and the context.

To make a safe electrode to be used with biological systems, you provide a controlled amount of current at a controlled voltage to the otherwise isolated side.  Note that that means an electrical circuit is formed between the supply and the otherwise isolated device. The voltage is typically a few volts, since there likely is a couple of opamps for amplifying and buffering the measurements, and very little current; typically under ten milliamps, most of that powering the signal isolators.  MOPP standard basically says that anything under 10mA is considered safe, as that is the amount of current MOPP allows to leak through to ground (say, via a human body; forming a second circuit through the ground).

Have you seen this being done in actual equipment, where there is provided a controlled amount of current at a controlled voltage to the otherwise isolated side?  I think most with such isolator may still accept the entire power supply voltage and current at the isolated input.

Also it is not hard to create a module which can work with all equipment? The module would simply provide isolation to the 2 differential inputs and one ground/common,reference? This means the voltage coming in would be the same voltage in the output. No amplification, only sure isolation with that very tiny amount (less than 10mA) of current controlled just to make the isolator operate.

Is there already a commercial unit that does this? Where you can put it between your electrodes and any equipments? If a company create this. It would be very useful, isn't it?

If no one has the foresight. What is the schematic for this circuit say using the ISO122 (capacitiv isolator) vs an optoisolator? What is more appropriate for the case or better?

[Andy Chee]

I think most with such isolator may still accept the entire power supply voltage and current at the isolated input.

Depends on the device.

For medical devices, they will sometimes use an isolated power supply to power the opamps, for example:

https://recom-power.com/pdf/Econoline/REC3.5-RW_R.pdf

Note the medical approvals and standards.  You can lookup the EN and UL numbers.

[Nominal Animal]

Have you seen this being done in actual equipment, where there is provided a controlled amount of current at a controlled voltage to the otherwise isolated side?

Sure.  A simple way is an isolated DC-DC converter driven by a current- and voltage-limiting supply.  You can find such even in USB high-speed isolators for the downstream port, limiting the current to 100mA until USB connection has been formed and the actual current limit desired discovered from the USB descriptor; then usually limited to 500mA.

For sensors and medical equipment, you often have a filter and a linear voltage regulator followed by a secondary current limiter (based on a dedicated chip; or an instrumentation amplifier, shunt resistor to measure the current, and a high-side P-MOSFET or low-side N-MOSFET to cut the power altogether) so that if the current draw is exceeded, instead of just voltage sagging the power is cut completely.

(That said, I'm only somewhat familiar with scientific sensors and such, and never worked on real medical equipment myself.)

I think most with such isolator may still accept the entire power supply voltage and current at the isolated input.

Most isolators do not provide any current to the isolated side, and only deal with the signal.  This includes USB isolators like ADuM3160, ADuM3166, ISOUSB211, and others; and isolation amplifiers like ISO122.

Voltage and current is typically provided by an isolated DC-DC converter, which uses a high-frequency transformer to pass current over the isolation gap.  Some input power (from a couple of percent to twenty percent, depending on the transformer; up to half for low-voltage ones like 3.3V and 5V ones) is lost as heat, but the rest is coupled to the other circuit.  The number of turns in the transformer dictates how the voltage over the primary side reflects on the voltage induced on the secondary side; and since the output power matches input power minus losses, the current similarly.  That is, by controlling the current and voltage on the primary side with a transformer having known number of turns (ratio), you can tell approximately how much current and voltage is induced in the secondary side.  It is easy to regulate to a fixed voltage with a linear regulator, and the current drawn by the rest of the circuit can be measured and connection stopped if it exceeds a set limit.

Also it is not hard to create a module which can work with all equipment? The module would simply provide isolation to the 2 differential inputs and one ground/common,reference? This means the voltage coming in would be the same voltage in the output. No amplification, only sure isolation with that very tiny amount (less than 10mA) of current controlled just to make the isolator operate.

It is not that simple, essentially because there is no transformer suitable for DC, nor an universal one-fits-all transformer for AC.  The geometry depends on the primary side current and voltage, as well as the turns ratio needed.  Plus, if the isolated side is connected to the main ground, it is no longer isolated: current can flow through to ground, making the isolation useless.

This is the difference between class I and class II isolation: class I has a safety capacitor between the grounds, reducing EMI generated during switching, and class II has the powered side completely isolated.  In both cases, the potential difference between the zero-volt rail on either side can be hundreds of volts without any extra current passing, but in class I the capacitor means that some leakage is possible, and can lead to "tingling" sensation.

Medical equipment uses MOPP (Means of Patient Protection) AKA IEC60601-1 standard compliant, class II isolated power supplies.  They're not that expensive; Mean Well ones seem to be common.  For example, Mean Well RPS-30: 30 watt class II isolated AC/DC supply, with several submodels providing one 3.3V, 5.0V, 7.5V, 12V, 15V, 24V, or 48V output).  Basically, if you short the output to ground, only 5-10mA will flow over the short.  This is not enough to stop the human heart.  (In pulses it could disrupt the rhythm of the heart causing arrythmia, though.)

Of course, if you put a human within the circuit, for example connect that power supply 0V to a needle in one hand, and the +V to a needle in the other hand, the current will pass through the human and possibly kill them.  We are fragile bags of conductive salty water, mostly, with a rubber-like somewhat resistive outer covering; and easy to kill.

Is there already a commercial unit that does this?

No magic bullet exists, no.

But the components needed to make your supply and your equipment thus safe are extremely common and cheap.

For example, you could start with an extremely common USB wall wart.  (Of course, for medical equipment you get something better and safer, especially with careful EMI checks so that it does not generate interference to other devices nearby.)

That gives you about 5V and ample power.  You then add current limiting that cuts the circuit for a set duration (say, one second), if the current exceeds some set limit.  They exist for USB (at 100mA and 500mA selectable limits), but you can construct one yourself using an instrumentation amplifier or analog current sense IC, a shunt resistor (under 1Ω –– the voltage drop over the resistor corresponds to the current over the resistor, U = I R, and the amplifier boosts that by a fixed factor), a comparator, a MOSFET, and some capacitors and resistors.  Not too complicated.

You then add an isolated DC-DC converter.  Because the voltage and current is limited on the primary side, the isolated secondary side is already limited to how much power it has available.  I would filter the output (using a Pi filter or capacitive multiplier) followed by a linear regulator (dropping a volt or two) to get a really nice, stable voltage, necessary for many sensor applications, and might even add a secondary current cut-off limiter (the same thing as on the primary side) for safety.

Of these, the isolated DC-DC converter is the most "expensive" component; at Mouser in singles, they cost between 5€ and 15€ apiece.  In the above cost-optimized scheme, I might use RECOM power RM-0505S, whose 5V output is unregulated (which is fine, because I'd filter and regulate the output to 3.3V anyway, assuming 3.3V needed for the medical electronics).  It itself is limited to 50mA on the isolated side, which is ample for powering signal isolators, but it has no minimum load, so it should work for the targeted 10mA max on the isolated side.

The main cost would be the AC-DC and isolated DC-DC converters, with the rest of the components being 'jellybeans' as Dave calls them.
In commercial medical use, the cost of safety and standards compliance testing would probably be much higher!

What is the schematic for this circuit say using the ISO122 (capacitiv isolator)

TI ISO122 is a precision isolation amplifier.  On both sides of the isolation barrier, you have a positive and a negative supply, and a ground reference.  On the isolated side, an analog input voltage (referenced to its own ground) causes a corresponding output voltage (referenced to its own ground), within the valid range; the input resistance is about 200kΩ, so very little current is drawn from the input pin.  On both sides, it may draw up to 7mA of current from the supply on that side.

You could use the power supply circuit I outlined above to power this (except with a dual-output isolated DC-DC converter, to get both rails; or with a circuit to generate the negative rail).  Then, all the microcontroller stuff would be on the side between the DC-DC converter and AC-DC supply, and use as much power as it needs (as long as the AC-DC supply can provide it).  Assuming you set the current cutoff on the isolated side to say 8-9mA, and on the non-isolated side (before the DC-DC converter) to say 17mA, you could prove that even if the ISO122 part was directly grounded, less than 10mA would flow to/from ground.

To pass medical safety tests, you'd of course have to do more design work, and consider and test for all kinds of fault conditions, which is what tends to make medical-grade electronics so much more expensive.  Without the testing and verification, their designs and implementation aren't that much more expensive, really.

[Zero999]

9V DC can kill. Did you hear the story of this person who tried using multimeter to test resistance of his skin. He tried to thin his skin then the current of the multimeter probe got bigger and stopped his heart and he got killed.

Yes I've heard the story and it's a load of nonsense. In order to kill, the current needs to flow through the heart, which won't happen with a 9V battery because the terminals are too close together and the resistance of human flesh is so high, it's highly unlikely for a lethal current to pass though the heart at such a low voltage.

EDIT:
I missed the part about the multimeter, so the battery terminal distance is irrelevant. I suppose the resistance of the human flesh being too high, whist still true, doesn't make much difference, since even the old multimeters limited the current, so as not to destroy the movement, when the probes were shorted together.

[CaptDon]

First off, I strongly call bullshit on that 9 volt story!!!! The old VTVM meters used a 1.5 volt battery for the ohms scale and the newer DVM's that use up to 10 volts also use a constant current circuit which attempts to pass 1 milliamp across the circuit under test and then at 1ma. the voltage read across the circuit is directly read out as ohms. They began using this higher voltage so sensible readings could be obtained when forward biasing semiconductors. The downside is 9 to 10 volts can now destroy the gates of logic level FET's and other devices. As for your BioMedical question, the chassis of all medical equipment and all metallic surfaces most show less than 300 milliohms (in some cases less than 100 milliohms) between the unit under test and the ground testing device that the unit is plugged into. In some cases the tester can also provide a substantial test current (10 amps to 30 amps) and measure voltage drop proving the ground connection is not feeble. Also, the 'front end' of E.C.G. (E.K.G. in Europe) is limited to no more than a 10 volt supply (often plus and minus 10) and as for shock prevention you missed the very obvious.....The inputs are high impedance FET preamplifier differential circuits usually with 10K resistors in series with each lead limiting available fault current to 1 milliamp or 2 milliamps if one lead was shorted to the +10VDC supply and any other lead shorted to the -10VDC supply. Also, our 120VAC operating room supply is actually 60VAC each side of center tap with the center tap going through a 2.2 meg resistor to earth ground and the imbalance monitored by a device referred to as a 'LEM' which will kill all O.R. outlets (until manually reset) with a leakage higher than 1 milliamp to earth ground. Each O.R. has its own 5KVA isolation transformer associated with this circuit and the O.R. receptacles. B.T.W., if you want to talk about current passing through skin, start thinking about Electro-Surgical-Units (ESU's) that can sear flesh to stop bleeding, seal wounds, or cut flesh like in removing tonsils. Most tonsils are removed using ESU's because it cuts and cauterizes at the same time, unlike a normal scalpel. Cheers mates!!

[Zero999]

The test voltage of all my digital multimeters is well under the 600mV or so required to forward bias a silicon junction, which I see as a welcome feature. It means I can measure resistors in circuit, as well as not damage my semiconductors. If I want to test silicon junctions, I use a the diode test function.

[m k]

the heart, stopping the regular beating and stopping blood flow.  Without an external shock provided by e.g. an defibrillator, it may not automatically regain its own rythm,

(just a note)

Heart has a tendency to beat.
There are fast and slow parts.

When heart stops to flat line it's worse than fibrillation.
Fibrillation, the fast part, means that heart is still, at least partially alive.

Defibrillator, like the name says, is to stop fibrillation.
After that heart's tendency to beat will take over.
So it's basically fingers crossed, actually it's the same with incredibly many cases with medical stuff. (excluding measurements of course)

[loop123]

Have you seen this being done in actual equipment, where there is provided a controlled amount of current at a controlled voltage to the otherwise isolated side?

Sure.  A simple way is an isolated DC-DC converter driven by a current- and voltage-limiting supply.  You can find such even in USB high-speed isolators for the downstream port, limiting the current to 100mA until USB connection has been formed and the actual current limit desired discovered from the USB descriptor; then usually limited to 500mA.

For sensors and medical equipment, you often have a filter and a linear voltage regulator followed by a secondary current limiter (based on a dedicated chip; or an instrumentation amplifier, shunt resistor to measure the current, and a high-side P-MOSFET or low-side N-MOSFET to cut the power altogether) so that if the current draw is exceeded, instead of just voltage sagging the power is cut completely.

(That said, I'm only somewhat familiar with scientific sensors and such, and never worked on real medical equipment myself.)

I think most with such isolator may still accept the entire power supply voltage and current at the isolated input.

Most isolators do not provide any current to the isolated side, and only deal with the signal.  This includes USB isolators like ADuM3160, ADuM3166, ISOUSB211, and others; and isolation amplifiers like ISO122.

First of all. Thanks a lot for your comprehensive explanations. I'd like to inquire about this ISO122 in particular first. You said "Most isolators do not provide any current to the isolated side, and only deal with the signal.". But in the following ISO122 specs. They are voltages in +Vs and -Vs in both sides.

https://www.ti.com/lit/ds/symlink/iso122.pdf?ts=1706601715056&ref_url=https%253A%252F%252Fwww.google.com%252F

Voltage and current is typically provided by an isolated DC-DC converter, which uses a high-frequency transformer to pass current over the isolation gap.  Some input power (from a couple of percent to twenty percent, depending on the transformer; up to half for low-voltage ones like 3.3V and 5V ones) is lost as heat, but the rest is coupled to the other circuit.  The number of turns in the transformer dictates how the voltage over the primary side reflects on the voltage induced on the secondary side; and since the output power matches input power minus losses, the current similarly.  That is, by controlling the current and voltage on the primary side with a transformer having known number of turns (ratio), you can tell approximately how much current and voltage is induced in the secondary side.  It is easy to regulate to a fixed voltage with a linear regulator, and the current drawn by the rest of the circuit can be measured and connection stopped if it exceeds a set limit.

Also it is not hard to create a module which can work with all equipment? The module would simply provide isolation to the 2 differential inputs and one ground/common,reference? This means the voltage coming in would be the same voltage in the output. No amplification, only sure isolation with that very tiny amount (less than 10mA) of current controlled just to make the isolator operate.

It is not that simple, essentially because there is no transformer suitable for DC, nor an universal one-fits-all transformer for AC.  The geometry depends on the primary side current and voltage, as well as the turns ratio needed.  Plus, if the isolated side is connected to the main ground, it is no longer isolated: current can flow through to ground, making the isolation useless.

This is the difference between class I and class II isolation: class I has a safety capacitor between the grounds, reducing EMI generated during switching, and class II has the powered side completely isolated.  In both cases, the potential difference between the zero-volt rail on either side can be hundreds of volts without any extra current passing, but in class I the capacitor means that some leakage is possible, and can lead to "tingling" sensation.

Medical equipment uses MOPP (Means of Patient Protection) AKA IEC60601-1 standard compliant, class II isolated power supplies.  They're not that expensive; Mean Well ones seem to be common.  For example, Mean Well RPS-30: 30 watt class II isolated AC/DC supply, with several submodels providing one 3.3V, 5.0V, 7.5V, 12V, 15V, 24V, or 48V output).  Basically, if you short the output to ground, only 5-10mA will flow over the short.  This is not enough to stop the human heart.  (In pulses it could disrupt the rhythm of the heart causing arrythmia, though.)

Of course, if you put a human within the circuit, for example connect that power supply 0V to a needle in one hand, and the +V to a needle in the other hand, the current will pass through the human and possibly kill them.  We are fragile bags of conductive salty water, mostly, with a rubber-like somewhat resistive outer covering; and easy to kill.

Is there already a commercial unit that does this?

No magic bullet exists, no.

But the components needed to make your supply and your equipment thus safe are extremely common and cheap.

For example, you could start with an extremely common USB wall wart.  (Of course, for medical equipment you get something better and safer, especially with careful EMI checks so that it does not generate interference to other devices nearby.)

That gives you about 5V and ample power.  You then add current limiting that cuts the circuit for a set duration (say, one second), if the current exceeds some set limit.  They exist for USB (at 100mA and 500mA selectable limits), but you can construct one yourself using an instrumentation amplifier or analog current sense IC, a shunt resistor (under 1Ω –– the voltage drop over the resistor corresponds to the current over the resistor, U = I R, and the amplifier boosts that by a fixed factor), a comparator, a MOSFET, and some capacitors and resistors.  Not too complicated.

You then add an isolated DC-DC converter.  Because the voltage and current is limited on the primary side, the isolated secondary side is already limited to how much power it has available.  I would filter the output (using a Pi filter or capacitive multiplier) followed by a linear regulator (dropping a volt or two) to get a really nice, stable voltage, necessary for many sensor applications, and might even add a secondary current cut-off limiter (the same thing as on the primary side) for safety.

Of these, the isolated DC-DC converter is the most "expensive" component; at Mouser in singles, they cost between 5€ and 15€ apiece.  In the above cost-optimized scheme, I might use RECOM power RM-0505S, whose 5V output is unregulated (which is fine, because I'd filter and regulate the output to 3.3V anyway, assuming 3.3V needed for the medical electronics).  It itself is limited to 50mA on the isolated side, which is ample for powering signal isolators, but it has no minimum load, so it should work for the targeted 10mA max on the isolated side.

The main cost would be the AC-DC and isolated DC-DC converters, with the rest of the components being 'jellybeans' as Dave calls them.
In commercial medical use, the cost of safety and standards compliance testing would probably be much higher!

What is the schematic for this circuit say using the ISO122 (capacitiv isolator)

TI ISO122 is a precision isolation amplifier.  On both sides of the isolation barrier, you have a positive and a negative supply, and a ground reference.  On the isolated side, an analog input voltage (referenced to its own ground) causes a corresponding output voltage (referenced to its own ground), within the valid range; the input resistance is about 200kΩ, so very little current is drawn from the input pin.  On both sides, it may draw up to 7mA of current from the supply on that side.

You could use the power supply circuit I outlined above to power this (except with a dual-output isolated DC-DC converter, to get both rails; or with a circuit to generate the negative rail).  Then, all the microcontroller stuff would be on the side between the DC-DC converter and AC-DC supply, and use as much power as it needs (as long as the AC-DC supply can provide it).  Assuming you set the current cutoff on the isolated side to say 8-9mA, and on the non-isolated side (before the DC-DC converter) to say 17mA, you could prove that even if the ISO122 part was directly grounded, less than 10mA would flow to/from ground.

To pass medical safety tests, you'd of course have to do more design work, and consider and test for all kinds of fault conditions, which is what tends to make medical-grade electronics so much more expensive.  Without the testing and verification, their designs and implementation aren't that much more expensive, really.

[Andy Chee]

First of all. Thanks a lot for your comprehensive explanations. I'd like to inquire about this ISO122 in particular first. You said "Most isolators do not provide any current to the isolated side, and only deal with the signal.". But in the following ISO122 specs. They are voltages in +Vs and -Vs in both sides.

https://www.ti.com/lit/ds/symlink/iso122.pdf?ts=1706601715056&ref_url=https%253A%252F%252Fwww.google.com%252F

Please examine Figure 24 in your document.  Note that there are two separate supplies Vs1 and Vs2.

Also examine Figure 22 & 23 in your document.  The HPR117 is an isolated power supply that provide current to the isolated side.

https://www.murata.com/products/productdata/8807039467550/tdc-hpr1xxc.pdf?1617679818000

[loop123]

First of all. Thanks a lot for your comprehensive explanations. I'd like to inquire about this ISO122 in particular first. You said "Most isolators do not provide any current to the isolated side, and only deal with the signal.". But in the following ISO122 specs. They are voltages in +Vs and -Vs in both sides.

https://www.ti.com/lit/ds/symlink/iso122.pdf?ts=1706601715056&ref_url=https%253A%252F%252Fwww.google.com%252F

Please examine Figure 24 in your document.  Note that there are two separate supplies Vs1 and Vs2.

Also examine Figure 22 & 23 in your document.  The HPR117 is an isolated power supply that provide current to the isolated side.

https://www.murata.com/products/productdata/8807039467550/tdc-hpr1xxc.pdf?1617679818000

So the HPR117 is not built in. You have to provide it when using the ISO122. Does it mean all circuit boards with ISO122 always uses HPR117 or equivalent isolated power supply? have you seen any ISO122 that doesn't use them?

[Andy Chee]

So the HPR117 is not built in. You have to provide it when using the ISO122. Does it mean all circuit boards with ISO122 always uses HPR117 or equivalent isolated power supply? have you seen any ISO122 that doesn't use them?

Correct.  In order to take advantage of the isolation feature of the ISO22, you must use an isolated power supply.

[Nominal Animal]

You said "Most isolators do not provide any current to the isolated side, and only deal with the signal.". But in the following ISO122 specs. They are voltages in +Vs and -Vs in both sides.

Yes, and this also applies to all the other isolators I mentioned (although they only need a local ground and a positive supply).
Like Andy Chee explained, you need an isolated power supply to power the isolated side of the device for the device to work at all.

The datasheet specs simply mean each side requires their own supply.  They have to be isolated, for the isolation to actually work.

Think of it as two completely separate circuits, with the device precisely straddling the isolation boundary; and internal features allow only the signal to be passed over the isolation barrier (via capacitive, optical, or magnetic coupling).  No electric current passes over the barrier, and the two sides are electrically completely separated.  (To within the limits of the isolation, of course.)
But to work, both sides of the device need to be powered.  Having power only on one side is not sufficient.

Isolated power supplies work on a very similar principle (typically using magnetic coupling via transformers), except that the signal itself carries enough power to be useful.  Most isolators don't do this; you need an isolated supply to pass useful power over the isolation barrier.  In all cases there is a conversion of electric current into something else that then passes over the isolation barrier that is then converted back into electric current.

Isolation transformers work with AC, typically mains voltage and frequency (and they differ depending on the mains voltage!).  The DC-DC ones need to create some form of AC (often high-frequency pulses, not sine waves) to be able to use transformers; higher frequency means a physically smaller transformer can be used.  Then, it has to be converted back to DC and optionally regulated on the other side.  Unlike AC-AC isolation transformers, the DC-DC ones only work in one direction.

Optically coupled isolated power supplies –– essentially LEDs powering a photovoltaic cell ("solar cell") –– are only used with very specialized devices (typically scientific experiments involving something that makes transformers and switchmode supplies unsuitable), but most of the power is lost; their efficiency is very low.  (On the positive side, the voltage and current tends to only have thermal noise, and the circuitry is utterly simple and straightforward.)
Some nuclear batteries work on a similar principle, with the radioactive decay (typically beta radiation) captured by a suitable "cell" that converts its energy to useful (but typically low) currents.  Other nuclear batteries (like those used on some space probes) are thermoelectric devices, turning a heat flow (from hot to cold) into electric current, the heat generated by nuclear decay.  Their efficiency is similarly low, because of the low efficiency in converting the available energy to an useful electric current.

Capacitive coupling (of alternating current) is not considered isolation at all, and is often used in low-power devices and some mains LED lights to limit current ("capacitive dropper").  In signal isolators, the capacitively coupled signal has very little energy –– the coupling within the device is such that it cannot pass much energy –– that is not enough to power even the receive-side circuitry.

[loop123]

So the HPR117 is not built in. You have to provide it when using the ISO122. Does it mean all circuit boards with ISO122 always uses HPR117 or equivalent isolated power supply? have you seen any ISO122 that doesn't use them?

Correct.  In order to take advantage of the isolation feature of the ISO22, you must use an isolated power supply.

What is your and others comments about the

https://www.ti.com/lit/ds/symlink/dcp010512db.pdf?ts=1706603932234&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FDCP010512DB%252Fpart-details%252FDCP010512DBP-U

DCP010512DBP
DCP01 Series, 1-W, 1000-VRMS Isolated, Unregulated DC/DC Converter Modules

It's in a circuit I'm evaluating that powers the ISO122. What happens if it's unregulated. All isolated power supply automatically have less than 50mA current so as not to affect the heart?

[Andy Chee]

So the HPR117 is not built in. You have to provide it when using the ISO122. Does it mean all circuit boards with ISO122 always uses HPR117 or equivalent isolated power supply? have you seen any ISO122 that doesn't use them?

Correct.  In order to take advantage of the isolation feature of the ISO22, you must use an isolated power supply.

What is your and others comments about the

https://www.ti.com/lit/ds/symlink/dcp010512db.pdf?ts=1706603932234&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FDCP010512DB%252Fpart-details%252FDCP010512DBP-U

DCP010512DBP
DCP01 Series, 1-W, 1000-VRMS Isolated, Unregulated DC/DC Converter Modules

It's in a circuit I'm evaluating that powers the ISO122. What happens if it's unregulated. All isolated power supply automatically have less than 50mA current so as not to affect the heart?

Depends on the circuit.  Unregulated could be perfectly ok.

https://g.recomcdn.com/media/Datasheet/pdf/.f-zadYxY/.t07de66f7699f06eb9220/Datasheet-252/REM10.pdf

[Nominal Animal]

DCP010512DBP
DCP01 Series, 1-W, 1000-VRMS Isolated, Unregulated DC/DC Converter Modules

It's in a circuit I'm evaluating that powers the ISO122. What happens if it's unregulated.

Unregulated means the output voltage can vary with load.  This one requires a load of 10% (of rated power, i.e. 100mW).  With lower loads, typically in unregulated isolated DC-DC converters, the voltage can go much higher, even twice the nominal voltage.  If the isolated side always loads the regulator by at least 10%, and does not require a very specific voltage –– ISO122 works fine with supplies being between ±4.5V and ±18V, as long as there is at least 3V or so headroom between the signal levels and the closer supply rail, and the supplies don't have too much noise –– it works absolutely fine; just as well as it would with regulated supplies.

This is typical with many circuits involving operational amplifiers and isolation amplifiers, too.  They don't need exact rail voltages to handle the signals they work with; they just need rail voltages with sufficient headroom to the signal voltage levels, with not too much noise on those rails.  For example, on the ISO122, the datasheet Figure 7 shows that on the isolated side, any noise in the rails below 1kHz is attenuated by 54 dB.  This means that if you have 500mV of under-1kHz ripple between the power rails, it shows up as 1mV of ripple at the same frequency in the signal output.  (Voltage amplitude output ripple is \$10^{\text{PSRR}/20}\$ of the supply ripple amplitude, when \$\text{PSRR}\$ is expressed in dB.)

If your device has supply filtering on the isolated side –– capacitors and inductors –– then those can significantly reduce any noise, without regulating the voltage rails to any particular voltage levels.  That would be absolutely fine for ISO122, excellent even.  It is only that when making your own devices, using linear regulators you get a specific output voltage (independent of the load, as long as the input voltage is within a suitable range) that also rejects noise on the input supply voltage.

All isolated power supply automatically have less than 50mA current so as not to affect the heart?

Oh no, they can go to many amperes of current.  RPS-30-05, for example, provides up to 6A at 5V DC (with maximum 80mV = 0.08V peak-to-peak ripple).

MOPP (Means of Patient Protection) limits the leakage current, the amount of current that may be passed from the isolated circuit to ground, for example through the patient via contact to the isolated circuit "ground" or 0V rail.

Oops!  Sorry!    I thought it was 10mA for DC, but it's actually just 50µA = 0.05mA DC in normal operation, and 100µA = 0.1mA DC in single failure condition.

[Nominal Animal]

My main unit has 3.95V and -11.95V output instead of 7.2V and -7.2V.

I do believe this was resolved in the other thread: a faulty 0V connection (C/white corroded, if I understood correctly).

This isolation unit can be used in any other bio-amplifiers, right? If the output plug can work. Or does it have to depend on the main unit ground and Vs -Vs to make it work?

On the output side, + (red) has to be about +12V with respect to C (white), and - (black) about -12V with respect to C (white), and provide at least 40mA of current.  The output is ±6V, referred to its signal ground at or very near to C.

On the input side, the sensor must provide a differential signal, and work from the ±7.2V power rails.  The isolation unit has a fixed gain of 10, so a 1mV sensor signal is 10mV at the output.

If these are fulfilled, any sensor can be used, as can any amplifier/data acquisition unit.

What do you think of the circuit? Is the isolation very good that you would make you life depend on it? It uses ISO122 and the DC converter dcp010512dbp. See last message for question about this, like is it very good?

The specs look absolutely fine to me.  My life has depended on much worse things at times.

As to the circuits, I'm just a hobbyist on the electronics side, and no BigClive or Dave.

What actually matters is whether the device has been certified for medical uses by a reputable laboratory.  Theory is one thing; practical testing and examination by professionals and their certification of the device is what makes a medical device safe to use.

In the circuits, there are many chips. Do you know which of them is the one where when you push it, a 500microVolt p-p 10 Hz would be displayed suppressing the electrodes?

No, but I guess LM741 is part of the twin-T oscillator generating the source 10Hz signal; it's output is pin 6 (top row, second pin from right), so you could probe that with respect to pin 4 (bottom row, rightmost pin), to see if you see the signal at all.

I can't see all the tracks on both sides, but I think the sinusoidal signal is then clipped by the diodes, divided by a voltage divider to the 50µVpp or 500µVpp range (1:1200 or 1:12000 or so), and finally buffered by the INA411AP before supplied to the ISO122 isolation amplifier.  If so, you should see the 50µVpp or 500µVpp 10Hz signal on the output pin of the INA411AP, most likely pin 6 (top row, second pin from right), as you can see the track going up towards the ISO122 on the component side.   (I can't tell at which stage the gain is set, but I think in the ISO122, which means the calibration signal is only 50µVpp on its input.)

(While it might feel better to have the calibration signal generated at the output, this way the signal actually goes through the isolation amplifier, basically the same route as the actual signal (except for the initial unity gain buffer op-amp, LF411CP).

Please identify all the components like what they do so I can more easily analyze them.

No!

If I were you, I'd start by drawing a copy of the tracks (on both sides).  You could use e.g. EasyEDA (free online) PCB for it; or, if you take pictures straight above, in Inkscape, on top of the picture.  Use a multimeter to check continuities; and better flip the solder side left-to-right, so you see the tracks as if the PCB was transparent.  Then you can start identifying all the components.   For example, because the sensor input connector input is connected to the LF411CP operational amplifier, and that is in unity gain configuration (OUT and IN- directly connected together), you can tell the LF411CP is the initial buffer for the signal.

[loop123]

I'd start testing and fixing it tomorrow. But I'm reading about active electrodes where the electrodes have amplifier. Here you don't need low impedance in the skin. Is it not the danger of shock or electrocution from dc source is due to low impedance. Then if the skin would remain high impedance and you use active electrodes. Then there is no more danger? Do active electrodes use isolation too or do they bypass it altogether because the skin target is already high impedance (remember huge current can only flow in low impedance).