question about galvanic and DC isolators - Page 1
Forum section: electronics / beginners

author: loop123
author_rank: Frequent Contributor
timestamp: January 28, 2024, 01:23:34 am
content: 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%252FCan 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?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 28, 2024, 02:31:58 am
content: Here you go:

author: loop123
author_rank: Frequent Contributor
timestamp: January 28, 2024, 07:46:49 am
content: Quote from: Andy Chee on January 28, 2024, 02:31:58 amI 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?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 28, 2024, 08:14:30 am
content: Quote from: loop123 on January 28, 2024, 07:46:49 amI 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.

author: selcuk
author_rank: Frequent Contributor
timestamp: January 28, 2024, 08:17:23 am
content: 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.

author: loop123
author_rank: Frequent Contributor
timestamp: January 28, 2024, 09:00:32 am
content: Quote from: Andy Chee on January 28, 2024, 08:14:30 amHere: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?

author: Nominal Animal
author_rank: Super Contributor
timestamp: January 28, 2024, 09:13:32 am
content: 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.

author: Andy Chee
author_rank: Super Contributor
timestamp: January 28, 2024, 09:25:22 am
content: Quote from: loop123 on January 28, 2024, 09:00:32 amFirstly, 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.

author: loop123
author_rank: Frequent Contributor
timestamp: January 28, 2024, 09:37:38 am
content: Quote from: Nominal Animal on January 28, 2024, 09:13:32 amHave 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?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 28, 2024, 09:58:18 am
content: Quote from: loop123 on January 28, 2024, 09:37:38 amDepends 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.pdfNote the medical approvals and standards. You can lookup the EN and UL numbers.

author: Nominal Animal
author_rank: Super Contributor
timestamp: January 28, 2024, 11:05:26 am
content: Quote from: loop123 on January 28, 2024, 09:37:38 amSure. 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.)Quote from: loop123 on January 28, 2024, 09:37:38 amMost 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.Quote from: loop123 on January 28, 2024, 09:37:38 amIt 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.Quote from: loop123 on January 28, 2024, 09:37:38 amNo 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!Quote from: loop123 on January 28, 2024, 09:37:38 amTI ISO122is 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'tthatmuch more expensive, really.

author: Zero999
author_rank: Super Contributor
timestamp: January 28, 2024, 12:20:48 pm
content: Quote from: loop123 on January 28, 2024, 01:23:34 amYes 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 theterminals are too close together and theresistance 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.

author: CaptDon
author_rank: Super Contributor
timestamp: January 28, 2024, 02:15:39 pm
content: 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!!

author: Zero999
author_rank: Super Contributor
timestamp: January 28, 2024, 05:16:48 pm
content: 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.

author: m k
author_rank: Super Contributor
timestamp: January 28, 2024, 07:00:37 pm
content: Quote from: Nominal Animal on January 28, 2024, 09:13:32 am(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)

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 08:03:28 am
content: Quote from: Nominal Animal on January 28, 2024, 11:05:26 amFirst 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%252FQuote

author: Andy Chee
author_rank: Super Contributor
timestamp: January 30, 2024, 08:13:00 am
content: Quote from: loop123 on January 30, 2024, 08:03:28 amPlease 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

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 08:32:10 am
content: Quote from: Andy Chee on January 30, 2024, 08:13:00 amSo 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?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 30, 2024, 09:23:42 am
content: Quote from: loop123 on January 30, 2024, 08:32:10 amCorrect. In order to take advantage of the isolation feature of the ISO22, you must use an isolated power supply.

author: Nominal Animal
author_rank: Super Contributor
timestamp: January 30, 2024, 09:40:57 am
content: Quote from: loop123 on January 30, 2024, 08:03:28 amYes, 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 allowonly the signalto 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 anisolated supplyto pass useful power over the isolation barrier. In all cases there is aconversionof 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.

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 09:43:11 am
content: Quote from: Andy Chee on January 30, 2024, 09:23:42 amWhat is your and others comments about thehttps://www.ti.com/lit/ds/symlink/dcp010512db.pdf?ts=1706603932234&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FDCP010512DB%252Fpart-details%252FDCP010512DBP-UDCP010512DBPDCP01 Series, 1-W, 1000-VRMS Isolated, Unregulated DC/DC Converter ModulesIt'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?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 30, 2024, 09:53:16 am
content: Quote from: loop123 on January 30, 2024, 09:43:11 amDepends on the circuit. Unregulated could be perfectly ok.https://g.recomcdn.com/media/Datasheet/pdf/.f-zadYxY/.t07de66f7699f06eb9220/Datasheet-252/REM10.pdf

author: Nominal Animal
author_rank: Super Contributor
timestamp: January 30, 2024, 10:33:56 am
content: Quote from: loop123 on January 30, 2024, 09:43:11 amUnregulated 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 needexactrail 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.Quote from: loop123 on January 30, 2024, 09:43:11 amOh 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 theleakage 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.

author: Nominal Animal
author_rank: Super Contributor
timestamp: January 31, 2024, 10:37:55 am
content: Quote from: loop123 on January 30, 2024, 09:59:51 amI do believe this was resolved in the other thread: a faulty 0V connection (C/white corroded, if I understood correctly).Quote from: loop123 on January 30, 2024, 09:59:51 amOn 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 sensorcan be used, as cananyamplifier/data acquisition unit.Quote from: loop123 on January 30, 2024, 09:59:51 amThe 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.Quote from: loop123 on January 30, 2024, 09:59:51 amNo, 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).Quote from: loop123 on January 30, 2024, 09:59:51 amNo!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.Thenyou 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.

author: loop123
author_rank: Frequent Contributor
timestamp: January 31, 2024, 01:58:56 pm
content: 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).

author: Nominal Animal
author_rank: Super Contributor
timestamp: January 31, 2024, 05:47:54 pm
content: Quote from: loop123 on January 31, 2024, 01:58:56 pmActive electrodes have a power input, reference voltage ("ground"), and a signal output (single-ended or differential).The amplifier is some kind of an operational amplifier – perhaps a JFET input precision low-noise opamp – which do not consume much power at all, typically well under a milliamp maximum. The input impedance is extremely high (megaohms), too. So, the main safety feature is limiting the amount of power supplied to the active electrode, with minimal allowed leakage (MOPP, 2×MOPP) through earth. That means, I believe, that devices you connect active electrodes to, have to have a medical-rated isolated DC supply for the active electrode.If you compare to the ISO-Z, the INA114AP is this active part, and it is on the electrode side of the isolation barrier, I believe, based on the images and the fact that cutting the resistor (to ground) makes it an unity gain buffer. So, you can think of active electrodes as having this part moved to the electrode itself, requiring a tiny bit (a milliamp or two) of current from a MOPP-rated isolated supply.

author: loop123
author_rank: Frequent Contributor
timestamp: February 01, 2024, 12:19:08 am
content: Quote from: Nominal Animal on January 31, 2024, 05:47:54 pmOne thing that always concerns me is ESD. It is more difficult if ESD from hands holding electrodes can damage the main equipment, so better if there is a fuse just before the main unit. Active amplifier in electrodes with isolated power supply can be a good fuse, isn't it. So the electrodes are destroyed instead of the main unit. Do you know if there is a special ESD fuse that can be used between electrodes and main unit.Also when say ESD from electrodes reach the ISO-Z. In your experience, would the ESD destroyed all the chips at at once (like EMF pulse) or would the damage be only from one chip? But then the current can still spread before it burns, isn't it?

author: Nominal Animal
author_rank: Super Contributor
timestamp: February 01, 2024, 02:37:22 am
content: Quote from: loop123 on February 01, 2024, 12:19:08 amUse ESD protection diodes.For example, if your sensor input range is ±6V, you might use aNexperia PESD7V0L1BSLAZbidirectional ESD diode between the input and your 0V rail. Then, any ESD pulse will be conducted to the 0V rail, with the voltage on the sensor input clamped to under ±10V (with respect to the 0V rail). That particular one does add about 20pF of capacitance to the input, so it can limit the sensor bandwidth.For example for USB, there are integrated chips likeST USBLC6-2, which protects both the data lines and the 5V USB from ESD, by shunting any excess voltage pulses to ground. And these are pretty cheap, well under 1€ apiece even in singles at Mouser.Toshiba has a nice application note about this:Basics of ESD Protection (TVS) Diodes.

author: Nominal Animal
author_rank: Super Contributor
timestamp: February 01, 2024, 03:46:35 pm
content: Quote from: loop123 on February 01, 2024, 01:05:01 pmNo idea.Quote from: loop123 on February 01, 2024, 01:05:01 pmWell, I myself have wondered about building an isolated sampling front-end based onTI ADS8681orTI ADS868816-bit ADC and anTI ISO7762digital isolator. The inputs are resistive with over 1MΩ impedance. The ADS consumes under 15mA, the signal isolator about 15mA assuming 3.3V signaling with SPI clock under 25MHz. The small signal -0.1dB bandwidth is only 2.5kHz (15kHz for -3dB), but it is 16-bit with 1Msps/500ksps sample rate (aggregate for '88, 62500 samples/sec per channel when all 8 channels used). For power, I think I'd use a Murata NTE0506MC, deriving the 5V analog supply using a TI LP2982IM5-5.0 and 3.3V digital supply using a TI LP2982IM5-3.3 (PDF datasheet), with a pi filter between the DC-DC and the LDOs.ATeensy 4.0microcontroller can collect statistics and even forward the 16-bit data to a host computer over USB. It has high-speed USB interface, and I've verified that it has no trouble providing data at 25+ Mbytes/s over USB Serial (in Linux), so the 16 Mbytes/s (1M 16-bit samples per second) would not be a problem at all. It also has ample memory, so one could use a cyclic buffer with say 0.2 seconds of samples (200k samples), which would mean no data would be lost even if the host program hiccuped a bit.I do not do biophysics myself, just sensors and such as a hobby, so I amnotclaiming the above would besafe. Ibelieveit could be made safe, but last word on that would always be with a certified laboratory in such stuff. I expect certification to cost upwards of 10k€/10kUSD, too.(I play a lot with Linux SBCs and appliances like routers, so I've been thinking about a 2-4 channel isolated probe where each channel is connected to a slow-but high-bit ADC, a fast but low-bit ADC, a pair of comparators with levels controlled from a shared DAC, with the digital data routed to UARTs and SPI. This would let one not only snoop on data, but also acquire information about the voltage levels, spikes, slew rates, and such.)Olimex sells somehobbyist EEG open-source stuffdeveloped atopeneeg.sourceforge.net,notmedical grade,with a big warning and no guarantees, so you can only use these on yourself at your own risk, and not others. You can take a look at theschematics for EEG-SMT, for example.

author: loop123
author_rank: Frequent Contributor
timestamp: February 01, 2024, 10:50:13 pm
content: Quote from: Nominal Animal on February 01, 2024, 03:46:35 pmI fixed it already. The noises came from a particular sine wave generator unit. Using others don't cause the same noises.There is not even a commercial ESD module to protect electrodes? Since there is also no stand alone commercial Isolated module that works with all amplifiers. Why don't you build them and sell them as finished products? I won't try building them in breadboard.Before you forgot about the circuit and this thread. I want to ask something. First I messed up the Molex socket by cutting and soldering which created a mess so plan to crimp this. Before I do. I want to know if it is better to have a separate power supply at the ISO-Z instead of getting it from the main amplifier. What noises can couple to DC? I know ADC can cause noises in power supply but it has no ADC.

author: Nominal Animal
author_rank: Super Contributor
timestamp: February 02, 2024, 04:42:26 pm
content: Quote from: loop123 on February 01, 2024, 10:50:13 pmIt's because to be able to sell for medical uses, you need that certification, and it is impossible to certify a device to work with all amplifiers or all electrodes.Quote from: loop123 on February 01, 2024, 10:50:13 pmThe amplifier requires so little power, it does not actually matter. It makes things more straightforward if the power is supplied by the main amplifier.Actually, I don't like the design of these devices much. Above, I tried to explain that I'd prefer the entire sampling front-end to be in the preamplifier: onlydigitalinformation would pass. Even with discrete components, say AD7766 and a precision amplifier frontend with high-impedance inputs, you could get 16-bit data from an isolated sampling frontend that only involves very low voltages and currents, with double or triple safety features.Quote from: loop123 on February 02, 2024, 07:15:48 amFromthis article:"The heartbeat has frequency from 0.1 Hz to 150 Hz. The electrical signal which is resulted from the heartbeat has amplitude of 100μV - 4 mV."So, could very well be designed that way.Quote from: loop123 on February 02, 2024, 01:54:55 pmBecause I was wrong; I looked at the ISO-Z upside-down (as if signal came in at the top, and came out of the jack near the power connectors).Quote from: loop123 on February 02, 2024, 01:54:55 pmIn that case, I would use something else. It looks fine in theory, but with the issues you're seeing, a self-made PCB could be made better and inherently safer (by limiting the power available to the isolated side, and with low-leakage ESD diodes).Quote from: loop123 on February 02, 2024, 01:54:55 pmTwo things happen: the button pulls the sensor input to ground via a resistor, and connect the clipped 10Hz signal to the output.Quote from: loop123 on February 02, 2024, 01:54:55 pmI don't know, and knowing the device isn't really certified, I wouldn't bother: the design is old, and we have better components now.A fully digital front-end wouldn't need a calibration signal generation. If needed or desired, a completely separate signal synthesizer (that could play back recorded signals for educational purposes) would be much better; and that's just a microcontroller-driven DAC with attenuation and opamp buffering.Quote from: loop123 on February 02, 2024, 01:54:55 pmI thought the red connector on the bottom of the board image is where the amplified signal comes out, and the red six-pin Molex connector at the top was the input where the electrodes are connected to. It's obviously the other way around, the Molex connects to the bioamplifier, and the electrodes to the connectors at the bottom.So, the electrode 0V reference is the white connector at the bottom (black wire), and the red and black are the differential signals. Based on the silkscreen, the button connection are4 1Normally 1 is connected to 3, and 4 connected to 65 26 3When pressed, 2 is connected to 3, and 5 connected to 6If so, the electrode positive and negative go to pins 1 and 4 of the switch, and from pins 3 and 6 through the two blue resistors to the inverting and non-inverting inputs of the INA114AP amplifier. (Note that the positive and negative get swapped here, and you should normally have excellent continuity with no noise between the right side of the two bottom blue resistors and the electrode input jacks, when the button is not pressed; and none when the button is pressed.)INA114AP is configured as an inverting amplifier with a gain of 10. Its output goes to the isolation amplifier ISO122P, crossing the isolation barrier between the electrode and the bioamplifier. ISO122P is in unity gain configuration. Its output goes to the inverting input of the final LF411CP buffer amplifier, also in unity gain configuration. The non-inverting input to LF411CP is from the blue DC offset potentiometer.That is the full electrode signal path. I would suspect the noise is in the tracks and the switch, before the first resistors.When the button is pressed, the electrode positive and negative are disconnected, and instead of the positive electrode, the inverting input to INA114AP is from the third pin of RP-1. The electrode negative connects to the solder point just above the push button contacts, and from there to the resistor closest to the green wire that connects it to the electrode positive: essentially, when you push the button (even without any power), you should see that resistor between the electrode positive and negative inputs, and that only.L7805CP regulates the positive supply from the bioamplifier to a 5V used by both the LED (the light gray 680 Ω resistor top center is its current-limiting resistor) and the DCP010512DBP DC-DC converter. The copper area on the other side is the common of its isolated output, and the component-side tracks are the derived +12V and -12V (unregulated) supply rails. These power the LM741CN in a twin-T oscillator configuration (as well as the INA114AP and the isolated side of ISO122P). I do believe you should see a large sine wave constantly on LM741CN pin 6 (top row, second from right). It is the wide track on the solder side, and although I can't see it fully, I think it comes up to the component side as the left one of the wide black tracks in the middle of the board. That is then connected to the diodes and an RC low-pass filter (two light brown resistors and two capacitors in the center) for each polarity. I think the second diode from the left is connected to one of the blue resistors at the bottom, which pass the electrode positive and negative signals when the button is not being pressed; and this is where the calibration signal gets injected.I don't see exactly where the oscillator signal gets attenuated to 500 µVpp = 0.5 mVpp = 0.0005 Vpp.If the 10Hz couples to the output signal always, then I'd look at the two bottom blue resistors and corrosion.For noise in the electrode signal, I'd check the tracks and the button switch before the blue resistors, and maybe replace the INA114AP (10€ at Mouser, in stock) and LF411CP (1.6€ at Mouser, in stock) opamps, as they're already socketed. Also clean and check the sockets for corrosion, too. I don't know if it is the picture angle, but the LF411CP does not look like it is fully seated to me.

author: loop123
author_rank: Frequent Contributor
timestamp: February 03, 2024, 02:34:26 pm
content: To add to the above tests injecting signal generator to the blue resistors and getting noise from the output of the unit. I removed the output from unit and I did the obvious and I tapped directly the pin 6 OUT of the INA114AP just before the ISO122P. There was NO noise! When I tapped the OUT and Gnd of ISO122P. There was noise! So the ISO122P is the culprit.

author: Nominal Animal
author_rank: Super Contributor
timestamp: February 03, 2024, 07:03:33 pm
content: Quote from: loop123 on February 02, 2024, 11:24:53 pmI'm broke (because I'm broken; burnout, depression, you name it) myself, so I get by with what I can make myself, plus a small order of components from Mouser every few months. You'd be surprised what you can do with aTeensy 4.0and ADC modules: it can easily sustain 25 Mbytes/sec over USB 2.0 High Speed (480 Mbit/s) to my Linux laptop over USB serial. (I still need to do a high-speed USB isolator, though; with the cheap ($10) ADuM3160-based isolators, transfer rate is limited to 1 Mbyte/s.)Quote from: loop123 on February 02, 2024, 11:24:53 pmNo, I mean you'd usesomethingto probe the signal with respect to the isolated side ground.Quote from: loop123 on February 02, 2024, 11:24:53 pmDude, use an image editing program! If you don't yet have any, installInkscape. It's free.Take pictures carefully perpendicular above the center of the board, on both sides. If you do this right, in the images the board will be rectangular.Create a new document in Inkscape, and import both images, each on their own layer. Mirror the solder side image, so their features match. You can show and hide each layer separately, and even make them partially transparent, so moving and scaling them until they match exactly only takes a few minutes.Make the layers fully opaque, and lock them. Then you won't accidentally affect the images anymore, but you can select which one (if either) is shown.Create a new top layer, and start drawing vector lines for the tracks. Use different colors for the key signals, with power rails the least noticeable: the signal path is the most important.Quote from: loop123 on February 02, 2024, 11:24:53 pmAre yousure? The connection could be incidental due to tracks on board, and not because the switch is wired so. Finding out the exact switch model (usually marked somewhere on it) might help find a datasheet or at least a pinout.Quote from: loop123 on February 02, 2024, 11:24:53 pmRP is often used prefix for resistor networks. It is a carefully matched bunch of resistors in a single package. Exactly how they are wired (they might even be separate resistor for each of legs, just carefully matched to have the same resistance!) depends on the part, and I can't read the markings from your images.To me, it looks like a Bourns 4608X-102-224 (marked 8X-2-224 with a dot/disc on bottom left, followed by a stylized B, followed by a four-digit manufacturing code YYWW, where YY is the last two digits of the year, and WW is the week number), which would make it four isolated 220000Ω = 220kΩ resistors. Leftmost pair of legs is one 220kΩ resistor, second pair another, third pair another, and the fourth and final pair another. The pairs are not connected together.Quote from: loop123 on February 02, 2024, 11:24:53 pmCurrent does not clump or pool; you need a circuit. The question is, how would a human become a part of the power supply circuit? The isolation is there to stop any circuits from forming through ground. When properly implemented (per MOPP), the isolated supply cannot leak enough current either.This NIH article describes how this can happen, in detail. It mostly concentrates on AC, though.Quote from: loop123 on February 02, 2024, 11:24:53 pmNo, I don't think so, because the signal being processed is less than one volt off from the 0V reference.Quote from: loop123 on February 02, 2024, 11:24:53 pmNo, do not. I was specifically referring to a separate voltage probe, assuming you have one, working to fix such circuits.

author: S. Petrukhin
author_rank: Super Contributor
timestamp: February 04, 2024, 10:58:43 am
content: It's not voltage kills a person, but current.You can watch how birds sit on high-voltage wires or how electrified hair takes on, having a charge of several kV.Protection in medical equipment is a huge topic that has its own standards.And not only insulation is used there, but also high-resistance circuits. If the sensor is connected via a 1M resistor, will the current in this circuit at a voltage of 230V not exceed 350uA this is 100 times less than the extremely dangerous current in the body of the body.Nevertheless, even such a small current can have a strong effect on the nervous system.

author: loop123
author_rank: Frequent Contributor
timestamp: February 05, 2024, 12:31:31 pm
content: Nominal Animal. I discovered the origin of the ripples. It's right there in page 10 of the datasheet. The LF711CP output filter after the ISO122P is supposed to make it disappear but the manufacturer used other values that makes them stay at certain frequencies.I believe you now that one must use Inkscape to get photos of both sides and combine them. It's painstaking to trace them by hand and using multimeter continuity tests a lot. But after more verifications. The following is the final schematic. I removed those irrelevant to make it clearer. I also check the 6 contacts of the calibration switch by continuity tests when I pushed it. Without pushing it. The -, + signal get into the blue resistors (legs 1,2 and legs 4,5 connected). After pushing it, legs 2,3 and legs 5,6 got connected diverting it to the blue resistors and into the INA114AP.Please take a look at this one and help (one last request) to figure out how the 0.5mV 10 Hz is created in the 741CN. So in the future, I could remove that part of the circuit or repair them when needed. Without it. The circuit is just plain simplicity. Also why are there diodes right after the blue resistors going to ground? It's not a rectifier diode. What is its purpose?Many thanks.