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If it's an analog sensor, you have to use something like an instrumentation amp to isolate it instead of an optocoupler.
Can you recommend a DC/DC converter?
Why can't you use a optocoupler with an analog sensor? I was thinking about using PS2506. http://www.cel.com/pdf/datasheets/ps2506.pdfThat opto is not a good choice due to its CTR varies greatly (200 to 2000%) with current, temperature and opto to opto. Look at the graphs in that data sheet.
It can be done but it is not easy as it requires a very carefully design circuit to do successfully. The best parts to do this is a lamp (light bulb) and a photo-resistive sensor (CDS) and a few Op-Amps. I've seen this done plenty of time in circuits built before the 1980's. Today, with cheap and readily available digital chips it is almost never done analog.
QuoteCan you recommend a DC/DC converter?What is the input Voltage, output Voltage and current (or power) you require. There are hundreds available and yes there can be a bit pricey.
Will an opto like the one I linked not work?
The sensor is an ECG sensor
Yep, jwatt laid out the CTR issue well.How accurate do you need the Analog signal to be on the isolated side? QuoteThe sensor is an ECG sensorIf this is going to be attached to a human body you need very good PS isolation and preferably devices that have a Medical rating. This does up the cost a bit.Since your sensor only requires a few mA would a battery be an option?
QuoteWill an opto like the one I linked not work?Not without VERY careful circuit design, that includes temperature compensation and per-device calibration.Look at this diagram from the data sheet:This tells me that, at 25C, 0.95V gives me a forward current of 0.05 mA, and 1.3V gives me a forward current of 80 mA. However, at 0C, those move up by at least 0.5 volts! Same thing at 50C, etc.The reason current transfer matters, is that you have to derive the output voltage using a voltage divider. If the supply voltage on the output is 5V, and there is a 50 Ohm current limiting resistor on the output, then a forward current of 80 mA means the total resistance should be (5/0.08 = 62.5) ohms, so the optocoupler will be 12.5 Ohm impedance. A forward current of 10 mA (from that diagram) means the total resistance should be (5/0.01 = 500) ohms, so the optocoupler will be 450 Ohms impedance. A forward current of 1 mA means the optcoupler will be 4950 Ohms impedance. And the voltage differential between the current limiting resistor and the optocoupler will be the same -- 50:12.5, 50:450, 50:4950, which also means that the lower the output current, the more precision your analog converter needs to be.An instrumentation amp with separately adjustable gain is a lot easier to deal with!HCPL-7840 is available in DIP and seems legit, although a little pricey at $6.80 in singles:http://www.digikey.com/product-detail/en/HCPL-7840-000E/516-1482-5-ND/669909Note that it contains an A/D converter powered from one end, and a D/A converter powered from the other end, with no galvanic contact in between!I know that AD has some cheaper ones, but likely only in surface mount these days.Digi-Key has a list of isolated board-mount DC/DC converters: http://www.digikey.com/product-search/en/power-supplies-board-mount/dc-dc-converters/4325599?k=isolated%20dc%20converterNote that these seldom give "the same voltage" out as in; rather, they give a fixed (or adjustable, specific) voltage out for a possible range of input voltages.The Recom RE-0505S seems to be the cheapest. If you can live with 5V in, 5V out, isolation, you can get it for $2.75 in singles: http://www.digikey.com/product-detail/en/REE-0505S/945-1654-5-ND/3461633
I think almost all isolation amps are dual supply, but given that you need a supply for your sensor, you could just use that supply for the amp, too.To me, it looks as if the iso amp and the DC DC converter I linked will solve your problem, for < $10 (plus some soldering.)