# Optoisolate two ICs sharing same power supply

I have two ICs. The former acts like an ADC, measuring the voltage in a power outlet and transmitting the result to the last. Both ICs requires 5 Volts to work.

I want to opto-isolate the channel between the two ICs to protect the last IC against peaks in the grid that the former IC is measuring.

Below there is a VERY simplified version of the circuit:

The power supply is the classic one (voltage transformer, regulator, capacitors, etc) and the input voltage is the same as the power line: 220 Volts RMS. The power line has a transformer too (not shown).

Is safe to both ICs share the same power supply? It is possible that in presence of a very large noise, the first IC affects the second in that common point?

If it is wrong... how to accomplish the isolation?

EDIT
IC1's datasheet

IC1 is a specific purpose IC to measure voltage (and current) from the power grid. The IC's datasheet has a example circuit that I pretend to use.

IC2 is a microcontroller that will receive the information from IC1.

The example circuit

Although the circuit does not use a transformer and voltage regulator, I intend to use them to power the IC.

The datasheet recommends to isolate the channel between ICs.

My focus right now is discover if it is possible to use the same supply for both ICs and yet opto-isolate this channel between them.

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 What is IC1 (datasheet)? The fact that it has an "IN_GND" pin might imply it has some defined behavior related to isolation from the mains it's measuring. – The Photon Jul 18 '12 at 22:16 – borges Jul 18 '12 at 22:23 YOu have no idea about EMC ingress measurement issues, UL/CE safety requirements, Power Line transients and Surge protection, power line monitoring techniques. Get an advisor to read my suggestions. – Tony Stewart Jul 19 '12 at 4:51

## 6 Answers

My focus right now is discover if it is possible to use the same supply for both ICs and yet opto-isolate this channel between them.

Notice that in the recommended circuit, the ground of the CS5463 is shown with a different symbol than the ground of the "serial data interface".

Also notice that the CS5463's ground is tied to line at the upper left.

If you connect your interface to the same ground as the CS5463 in this configuration, your entire circuit will be riding the 120 V line voltage.

If you do not connect your interface circuit to the same ground as the CS5463, then you won't be able to use the same power supply to power the two subcircuits.

If you want to power your circuit from the same supply as the CS5463, you should use the other recommended circuit from the datasheet:

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If the two sides of the opto-isolation share ground and/or +5V power, the effect of the opto-isolation is completely negated.

I can not judge whether this is safe or not, but judging from the fact that you included an opto-isolator in the first place I think that sharing power and groundf is a bad idea. use two separate powers, and don't connect the grounds.

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Answer: not safe.

Question is missing the context. I assume the safety meant to be the user's safety. Some original design involved opto-insulated group connected to mains. The next designer or hacker tries to reuse existing power source and concerned about electrical shock.

Certainly not safe. Do not galvanically connect anything on insulated side to mains.

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If the goal is to protect IC2 from peaks on the input I would use a resistor to limit the current and a zener.

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Q Is safe to both ICs share the same power supply?

A Maybe not. Depends on ground faults and miswired outlets. Does it meet UL CE safety requirements for line connect to electronics ground? No Supplementary design details are missing. If it has human access to signals or signal ground, even with ESD it must have an earth connection, even if a floating apparatus. i.e. a 3 pronged plug. OTherwise double Insulation is needed.

You ought to protect the electronics from easy failure too with surge suppressor. how you manage to protect users from stray lightning surges at say +/- 1500V 0-pk 50uS. Your resistors also need to be rated for 3kV and preferably more. YOu could consider a capacitive network AC coupled and not connect Neutral to DC ground with ground current limited at 1200Vac for Hipot test and leakage current for safety tests limited to 0.5mA on any filter caps to ground.

If you are not planning to get UL CE certification, quit now and change to a low voltage transformer output. That's a safe plan.

Q It is possible that in presence of a very large noise, the first IC affects the second in that common point?

A Yes but the 2nd IC does nothing useful at present. Common mode and stray RF noise will plague your design unless you assume there will be interference and block it.

Q If it is wrong... how to accomplish the isolation?

A high impedance differential measurement like a DMM is a better approach with CMMR , LPF Surge protection, OVP, Hipot protection, leakage to ground and CM filter covered into the design. CM ferrite choke, Differential Instrument Amp or reliable Op Amp with small RF caps to ground or not is your choice if you want 3 prong plug with MOV after a 3KV series cap or resistor before divider to scale down to desired level. Is this bipolar supply(+/-V) then bias to V+/2.

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 Unfortunately, I understood almost nothing of what you said. Please, see my edit. – borges Jul 19 '12 at 1:33 If you do not understand my comments, then understand this, Buy a LOW VOLTAGE AC adapter and design around the low voltage signal which may be rectified to power the unit. Use a DMM to calibrate the unit. – Tony Stewart Jul 19 '12 at 4:43

You need an isolating transformer to pass power between isolated circuits. Nothing else will cut it.

You could use a large transformer with the 50Hz mains, or perhaps some sort of switch mode power supply circuit.

Even a 5V boost/buck type circuit could work if you find a suitable isolated low power transformer.

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