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I have a following issue, which I didn't see in IC books: I would like to know if it is possible to send analog signals (sin waves) between two power domains in an integrated circuit. The issue looks as follows:

enter image description here

I have never had a problem with sending digital signals between two different power domains as it is done through level shifters / level converters. However, I never thought about sending analog signals between two power domains. Is it done using protection diodes as below?

enter image description here

Of course, signal send from 3.3 V domain is limited to 2.5 V. This sinusoidal 3.3 Vpp emphasizes that signal is send from 3.3 V domain to 2.5 V domain and that maximum signal that can appear on the line is 3.3 Vpp.

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  • \$\begingroup\$ If they are "common grounded", yes. But you should limit you signal swing to the input range of your receiver. \$\endgroup\$ – diverger Nov 3 '14 at 15:39
  • \$\begingroup\$ @diverger Right. Of course signal send from 3.3 V domain is 2.5 Vpp. I added the comment at the end of my main message. Thank you. \$\endgroup\$ – Tako Nov 3 '14 at 15:50
  • \$\begingroup\$ Yes that looks fine. Depending on the output impedance of the high voltage side you might want some resistance in series to limit the current into the diodes. (You also need to check that the 2.5 V supply can sink an over voltage.. sometimes a resistor to ground is added.) \$\endgroup\$ – George Herold Nov 3 '14 at 16:03
  • \$\begingroup\$ Hmm, I think I had an eclipse of the mind at the end of a work day... Of course I can connect two power domains using capacitor + voltage divider. @GeorgeHerold , I understand the issue about resistor to limit the current into diodes - this is somewhat ESD protection, but I do not understand the part about that th3 2.5 V supply can sing an over voltage and where the resistor to ground should be added. It should be just added from the signal line to the ground? If so, it should be very big resistor. \$\endgroup\$ – Tako Nov 3 '14 at 16:13
  • \$\begingroup\$ Well most power supplies are designed to source current (if they are positive voltage.) If the diode going to the positive rail conducts, then you are asking the power supply to sink current. Most of the time this is not an issue... there are other drains on the supply, but it is something to be aware of. And yeah I try to scale the signal on the high voltage side, that keeps it all linear too. \$\endgroup\$ – George Herold Nov 3 '14 at 16:24
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You don't say so I am assuming that the 3.3V domain signal is a cell that you can't touch and scale back. Otherwise the right solutions is ... don't send a 3.3V signal into a 2.5V domain.

Using clamping diodes is problematic. Do you know how the output driver will respond to having it's load impedance modulated like that? What about the parasitic capacitance of the diodes? What is the impact on interference from you dumping a signal onto the rail?

It would be easier (again assuming that you can't touch the 3.3V output) to convert this signal locally. Design a simple emitter follower in NMOS (which gives you a gain of 0.8) using the 3.3V transistors and you will get an easily scaled voltage that is only two components and gets you very close 2.65 V. You could probably degenerate the source current a bit more for a nice linear inbounds conversion. The drive will be asymmetrical but again you don't say what the speed requirements are.

Make sure you have a good path back to the driver for image current flow.

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  • \$\begingroup\$ I see. But sometimes there is a need for such connection. For example see Fig. 32 and Fig. 49 here: link . The input signal of the PGA1 is 10Vpp while the PGA2 can support input signal equal to its power domain that is 3.3 V. These two PGAs are connected through low-pass filter LPF and a capacitor (C10 in Fig. 49). \$\endgroup\$ – Tako Nov 4 '14 at 16:59

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