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I'm familiar with using CMOS switches ("transmission gates") for switching analog signals, but you can't use them for switching voltages beyond their supply rails. (So if you have a single supply circuit, and a voltage signal that's been AC-coupled and to a ground reference, you cannot use a CMOS switch because the voltage of the signal drops below ground, which is the negative supply for the switch.)

However, there are some analog switches that will switch ±25 V signals with a power supply of 0 V and +3.3 V. How do they work? Is it simple to build a circuit like this out of discrete parts?

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up vote 7 down vote accepted

How do they work?

They have integrated charge pumps to create internal supply voltages that are greater than ±25 V.

Is it simple to build a circuit like this out of discrete parts?

Possible? Yes. Simple? Not particularly.

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Would it be worth mentioning that the elevated supply rails don't need to source or sink any significant amount of current and, as a consequence, the charge pumps can be very small? – supercat Feb 20 '14 at 16:15
Making the circuit out of discrete parts gets complicated, but a charge pump can be two diodes, two caps, and a square wave from a microcontroller. Integrated standalone charge pumps are also available, if the beyond-the-rails parts don't meet OP's specification. – markrages Feb 20 '14 at 20:47
Wouldn't this produce at least small amounts of switching noise? – endolith Feb 20 '14 at 20:51
All charge pumps make switching noise. If you roll your own you can set the frequency or filter it or postregulate it. Certainly switching noise is a design consideration. – markrages Feb 20 '14 at 20:57
Some, not all, of those chips seem very explicit on the charge pump properties and what start-up and noise considerations are involved. – Spehro Pefhany Feb 20 '14 at 21:15

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