How do "beyond the rails" analog switches work?

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?

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.

• 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? Commented Feb 20, 2014 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. Commented Feb 20, 2014 at 20:47
• Wouldn't this produce at least small amounts of switching noise? Commented Feb 20, 2014 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. Commented Feb 20, 2014 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. Commented Feb 20, 2014 at 21:15

markrages wrote: "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."

And a potentiometer if you want a variable-voltage charge pump. It acts as a voltage divider, it's placed across the pump diode and just upstream of the pump capacitor. If you want the charge pump to quick charge just as it's starting up, put a momentary switch between the output of the pump diode and the input of the storage diode.

https://i.sstatic.net/9YFp7.png

• Beware, the question is from 2014, so you might not get traction from the OP. Commented Jun 22, 2020 at 18:22