How to make Op Amp Gain Switch Solid state. (vs jumper)
Circuit modified.   sources commoned and R&amp; D added. May well work without R3 - see below.
R3 value not shown but could be 10M or higher if environment allows.
R2 can be lower if desired.
Needs thinking but workable (I believe). It's the sort of circuit that can be made to work entirely reliably but which 'has things to think about along the way'.
The main point of unusualness is that the mode with two gates is "floating". R2 turns both FETs off when no external drive is applied.
My experience has been that applying external drive to the floating gate pair tends to cause them to assume the expected on state BUT there is no definite formal path to get there. Adding R3 allows the gates-node to be ground referenced but to float until driven. Once on, it works as expected.   

Note: Almost all circuits are "designable". This one is too - but to be fully formally designed needs a few parts which in practice can be omitted. That does not mean that the end result need be unreliable - just that the factors at work that make it work with fewer parts than may be expected take more thinking about than usual. 



Place a resistor in parallel with R6 and in series with 2 s 'jellybean" N channel MOSFETS in series opposed connection. ie
D1 ground
S1 S2 connected
G1 G2 connected 
D2 to R6b
Drive G1-G2 via a high = on, ground = off signal.
Note resistor biasing to 'persuade' floating node to be ground referenced.
Diode drive provides high or O/C drive. 

Adjust R5 and R6a-R6b values to suit.
If R6 = say 12k then a MOSFET Rdson of 100 milliohms per FET = resistance error of +0.2 Ohm/12k ~= 0.002% - so temperature and device variations "not an issue".

Cost in modest volume manufacturing in China = $US0.01/MOSFET + assembly and PCB area costs. 

The circuit utilises the fact that MOSFETs are 2 quadrant devices. Each device is on bidirectionally when gate is positive relative to source regardless of D-S polarity.
Two devices are required in series opposition due to body diodes which otherwise would provide reverse conduction when the FET was off.
For small amplitude signals where V_R6 was &lt;&lt; 0.6V one might "get away with" a single MOSFET, but this is an imperfect solution as diodes still conduct (at 'exponentially' lower currents) at lower voltages. 







simulate this circuit &ndash; Schematic created using CircuitLab