What Happens to Op-Amp Gain When Switch is On?

[foobag]

http://img684.imageshack.us/img684/8362/opamp2.png

Lets say we are given a non-inverting Op-Amp

The gain of this circuit would be A = 1 + (R1+R2)/R3 if the switch is off.

What happens when the switch is turned on? It shorts the R1 and R2 resistors am I correct? Does this mean that R1 and R2 will behave as if they are in parallel, and decrease the overall gain of this amplifier?

Appreciate your help!

 

[vk6kro]

As you have drawn it, only R2 is shorted.

So, the gain is A = 1 + (R1)/R3.

 

[foobag]

ah ok, thank you for your help!

 

[uart]

As the circuit is drawn it doesn't work (in any sensible way) when the switch is open as there is no negative feedback.

Where two wires cross on a circuit diagram you MUST place a "junction point" (filled in dot) if the two wires are connected at that point. Otherwise there is NO connection. This is the standard convention.

 

[vk6kro]

Another convention is to draw a hump where there is a crossover that doesn't touch.

Yet another is draw a broken line where it passes under another wire.

All three methods are used but I like the solid circle method as it is unambiguous.

In this case, there is apparently a connection at each side of the resistor with the switch across it.

 

[uart]

Another convention is to draw a hump where there is a crossover that doesn't touch.

Yeah but that is somewhat outdated now. The most common convention (which is used almost universally in circuits and schematic software) is the use of the junction point "dot".

btw UART, where are you in oz? Up very late unless you are in WA?

Yeah I'm just up late.