I am not quite understand this configuration here, can you explain it a little bit more please, thanks!The mux/switch selects the tap point off the resistor string that goes to the minus (-) input which determines the gain. That configuration avoids any effect the mux/switch resistance has on the gain.
Below is a simplified schematic of the scheme.I am not quite understand this configuration here, can you explain it a little bit more please, thanks!
Thanks crutschow, this is a little neat configuration I haven't come crass before, thank for showing that to me. It's going to be very useful.Below is a simplified schematic of the scheme.
Can you explain more about the use of this resistor?Also, a 1 Meg resistor from output to inverting input would prevent the gain from being open loop if the switch if break before make.
Some analog multiplexers have make-before-break switching. This means that, for a brief instant during switching, the "wiper" on the switch will be connected to the input it is leaving AND the input it is going to. In this case, feedback always exists between output and input. This is probably the best choice, if one can be found.Can you explain more about the use of this resistor?
Do you refer a circuit like this?Other MUXes have break-before-make switching. This means that, for a brief instant during switching, the "wiper" on the switch will be connected only to the op amp -input, so there is no feedback path. The -input will have no current path to ground, so the output will have a "glitch" as the output attempts to go to one of the rails.
I used LTspice which is a free program from Linear Technology that many on these forums use.Thanks Crutshow. A nicely detailed response. What cct simulator have you used here?
That is a very simplified diagram of an op amp. It doesn't really show the open-input condition being referred to. When one of the inputs of an op amp is open-circuit then there is no place for the bias current to go and the input will drift in voltage. As soon as that occurs, the high open loop gain of the op amp will drive the output to one of the rails. If you add a high resistance from the output to the negative (-) input, then there is always a path for the bias current and the output won't saturate if the negative feedback path is momentarily interrupted.Thanks, that is completely new to me.
This part is not quite clear to me.
Do you refer a circuit like this?
Then the output will be V+ or V- which depends on tapping point?
How much bandwidth do you need?So I just one or two switch, the mux I found are usually 8 channels, can I do that with transistor or something?
The gain I'm interested in are unity gain and x10
I have not thought of that part yet, but I would like to start with an easy one, say 1KHz.How much bandwidth do you need?