Guard rings for ic input

What are these? Is it like some 'physical ring' where you out a wire through it or just a connection to make both input to the op-amp common? How does this really help? Seems like you're making a source common to both like gnd so if the gnd is noisy it becomes common for both and rejects it?

Here's a chip i was considering using, a JFET due to the low input noise.
http://focus.ti.com/lit/ds/slos182b/slos182b.pdf

Page 64 talks about using guard rings and driving them from a low impedance source at the same voltage level as the common mode input.
 

berkeman

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Pretty cool. The guard rings are to prevent leakage currents on the PCB surface from interfering with these super-high-Zin inputs. You drive the guard rings to the same voltage as the input pins are at, so there is no electical potential to drive the leakage currents.

I've seen a similar technique for reducing input capacitance -- it's called bootstrapping. Pretty handy in some situations that need higher bandwidth, but often difficult to get stable.
 
so this 'guard ring' is an electrical connection as in a physical trace connected to both inputs and have a common gnd or source to drive it? Doesn't this interfere with the signal Vin then? This is somewhat confusing..
 

berkeman

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No. The rings surround the IC input pads, but are not connected. The rings are not grounded (unless the CM input voltage happens to be ground, which it wouldn't be in the general case). The rings are driven as shown on page 64 to the CM voltage of the input pins. This minimizes/eliminates any electric field bewteen the input pins and the guard rings, which minimizes any surface leakage currents on the PCB to/from those input pins.
 
so if i were to do a pcb board, the decal would have the regular mask for the chip, and then some rings surrounding the pads (driven by common mode voltage) with room enough to allow traces to be connected to the chip?
 

berkeman

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so if i were to do a pcb board, the decal would have the regular mask for the chip, and then some rings surrounding the pads (driven by common mode voltage) with room enough to allow traces to be connected to the chip?
Basically, you need to surround the input pins and whatever other metal connects to them on the PCB with the guard rings (guard traces really). So in the case of DIP packages, you would run guard traces around the input pins, and run parallel on the sides of each of the input traces, back to where the input signals are originating (like where a sensor plugs into the PCB). Where the metal of the signal lines is exposed (like at the DIP IC pin going through the plated through hole of the PCB, the guard trace should be likewise exposed without soldermask on top of it. Where the input traces are under soldermask in their run from the source, the guard traces should likewise be under the soldermask, running parallel to the input trace.

The objective is to keep the input trace/pin/etc. metal from having a creepage path to some other metal that is at a different electrical potential.
 

berkeman

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Oh, I forgot to mention -- These guard ring/trace structures need to be on any PCB layer where the input metal is exposed. So for DIP packages and a 2-layer PCB, that would be on the top and bottom layers (watch out for solder bridge issues for exposed metal stuff at close spacing, though, when wave soldering). For SMT packages on the top side of a PCB without vias on the signal lines, the guard rings/traces would just be on the top layer. For DIP packages or vias on the input lines and a 4-layer PCB, you will need to look at the inner layers too, and add guard rings around the inner layer vias.
 

chroot

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I don't usually hear them called "guard rings," I usually hear them called "gaurd traces." Normally, you just draw a dummy trace on either side of the signal traces, and drive those dummy traces with whatever voltage is least likely to cause interference (the common mode voltage is obviously a good choice). You drive it with a low impedance, hoping that any crosstalk or EMI will couple more into the low-impedance path than the very high-impedance path at the chip's input pin.

- Warren
 
ok thanks, just run outer traces along the input trace (and at the same side of trace -with/without soldermask depending on input trace) and drive them with a low impedance source.

what voltage do you typically drive it with? when someone mentions common mode voltage how can you know the common mode voltage without having even look at it or choose the proper one?? +5v, +12v..??

thanks
 

berkeman

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Look back on that page 64 in the datasheet. You want to drive the guards with the same voltage that is present on the inputs. So you tap off of your circuit at whatever place is appropriate (wherever you can get a copy of Vin), and buffer it if needed.

In the 2nd figure, the two inputs will be near ground, so you just tie the guard to ground. In the 3rd figure, the follower arrangement lets you drive the guard with just the output of the opamp.

Quiz question -- in the first figure, what should R3 and R4 be in terms of R1 and R2?
 
it says r3/r4 = r2/r1
i suppose if you want to change the gain just make r4/r1 smaller
 

berkeman

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it says r3/r4 = r2/r1
i suppose if you want to change the gain just make r4/r1 smaller
Yes, but why is that ratio important? What are you trying to do with that extra voltage divider that supplies the guard voltage....?
 
well the voltage divider supplies to both guard, keeps it balance for high common mode. also it seems it uses output of op amp to drive the guard which has low impedance output so if you made r4/r1 small you can get a low impedance source to drive the guard ??
 

berkeman

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well the voltage divider supplies to both guard, keeps it balance for high common mode. also it seems it uses output of op amp to drive the guard which has low impedance output so if you made r4/r1 small you can get a low impedance source to drive the guard ??
Mostly correct, but what I was driving at was to have you see how that other voltage divider is just making a copy of the voltage that is being fed back to the - input on the opamp. That is how you are matching the common-mode voltage that is at the inputs, by making another copy of it in parallel to use.
 
i see what you mean
thanks
 

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