Input impedance on an instrumentation amp

How about biasing one input to say 1V with a resisitve divider and having the other open ?

Rene

The bias current paths are usually provided by the the feedback and input resistors.

Do you have a schematic ?

Graham

Not in the case of high-Z instrumentation amplifiers, Graham, where the feedback resistors come after the input buffers (see the AD623 datasheet).

Zach, are the batteries really floating? I mean, is one end of your battery-stack grounded, etc? In that case you don't need to provide another DC bias path, that's provided by the cells.

Also, 100k, etc., bias path resistors are unreasonable, since the AD623 has only 17nA of bias current. I don't know your battery / system-power setup, but if the system power is ever off while the electronics is connected to the battery cells, you can have the awkward situation of powering the system through the 4052 CMOS switch or the AD623's protection diodes, etc. You may want series resistors for the battery-cell connections. Also, there are more advance MUX types that disconnect themselves when the power is off, or if the input voltages exceed their supply rails.

Zach, You must provide a path for bias current. However, the value of the resistor value to ground depends on the output error that you can tolerate due to input bias current. The output error will be approcimately equal to Ib X Rbias X Ac, where Ib is the bias current, RBias is the summ of the bias resistors, Ac is the closed loop gain. Regards, Jon

Thank you to everyone that responded.

Let me answer some of the questions:

Rene, I am not sure I understand your suggestion. This is a differential system so both inputs on the amp are inputs.

Graham, I am just copying the schematic on page 14 of the datasheet figure 47c.

Win, yes the batteries are completely isolated and need to stay that way.

I wired up a test circuit and found that things start to bounce around with ~300k ohms per DC bias resistor. There must be another trick to get higher input impedances when working with a fully isolated input.

Good point on the power off condition. I will start looking for a more advanced analog switch. Any suggestions?

Thanks,

Zach

One other data point...

I was looking at some of TI's in-amps and noticed on their data sheets they only use a DC bias resistor on one input (not both). So, I tried that on my test circuit and it seems to work. Just a 100k from the + input to gnd. Any reason I shouldn't do this?

Are the batteries floating? If so, just a little static and you'll likely exceed the common-mode range of the InAmp. Can you guard the batteries by having a high value (say 100k) resistor from one battery terminal to gorund?

According to the data sheet, when measuring

The recommended circuit uses two

You should only need one resistor to ground, to guarantee a safe common-mode voltage and to provide a few nanoamps of bias current.

Mots InAmps have input buffer amps, so their input impedance is REALLY up there. See the spec sheet and calculate how high you can safely go with the resistor or resistors.

Bounce is probablyl caused by noise and Ac pickup. You may want to add a low pass filter at the input, say a 1 megohm resistor in series with each input and a 5uf poly capacitor to ground. That wil roll off a whole heck of a lot of 60Hz and random noise.

The AD623 looks similar enough to the AD620. We typically use 10Meg bias resistors on both inputs of our circuits using the AD620 and AD621. The bias current is low enough that even with 5% bias resistors it doesn't present a problem at the output.

HTH, Tom

The AD620/1 are dual supply and have .5nA of input bias current. The AD623 uses a single supply has 17nA of input bias current (34 times as much current). If you take your 10meg resistors and divide them by 34, you get ~300k ohms. Which is about the max I could use and still get readings without significant errors. I'm still not sure if I need resistors on both inputs. My test circuit seems to work fine with just one, but I'd hate for something bad to happen under different operating conditions.

Thanks,

Zach

The battery itself constitutes a very low ohmic serial resistance for almost all frequencies including DC. So you need only a single resistor, or you can just ground 1 terminal. There will be only the bias current flowing through the battery. Why do you use a differential measurement, IMHO you can do as well with a single ended ADC + opamp.

In article , wrote: ]snip]

Battery voltages move slowly, so if you can accept an update rate of no more than about once each 30 seconds then a truly isolated scanner can be built with a relay-based flying capacitor system.

RL1a | Cell+ ---/\\/\\-----o\\ o-+---->ADC R 470k \\ | o | C === 0.1u | o R 470k / | Cell- ----/\\/\\----o/ o----+->0v RL1b |

A 4-channel scanner would require 4 2PCO relays, wired as above, with all 0v and ADC outputs bussed together.

To measure a cell voltage, energise that relay to swing the capacitor over to the ADC, measure the voltage, put the relay back. With a 0.1uF cap, each relay should be energised for no more than about 10mS.

Once each C has been charged for the first time the load presented to each cell will be the leakage of the C.