load cell to AD620 op amp, circuit not working, is chip dead?

[moorea211]

Hi,

I'm building a machine that will utilise a load sensor from a digital weighing scale. The load sensor has a built in wheatstone bridge, and I'm using an AD620 op amp to get the signal up to a readable level. The first circuit is for this. There is a resistor network to provide a 3V reference voltage, with a decoupling capacitor.

The gain is set by the resistor on the far left.

I've checked all the input voltages, and all component values, (and my calculation for gain) and checked the wiring (breadboard) many times.

I get no output from pin 6 with respect to 0V.

Am I daft? Is this circuit no good for this chip? Or is the chip dead? I had a feeling it's my circuit, so I built the one below, with a small gain and a fixed input voltage, and the - input held to 0V. Still no output!

Attached are the circuits (apologies for slightly non standard symbols, I'm in UK, we draw them differently here!), and some pages from data sheet.

No idea what is wrong, but someone here will spot it instantly, I'm sure!

Thanks, Richard B

View attachment 41125

View attachment 41126

View attachment 41127

 

[moorea211]

A couple more relevant bits of datasheet...

Richard B

View attachment 41128

View attachment 41129

 

[Hero999]

I think the input voltages are out of range. The chip only works when the inputs are 1.9V above the negative supply and 1.2V below the positive supply.

 

[moorea211]

Thanks, Hero999,

I tested the outputs from the load cell, with a 3V supply voltage, the sense outputs are both 1.5V, with a maximum change of 1mV under physical load.(Which is all as expected from a wheatstone bridge) Obviously the sense voltage is below the figure of 1.9V for chip input, but below the (3V - 1.2V =1.8V) too, so I'll up the supply voltage to chip and sensor 4-6V.

Then I'll let you know what happened!

Thanks, your clarity and experience are welcome,

Richard B

 

[Hero999]

Increasing the positive power supply voltage won't make any difference. You need a negative supply voltage, if the input is below 1.9V.

 

[moorea211]

Okay, I think...

If I increase the excite voltage to 4V so the sense voltages are approx. 2V, would that work? Not sure at all how I'd make a negative supply. 15 years ago, when I last used any electronic knowledge, I'd have known, but it's amazing how much gets forgotten if you don't use it.

Haven't been able to test the chip yet. Do I need a negative supply to do the test you described? or can I just use 0V?

Richard B

 

[moorea211]

Just a thought: if the reference voltage is 3V, the +supply is 6V, then isn't the negative supply then equal to -3V, relative to the reference voltage? Or am I missing something? (Probably.)

Richard B

 

[moorea211]

I wrote the wrong value for the gain resistor in the first drawing, I've updated that and added more information.
Odd anomaly: if I wire it up as shown, with the excite  voltage (4V as tested when not connected to anything) taken from the bias network, it then reads 0.41V... Mystified as to why. The sense voltages are both 0.205V. If I put the positive lead from the sensor (excite+) and jam it it under the third 1.5V battery in my 6V supply, it gives me 4.5V (obviously,) with sense wires at 2.25V.

Why should this happen? It's totally beyond me.

Can anyone help, i think my brain is evaporating...

Richard B

View attachment 41132