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I am currently using an AD620 to amplify the signal coming from my load cell. I am supplying it with single supply of 5V and 0V. Below is my preliminary circuit.enter image description here

I am feeding half the supply to the reference pin so i am getting an output of 2.5V to 3.7V. I also tried feeding the reference pin of 1.5V and I am getting an output of 1.5V to 3.7V as expected from the information provided from the datasheet. What can i do to have a 0 to 5V rail-to-rail swing? Do i have to feed this output to a rail-to-rail op amp like an AD822?

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I have modified it by using an AD822 "rail-to-rail" op amp to subtract a voltage so I could scale down my output from the AD620. is there something wrong with this? Because I am having some inaccuracies regarding my AD822 output.

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  • \$\begingroup\$ Note that the 741 op-amp will not work with a supply as low as 5V. You'll need to use a different op-amp chip, one that is specified to work at 5V. \$\endgroup\$ – John Honniball Dec 18 '14 at 9:49
  • \$\begingroup\$ as of now, I replaced that 741 with a simple voltage divider with a pair of resistors to have a voltage reference to my AD620. but as expected it is unstable, what would you recommend to have a voltage ref of about 1.5V? \$\endgroup\$ – Rhonald Rei Pahayac Dec 18 '14 at 11:46
  • \$\begingroup\$ For a voltage reference (as opposed to the virtual ground shown in your schematic) I'd suggest the TL431. It's an adjustable band-gap reference diode, very commonly used and very cheap. \$\endgroup\$ – John Honniball Dec 19 '14 at 9:30
  • \$\begingroup\$ ok thanks @JohnHonniball I will try that one. i am thinking on using it to have a stable excitation voltage of 5V to my load cell and 5V reference voltage for my ADC. and also for the 1.5V reference voltage for the amplifier. is that part suitable for this purposes? \$\endgroup\$ – Rhonald Rei Pahayac Dec 20 '14 at 15:36
  • \$\begingroup\$ You'll need two TL431s if you want both a 5V reference and a 1.5V reference. But thinking about it, I don't think the TL431 can go below 2.5V. So, OK for the 5V but no good for 1.5V. \$\endgroup\$ – John Honniball Dec 20 '14 at 18:10
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Yes. What you need is a difference amplifier with gain. You can do this in a single stage, but I think you're better off doing it in two stages. First, create a voltage of 1.5 volts, and subtract that from your AD620 output. You can do that with a simple difference amplifier: one op amp and 4 equal resistors. You will now have a useable voltage swing of 0 to 2.2 volts. Then you make a non-inverting amplifier with a gain of a little more than 2, and boost your 0 to 2.2 to a 0 to 5 volt signal.

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  • \$\begingroup\$ so i have to set my AD620 reference pin to 1.5V. i think i will research on this. thanks! \$\endgroup\$ – Rhonald Rei Pahayac Dec 18 '14 at 5:47
  • \$\begingroup\$ No, use the existing 2.5 volts. You say you're getting a minimum output of 1.5. If you subtract 1.5 from this, you'll get a minimum output of 0, which I presume is what you want. \$\endgroup\$ – WhatRoughBeast Dec 18 '14 at 21:51
  • \$\begingroup\$ I use the 1.5V reference to have a minimum output of 1.5 and also tested the 2.5V reference and have a minimum output of 2.5. But i also get voltages a little lower than my reference. Is that a problem? Does setting my reference voltage to 1.5V MUST mean that my output should start at 1.5V? \$\endgroup\$ – Rhonald Rei Pahayac Dec 20 '14 at 15:40
  • \$\begingroup\$ No. What I mean is, with whatever reference you use, you must measure the AD620 output at zero load, then subtract a little less than that from the AD620 output to get a zero load output slightly more than zero volts. And the reference must be such that the AD620 output is not pegged at whatever the minimum is with zero load on the cell. This is possible if you set your reference too low. Of course, the higher the reference, the less span you have to work with before the output becomes too high for the AD620 to drive it. And don't go too close to either extreme, either. \$\endgroup\$ – WhatRoughBeast Dec 20 '14 at 16:18
  • \$\begingroup\$ I have made the circuit and set the AD620 REF pin to 1.5V. Would it be right if I also use that ref voltage as the one to subtract to my AD620 output(tied to the inverting pin of the differential amp)? Because thats what I have done to my circuit, and then after I subtracted it, i amplified it using a non-inverting amp and now i get something like 10mV at zero load, which i think is ok. \$\endgroup\$ – Rhonald Rei Pahayac Dec 21 '14 at 14:43

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