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I've been using the AD620 to offset and scale up a tiny AC signal for my project.

enter image description here

The AD620 is set up in the following way:

  • Rg= 6.8kOhms (between pins 1 and 8)
  • Vin = 0.123V rms AC signal (pins 2 and 3)
  • REF = 2.5V (pin 5)
  • +Vs = 5V (pin 7)
  • -Vs to Ground (pin 4)

Even though the gain is supposed to be G=49.4k/Rg+1=8.26 and I'm supposed to see an output (pin 6) of: V=0.123*8.26=1.02 V-rms, I get no AC output signal at all.

What I'm getting is a DC 1.6 Volts signal (which, I suppose, comes from Vref).

Do you guys have any idea of what's going on?

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    \$\begingroup\$ There's a schematic button on the editor toolbar. Hit it and add a schematic. Double-click to edit component properties. We don't do "y'all" greetings on this site. It's for Q&A and isn't the forum style you may be used to. See the Help Center and read some other posts. Welcome to EE.SE. \$\endgroup\$ – Transistor Jul 7 '16 at 23:49
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    \$\begingroup\$ What is the common mode voltage of the input signal? You are using AD620 very near its minimum supply voltage limit and you will need to have Vicm between 1.9 and 3.8 V to get normal operation. \$\endgroup\$ – The Photon Jul 8 '16 at 0:45
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    \$\begingroup\$ The AD620 needs a bipolar power supply of +/-2.3 volts to +/- 15 vdc. This is in the spec sheet for this part. Please read these fine details. They DO make a difference. \$\endgroup\$ – user105652 Jul 8 '16 at 1:59
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    \$\begingroup\$ Switch to the pin compatible AD623 \$\endgroup\$ – Scott Seidman Jul 8 '16 at 2:49
  • \$\begingroup\$ Agree with The Photon. You don't show what the common-mode voltage is in your schematic. I bet it is that. \$\endgroup\$ – Vince Patron Jul 8 '16 at 4:59
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With a single 5-volt supply, you have barely enough voltage to operate. The maximum allowed on an input is Vs - 1.2 or in your case 3.8 volts. the minimum voltage is -Vs +1.9 or 1.9 volts. Therefore your input voltage (both pins) must stay within this range. The output must be between 1.1 and 3.8 volts (2.7 volt swing) as well, so you don't have enough swing for your output p-p voltage of 2.9 volts at 1.02 volt rms.

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Here's what looks to be one of your main problems: -

enter image description here

Your circuit implies that you expect to amplify a floating source and this won't work.

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  • \$\begingroup\$ This can certainly be part of the issue, but the power scheme certainly doesn't match the chip \$\endgroup\$ – Scott Seidman Jul 8 '16 at 10:58

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