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I tried the circuit shown below on breadboard and found that when I connect both the IC's input pin to the node 1, the op amp output is giving a different value. I believe this is due to the imbalanced input resistance on the IC (U3 & U6) pin. The AD736 has a very high input impedance. Is there a way to isolate the loads of the op amp from each other so that the correct voltage will be seen at each input pin of IC.

Edited circuit enter image description here

Original circuit enter image description here

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  • \$\begingroup\$ What is the box labelled "AD8302" supposed to be? It doesn't have enough pins to actually be an AD8302. Why is its output connected directly to its input? \$\endgroup\$ – The Photon Aug 31 '16 at 16:36
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    \$\begingroup\$ Your problem is probably connecting to the "AD8302" output, which is probably not high impedance. What problem is your circuit actually supposed to solve? \$\endgroup\$ – The Photon Aug 31 '16 at 16:39
  • \$\begingroup\$ U3 is improperly connected for a non inverting buffer. It's output is shorted to U2 out and seems pointless. \$\endgroup\$ – Sunnyskyguy EE75 Aug 31 '16 at 16:55
  • \$\begingroup\$ @ThePhoton. Im sorry for the confusion. I have edited the question to make it easier for the discussion. Right now,I'm giving a sin input voltage through U2 and the output of U2 is fed into U3 and U6. U6 is used to measure the DC voltage of the U2 output. And, U3 is another inverting unity-gain amplifier to present the actual input voltage. The problem is, when i connect the output of U2 to both the input pin of U3 & U6, the output value of U2 is changing and not the desired value. Im looking to isolate the loads of the U2 to avoid this problem. \$\endgroup\$ – rajk Aug 31 '16 at 17:18
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    \$\begingroup\$ Don't edit the question like you have because now you make the comments and answer seem stupid. Do you want people to feel that way when they are trying to help you? Add the corrected circuit as an addendum and offer some apologies for screwing up. \$\endgroup\$ – Andy aka Aug 31 '16 at 17:20
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U3 is improperly connected for a non inverting buffer. It's output is shorted to U2 out and seems pointless.

You can disconnect these outputs from U2out to U3-in, but what you do with U3out is unknown.

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