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I come from a non-electronics background and I'm interested in understanding the operation of the following circuit configuration. The drawing is very crude, but this is the representation that I have available:

enter image description here

I do recognize some elements, like the resistance and the ground (lower pointing triangle.) The left-pointing triangle seems to be an op-amp from a quick search in Google. Besides the translation, I'd like to have a clear idea of how this circuit would work.

Could you please help me out decoding this?

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    \$\begingroup\$ Where did you find this? \$\endgroup\$
    – user253751
    Oct 11, 2022 at 11:08
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    \$\begingroup\$ The circuit image in the question makes little sense to me. \$\endgroup\$
    – Andy aka
    Oct 11, 2022 at 13:04
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    \$\begingroup\$ Not enough context to suggest an application of this circuit. But "V" must be an input, which by default, makes "I" an output. The left-pointing triangle might indeed be a standard opamp. It might also be a transconductance-type amplifier (which has voltage inputs, and current output). \$\endgroup\$
    – glen_geek
    Oct 11, 2022 at 13:39

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It would have been nice if you had labelled the circuit. If you are new to this, generally you label nodes at the intersection of wires connecting different components.

Here's how the circuit would work. Assume voltage at non-inverting lower terminal of OP-amp be \$V_a\$ and at the upper terminal \$V_b\$. \$V_b\$ is connected to ground so current will flow from current source I through the resistor R.

Attached is a circuit diagram assuming some values for V,I and R. You have to rotate this circuit 180 degrees to get your circuit diagram.

enter image description here

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  • \$\begingroup\$ Thanks for your answer. Could you mind explaining the use of the resistance in this type of circuit? \$\endgroup\$ Oct 17, 2022 at 14:16
  • \$\begingroup\$ This circuit is peculiar because current flowing in resistor R is not I. if you increase I in your diagram, the current flowing in the reisistor R is not changed , it is somewhat lower in mA range. This circuit seems to give constant current in resistor R with variation in V and I so it can be used as current source or a crude current regulator. This behavior has something to with op amp characteristics. \$\endgroup\$
    – Amit M
    Oct 17, 2022 at 15:25

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