IMHO, since current doesn't flow into the op amp, the best way is to look at it as a voltage divider formed by Ri and Rf. The middle point and Vi are fixed, and Vout is allowed to vary. Current flows through it in one direction. Therefore, when Vi is below ground, Vout is above ground. And when Vi is above ground, Vout is below ground.
What I'm trying to understand is that why Vout decreases as the resistance of photoconductive cell increases, and why Vout increases as the resistance of photoconductive cell decreases. I'm looking at it in intuitive terms. Thank you.
Hi
Please use this attachment and please help me. Thank you.
An op-amp tries to make differential voltage across its input terminals zero which also means that no input current flows into the op-amp.
In this case of Figure 12-22, it's easier to understand what is really happening. As Vout = (-Rf/Ri)Vin, therefore when value of Ri is low, the Vout is made more negative so that 'positive charge carriers', i.e. conventional current, are pulled away from the inverting input terminal of the op-amp with greater force and no charge carriers enter into the op-amp.
Q: The circuit on the left uses inverting configuration whose gain is given as Vout/Vin = -Rf/Ri.
In the given case Vin is negative therefore Vout = (Rf/Ri)|Vin|.
I'm assuming that the relation for Vout given above is correct. When the resistance of photoconductive cell decreases (which in this case is Ri), the Vout increases. Conceptually, I don't get why an op-amp would increase its Vout as Ri decreases.
Likewise, when Ri increases, Vout will be decreased. Why?
Regards
PG
The op amp wants to maintain 3V voltage at the red point.
Do you think my connections are alright? Kindly let me know.
I'm assuming the connections are alright. The current can flow along two different paths from the yellow point, either toward the red point or toward the ground (or, black point). I think that the current gets divided to flow along two different paths at any time. In other words, some of the current flows from yellow point toward the black point and some of the current flows from yellow point toward the red point (or blue point). Do I have it correct?
That configuration will work, but it is a bit clunky.I believe this is the way I would need to power up a dual polarity op-amp. Is this correct? In other words, a dual-polarity power supply can be made using two 7805 regulators.
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