Hi to you guys, While i was perusing in an Albert Malvino's textbook to learn more about integrators,i stumbled over a problem(see photo) in the troubleshooting section in page 508 of the book. The question reads like this.-Somebody mistakenly reverses the battery of figure 16-27b.What happens? I describe the circuit in case the photo appear confusing. The circuit is a voltage follower with a 50 ohm resistance,1 milliamp sensitivity full scale connected in the output of the opamp(presumely a 741)and fed back to the inverting input of the op amp.There is also a thermistor connected inbetween the inverting input and ground.On the other hand,a 10 volt battery drives the non inverting input with the positive tag connected to that same non inv.inputand the negative to ground.What happens if the battery is connected the wrong way round?Awaiting your comments thanks!
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Casse-tête!! Malvino has *many* textbooks! p508 in which one?
You keep saying "integrator", here and in another thread. But you seem to be wrestling with a voltage-to-current converter, what Malvino calls "Voltage-Controlled Current Source".
Anyway, if the battery is backward, the thermistor is bi-directional, and the opamp is powered to swing both ways, the meter swings backward and hits the zero-peg.
You could predict this from the base equation of the VCCS. I=V/R. If V is negative, I is negative.
{for non-French folks: "casse-tête" is 'jigsaw puzzle, brainteaser, difficult problem, headache'.}
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