you are right , A1 A2 are voltage follower, so their inverting terminal should get shorted with output terminal and we give input signal to non-inverting terminal... however those two buffer diagram plus minus sign should get altered , then it's ok
In the circuit of Figure 1, R1 = 1 kΩ, RF =4.7 kΩ, RA=RB=RC=100 kΩ, Vdc = +5 V, and op-amp supply voltages = 15V. The transducer is a thermistor with the following specifications: RT= 100 kΩ at a reference temperature of 25⁰C, temperature coefficient of resistance = -1 kΩ/⁰C. Determine the output voltage at 0⁰C and at 100⁰C. MAM PLZ SOLVE THIS FOR ME IT IS VERY IMP FOR ME
Thanks for the information mam, even if the unity gain amplifier are wrongs your explanation is absolutely easy to understand.
Awesome explanation mam
Thanks a lot
excellent video very informative and helpful
Glad you liked it
i think that unity feedback should have negative feedbacks?? and also Vab=Va-Vb(by standard books)
Why is A1 and A2 connected as positive feedback?
non inverting configuration should be used its mistake of drawing figure
Vab=va-vb or vb-va
@@sindhujagalam3007 By definition must be: Vab = Va-Vb
you are right , A1 A2 are voltage follower, so their inverting terminal should get shorted with output terminal and we give input signal to non-inverting terminal...
however those two buffer diagram plus minus sign should get altered , then it's ok
what if i place ldr at place of thermister ???
Very nice mam
Buffer stage input terminals must be swapped.
Ghub bhalo
In the circuit of Figure 1, R1 = 1 kΩ, RF =4.7 kΩ, RA=RB=RC=100 kΩ, Vdc = +5 V, and
op-amp supply voltages =
15V. The transducer is a thermistor with the following
specifications:
RT= 100 kΩ at a reference temperature of 25⁰C, temperature coefficient of resistance = -1
kΩ/⁰C. Determine the output voltage at 0⁰C and at 100⁰C.
MAM PLZ SOLVE THIS FOR ME IT IS VERY IMP FOR ME
do you have the solution?
schematic is wrong. Positive feedback will result in amplifier overflow...