# Consider the amplifier is nulled at 27^oC. Calculate the output voltage , if the input voltage is 6.21mv dc at 50^oC. Assume LM307 op-amp with specification: △Vio/△T=30µV/^oC ; △Iio/△T = 300pA/^oC; VS =±15v.

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Consider the amplifier is nulled at 27^oC. Calculate the output voltage , if the input voltage is 6.21mv dc at 50^oC. Assume LM307 op-amp with specification: △Vio/△T=30µV/^oC ; △Iio/△T = 300pA/^oC; VS =±15v.

(a) +0.53v or -0.68v

(b) +0.52v or -0.78v

(c) +0.54v or -0.90v

(d) +0.51v or -0.86v

I have been asked this question during an internship interview.

The doubt is from Thermal Drift topic in division Operational Amplifier Fundamentals of Linear Integrated Circuits

by (39.8k points)
The accurate answer is (d) +0.51v or -0.86v

Delve into insight with: Change in temperature △T = 50^oC-27^oC = 23^oC.

=> Error voltage, Ev =[1+(RF/R1)]×(△Vio/△T)×△T + RF×(△Iio/△T)×△T = [1+(100kΩ/1kΩ)]×(30µv/1^oC)× 23^oC + 100kΩ×(300pA/1^oC)× 23^oC = 0.06969+ 6.9×10^-9

=> Ev= 0.0704 = 70.4mv.

For an input voltage of 6.21mv dc, the output voltage,

Vo=-(RF/R1)×Vin±Ev = -(100kΩ/1kΩ)×6.21mv±70.4mv = +0.69v or -0.55v.

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