Question

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

Answer 1

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.