Transistors Q1 and Q2 are identical and β >> 1 in the circuit shown in the figure below. The output voltage is (Vt = 0.026 V):

\(2.3 \ {\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)

\(2 \ {\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)

\({\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)

\(4.6 \ {\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)

Transistors Q1 and Q2 are identical and β >> 1 in the circuit shown

1.	Transistors Q1 and Q2 are identical and β >> 1 in the circuit shown in the figure below. The output voltage is (Vt = 0.026 V):
A.	\(2 \ {\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)
B.	\({\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)
C.	\(2.3 \ {\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)
D.	\(4.6 \ {\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)
Answer» C. \(2.3 \ {\log _{10}}\left( {\frac{{{\upsilon _2}}}{{{\upsilon _1}}}\frac{{{R_1}}}{{{R_2}}}} \right)\)