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Figure Q1 shows the circuit diagram of a "Logarithmic Amplifier". In this circuit the feedback resistor of a normal inverting amplifier is replaced by a silicon diode. Since the current flowing through the diode increases exponentially with the voltage across the diode, the voltage across the diode will be proportional to the logarithm current flowing through it. Hence the output of the circuit is proportional to the logarithm of the input voltage. Q1 D1 VN RIN A1 VOUT RM Figure Q1 The current flowing through a diode is given by the Shockley Diode Equation: where I, is the current flowing through the diode q is the charge on the electron (given at the back of the examination sheet) Vp is the voltage across the diode K, is Boltzmann's constant (given at the back of the examination sheet) T is room temperature (given at the back of the examination sheet) Given that I,>>I; an approximate expression for V, as a function of I, is

Figure Q1 shows the circuit diagram of a "Logarithmic Amplifier". In this circuit the feedback resistor of a normal inverting amplifier is replaced by a silicon diode. Since the current flowing through the diode increases exponentially with the voltage across the diode, the voltage across the diode will be proportional to the logarithm current flowing through it. Hence the output of the circuit is proportional to the logarithm of the input voltage. Q1 D1 VN RIN A1 VOUT RM Figure Q1 The current flowing through a diode is given by the Shockley Diode Equation: where I, is the current flowing through the diode q is the charge on the electron (given at the back of the examination sheet) Vp is the voltage across the diode K, is Boltzmann's constant (given at the back of the examination sheet) T is room temperature (given at the back of the examination sheet) Given that I,>>I; an approximate expression for V, as a function of I, is

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Neglecting any input errors of Al (i.e. using a perfect opamp calculation) derive an expression for VoUT as a function of a positive value of VIN showing that the output voltage has a logarithmic dependence on Viy. State clearly any assumptions you make Q1 (a) (i) (ii) What would happen if Vy was negative? (Hint: Remember that D1 is a diode) (b) Re-draw the circuit showing all sources of DC error. From your answer to part b) derive an expression for the DC error in input voltage due to the input voltage and current errors of Al. (c) Al is one of the opamps listed in table Q1. Using the expression derived in c) choose a value of Ry and carefully explain your choice (d) Орamp Vos (uV) Ib (pA) los(pA) OPA189 300 600 ICL7652 5 30 30 ADA4530-1 40 0.02 0.02 Table QI The maximum permissible current through the diode is 1mA. Given that Vys 10V, choose a value for RN (e) Al is one of the opamps listed in Table Q1, and given your value of RM fror d), define a figure of merit for Al and hence choose the best opamp for Al.1: (f) (g) What is the ratio of the largest to the smallest input voltages which the circuit can reliably accept? (i) (ii) If Is = 0.5pA what range of output voltages does this correspond to?'!
Transcribed Image Text:Neglecting any input errors of Al (i.e. using a perfect opamp calculation) derive an expression for VoUT as a function of a positive value of VIN showing that the output voltage has a logarithmic dependence on Viy. State clearly any assumptions you make Q1 (a) (i) (ii) What would happen if Vy was negative? (Hint: Remember that D1 is a diode) (b) Re-draw the circuit showing all sources of DC error. From your answer to part b) derive an expression for the DC error in input voltage due to the input voltage and current errors of Al. (c) Al is one of the opamps listed in table Q1. Using the expression derived in c) choose a value of Ry and carefully explain your choice (d) Орamp Vos (uV) Ib (pA) los(pA) OPA189 300 600 ICL7652 5 30 30 ADA4530-1 40 0.02 0.02 Table QI The maximum permissible current through the diode is 1mA. Given that Vys 10V, choose a value for RN (e) Al is one of the opamps listed in Table Q1, and given your value of RM fror d), define a figure of merit for Al and hence choose the best opamp for Al.1: (f) (g) What is the ratio of the largest to the smallest input voltages which the circuit can reliably accept? (i) (ii) If Is = 0.5pA what range of output voltages does this correspond to?'!
Figure Q1 shows the circuit diagram of a "Logarithmic Amplifier". In this circuit the feedback resistor of a normal inverting amplifier is replaced by a silicon diode. Since the current flowing through the diode increases exponentially with the voltage across the diode, the voltage across the diode will be proportional to the logarithm current flowing through it. Hence the output of the circuit is proportional to the logarithm of the input voltage. Q1 D1 RIN A1 VOUT RM Figure Q1 The current flowing through a diode is given by the Shockley Diode Equation: =. where I, is the current flowing through the diode q is the charge on the electron (given at the back of the examination sheet) Vp is the voltage across the diode K, is Boltzmann's constant (given at the back of the examination sheet) T is room temperature (given at the back of the examination sheet) Given that I,>>I; an approximate expression for V, as a function of I, is
Transcribed Image Text:Figure Q1 shows the circuit diagram of a "Logarithmic Amplifier". In this circuit the feedback resistor of a normal inverting amplifier is replaced by a silicon diode. Since the current flowing through the diode increases exponentially with the voltage across the diode, the voltage across the diode will be proportional to the logarithm current flowing through it. Hence the output of the circuit is proportional to the logarithm of the input voltage. Q1 D1 RIN A1 VOUT RM Figure Q1 The current flowing through a diode is given by the Shockley Diode Equation: =. where I, is the current flowing through the diode q is the charge on the electron (given at the back of the examination sheet) Vp is the voltage across the diode K, is Boltzmann's constant (given at the back of the examination sheet) T is room temperature (given at the back of the examination sheet) Given that I,>>I; an approximate expression for V, as a function of I, is
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