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The circuit you've provided is indeed a Howland Current Source. It's designed to provide a constant current output (iout) relatively independent of the load resistance (RL). The equation for the output current is: iout = Vin/R1 Given that R1 = R2 = 10kQ: 1. Vin=2V: iout 2V/10kQ = 0.2 MA = 2. Vin = -2V: iout -2V / 10kQ = -0.2 MA = Therefore, for the given values of R1 and R2: When Vin = 2V, iout = 0.2 MA •When Vin = -2V jout = -0.2 mA. As you can see, the output current changes. direction when the input voltage changes polarity. but its magnitude remains the same. This is a characteristic of a well-designed Howland current source. R1 Vin R₁ R₂ V₁ + RL R₂ |iout= Vin/R1 ол

The circuit you've provided is indeed a Howland Current Source. It's designed to provide a constant current output (iout) relatively independent of the load resistance (RL). The equation for the output current is: iout = Vin/R1 Given that R1 = R2 = 10kQ: 1. Vin=2V: iout 2V/10kQ = 0.2 MA = 2. Vin = -2V: iout -2V / 10kQ = -0.2 MA = Therefore, for the given values of R1 and R2: When Vin = 2V, iout = 0.2 MA •When Vin = -2V jout = -0.2 mA. As you can see, the output current changes. direction when the input voltage changes polarity. but its magnitude remains the same. This is a characteristic of a well-designed Howland current source. R1 Vin R₁ R₂ V₁ + RL R₂ |iout= Vin/R1 ол

Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter16: Inductance In Ac Circuits
Section: Chapter Questions
Problem 1PP: Inductive Circuits Fill in all the missing values. Refer to the following formulas:...
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  1. If you replaced RL in the current source circuit with a 0.1μF capacitor and passed a current iout through it equal to the values calculated in part Attached, what slope (dv/dt) would you expect the capacitor voltage to have in each case?

 

The circuit you've provided is indeed a Howland Current Source. It's designed to provide a constant current output (iout) relatively independent of the load resistance (RL). The equation for the output current is: iout = Vin/R1 Given that R1 = R2 = 10kQ: 1. Vin=2V: iout 2V/10kQ = 0.2 MA = 2. Vin = -2V: iout -2V / 10kQ = -0.2 MA = Therefore, for the given values of R1 and R2: When Vin = 2V, iout = 0.2 MA •When Vin = -2V jout = -0.2 mA. As you can see, the output current changes. direction when the input voltage changes polarity. but its magnitude remains the same. This is a characteristic of a well-designed Howland current source.
Transcribed Image Text:The circuit you've provided is indeed a Howland Current Source. It's designed to provide a constant current output (iout) relatively independent of the load resistance (RL). The equation for the output current is: iout = Vin/R1 Given that R1 = R2 = 10kQ: 1. Vin=2V: iout 2V/10kQ = 0.2 MA = 2. Vin = -2V: iout -2V / 10kQ = -0.2 MA = Therefore, for the given values of R1 and R2: When Vin = 2V, iout = 0.2 MA •When Vin = -2V jout = -0.2 mA. As you can see, the output current changes. direction when the input voltage changes polarity. but its magnitude remains the same. This is a characteristic of a well-designed Howland current source.
R1 Vin R₁ R₂ V₁ + RL R₂ |iout= Vin/R1 ол
Transcribed Image Text:R1 Vin R₁ R₂ V₁ + RL R₂ |iout= Vin/R1 ол
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