Creating a 1mA AC Current Source with Noninverting Op-Amp Configuration

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Discussion Overview

The discussion revolves around designing a circuit using a noninverting op-amp configuration to create a 1mA AC current source. Participants explore the theoretical and practical aspects of implementing this design, including the implications of load resistance on current output.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Exploratory

Main Points Raised

  • Some participants express uncertainty about how to start the design process for the current source.
  • Questions arise regarding the "virtual ground" property of op-amps and its application in this context.
  • Concerns are raised about how external load resistance affects the current output, with some participants noting that it complicates achieving a constant current.
  • One participant identifies that the current flowing into the op-amp inputs is essentially zero, which is a key point in understanding the circuit behavior.
  • Another participant confirms that the voltage at the negative input will equal the voltage at the positive input, leading to a relationship between input voltage and output current through resistors.
  • Participants discuss specific resistor values and input voltages to achieve the desired output current, while noting that the output current is dependent on load impedance.
  • There is acknowledgment that if the load impedance is too high, the op-amp may saturate, affecting the output current.

Areas of Agreement / Disagreement

Participants generally agree on the principles of op-amp operation and the relationship between input voltage and output current. However, there remains uncertainty regarding how to maintain a constant output current independent of load resistance, indicating that the discussion is not fully resolved.

Contextual Notes

Participants mention the limitations of the reference book used, indicating that it does not provide sufficient clarity on using op-amps as current sources. There is also an implicit assumption that the op-amp will not saturate under normal operating conditions.

hogrampage
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Homework Statement


Design a circuit based on the noninverting op-amp configuration that functions as a 1mA ac current source.

Homework Equations


Gain = Vout/Vin = 1 + R2/R1

The Attempt at a Solution


I really don't know how to start.
 

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hogrampage said:

Homework Statement


Design a circuit based on the noninverting op-amp configuration that functions as a 1mA ac current source.

Homework Equations


Gain = Vout/Vin = 1 + R2/R1

The Attempt at a Solution


I really don't know how to start.

What is the "virtual ground" property of opamps? And how can you use that to help in this problem?
 
Well, the current flowing into the +/- inputs is essentially 0. I understand the basics of the op amp (how to derive gain, vout, etc.), but I have never encountered a problem asking to use one as a current source. And, the book (Microelectronic Circuits & Devices, 2nd, Horenstein) is very vague, unfortunately.

If I have a different resistance connected to Vout (not R1 or R2), then it will change the current, won't it? That's where I am getting lost. I don't know how to make the current independent of whatever load resistance is connected.
 
hogrampage said:
Well, the current flowing into the +/- inputs is essentially 0. I understand the basics of the op amp (how to derive gain, vout, etc.), but I have never encountered a problem asking to use one as a current source. And, the book (Microelectronic Circuits & Devices, 2nd, Horenstein)is very vague, unfortunately.

If I have a different resistance connected to Vout (not R1 or R2), then it will change the current, won't it? That's where I am getting lost. I don't know how to make the current independent of whatever load resistance is connected.

The property you mention is not the "virtual ground" property of opamps. Try again?
 
V+ = v-.
 
hogrampage said:
V+ = v-.

Bingo!

So what will the voltage at the - input always be? And so what will the current be through R1? And since the current into the - input is zero as you already pointed out, what will the output current be?
 
iout = iR1 = iR2 = \frac{Vin}{R1}

So, I could choose Vin = 1V, R1 = 1kΩ, and R2 = whatever?
 
Last edited:
hogrampage said:
iout = iR1 = iR2 = \frac{Vin}{R1}

So, I could choose Vin = 1V, R1 = 1kΩ, and R2 = whatever?

Yep!

The Vout will depend on the load impedance. If the load impedance is too large, the output of the opamp will saturate (like if Rload = 10kOhms, and the supplies to the opamp are +/-5V). But as long as the opamp doesn't saturate, the output current will be 1mApp.
 
Ah, okay. That makes sense.

Thank you so much for your help! I really appreciate it.
 
  • #10
All I did was ask questions... :smile:
 

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