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

AI Thread Summary
To design a 1mA AC current source using a noninverting op-amp configuration, understanding the "virtual ground" property is crucial, where the voltage at the inverting input equals the non-inverting input. The current through R1 can be calculated as iout = Vin/R1, allowing for the selection of input voltage and resistance values to achieve the desired output current. The output current remains stable as long as the op-amp does not saturate, which can occur if the load impedance is too high. The discussion highlights the importance of managing load resistance to maintain the current level. Overall, the conversation emphasizes the relationship between input voltage, resistance, and output current in op-amp circuits.
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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