The circuit shown in the paper can provide a maximum current of only 10 mA. There are also limitations in the minimum and maximum voltage output.
I made a similar circuit many years ago. It stopped working due cracked wires. I completely disassembled the circuit and made from scratch but with the same operational amplifier IC and the same IC socket. I usually use wire wrap. However, the IC socket had short legs because its not a wire wrap socket. Thus I had to user a soldering iron. I also used a 67 kohm potentiometer from an old VCR (video cassette recorder).
You do not need to user a 14 pin IC socket that I used. You can purchase an 8 pin wire wrap IC socket.
You can see the circuit operation in this video:
Designing the Circuit
I have drawn the circuit in https://easyeda.com online software.
The circuit below is showing a simple opamp voltage follower configuration with a very high input resistance.
Rv can be of any potentiometer value from 10 kohms to 1 Megohms. Reducing Rv to just 1 kohm will a reasonable amount of current from power source and will waste power. However, there is still a small amount of power consumed by the operational amplifier IC even when the output current is zero.
The 4.5 V power source is the minimum that you can supply to a general purpose opamp.
If the power supply is raised to 15 V and opamp saturation voltage is 3 V, the maximum opamp output voltage will be 12 V.
The maximum LED current will equal to:
IledMax = (Vo - Vled) / Rd = (12 V - 2 V) / 1000 ohms
= 10 V / 1000 ohms = 10 mA
Ro is used for short circuit protection to prevent damage to the opamp IC.
The maximum opamp output current at 12 V output will equal to:
IopampMax = Io + Iled = 12 V / 1000 ohms + 10 mA = 22 mA
This is twice the maximum opamp output current. However, this occurs only when the output is shorted and Rl is zero.
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