Possible Adaptations:

1) Adjust R3 - You can easily produce a different current by changing the resistance of R3. The voltage drop over R3 will remain constant (due to the op amp), so the current can be calculated as before from ohms law: I = 1.36V/ R3. The upper limit of the current is determined by the power tolerance of R3. When R3 gets smaller, the current is larger, but you may burn out your resistor if you exceed the resistor’s power rating. If you want to easily adjust the current out, R3 could be replaced by a potentiometer.

2) Adjust the voltage divider - Another way to change the current is to change the voltage from the voltage divider. Adjusting R1 and R2 augments the voltage in the Ohm’s law calculation. However, the larger the voltage drop over R3, the smaller voltage drop is available over the load. Remember that your power supply a limited voltage (15 V in our circuit).

*Note that the transistor must have at least about 0.2 V drop from the collector to the emitter terminal (corresponding to the point that connects to the R3 and the point connected to the load). Without this drop, the transistor no longer functions properly and the current sags to almost 0 mA. One can see this sag begin around 1.4 k.

Possible Applications:

The current source can be useful in lighting up LEDs as they need only a small current and about a 0.6 V drop across.

1) Adjust R3 - You can easily produce a different current by changing the resistance of R3. The voltage drop over R3 will remain constant (due to the op amp), so the current can be calculated as before from ohms law: I = 1.36V/ R3. The upper limit of the current is determined by the power tolerance of R3. When R3 gets smaller, the current is larger, but you may burn out your resistor if you exceed the resistor’s power rating. If you want to easily adjust the current out, R3 could be replaced by a potentiometer.

2) Adjust the voltage divider - Another way to change the current is to change the voltage from the voltage divider. Adjusting R1 and R2 augments the voltage in the Ohm’s law calculation. However, the larger the voltage drop over R3, the smaller voltage drop is available over the load. Remember that your power supply a limited voltage (15 V in our circuit).

*Note that the transistor must have at least about 0.2 V drop from the collector to the emitter terminal (corresponding to the point that connects to the R3 and the point connected to the load). Without this drop, the transistor no longer functions properly and the current sags to almost 0 mA. One can see this sag begin around 1.4 k.

Possible Applications:

The current source can be useful in lighting up LEDs as they need only a small current and about a 0.6 V drop across.

<p>What about replacing PNP with NPN? I think it'll not make any difference...</p>

<p>I guess I'm a little late to the party, but:</p><p>The sentence: "Connect a load of your choosing with a resistance of less than 1.4 k to the emitter terminal of the transistor (the nonconnected terminal) and to ground as shown in the circuit diagram." misstates the transistor terminal as being the emitter, when in fact it is the collector. The choice of the op amp is not arbitrary, since the inverting input is held near the V+ rail and for many op amps (esp. older ones) the rails are off limits and can result in destruction at worst and invalid operation at best.</p>

<p>Hi ,can this be considered constant current and constant voltage ?</p>

<p>Very interesting. I wonder if it could be possible to decouple the power source needed by the OA and R3 so R3 and R1 would not share the same power supply. Why? Because I have a power supply sourcing the load that may vary between 0.8V and 9V but I want to control the current going thru R3 (as in the schematic) to a constant current value, not limiting the current but keeping it constant. So one PS (stable) would source R1 for the OA and another one (could vary from 0.8V to 9V) would feed R3 for the load.</p>

So is the op amp and the transistor mirroring the current the voltage divider does? I do not have any LF411 Operational Amplifiers but I do have a number of TL082 ICs that are used in the schematics. I may have to breadboard this circuit up using one of them to try to better understand what is happening here.<br> <br> If I do I'll be sure to use my dual adjustable supply.<br> <br> <a href="https://www.instructables.com/id/Dual-POS-NEG-Power-Supply/" rel="nofollow">https://www.instructables.com/id/Dual-POS-NEG-Power-Supply/</a><br> <br> Which I think works well for op amp experimenting.

The main purposes of the op amp are 1) to provide a stiff voltage (that is not effected by the load's resistance, as it would be if it were just a voltage divider). And 2) set the voltage drop over R3, so that the current is fixed. Using another power supply instead of a voltage divider for the purpose 1 would definitely work. I'm not sure how you could replicate the 2) purpose of the op amp without getting some current sag for large loads. Is that what you were asking?

No. I was asking if the op amps were using the divider as a voltage reference.

This current source has the nice feature of a grounded output. It has the disadvantage of being only as good as the 15 volt rail is regulated. Separate regulation for the input to the voltage divider might be a way to address this.

Thanks russ_hensel. That is a good augmentation idea.