Constant Current Source With Operational Amplifier




This instructable will show you how to make a current source with negligible current sag for loads with a resistance of less than 1.4k. In our example using a power supply, the current provided is 9.1 mA but can be adjusted. The advantage of a current source is that it provides a constant amount of current that is independent of the load’s resistance. Further, this circuit has the load grounded, which eliminates the danger of having two exposed connections with floating voltages.

Materials Used:
One LF411 8 pin Operational Amplifier (Op Amp) ~$4
One PNP transistor model 2N3609 ~$2
Three resistors of values 10k, 1k, 150Ω  <$1
A load of your choosing (3 LEDs in the above picture)
Power Supply. In this example, we used a +15 V and -15V commercial power supply.*

*Note: We were also able to power this circuit with two 9V batteries (one for each the + and the -), but circuit produced a current of only 5.4mA with a similar maximum load. Also, the negative terminal battery became very warm quickly, so the power supply is highly recommended. Power supplies can be made by converting an ATX power supply from an old computer such as this instructable by abizar.

Step 1: Assible the Circuit

Assemble materials and connect as shown in the circuit diagram. Note that this assembling is easiest using a breadboard.

The Op Amp has 8 pins; the first pin is designated by a small, indented circle. The rest of the pins are enumerated around counter-clockwise. The connections shown to the V- terminal is pin 2, V+ is pin 3, and the point of the triangle connected to the transistor is pin 6 . In addition, pin 4 should be connected to -15V and pin 7 to +15V (not shown in circuit diagram).

Step 2: Connect a Load

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. As long as the load’s resistance meet the requirements, 9.1 mA will flow through the load.

Congratulations. You have built a current source.

Step 3: How It Works:

This circuit achieves its constant current through the use of the Op Amp and the transistor. The first two resistors (R1 and R2) are used as a voltage divider. The voltage between them is given by
Vout = Vin * R2R1 + R2
where Vin is from the power supply. From here, the Op Amp has negative feed back (ie, the output is connected back to the input), and balances the V+ and V- terminals to approximately the same potential. In other words, the voltage at the V- terminal will also be the Vout from the voltage divider. Thus, the voltage drop across R3 is found to be the difference of 15 - 13.64 = 1.36V. This voltage drop produces a current given by Ohm’s law of 1.36V / 150Ω, giving a current of 9.1 mA. Even with the addition of a negligible amount of current from the output of the Op Amp (a factor of 100 smaller), the current is constant through the transistor, so the output to the load is 9.1 mA.

Step 4: More

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.



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    13 Discussions


    Reply 4 months ago

    Hi... I think you can replce the R1,R2 divider with a non-inverting amplifier whose input is a DAC output voltage.


    9 months ago

    The formula for Vout is wrong, as it misses a division sign ("/")


    1 year ago

    can i make it with IC 741 OP-amp


    2 years ago

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


    2 years ago

    I guess I'm a little late to the party, but:

    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.


    4 years ago on Step 4

    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.


    6 years ago on Introduction

    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.

    If I do I'll be sure to use my dual adjustable supply.

    Which I think works well for op amp experimenting.

    2 replies

    Reply 6 years ago on Introduction

    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?


    Reply 6 years ago on Introduction

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


    6 years ago on Introduction

    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.

    1 reply