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.
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
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.
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.