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).
<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: &quot;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.&quot; 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.

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