Laser diodes are quite useful in some ways especially in trigger/alarm systems such as triplight mechanisms or even in DIY toys if you just want to play with it.
These two terminal devices are generally much more expensive than LEDs although technically speaking, they are both light emitting diodes. They just emit light in different ways. In some microcontroller projects, it can be used as an input/trigger to a photoresistor/phototransistor and in some cases, one may opt to use a supply that is not directly connected to the other components especially some delicate ICs or microcontrollers that cannot provide enough supply current.
In this instructable, we are going to make a very simple module for any laser diode with a maximum forward voltage of about 4.5 Volts, supply it with a maximum of 20mA current (just like a typical LED) from an unregulated DC supply of 6V-18V or battery.
For questions, concerns, clarifications, corrections etc.
email me at: email@example.com ***************************************************************************************************
Step 1: Know Your Diodes(LED, Zener and Laser Diodes)
Laser diodes come in different specifications just like LEDs and are generally defined by their dissipated power and forward voltages. In this case, we are limiting our scope to laser diodes of up to 4.5V forward voltage(sometimes called the turn-on voltage) and the reason will be explained later on as we discuss the simple schematic for our module.
Why create a module? Well, just like LEDs, laser diodes when supplied with too much voltage and current, DO BLOW UP as well as other things related to electronics. The goal is pretty simple. We just need to provide what is necessary for the laser diode to function properly and can last a long time.
Diodes in general has a turn-on voltage that we must supply before conducting current and ideally, this turn-on voltage remains constant regardless of the supply current. Practically, we just need to provide a supply voltage equal to its forward voltage to turn it on and the supply current... well its a different kind of story.
In common diodes(those black and white resistor like devices), the supply current can be neglected in analysis since they are generally used as rectifiers... and we don't generally mind what is the actual amount of current passing through it as long as we don't exceed the maximum current specifications. Meanwhile, in zener diodes, we sometimes need to minimize the current supplied through it since it wastes a lot more power as compared to the common ones. Why? because we are usually taking advantage of its breakdown reverse biased voltage rather than its forward voltage. Zener diodes have typical breakdown voltages of 5 Volts and above and breakdown voltages are pretty much the most important data you want to find in its datasheet (ideally, breakdown voltages across a zener diode remain constant regardless of the circuit configuration). Having the need to supply a much higher voltage for a zener diode means a higher power dissipation for a given amount of supplied current.
Lastly, LEDs and Laser diodes are a little bit interesting (this time we may assume that LEDs and laser diodes are the same thing). Aside from a forward voltage requirement, they also require enough current to light up properly (not too dim and not too bright as they may heat up). Typical LEDs require 20mA of current to properly light up (don't ask why... a lot of youtube videos that i've watched tell the same thing and looking at some LED datasheets can confirm this). That is why we often hear the term current limiting resistors when creating a circuit that has LED components. For example, when we want a 9 Volt battery to power our LED with a forward voltage of let's say 2V, we don't connect them directly in parallel but rather, we use a current limiting resistor. The value of the resistor is calculated as follows:
R = (9V - 2V)/20mA = 350 ohms [this makes the current passing through the circuit exactly equal to 20mA]
Increasing the resistor value would then decrease the supplied current thus less brightness for the LED output. The reverse is true but mat heat up the LED. If the LED is defined by its maximum power consumption, then the ideal supply current can be calculated by dividing the maximum power to its foward voltage.
Step 2: Module Design
Creating a simple module means we SHOULD NOT only cater one specific model of a laser diode but our module must be able to drive a wide variety of different laser diode specifications available in the the market. Here we have to keep in mind that we have to supply the necessary voltage and currents to properly operate the diode. In this case, i assumed that typical supply currents for the laser diode is 20mA. This also depends on the laser diode already available at hand when making this project. For example you might want to test your diode first if what value of supply current would decently light it up given a supply voltage equal to its forward voltage or you may look directly at its datasheet. In my case, I have a 3.3V laser diode and i didn't have a copy of the datasheet at hand therefore i tested it by initially supplying a 10mA current using a current limiting resistor (potentiometer) from a 9 Volt DC source. Well, it didn't seem to be a right idea since the output light was very dim. I increased the current supplied to 20mA and it lighted up just good where i can see a solid red dot as the output. I left it on for about an hour. It doesn't seem to heat up and thats when i decided to supply a constant 20mA current for the module that I wanted to make.
In the circuit schematic as shown in the image above, i used a very simple current mirror transistor configuration to supply the needed 20mA current. Ideally, transistors Q1 and Q2 should be identical. We all know that the collector current(Ic) of the transistor is dependent on its Vbe(base emitter). So if two identical transistors have the same Vbe, then they both have the same collector current. We can then say that the Ic on the left side of the transistor configuration is mirrored to the right side.
Transistor Q1 is diode connected and in series with a current limiting resistor. So if we want to have a 20mA current at the right side(Ic of transistor Q2) of the configuration to supply the diode, we just need to adjust the current on the left side to 20mA. In order to adjust the current on Q1, we need to calculate the needed value for the resistor. We then assume that the voltage across the diode connected transistor(Q1) is equal to its Vbe(typically 0.7 Volts). So by KVL, we have:
Vee = 0.7 + RIc
where Vee is the voltage at the emitter of both Q1 and Q2. In this case, I am using a 5V linear regulator to supply the transistor configuration therefore Vee = 5V. We want Ic to be equal to 20mA.
Therefore R = 215 ohms
I didn't have an available 215 ohm resistor so i decided to replace it with a 220 ohm resistor. The supply current would then be: Ic = (Vee - 0.7)/220 = 19.545455mA. A little bit less than 20mA but it doesn't give a noticeable decrease in brightness.
I mentioned earlier that we can only cater diodes with up to 4.5V as forward voltage. This is mainly because I am using a 5Volt regulator prior to the transistor stage. Connecting a laser diode with a 5V forward voltage may force the transistor Q2 to operate at saturation since there is no more voltage left for its Vec. So a maximum of 4.5V forward voltage for a diode ensures that the transistors are in forward active region and current adjustments are still possible. If you want to use diodes with much higher forward voltages, you may replace the regulator with a 9V one, a 12 Volt one or even a variable version like the LM317. You just have to calculate the value of the resistor over again.
We can supply a battery from 6V - 18V at the input of the regulator but personally, i would limit it to 9V as the regulator heats up faster with higher supply voltage. In that case, you may use a heatsink to dissipate the heat.
Step 3: Wire the Circuit
We have the following needed materials/tools:
- 2pcs 10uF capacitor 16V
- Laser diode - LM7805 linear regulator
- 2pcs 2N3906 PNP transistor
- 220 ohm resistor
- small pcb board
- soldering iron
Construct the circuit as shown in the design(or in your modified design). A breaboard is essential if you want to test it first and later, solder it in a small pcb if you like.
Step 4: Testing the Circuit
Here I am using an old set of 1.5V batteries cells, 6 all in all to form a 9V battery.
Measured parameters are shown in the image above.