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FrozenGate by Avery

Clarification on DDL Drivers

Joined
Apr 4, 2016
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Hello Everyone,

I've been researching the DDL laser driver and would like some help in understanding the circuit. I intent to use it to build a 690nm laser with a 10mW output (I'd like to start small and happen to have this diode around), preferably using 5 volts. I figured I'd start by stating what I do understand before going onto my question. If I'm posting this in the wrong forum I apologize. I'm new here and the amount of content is overwhelming.

Concerning the components of the DDL circuit attached below, I understand that:

-The diode is to protect the laser diode in case the batteries are placed in reverse
-The Capacitor is to smooth out the current going into the laser diode
-The potentiometer is to adjust the resistance of the system ( the greater the resistance, the less the current).

Additionally, I understand that you can test the current output by using a dummy load instead of the laser driver if you do not want to burn it (other picture). the diodes are chosen to have a voltage drop close or at the operating voltage of the laser diode. The 1-ohm resistor is to use Ohm's law to figure out the current flowing thru it.

My questions are:


-Is there a calculation for determining the capacitance to use?
-Why do we have fixed resistors if we have a potentiometer?
-How do we calculate the fixed resistor values?
-When choosing a voltage to supply to the LM317, is it 3V + Operating voltage of the laser driver? I'm not sure why 3V, but from what I can understand the LM317 requires 3V.

Again, I apologize if I've done something wrong and thank you for your time.
 

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-Is there a calculation for determining the capacitance to use?

The lm317 datasheet (ask google) will give some recommendations. Anything between 0.1 and 10uF would be fine though. This value is not critical.

-Why do we have fixed resistors if we have a potentiometer?

To set a minimum resistance, and therefore a maximum current. This prevents over-current from turning the pot too far.

-How do we calculate the fixed resistor values?

As mentioned above, this depends on what you want the maximum current to be. Ohm's law again. Vref in this case is held at the reference voltage of 1.25V, so 1.25V/(resistance)=current.

-When choosing a voltage to supply to the LM317, is it 3V + Operating voltage of the laser driver? I'm not sure why 3V, but from what I can understand the LM317 requires 3V.

Not quite. The LM317, like all linear regulators, has what is called a minimum dropout voltage. That is to say the supply voltage needs to exceed the load voltage by at least this value. The minimum dropout voltage depends on temperature, CC/CV operation, drive current, and internal pass-transistor topology. For most cases using the lm317 as a constant current driver, you'd want at least 3.5V above the load voltage. Again, the datasheet will give some more precise values. A red laser diode is typically 2.5-3V or so.

This leads me to MY question. Why 5V drive? That clearly won't work with this driver.

If the 5V is a regulated 5V, it would be advisable to simply use a current limiting resistor instead of a driver, given that this is low power. This would allow operation from 5V and obviously greatly reduce the complexity of the circuit.
 
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The actual required dropout of this circuit is quite high: Per the datasheet 3 volts between input and output, PLUS the 1.25 volts across the resistor between out and adjust, so a total drop of 4.25 volts to operate within specified parameters.

It will -usually- work with quite a bit less, but that is by no means guaranteed.

In all honesty i don't think this is a good design for practial application anymore. DDL obviously designed it to make use of a component available from any electronics store which is convenient, but performance wise it's not attractive to use any longer on todays batteries and laser diodes.
 
I absolutely agree - there are substantially better driver circuits out there. However, it's hard to beat this one for simplicity, and it's not such a bad beginner's project.

Per the datasheet 3 volts between input and output

That's after adding a safety margin. Look at page 9 here and you'll see it's less than 2.25V the vast majority of cases. 2.25+1.25 is 3.5. It would be under 2V (+1.25) for the low current OP would potentially be using.
 
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Thank You for replying everyone. It's been very informative and insightful.

To answer Cyp, I would like to power it from the 5 volt source on my Arduino. I'm interested in exploring the idea of using the arduino to control the laser. As such I'd like to see if I can get 35mA of current to power one of my more low powered laser diodes using 5 volts.

Would anyone have any recommendations of what type of driver I should look into then?
 


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