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Old 12-25-2011, 10:03 PM #1
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Default Can someone explain driver voltage?

So as simple as this is, I searched, and couldnt find and info that seemed to make sense.

So the diode I am talking about here is the 10mW 635nm here which has a voltage range of 2.3-2.6 volts

I am planning on using a rckstr driver which has a voltage drop of 2.25 volts.

But this would require a range of 4.55-4.85 volts, which would require a LM317 or similar circuit, just to get the right voltage to the driver.

DTR says

Quote:
Originally Posted by DTR View Post
Yep you need to look up how to set the driver on a test load or with the P3 I believe you can test the current on the output side via a multimeter as well. You don't need to worry about voltage to the diode. The only voltage you need to worry about is that your power source can supply more than 6V to the driver.

Did you try a search for how to set the current on the P3?

That seems odd, can someone please clarify?


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Old 12-25-2011, 10:36 PM #2
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Default Re: Can someone explain driver voltage?

Quote:
Originally Posted by Jacob32123 View Post
I am planning on using a rckstr driver which has a minimum voltage drop of 2.25 volts.
fixed
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Old 12-25-2011, 10:48 PM #3
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Default Re: Can someone explain driver voltage?

Put more work in it, Alex. Or at least sarcasm or few bad jokes. It ain't you.

Jacob,

You should not worry about the voltage of the diode.
Drivers are constant current sources, which means that they will provide exact current to the diode, and voltage will vary - it will try to supply as much voltage as the diode requires at given current.

In other words, you set current - voltage magically sorts itself out.

If your diode requires 30 mA (I'm guessing, no idea), then you set your resistance between OUT and ADJ pins of your DDL driver to corresponding value to get the needed current, and power the driver with enough voltage to cover both diode's voltage drop and driver's voltage drop. If diode says it has a voltage drop of say, 2.5 V, and driver also needs say 2.5 V, that means that 5 V supply will work exactly the same as 10V supply regarding the optical output power.
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Old 12-25-2011, 11:38 PM #4
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Default Re: Can someone explain driver voltage?

The more extra voltage you throaw at it the more waste heat generated though. P = I E (Watts = Amps x Volts), so if your current is stationary, throwing more volts at it increases the wattage being dissipated (not the wattage of the diode's light output).

It doesn't increase the optical wattage because light power generated is a function of current for diodes, so pumping more watts via voltage into a diode just increases the waste by making it run less efficiently.

Cover your voltage drops by enough to compensate for battery sag and you'll be fine.

Semi-OT; I've been out of the laser hobby a while, and would appreciate a link to (or explanation of) the various Vf's (forward voltages) of the "modern" selection of LDs. Last time I was in the laser hobby, the 405, 445, 510, and 635 diodes were not even invented yet!
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Old 12-28-2011, 05:48 AM #5
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Default Re: Can someone explain driver voltage?

dose that mean I can use a driver that puts out 4 volts wiith a diode that says it need 3 volts?
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Old 12-28-2011, 07:50 AM #6
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Default Re: Can someone explain driver voltage?

Quote:
Originally Posted by wolfman View Post
dose that mean I can use a driver that puts out 4 volts wiith a diode that says it need 3 volts?
Driver doesn't output 4 Volts, driver outputs current, and when you hook up a diode on it, then it supplies the voltage needed exactly to run that amount of current through it.

If you want to put it really bluntly...
Yes.
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Old 12-28-2011, 05:49 PM #7
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Default Re: Can someone explain driver voltage?

Quote:
Originally Posted by Jacob32123 View Post
So as simple as this is, I searched, and couldnt find and info that seemed to make sense.

So the diode I am talking about here is the 10mW 635nm here which has a voltage range of 2.3-2.6 volts

I am planning on using a rckstr driver which has a voltage drop of 2.25 volts.

But this would require a range of 4.55-4.85 volts, which would require a LM317 or similar circuit, just to get the right voltage to the driver.

DTR says

Quote:
Originally Posted by DTR
Yep you need to look up how to set the driver on a test load or with the P3 I believe you can test the current on the output side via a multimeter as well. You don't need to worry about voltage to the diode. The only voltage you need to worry about is that your power source can supply more than 6V to the driver.

Did you try a search for how to set the current on the P3?



That seems odd, can someone please clarify?
The rkcstr driver is a linear driver, so as long as your supply is at least 2.5 volts above your diode demand, then it will work.

Even with a red build, it is better to use 2 X Li-Ion's though. Because the voltage of the batteries will drop to as low as 6 volts before you need a charge...
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Old 12-28-2011, 06:30 PM #8
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Default Re: Can someone explain driver voltage?

I think a lot of people describe DDL drivers incorrectly. Correct me if I'm wrong, there are two things that I hear constantly, and don't think are accurate:

1) That the drivers "increase current" until the drop across the resistor is 1.25V. I'm pretty sure they actually "increase voltage" at the OUT pin, until the laser diode's IV curve results in a current that produces a drop across the resistor of 1.25V. A technicality perhaps, but these are inherently Voltage regulators, so I'm pretty sure they are actively playing with Voltage at the OUT pin, not Current (current regulation is a byproduct of the Laser Diode's IV curve).

2) Early on, people would sometimes draw incorrect diagrams of DDL drivers where the Laser Diode was connected to the OUT pin. Then someone would jump in and correct them by saying that the current is actually output from the ADJ pin. The way we wire these up, it certainly looks like the current is leaving the ADJ pin, but this is also technically inaccurate. The current is output from the OUT pin, flows across the resistor, gets its Voltage measured at the ADJ pin, and then continues on it's way through the Laser Diode.

The DDL driver explained:

- The LM317 (or LM1117, 1084, 1085, etc) ICs want to maintain a Voltage Drop between the ADJ and the OUT pins of exactly 1.25V. That's their goal.

- But you've put a resistor between those pins, which means the Voltage Drop it measures won't hit 1.25V until the current increases to 1.25 / Resistor Ohms.

- To reach its goal, the IC starts increasing Voltage at the OUT pin, hoping that this will help it achieve it's target of a 1.25V drop across the OUT and ADJ pins.

- Remember, current is flowing out of the OUT pin, across the resistor, gets its Voltage measured at ADJ, and then continues to pass through your laser diode.

- Here's the *magic* part. Since your Laser Diode has an IV curve, that Voltage your driver sets at the OUT pin determines where on your Laser Diode's IV curve we land in terms of Current.

- This current is what interacts with the resistor value, to determine the Voltage Drop between ADJ and OUT. Remember, the IC wants that voltage drop to be 1.25V.

- If heat causes the IV curve to shift a bit, your driver's OUT pin just adjusts it's Voltage up or down. This causes us to land at the same constant current, on the newly shifted IV curve.
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Last edited by rhd; 12-28-2011 at 06:33 PM. Reason: Typos
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Old 12-28-2011, 08:19 PM #9
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Default Re: Can someone explain driver voltage?

In the circuit configuration used in the DDL driver, which is a current limiting configuration, the LM317 limits current. The laser diode regulates it's own voltage by acting like a zener diode. Any excess voltage supplied by the LM317 IC is converted to heat within the IC and is absorbed by whatever heatsink is supplied. I didn't read every bit of rhd's post above, but this is how I understand things to work.
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Old 12-28-2011, 09:09 PM #10
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Default Re: Can someone explain driver voltage?

The LM317 implements a bandgap voltage reference, more specifically the Brokaw variety.

You can look at the circuit diagram to understand how it works. There are only two things you should assume about the LM317 or any other voltage regulator of that type:
  1. It provides 1.25V between OUT and ADJ.
  2. You must provide Vin with Vout + Vdropout for the voltage regulator to work.
That is it.

Don't assume any other LM317 effects other than the above. Everything else is just a result of the enforced conditions above. The parameters and side-effects of the device, such as amperage limits, heat, etc. are functions of the physical limitations of the materials, not the assumed operation of the device.

The LM317 will:
  • Increase the amount of current out of Vout in order to force the voltage drop across a resistive load between Vout and Vadj to 1.25V.
  • In a constant current arrangement: increase the voltage enough at Vadj to cause current to pass through a diode in order to allow current to pass from Vout through the diode to ground (assuming adequate Vin)
  • Filter out noise at the input in order to maintain 1.25V between Vout and Vadj.
  • Drop excess voltage if output voltage can't increase the Vout voltage any more (such as a diode with a fixed current)
  • Virtually anything to satisfy the two conditions expressed above.
  • Or die trying.
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