Some questions about my first laser build

[Greenhorn]

I have very limited knowledge of electronics, but I am a hobbyist and and am very eager to build a 405nm laser.  I already bought two PHR-803T diodes from eBay.  After hearing people talk about drivers, about voltages, etc, I'm confused.  Here are some issues I'm having.

1.  People say that you need 9 volts, but I thought a PHR-803T took 5.5-6 volts.  Why?

2.  What exactly is a driver?  Is this just a chip with resistors?

3.  What exactly is voltage drop, and how does it apply to current?  As far as I understand, the way to figure out how much resistance you need for a certain current is to divide the difference between the diode's voltage drop and battery voltage by the current you want (in amps).  Why is this?  What significance does the difference in VD vs. original voltage hold?  And what is the voltage drop of a PHR-803T?

4.  Is a resistor enough to keep a diode from possibly dying out (assuming that you don't mix batteries, or determine ohms needed by alkalines and then use lithiums)?  Do you NEED a voltage regulator (which I don't know anything about).

5.  If I use a pot to find the optimal resistance for my wanted current, will the current then be different if I solder a permanent resistor onto it?  In other words, will the pot lower the voltage or something, and mess up my readings?

I have ordered a metered 10-turn 0-10k ohm resistor, and two 405nm diodes and focus modules.  Is this all I need to build a reliable barebones laser?  Will four alkaline AA batteries be fine?

 

[chopper]

I have very limited knowledge of electronics, but I am a hobbyist and and am very eager to build a 405nm laser.  I already bought two PHR-803T diodes from eBay.  After hearing people talk about drivers, about voltages, etc, I'm confused.  Here are some issues I'm having.

1.  People say that you need 9 volts, but I thought a PHR-803T took 5.5-6 volts.  Why?

2.  What exactly is a driver?  Is this just a chip with resistors?

3.  What exactly is voltage drop, and how does it apply to current?  As far as I understand, the way to figure out how much resistance you need for a certain current is to divide the difference between the diode's voltage drop and battery voltage by the current you want (in amps).  Why is this?  What significance does the difference in VD vs. original voltage hold?  And what is the voltage drop of a PHR-803T?

4.  Is a resistor enough to keep a diode from possibly dying out (assuming that you don't mix batteries, or determine ohms needed by alkalines and then use lithiums)?  Do you NEED a voltage regulator (which I don't know anything about).

5.  If I use a pot to find the optimal resistance, will it be different if I solder a permanent resistor onto it?  In other words, will the pot lower the voltage or anything?

I have ordered a metered 10-turn 0-10k ohm resistor, and two 405nm diodes and focus modules.  Is this all I need to build a reliable barebones laser?  Will four alkaline AA batteries be fine?

I'm no expert as some on here are, but I'll do my best to answer your questions:

1. Most (if not all) drivers require some voltage to operate. You need to supply diode voltage + driver voltage.
(i.e. a rckstr driver requires ~2.25V and an phr803T requires ~5V so the minimum voltage you'd want is 2.25V+5V=7.25V a little more is ok, but rckstr suggest no going over the min more than 2V or over a total of 12V)

A boost driver like drlava's flexdrive will boost the voltage to that required by the diode so all that is required to supply is anything within the specs of the driver...

2. A driver is a circuit to control the current supplied to the diode. A diode will draw more current than it needs if left unregulated, so much that it will commit suicide , for this reason a driver circuit is required to regulate the current. As mentioned above, some drivers will regulate voltage too. There are other reasons to have a driver that are important but more technical dealing with the quality of the supplied power. They minimize spikes and noise etc...

3. Voltage drop is the voltage 'consumed' in a particular component of a circuit. For instance a resistor fallows Ohm's law, V=I*R (V=voltage I=current R=resistance) so the voltage drop across a resistor is the current times the resistance. I remember somewhere from my Physics 2 class something like Kirchoff's law states that the sum of all voltage drops/gains(battery...) in a circuit will equal zero. So a circuit with a 1V battery and a 1Ohm resistor goes like:

V(batt) + V(resist) = 0 but we know Ohms law so V(resist) = I*R = I*1ohm (this will be negative because it is a v drop not gain), V(batt) = 1V (positive because it is a gain)
=> 1V - I*1ohm = 0
=> 1V = I*1ohm
=> I= 1V/1ohm = 1 A

From what I have found in the forum a phr 803T has a voltage drop of about 5V

4. I believe a resistor alone is not sufficient (correct me if I'm wrong)

5. The pot is just a variable resistor. A fixed resistor is no different than a pot set at the same resistance as the fixed resistor.

Lastly, you'll need some more components like a voltage/current regulator IC like a LM317.

I think the DDL driver circuit might be about the simplest you can get, see this thread:

http://www.laserpointerforums.com/forums/YaBB.pl?num=1185701612

 

[Greenhorn]

Thanks for the response.  I have a few more questions now.

3.  Voltage drop is the voltage 'consumed' in a particular component of a circuit.  For instance a resistor fallows Ohm's law, V=I*R (V=voltage I=current R=resistance) so the voltage drop across a resistor is the current times the resistance.  I remember somewhere from my Physics 2 class something like Kirchoff's law states that the sum of all voltage drops/gains(battery...) in a circuit will equal zero.  So a circuit with a 1V battery and a 1Ohm resistor goes like:

V(batt) + V(resist) = 0  but we know Ohms law so V(resist) = I*R = I*1ohm (this will be negative because it is a v drop not gain), V(batt) = 1V (positive because it is a gain)
=> 1V - I*1ohm = 0
=> 1V = I*1ohm
=> I= 1V/1ohm = 1 A

That's a little confusing  :-[  I understand some of it.

I'm not even sure how to ask this question.  It's a mess in my head.  I guess my question is, basically, what is OHM as it relates to current?  What does a resistor do, exactly?  Does it actually cause a voltage drop?  How does one compensate for this? Do you need to know the BATTERY'S current to calculate this stuff?

Also, what information do you need to determine what current a certain resistor will give you?  Do you ONLY need the diode's voltage drop and the batteries' voltage?  If it's as simple as that, I won't ask WHY, I'll just accept it.  ;D

One more thing, how does the DIODE'S voltage drop and the resistor determine current?  When the current passes through the resistor, it hasn't gotten to the diode yet, so how does the diode affect anything at that point?  I hope I'm not being too confusing.

From what I have found in the forum a phr 803T has a voltage drop of about 5V

I'm confused.  You said that the diode requires 5 volts, but the VD is also 5?  If the needed ohm is measured by dividing the difference between voltage and VD by the required current, you can't divide by zero.

Lastly, you'll need some more components like a voltage/current regulator IC like a LM317.

Doesn't a resistor regulate current?  Or does it only lower the current based on the voltage it receives?  What I mean is, if there is a voltage spike, would this cause a current spike also, even if there was a resistor?

Finally, tell me if this is about right:  An LM317 regulates voltage, keeping spikes from killing the diode.  With regulated voltage, a resistor will then regulate the current.

Is this right, and is this all I need for a good, basic driver?  Will either of these components create voltage drop, which would require an extra battery or something?


Sorry for the random, confusing questions, but just two weeks ago I had no clue you could BUY laser diodes, much less high-power laser diodes, and my knowledge of electricity was almost zero.  I still don't know much more than that, but I'm learning.

 

[originalbigT]

a resister alone is not a good route, it's like working under your car using just the car jack.A driver is like the jack stand, it elimnates any variables the could kill your diode. Think of voltage as water pressure and current as the amount off water being pushed, your resister is like putting a smaller hose on the end. It will reduce the amount of water coming out, but if the pressure going in goes up, so does the pressure coming out. As batteries die, they sometimes jump up in current, if it jumps enough, you can get too much out of the resister and ruin your diode. A driver makes sure that regardless of what the battery does, your laser diode will never get too much current. This may be an over simplification, but it gives you a basic idea. Hope this helps. I'm new to this too and have killed my share of laser diodes, take my advise and get a driver. They're easy to get here and are great quality from what I've read.

 

[Greenhorn]

a resister alone is not a good route, it's like working under your car using just the car jack.A driver is like the jack stand, it elimnates any variables the could kill your diode. Think of voltage as water pressure and current as the amount off water being pushed, your resister is like putting a smaller hose on the end. It will reduce the amount of water coming out, but if the pressure going in goes up, so does the pressure coming out. As batteries die, they sometimes jump up in current, if it jumps enough, you can get too much out of the resister and ruin your diode. A driver makes sure that regardless of what the battery does, your laser diode will never get too much current. This may be an over simplification, but it gives you a basic idea. Hope this helps. I'm new to this too and have killed my share of laser diodes, take my advise and get a driver. They're easy to get here and are great quality from what I've read.

But as a hobbyist, I want to make the driver myself. It'll give me a sense of accomplishment, and help me learn about electricity.

I thought the resistor's purpose was to regulate current. But you're saying that it doesn't?

From what little I understand, isn't a driver basically just a chip that has a voltage regulator and current regulator?

 

[originalbigT]

no a resister does not regulate as it's simply a component with a set resistance. So if current going goes up, so does the current coming out. If you want to build your own and better understand what's involved and what it does then read the basic faqs before posting thread. It gives you all the info you need and a link to build a driver with a schematic and parts list.

 

[originalbigT]

just follow the link Chopper gave you to build a DDL driver

 

[chopper]

You don't need to know the total circuit calculation that confused you, if you want to know more about that you can find a physics text on electricity and magnetism or a book on circuits.

You should know Ohm's Law V=I*R.  This is valid for resistors.  a better way to right it is I=V/R.  So given a voltage from a battery for example the current allowed through a resistor will be the voltage divided by the resistance in ohms.  So if the voltage changes, the current changes.  One definition of an Ampere unit of current is 1 Ampere (A) is equal to 1 volt (V) per ohm (captial omega).  Parentheses are abbreviations for the units.

A resistor is a component that opposes the flow of electricity.  More technically (Usually they are made of a material where the flowing electrons run into obstacles where some of there energy is absorbed.)
One method often used to describe and understand the flow of electricity is like a hose of flowing water.  The voltage is like the water pressure.  A resistor is like a kink in the hose.  Current it like the amount of water flowing.  The kink lowers the pressure coming out of the kink (resistor lowers voltage or has a voltage drop, if you have a simple circuit you can measure the voltage across(a probe on each end) a resistor to find the voltage drop).  This decrease in pressure limits the amount of water flowing out of the kink in the hose.  If you increase the pressure source of the water flow more water will flow through the kink (increased voltage increases current through a resistor).

A diode has a voltage drop, but doesn't limit current the same way that a resistor does.  It works much differently and would be a discussion in solid state physics to try and understand better.  I don't know the specific relationship between current and voltage (drop) across a diode.  I just assume the diode does not limit current and would allow as much as a battery could supply if left unregulated.  The voltage drop of the diode is the voltage needed to allow current to flow through, again a detailed explanation requires knowledge of solid state physics and semiconductor devices.

Think of a diode like a one way water valve that will not let water flow in one direction, but will let water flow in the other direction, but only when the pressure is above a certain threshold.  Interestingly, if put a kink into the hose that had enough pressure to let water flow through the valve, regardless if the kink is before or after the valve, the pressure that the valve will feel will be lowered.  If the kink is too tight, the pressure will drop below the point where the valve will allow water to flow.  Again this works the same in a circuit, a resistor before or after a diode will have the same limiting effect on the available voltage to the diode.  A complete driver circuit also has this voltage reducing effect hence the need for a supply voltage equal to the sum of the voltage drop of the driver and the diode.

Using the water hose metaphor, a battery is like a water tank.  It can let as much water flow as the hose attached will let flow.  So a battery will try to provide as much current as the circuit will let flow.  I say it will try because there is a maximum amount of current that can flow (you don't need to know this: I think it s due to the battery's internal resistance).


When I say the phr803T requires 5 volts I mean that it will have a voltage drop in the circuit of 5 volts.  This means that if the total voltage (drop) of each component (the sum of all voltage drops) in the circuit is less than the supplied voltage (battery voltage) the diode will not let current flow properly.

To see this effect you can set up an experiment with a DDL driver and a dummy load (a circuit made to simulate the voltage drop of a laser diode in order to set current output from a driver).  My 803T dummy load has a voltage drop of about 5V and I think the DDL has a voltage drop of about 3V.  Now, my dummy load has a 1 ohm resistor on it.  If I measure it's voltage drop I can figure out the current flowing real easily.  Using Ohm's law I = V/R so I = (measured voltage drop) / 1 ohm so basically the current in A is equal to the voltagte in V.  If I supply the circuit with say 9V I can adjust the current with the driver and measure say .100V across the 1 ohm resistor.  This means that .100 A = 100mA is flowing through the circuit.  If I supply the circuit with something less than the total voltage like 6V, the voltage drop measured on the 1 ohm resistor will be 0.  Now I can measure the voltage across the diodes in the dummy load and measure ~3V which is not enough to let any current flow.  This implies an interesting idea if you are unfamiliar.  A voltage difference does not imply current flow.  This makes sense, if you have a AA battery and measure it's voltage without a circuit attached it should say 1.5V, but not current is flowing.

An LM317 is a component that will either regulate voltage or regulate current, but only one at a time.  The circuit design determines which it will do.  In the DDL driver circuit the LM317 is being used as a current regulator as the voltage is pretty well regulated (from going too high, not too low) by the batteries.  The first few posts in the DDL driver thread discuss this I think.

The DDL is a well tested driver circuit.  The resistor values in it are what tells the LM317 what current to allow.  This is what the pot (potentiometer = variable resistor) is used for.  You can make a fixed current DDL by using a fixed resistor instead of a pot.

Ok I said a lot.  I hope I don't confuse you , but there's alot to take in.  Questions are welcome.  

 

[Greenhorn]

Thanks, that was VERY good information, but I still have some questions.

I don't understand how the resistor value determines how much current the LM317 allows.  I always thought that the POT is what controls current - the higher the resistance for a given voltage, the lower the current, right?  So what does the LM317 do?  If the pot DOESN'T control current, what DOES it do?

I always thought that a resistor only affects current, but it actually creates a voltage drop?  How do you determine how much voltage drop a resistor has without some kind of expensive equipment?

In the DDL schematic, if you already have a pot, why does it also need the second 4 ohm resistor?

BUT, all that aside, from what I understand, an LM317, plus a resistor that will tell the LM317 how much current to allow, is basically all you need for a barebones driver.  A voltage regulator of some kind, like a capacitor, is helpful too.  Right?

Finally, what does the silicon diode do?
EDITED TO ADD: Does it slightly alter the voltage to get the optimal voltage to reach the laser diode? Just a guess.

Ugh, so many questions, and I don't know enough to know what or how to ask.  I kinda feel dumb.

 

[Ace82]

Greenhorn, you might want to consider buying pre-manufactured drivers. If you want a smaller build and less battery power, you could go with this booster driver:
http://www.laserpointerforums.com/forums/YaBB.pl?num=1209418066/0#0

Or this one is the one I prefer, but that’s because I’m big on battery power and I don’t mind larger builds:
http://www.laserpointerforums.com/forums/YaBB.pl?num=1206947255/0#0

And you should be able to find out all you need here:
http://www.laserpointerforums.com/forums/YaBB.pl?board=b_repairs

 

[chopper]

Thanks, that was VERY good information, but I still have some questions.

I don't understand how the resistor value determines how much current the LM317 allows.  I always thought that the POT is what controls current - the higher the resistance for a given voltage, the lower the current, right?  So what does the LM317 do?  If the pot DOESN'T control current, what DOES it do?

I always thought that a resistor only affects current, but it actually creates a voltage drop?  How do you determine how much voltage drop a resistor has without some kind of expensive equipment?

In the DDL schematic, if you already have a pot, why does it also need the second 4 ohm resistor?

BUT, all that aside, from what I understand, an LM317, plus a resistor that will tell the LM317 how much current to allow, is basically all you need for a barebones driver.  A voltage regulator of some kind, like a capacitor, is helpful too.  Right?

Finally, what does the silicon diode do?
EDITED TO ADD:  Does it slightly alter the voltage to get the optimal voltage to reach the laser diode?  Just a guess.

Ugh, so many questions, and I don't know enough to know what or how to ask.  I kinda feel dumb.

***Disclaimer: This is a metaphor for how this works, I haven't studied the internal workings of the LM317 in detail.***

The LM317 set up as current regulating circuit has the output current coming out of the V-out pin. A tap of this output is sent back to the adj pin after the 'current setting resistor'. This 'current setting resistor' could be a fixed resistor, a pot (a resistor which lets you vary the resistance value) or a combination of the two. The LM317 'looks' to see what resistance it 'sees' and determines the amount of current to deliver out. Since the tap after the resistor is off of the delivered current, the LM317 is always checking to see if it is sending the right current. You are correct the higher the resistance correspond to a lower current. To calculate the delivered current: I-out = 1.25 V / Resistor value. As I mention you could use only a pot to make a variable current regulator, but if you turn the pot to a value less than .625 ohms it might try to push more current than it is rated for, 1.5A. For lasers we are working with a desired current range of roughly 500mA down to 10mA (this is probably wider than the usable range for us). By placing a fixed resistor in series with the pot, there is a set maximum current if the pot is set for a 0 ohm resistance. I put a 5 ohm fixed resistor ( two 10 ohm resisters parallel to each other) with a 25 ohm pot in series. So my resistor range is from 5 ohms to 30 ohms. If I have the pot turned to 0 ohms, the LM317 sees 5 ohms and outputs about 250mA ( if you wanted 500mA the calculation yields a required resistance of 2.5 ohms which could be achieved by placing 4 10 ohm resistors in parallel). My minimum current would be at the pot turn all the way to 25 ohms plus the 5 ohm fixed resistor (30 ohm total) gives me a minimum current of about 42mA. If I wanted less current than that I would need to increase the fixed resistance (decreasing the maximum current) or increase the pot's maximum resistance (widening the range of available current settings).

The voltage drop of a resistor is dependent on the current. Ohm's Law V=I*R so if we have a current of 1A and a 1ohm resistor it has a voltage drop of 1V. Same current with a 2 ohm resistor would lead to a 2V drop. I know this is confusing since we don't usually look at it this way. This however is how the LM317 looks to see what current it should be setting. The LM317 adjusts the current so that the voltage drop across the 'current setting resistance' is 1.25V. This is where the LM317 current equation comes from.

If you read the first post in the DDL driver thread he basically says that battery voltage is pretty much already regulated minus some spikes and noise so no real voltage regulator is used. A capacitor is not a voltage regulator. It's more of a filter to smooth out the voltage supplied helping to minimize peaks etc...

I believe the silicon diode is there to prevent reverse polarity on the power supply.

 

[Greenhorn]

Wow, this is pretty much exactly what I wanted to know. I can't even really think of any more questions to ask now (I'll probably think of a few more eventually). Thanks a lot for your help!

***Disclaimer: This is a metaphor for how this works,  I haven't studied the internal workings of the LM317 in detail.***

The LM317 set up as current regulating circuit has the output current coming out of the V-out pin.  A tap of this output is sent back to the adj pin after the 'current setting resistor'.  This 'current setting resistor' could be a fixed resistor, a pot (a resistor which lets you vary the resistance value) or a combination of the two.  The LM317 'looks' to see what resistance it 'sees' and determines the amount of current to deliver out.  Since the tap after the resistor is off of the delivered current, the LM317 is always checking to see if it is sending the right current.  You are correct  the higher the resistance correspond to a lower current.  To calculate the delivered current: I-out = 1.25 V / Resistor value.  As I mention you could use only a pot to make a variable current regulator, but if you turn the pot to a value less than .625 ohms it might try to push more current than it is rated for, 1.5A.  For lasers we are working with a desired current range of roughly 500mA down to 10mA (this is probably wider than the usable range for us).  By placing a fixed resistor in series with the pot, there is a set maximum current if the pot is set for a 0 ohm resistance.  I put a 5 ohm fixed resistor ( two 10 ohm resisters parallel to each other) with a 25 ohm pot in series.  So my resistor range is from 5 ohms to 30 ohms.  If I have the  pot turned to 0 ohms, the LM317 sees 5 ohms and outputs about 250mA ( if you wanted 500mA the calculation yields a required resistance of 2.5 ohms which could be achieved by placing 4 10 ohm resistors in parallel).  My minimum current would be at the pot turn all the way to 25 ohms plus the 5 ohm fixed resistor (30 ohm total)  gives me a minimum current of about 42mA.  If I wanted less current than that I would need to increase the fixed resistance (decreasing the maximum current) or increase the pot's maximum resistance (widening the range of available  current settings).

The voltage drop of a resistor is dependent on the current.  Ohm's Law V=I*R so if we have a current of  1A and a 1ohm resistor it has a voltage drop of 1V.  Same current with a 2 ohm resistor would lead to a 2V drop.  I know this is confusing since we don't usually look at it this way.  This however is how the LM317 looks to see what current it should be setting.  The LM317 adjusts the current so that the voltage drop across the 'current setting resistance' is 1.25V.  This is where the LM317 current equation comes from.

If you read the first post in the DDL driver thread he basically says that battery voltage is pretty much already regulated minus some spikes and noise so no real voltage regulator is used.  A capacitor is not a voltage regulator.  It's more of a filter to smooth out the voltage supplied helping to minimize peaks etc...

I believe the silicon diode is there to prevent reverse polarity on the power supply.

 

[chopper]

Glad to help if I can!

 

[Greenhorn]

I'll just make sure I'm getting this right.

So the LM317 keeps the voltage drop of the resistor at 1.25v no matter the resistance, so in the equation I = V/R, the only way to change I is to change R (because V stays the same.) Correct? And the permanent resistor keeps the current from reaching dangerous levels, no matter where you set the pot?

From the schematic, it looks like the IN lead of the LM317 is (of course) connected to a battery, with the OUT lead leading to a pot and a permanent resistor (does it matter which comes first?), which then goes into both the ADJ lead and into the laser diode, right? Maybe this is a dumb question, but why does the current have to go into the ADJ lead? Why can't it just continue on into the diode?

I guess I'll have to learn more about series and parallel, too. If you put resistors in series they add up each other's resistance, but if you put them in parallel they reduce it, right? Two 5ohm resistors in series have 10ohm, but two 5ohm in parallel have 2.5?

 

[chopper]

I'll just make sure I'm getting this right.

So the LM317 keeps the voltage drop of the resistor at 1.25v no matter the resistance, so in the equation I = V/R, the only way to change I is to change R (because V stays the same.)  Correct?  And the permanent resistor keeps the current from reaching dangerous levels, no matter where you set the pot?

From the schematic, it looks like the IN lead of the LM317 is (of course) connected to a battery, with the OUT lead leading to a pot and a permanent resistor (does it matter which comes first?), which then goes into both the ADJ lead and into the laser diode, right?  Maybe this is a dumb question, but why does the current have to go into the ADJ lead?  Why can't it just continue on into the diode?

I guess I'll have to learn more about series and parallel, too.  If you put resistors in series they add up each other's resistance, but if you put them in parallel they reduce it, right?  Two 5ohm resistors in series have 10ohm, but two 5ohm in parallel have 2.5?

Right, the fixed resistor sets a minimum current.

It does not matter what order the pot and resistor are in.

Bettery plus to Vin. A few pics and schematics can be found in the DDL thread.

The feed needs to also go to the adj pin so the LM317 can measure the voltage drop across the resistor(s) ion order to keep the set curent.

Series (connected end to end) resistors
R(total) = R1 + R2 + R3 + ...

Parallel resistor ( all connected to common point and end at another common point):
1/R(total) = 1/R1 + 1/R2 + 1/R3 + ...

Say you have 4 ten ohm resistors.

In series :  (+ )-0- -0- -0- -0-(-)

Rt = 10 + 10 + 10 + 10 = 40

In parallel:
__ (+)_    1/Rt = 1/10 + 1/10 + 1/10 + 1/10 = 4/10
|  | |  |    => Rt = 10/4 = 2.5
0 0 0 0
|  | |  |
--(-)---

Say you have a series of  two in paralell  with two more in parallel:

(+)-=--=-(-)   Each pair in parallel, 1/R = 1/10 + 1/10  =2/10  R=5
  The series of the two parallel sets: R= 5+5 = 10
So with 4 resistors of the same resistance connected in a symmetric series parallel combination  you end up with the same resistance as one of the resistors.