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FrozenGate by Avery

Please help! :)

CynicalBrad said:
I do believe Rachel is saying each of the four diodes is being ran from 3xlm317's for a total of TWELVE LM317's.

Essentially 4 separate laser/driver combos being ran from one input power source.

Code: 
(+)           [battery]           (-)
 |                                 |
  >           3xLM317&LD          <
 |                                 |
  >           3xLM317&LD          <
 |                                 |
  >           3xLM317&LD          <
 |                                 |
  >           3xLM317&LD          <

Is that about right?
(Too lazy to make a proper diagram of what I believe is being done by the OP Rachel.)
Click to expand...


Hi brad

Yes that is correct :p I have a picture I can upload if needed but yes you know what I've done :)
 





I would check all the connections to the fourth non functioning driver/diode setup and ensure all the connections are good and try to take measurement both on the input side of the non functioning set aswell as setting up a testload and measuring current on the output side of that one.

Also have to accept the chance that the diode may have died as sometimes these things can be picky and die for seemingly no reason at all.
Really hope it is just a bad connection going unnoticed somewhere though.
Have to also admit that pictures of the entire build as is wouldn't hurt either so some other eyes can look over it and see how it is set up.
 
CynicalBrad said:
I would check all the connections to the fourth non functioning driver/diode setup and ensure all the connections are good and try to take measurement both on the input side of the non functioning set aswell as setting up a testload and measuring current on the output side of that one.

Also have to accept the chance that the diode may have died as sometimes these things can be picky and die for seemingly no reason at all.
Really hope it is just a bad connection going unnoticed somewhere though.
Have to also admit that pictures of the entire build as is wouldn't hurt either so some other eyes can look over it and see how it is set up.
Click to expand...

Hi brad,

No I've only ran it twice. So I have no reason why it wouldn't have blown! Really hope not, but I don't think it should have. I've been really careful with them so I would be very confused! I haven't checked the connection yet, but will tonight or tomorrow.

I've got a old picture from the prototype I made for my bread board. Exactly the same circuit so should be easy for you to read :).

I'll find a server host and the. Upload my pic through that link for you. Bare with me :)
 
I'm seeing two issues and one potential issue in the photo.

Issue 1: It looks like the regulator OUT pins are connected to the laser diode anodes--power to the laser diodes should be connected to ADJ pins of the 317 regulators. In the configuration in the photo current to the laser diodes will be limited only by the overcurrent protection of the regulators. This may not be a problem due to issue #2: A 7.4 volt battery really isn't enough voltage for a buck regulator using 317's. The circuit would only work when the battery is fresh off the charger with a surface charge of 8.0-8.4 volts. The potential issue is the capacitors on the regulator outputs. These will help keep ripple to the laser diodes low but they're dangerous in that if there is even a momentary disconnection of the laser diode these cap's will charge up to full battery voltage (minus a volt or so for regulator losses) and when the laser is reconnected these will send a discharge pulse of current that can easily kill the laser.
 
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WizardG said:
I'm seeing two issues and one potential issue in the photo.

Issue 1: It looks like the regulator OUT pins are connected to the laser diode anodes--power to the laser diodes should be connected to ADJ pins of the 317 regulators. In the configuration in the photo current to the laser diodes will be limited only by the overcurrent protection of the regulators. This may not be a problem due to issue #2: A 7.4 volt battery really isn't enough voltage for a buck regulator using 317's. The circuit would only work when the battery is fresh off the charger with a surface charge of 8.0-8.4 volts. The potential issue is the capacitors on the regulator outputs. These will help keep ripple to the laser diodes low but they're dangerous in that if there is even a momentary disconnection of the laser diode these cap's will charge up to full battery voltage (minus a volt or so for regulator losses) and when the laser is reconnected these will send a discharge pulse of current that can easily kill the laser.
Click to expand...

So I've wired up from the lm317 incorrectly then?

And I could only get a 7.4V. I would go 8V but I can't find a hobby pack of that voltage.

Would taking th caps of be better then or still leave them on ?
 
So I've wired up from the lm317 incorrectly then? Yes, google "LM317 current regulator" and you'll get the basic schematic. Take a look and you'll see that when connected as a current regulator the load is connected to the ADJ pin, not the OUT pin.

And I could only get a 7.4V. I would go 8V but I can't find a hobby pack of that voltage.Your 7.4 volt battery will put out between 8 and 8.4 volts when fully charged (I'm assuming we're talking about a Li ion battery) which is just at the edge of being `enough'. You might want to go to a 3 cell battery. They're usually rated 11.1 volts. The LM317 regulators will generate considerable heat so your duty cycle may be limited unless you heatsink them.

Would taking th caps of be better then or still leave them on ? I would say to leave them, but make them small ones. 0.1 uF should be fine. A 1kohm resistor in parallel with each capacitor isn't a bad idea. Just make sure the connections to the laser diodes are solid
 
WizardG said:
So I've wired up from the lm317 incorrectly then? Yes, google "LM317 current regulator" and you'll get the basic schematic. Take a look and you'll see that when connected as a current regulator the load is connected to the ADJ pin, not the OUT pin.

And I could only get a 7.4V. I would go 8V but I can't find a hobby pack of that voltage.Your 7.4 volt battery will put out between 8 and 8.4 volts when fully charged (I'm assuming we're talking about a Li ion battery) which is just at the edge of being `enough'. You might want to go to a 3 cell battery. They're usually rated 11.1 volts. The LM317 regulators will generate considerable heat so your duty cycle may be limited unless you heatsink them.

Would taking th caps of be better then or still leave them on ? I would say to leave them, but make them small ones. 0.1 uF should be fine. A 1kohm resistor in parallel with each capacitor isn't a bad idea. Just make sure the connections to the laser diodes are solid
Click to expand...

Okay thanks wizard I will have a look st the wiring schematic, I took my schematic of a basic drawing so that's probably why it's wrong. Thankful it hasn't damages my LD tho.

The 7.4v is a lithium polymer battery. It works in the midium voltage of the LD data sheet, minus the lm317 share of voltage (3v) I don't want to go 11V coz I feel that that would be too high and I don't want to blow them.

How do I wire the resistors ? Parallel how? To each laser or capacitor ? I'm confused here :p

Thanks
 
The 1k resistors would go in parallel with the capacitors. One for each capacitor. They're just to bleed off any charge in the capacitor when the circuit is powered off. Many here on the forum can tell you tales of woe from connecting a laser diode to a driver they had just tested......without discharging the capacitors in the driver first.
 
WizardG said:
The 1k resistors would go in parallel with the capacitors. One for each capacitor. They're just to bleed off any charge in the capacitor when the circuit is powered off. Many here on the forum can tell you tales of woe from connecting a laser diode to a driver they had just tested......without discharging the capacitors in the driver first.
Click to expand...

Okay thanks wizard. I'll make sure I do that.
So they just got across from the positive side then go over to the negative side of the laser ? Just like the capacitor is on the picture ?

On the picture the capacitor goes across the positive and negative before it goes to my LD do I do the exact same with the resistor ? Am I understanding what you're saying correctly ?

Thanks

Rach
 
Rachel said:
Okay thanks wizard. I'll make sure I do that.
So they just got across from the positive side then go over to the negative side of the laser ? Just like the capacitor is on the picture ?

On the picture the capacitor goes across the positive and negative before it goes to my LD do I do the exact same with the resistor ? Am I understanding what you're saying correctly ?

Thanks

Rach
Click to expand...


Yes, sounds like you're understanding correctly.

Resistor in parallel with the capacitor, before the LD.
 
diachi said:
Yes, sounds like you're understanding correctly.

Resistor in parallel with the capacitor, before the LD.
Click to expand...

Thanks Diachi

What wattage would I need to cover this with ? I'd say 1W would be okay ?

And would I need a diode going in there so there is no short circuiting from the resistor or will the resistor not do that ?

Thanks again,

Sorry for the silly questions :p :D !!!!
 
Rachel said:
Thanks Diachi

What wattage would I need to cover this with ? I'd say 1W would be okay ?

And would I need a diode going in there so there is no short circuiting from the resistor or will the resistor not do that ?

Thanks again,

Sorry for the silly questions :p :D !!!!
Click to expand...


0.25W should be enough. The calculation is easy, I suggest figuring out how to calculate resistor power ratings... At 1Kohm and <5V going to the diode (resistor also sees 5V as it is parallel) you'll have <5mA flowing through the resistor. That works out to <0.025W dissipation in the resistor. 0.25W is more than enough.
 
The cap would only be charged to a hazardous level if you tried powering the circuit without a load connected. Otherwise, a bleeder doesn't help anything here. The diode itself clamps the cap voltage to operating voltage or below.

1K is very low for a bleeder anyway. If you've got 10µF (any more than 10µF isn't necessary for lm317), the cap is discharged to <14% original voltage in 20ms. 100k would waste 99% less power and discharge to <14% in 2 seconds, which is plenty fast enough. 100µF with 10k would waste 90% less power and still discharge in 2s.

1k will work, but I personally try to be a little more practical than "pick a number, any number"

7.4V isn't enough. The minimum dropout depends on load current and junction temperature, but is usually around 3.5V for what we use them for. You're saying 4.8V from the diode for a total of 8.3V minimum to be in regulation. It's even worse when you take those anemic wires and internal battery resistance into consideration. It will turn on at 7.4V, but it will not be in regulation. The current will be lower than your setpoint, and the regulator is acting as a resistor here. Switch to 11.1V.
 
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diachi said:
0.25W should be enough. The calculation is easy, I suggest figuring out how to calculate resistor power ratings... At 1Kohm and <5V going to the diode (resistor also sees 5V as it is parallel) you'll have <5mA flowing through the resistor. That works out to <0.025W dissipation in the resistor. 0.25W is more than enough.
Click to expand...

Thanks for the calculations in that, I think I've seen this before when I tried to work out what resistor I needed! It just slipped my mind :D
 
Cyparagon said:
The cap would only be charged to a hazardous level if you tried powering the circuit without a load connected. Otherwise, a bleeder doesn't help anything here. The diode itself clamps the cap voltage to operating voltage or below.

1K is very low for a bleeder anyway. If you've got 10µF (any more than 10µF isn't necessary for lm317), the cap is discharged to <14% original voltage in 20ms. 100k would waste 99% less power and discharge to <14% in 2 seconds, which is plenty fast enough. 100µF with 10k would waste 90% less power and still discharge in 2s.

1k will work, but I personally try to be a little more practical than "pick a number, any number"

7.4V isn't enough. The minimum dropout depends on load current and junction temperature, but is usually around 3.5V for what we use them for. You're saying 4.8V from the diode for a total of 8.3V minimum to be in regulation. It's even worse when you take those anemic wires and internal battery resistance into consideration. It will turn on at 7.4V, but it will not be in regulation. The current will be lower than your setpoint, and the regulator is acting as a resistor here. Switch to 11.1V.
Click to expand...

Thanks for your comments cyparagon, I will move my battery to 11.1V in this case. I hope it doesn't damaged the LD.

Just to make you all I aware the LD I am using is the 08 N1ch!@. Dont know if u guys have any experience of this kind of LD but it is quite powerful :)

In sum I have a 10uF cap. So would I need a resistor in parallel with the cap still or is there not a need anymore ?

If I did need one, I found it hard to understand which type I did need and why :D sorry xxxx!!

What would you recommend

Rach
 
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