how does it add up ?

OK i have been trying to figure this out for at lease 4 hours straight and its driving me nuts ! :banghead: How is it that for a red laser you need a power supply of 7.2 when the diode needs 3, 1.2 for long life between battery changes and the lm317's dropout voltage is 1.2. Thats only 5.4, wheres the rest of it? On ROG8811 it says that its 7.2 b/c of

Voltage needed by red LD 3v

Voltage needed by regulator to make it work 3v

Additional voltage to get long life between battery changes 1.2v.

Add that up and you get 7.2volts (his words)

Where did the 3 volts for the 317 come from, i never heard of 3v being needed for the 317 to work. I have heard of 317 adj being able to put out 1.2v. So that + 1.2 dropout = 2.4v; that would be the min voltage you could set right ? Maybe there is no dropout V. I am just really tired of trying to figure this out and its late on top of that. Could someone just tell me where i went wrong or what i'm missing.

Thank you and sorry if i said anything noobish.

The 3 Volts for the LPM317 is a conservative number it is
closer to 2.5 Volts.
That is the basic Voltage drop across the LM317's internal
electronic circuit. Then you add the other voltage dropping
factors.
Electronic circuits are never run on the "edge" to be efficient..


Jerry

You can contact us at any time on our Website: J.BAUER Electronics

lasersbee said:

The 3 Volts for the LPM317 is a conservative number it is
closer to 2.5 Volts.
That is the basic Voltage drop across the LM317's internal
electronic circuit. Then you add the other voltage dropping
factors.
Electronic circuits are never run on the "edge" to be efficient..


Jerry

Click to expand...

Doesn't answer his question. :can:

Anyway -- as for the 2.5V - 3V nominal value -- that's just the Vdrop across the regulator.

The Vdrop is how much voltage the device (in this case, the regulator) takes 'off' the input voltage.

Simple Example: I put 6V through a 317, I'll get 3.5V out. 2.5V of the 6V I put in will be dropped ('lost') across the 317.

Making sense now? Another example: if I put 2.5V through the same 317, I'll get nothing out, because it'll take everything. Once you go below the dropout voltage, the output drops out.

Thing is, not all 317s will have an exact voltage drop of 2.5V; and as lasersbee pointed out, some may have a higher voltage drop. The 3V figure you've been quoted is to just allow for some extra headroom, if the 317 needs it. Most of them don't (I haven't had one need more than 2.7), but it's for the off-chance that it does.

As for the 1.2V figure -- that's only relevant when the LM317 is being run in voltage regulation mode. In this application, you're using it in current regulation mode. What it means is that by varying the values of the two resistors, you can adjust the voltage linearly down to 1.2V. You can't adjust the output voltage to any lower than 1.2V using the control resistors alone; at that point you'll have to start dropping the 317's input voltage.

In constant current mode, you just need to have enough left over to meet the laser diode's Vf. In most cases, it's a non-issue.

The whole '1.2V headroom' thing comes in here -- it gives the 317 some room to breathe. As the battery discharges, the voltage drops, and the headroom just allows for a further discharge (and by extension, lower voltage) before a recharge is required.

Still confused? Just ask. Hope this helped.

Oceansoul said:

Doesn't answer his question. :can:

Click to expand...

mrp1232 said:

Additional voltage to get long life between battery changes 1.2v.

Where did the 3 volts for the 317 come from, i never heard of 3v being needed for the 317 to work. I have heard of 317 adj being able to put out 1.2v. Could someone just tell me where i went wrong or what i'm missing.

Thank you and sorry if i said anything noobish.

Click to expand...

perhaps I didn't read that correctly....:thinking:

What was asked was where did the 3Volts (drop) come from..
above any additional voltage drops..


EDIT

Oceansoul said:

As for the 1.2V figure -- that's only relevant when the LM317 is being run in voltage regulation mode. In this application, you're using it in current regulation mode. What it means is that by varying the values of the two resistors, you can adjust the voltage linearly down to 1.2V. You can't adjust the output voltage to any lower than 1.2V using the control resistors alone; at that point you'll have to start dropping the 317's input voltage.

Still confused? Just ask. Hope this helped.

Click to expand...

IIRC you may have that explanation reversed...:undecided:


Jerry

You can contact us at any time on our Website: J.BAUER Electronics

lasersbee said:

perhaps I didn't read that correctly....:thinking:

What was asked was where did the 3Volts (drop) come from..
above any additional voltage drops..


Jerry

You can contact us at any time on our Website: J.BAUER Electronics

Click to expand...

You're not very good at this, are you?

Where did the 3 volts for the 317 come from, i never heard of 3v being needed for the 317 to work. I have heard of 317 adj being able to put out 1.2v. So that + 1.2 dropout = 2.4v; that would be the min voltage you could set right ? Maybe there is no dropout V. I am just really tired of trying to figure this out and its late on top of that. Could someone just tell me where i went wrong or what i'm missing.

Click to expand...

I don't see any explanation of exactly why there's a voltage drop, which, funnily enough, is what he wanted to know.

He's already worked out that there is a voltage drop. No further explanation needed.

Nor do I see an explanation for the 1.2V figure. :undecided:

And for the OP -- PM or post likewise if you still want help on this matter.

Oceansoul said:

You're not very good at this, are you?

Click to expand...

:huh:
:wtf:
:thinking:

EDIT....
Referring to post #13 may bring you up to speed...


Jerry

You can contact us at any time on our Website: J.BAUER Electronics

Simple Example: I put 6V through a 317, I'll get 4.5V out. 2.5V of the 6V I put in will be dropped ('lost') across the 317.


Just my 2 cents worth but i think the maths here is wrong also.

blue led said:

Simple Example: I put 6V through a 317, I'll get 4.5V out. 2.5V of the 6V I put in will be dropped ('lost') across the 317.


Just my 2 cents worth but i think the maths here is wrong also.

Click to expand...

Yeah... If you put 6V in, you'll get 3.5V out if 2.5V is dropped

6-2.5 = 3.5

So 1.2v is used by the 317 in voltage regulation mode and close to 3v for current mode Oceansoul ? I thought it was only 1.2 and not 3 for the 317, i did not think about that it is not being used for voltage regulation. Also so voltage drop and how much the 317 uses is the same thing.

Doesn't the lm317 drop ~2.5-3v plus the 1.25v over the resistors?

O never thought about that ether Bemwv. Quote by Benmwv - Doesn't the lm317 drop ~2.5-3v plus the 1.25v over the resistors?

Wait i think i got this one if theres a 1.25v drop across the resistors and if the Vdrop is close 1.2 for the 317 than thats 2.5 and with the Vdrop is not perfect in every 317 thats 3v. i could be wrong.

Yes, benmwv. The ~2.5V is for voltage regulation mode only. In current regulation configuration, you have the 1.25V drop of the current sense resistor. Mister Rog is correct. The minimum dropout depends on case temperature and load current, but it is somewhere between 1.2V and 2.5V. For a typical red, you can expect about 1.75V. Google the datasheet for graphs.

As for the battery, you can't take the full-cell voltage as a design voltage, because the cell only hangs out at that voltage for a minute or two. All cells drop voltage as they discharge. As for how much, It really depends on the types of batteries you're using.

3V: laser diode
1.75V: regulator
1.25V: current sense resistor
X: headroom

Unless you can guarantee your supply voltage will be at least 6V all the time, add some headroom.

Yes, that is right. I need 5.6V to get in regulation with a red diode and a LM317. You will need always two 3.0/3.7V cells for the LM317.

Ok i think i'm getting this but i have just a few other questions. First is the voltage drop and the voltage need by the regulator the same thing ? Also does the 317 use 2.5~3v for both constant current regulating and voltage regulating.
thanks you guys

mrp1232 said:

Ok i think i'm getting this but i have just a few other questions. First is the voltage drop and the voltage need by the regulator the same thing ? Also does the 317 use 2.5~3v for both constant current regulating and voltage regulating.
thanks you guys

Click to expand...

Yep, the Vdrop is the same regardless of how it's being operated. The voltage required by the reg is also the Vdrop.

The LM317 can't be operated in its two modes simultaneously -- you force it into CV or CC mode depending on how you've laid out the two control resistors.

The 1.2V figure is for voltage regulation mode only -- as in, the LM317 can run at constant voltage mode for any voltage between 1.2V and its maximum operating voltage (which is something around 35V).