Small Modification to LM317 Drivers

[LaserFerret]

Anytime someone asks how to make their own laser driver or how to build a laser This Link gets passed along with this image:



But I think there's a bit of a flaw with it, when the diode is placed where it is inserting the batteries backwards will short and fry the LM317, and would still most likely put voltage reverse biased into the diode, possibly damaging the diode and most likely destroying the LM317.

But if the diode is simply moved to the position shown in this modified picture I made these problems should be avoided.



Anybody else notice this problem? Of course you don't have to make your schematic look like ROGs, and ROGs schematic is just an example of this sort of driver, not the definitive layout.

 

[jib77]

Not exactly correct ... the diode in the first pic will conduct in a reverse polarity situation instead of the LD because its a lower resistance than the Laser Diode, and being in parallel it does not load the circuit (except for its tiny reverse leakage current) unless you insert the cells incorrectly.

The way you have it also works but you will always load your circuit, dropping the battery's input to the IC by the diode's forward voltage. Very inefficient.

 

[random person]

Not exactly correct ... the diode in the first pic will conduct in a reverse polarity situation instead of the LD because its a lower resistance than the Laser Diode, and being in parallel it does not load the circuit (except for its tiny reverse leakage current) unless you insert the cells incorrectly.

The way you have it also works but you will always load your circuit, dropping the battery's input to the IC by the diode's forward voltage. Very inefficient.

thats what she said...


sorry.. couldn't help myself... yup hes correct. Not good burn power over that diode if you don't have to. Also causes a voltage drop that is unwanted if your power source is limited.

 

[rhd]

thats what she said...

I don't get it... ?

But yes, I do agree with the analysis above. You don't want to be dropping voltage for no reason.

That said, I could see some merit to moving the diode to BEFORE the IC, while still keeping it in parallel.

 

[SOG]

Somehow I keep thinking the Diode is like an LED, it's a Semi-conductor... so battery is not supposed to be able to pass it form the other way around... so how does it get burn by installing the battery incorrectly?

Does install another piece of Semi-contuctor will work ?

 

[123splat]

SOG,
Well, sorta... A LED is a Diode (that emits light, Light Emitting Diode, LED). The reversed diode will not pass voltage below it's breakdown voltage level, so it will BLOCK voltage from a reversed battery. But forward biased, you have to overcome the diode's forward voltage drop to conduct, that is where you loose the voltage using it for reversed battery protection (it is a trade off, you have to decide if the loss is worth the benefit).

RTD, If you install a reversed battery before the E-reg, you will get a short to ground, if you reverse the battery, but you may burn the 1A diode open before the battery shuts down, so no real protection... If you take out the reversed diode in the output side, you loose LASER diode protection. I'd rather fry a regulator chip than puddle my LASER diode. But it's up to you. B.T.W. I don't use the 317 regulators for LD's. I'm part of that group looking for the ideal single battery driver (Go BOOST!!, or, Low Drop Out pass regulator).

 

[rhd]

RTD, If you install a reversed battery before the E-reg, you will get a short to ground, if you reverse the battery, but you may burn the 1A diode open before the battery shuts down, so no real protection... If you take out the reversed diode in the output side, you loose LASER diode protection. I'd rather fry a regulator chip than puddle my LASER diode. But it's up to you. B.T.W. I don't use the 317 regulators for LD's. I'm part of that group looking for the ideal single battery driver (Go BOOST!!, or, Low Drop Out pass regulator).

I usually use a 1N5404, so I wouldn't be too concerned about it frying first, as most of the battery combos I build linears with can't supply much more than 3A in the first place. That said, I usually don't use a protective reverse polarity diode in the first place. I just don't put my batteries in incorrectly

 

[123splat]

Best practice, he, he, he...

 

[qumefox]

I'd also rather lose the regulator than have a li-ion (or two) spontaneously combust in my hand from having their output shorted.

I think with as many electronics guru's as we have around here.. if there was a better way of wiring the DDL, it would have come out by now. It is a fairly straightforward circuit after all.

 

[random person]

Sorry, just for confirmation... is any damage done to the laser diode if wired in reverse? Anyone know the reverse breakdown voltage for LDs?

Yea, you wouldn't want to put the diode before the IC. That would fry the diode and then there would be no protection for the LD.

 

[qumefox]

Depends on the laser diode. It is possible to kill them by wiring them backwards, but it really just depends on how much current you try to push through them before you realize what's wrong.

 

[LaserFerret]

Not exactly correct ... the diode in the first pic will conduct in a reverse polarity situation instead of the LD because its a lower resistance than the Laser Diode, and being in parallel it does not load the circuit (except for its tiny reverse leakage current) unless you insert the cells incorrectly.

The way you have it also works but you will always load your circuit, dropping the battery's input to the IC by the diode's forward voltage. Very inefficient.

You said the exact problem in your message, "the diode in the first pic will conduct in a reverse polarity situation instead of the LD because its a lower resistance than the Laser Diode"

Thus, shorting the circuit. Then you fry your batteries and your LM317.

While you may avoid the voltage drop from the diode by putting it in parallel when the batteries are inserted incorrectly you short the circuit. The .7 volt drop the diode makes will only drop the input voltage to the IC to 6.5 volts when using two 3.6v batteries such as two CR123As or two 18650s, and the circuit will still run perfectly well on 6.5 volts, needing only 6 volts to run.

As RHD said it's best to simply learn to never install your batteries incorrectly. But if you happened to do so with the original circuit you would still be damaging components, and while the second circuit makes a voltage drop it will not complete the circuit and will not short anything in the case of backwards batteries, plus the voltage drop is negligible and will not prevent the driver from operating correctly.

 

[lasersbee]

I often use an inline Diode for reverse voltage protection
on some circuits but I use a schottky diode that only has
a 0.3V drop...


Jerry

 

[rhd]

You said the exact problem in your message, "the diode in the first pic will conduct in a reverse polarity situation instead of the LD because its a lower resistance than the Laser Diode"

Thus, shorting the circuit. Then you fry your batteries and your LM317.

I don't think that's an accurate statement of how the DDL circuit would behave in such a situation.

Rather, the protective diode will probably just drop its 0.7V, leaving the IC to dissipate the rest of the voltage as heat. Not great long term, but the IC can handle it for a few seconds no doubt.

However, I'm not actually sure how the IC will handle the reverse itself. It might block the current on its own. After all, there is a reason we have separate negative regulating ICs - because these don't simply work the same in reverse.

 

[LaserFerret]

I really don't see a downside to putting the diode inline except for the voltage drop, which is minimal and with good diodes as laserbee said the voltage drop can be very minor.

I will make the edited driver just as soon as I get my potentiometers in, and observe if there are any performance issues with the diode placed in series.

 

[jib77]

Just make sure you get one with a forward continuous current greater than the max current you are planning to provide.

0.3-0.5V may not seem like much but it might be the difference between being able to run a 445 from two cells instead of 3. Remember the LM317 requires 3V overhead ... so your nominal Vin of 7.4V just became 6.9 which means only 3.9V available for your load.