2 Stage Circuit: Stepdown/DD & Questions about Optics

Originally posted by IsaacHayes:
<pre>
+++...........Q1.......|
Bat | | |
Bat |.ZDiode...|..R1...|
Bat |
Bat |
---..........LS-LED.....
|capcitor|
</pre>
Q1 = transistor (pnp, npn? I forget)

Zdiode = Zenier Diode, reverse voltage breaks down ~3.5v
R1= Suficient value to drop 6v down to 3.5 for the LS.
Capacitor is for a neat fade out effect when shut off.

So when the voltage is above 3.5, the Zenier diode breaks down and applies voltage to the Transistor to shut it off from DD the LS. Then the current continues through the limiting resistor. When the voltage is below 3.5, the diode doesn't break down and the transistor contiunes to DD the LS.

Will this work?

Click to expand...

<font size="2" face="Verdana, Arial">No.

Do a search in the past posts on "LM334" for a better idea.

Ok. Now I can for get that idea. But if I'm going to do some small circuit, It must be using radioshack parts. So should I just forget using a stepdown circuit and just use resistors? or is there a tried & tested schematic out there? I haven't messed with electronics in about 4 years so I just about forgot everything.

What about the optics portion?

I just purchased what will be the host of a 1W Cyan LS. It should look pretty cool and futuristic, hopefully it will have a nice scifi look to it with the LS. It should arrive in the next week.
Now I must get the other parts ready. But I need to figure out what are best for my plans..

1st off is the electronics for it. It will be powered by 4 AA's. So I need to step down the voltage with a resistor. But then once the voltage drops below the LS rated voltage, would it be best if the resistor was no longer used, and then DD the LS?

If so, this circuit should do the trick right?
Or is there no advantage to removing the resistor once the voltage drops below ~3.5v?
<pre>
+++...........Q1.......|
Bat | | |
Bat |.ZDiode...|..R1...|
Bat |
Bat |
---..........LS-LED.....
|capcitor|
</pre>
Q1 = transistor (pnp, npn? I forget)
Zdiode = Zenier Diode, reverse voltage breaks down ~3.5v
R1= Suficient value to drop 6v down to 3.5 for the LS.
Capacitor is for a neat fade out effect when shut off.

So when the voltage is above 3.5, the Zenier diode breaks down and applies voltage to the Transistor to shut it off from DD the LS. Then the current continues through the limiting resistor. When the voltage is below 3.5, the diode doesn't break down and the transistor contiunes to DD the LS.

Will this work?
I'm thinking though that as the batteries approach being 3.5 volts, the LS will dim, and then when they go below 3.5 the LS will brighten up all of a sudden when it goes into DD mode. Am I correct?
Is there a better way to regulate this?

Secondly, I want the Optics for this to be as tight of a beam as possible. Unfortunatly in this cool looking flashlight body, I can't fit 30mm optics. Is a HD + NX05 the way to go?

Thanks for the help.

Lots of people order parts from these guys: http://www.digikey.com/

Shipping is $5. or free for orders over $25.

They'll send you a free catalog to study.

Your initial scheme will work OK and you will get decent performances. A typical LS will require no more than 3.5V forward volltage, thus giving you 1.5V to be regulated by the transistor. One suggestion though is to move the R1 right-hand lead from the LS positive to battery negative. A trade off for this kind of simple regulator is you only get ~60% efficiency(3.5/6), so the transistor will become fairly hot during operation, and you do need to select high power-rated transistors (use at least 1W--- 350mA*1.5V=0.525W, new batteries will give even higher voltage)

If you go with the current scheme, you will need a PNP type transistor, but I would suggest reverting the polarity to NPN as it's easier to make NPN-type transistors, and NPN's will be cheaper than similarly rated PNP's.

Originally posted by jeff1500:
Lots of people order parts from these guys: http://www.digikey.com/

Shipping is $5. or free for orders over $25.

They'll send you a free catalog to study.

Click to expand...

<font size="2" face="Verdana, Arial">I just ordered from them. There is handling fee of $5 on orders under $25 but the shipping is still extra. So on a $10 order you would pay $10 for parts, $5 for handling and $? for shipping. Ugh.