October 28, 2016 at 11:44 pm #20568nflourishParticipant
Hey folks. So my son went to a Maker Faire where he was enticed into a soldering workshop in which, you guessed it, they were using the Larson Scanner kit! He got it for free! I was astounded how well he did. He’s 5.6 years old. Yes it works. Yes he did 70-80% of the construction.So then he wants to use it in his halloween costume. The long story short is, for it to work with the GB Proton Pack I’ve built for him, it needs to run off of a 9v battery.I did some putzing around trying to figure out what the voltage and current requirements for the Larson Scanner are… but found A) that my multi-meter’s DVA mode is hosed (probably a bad fuse), and B) the controller chip for the kit has documentation that reads close to greek for me.Sooooo … I explored what it takes to bring 9v down to 3v and saw some hack solution to just add appropriate resistance. Not knowing the current, I was guessing so I started very high and dialed the resister value back down and down until it started working. Sort of. The scanner does scan, and the control button does control _speed_ … but on the 9v battery with resistance, it does not seem to have the ability to switch between high and low brightness… which is honestly fine. But confusing.I am a very green circuit guy. Physics classes and digital logic classes in college over 10 years ago are just enough foundation to fall off of. Honestly I had never soldered a circuit onto a PCB until this week… my son did it before I did. I’d only ever used solderless breadboards.So give it to me straight: what is the quickest hackiest and yet still valid way to run the Larson Scanner off of 9v? And then, what is the most correct way to run the Larson Scanner off of 9v?Thanks for the education!Nick
October 29, 2016 at 12:53 am #22641Windell OskayKeymaster
- This topic was modified 2 years, 11 months ago by Windell Oskay.
The microcontroller on the Larson Scanner is rated for an absolute maximum of 5.5 V. Any more than this can permanently damage it, and potentially create a hazardous situation. Chips have been known to overheat or even explode if exposed to too much voltage. And, the resistors in the circuit are configured to give a safe level of current at 3 V.Resistors alone cannot reduce the voltage from the battery in such a way that there is a guaranteed safe voltage level. (This is because as the current drawn goes to zero, so does the voltage across the resistor– effectively, the full 9 V goes across the microcontroller at turn-on time.)The *safe* way to run the Larson Scanner would be to give it its own 3 V power supply– from a separate pair of AA or AAA batteries, or a CR2032 coin cell if need be. This can be very compact, if that is the issue.Another method would be to use a set of diodes to drop the voltage. A small signal diode (such as a 1N914) drops about 0.7 V. You could use 8 of them in series to drop the 9 V to about 3.4 V. Or, you could use three red LEDs, each of which drops about 1.8 V, to drop 5.4 V, giving you about 3.6 V. This would also have the side effect of giving a few more red LEDs for the costume.October 29, 2016 at 10:41 pm #22642nflourishParticipant
A) Awesome information. Thank you! This might explain why one of the red LEDs on the scanner has suddenly stopped working! But, if a single LED has gone, it does not seem like that suggests the IC is damaged… because one would have to assume if the IC fried, none of the LEDs would work? I won’t know until ~maybe~ past Halloween because I can’t really disassemble the Larson scanner part of the pack. You can see it in this video pretty clearly…
Hadrian Ghost Buster #1B) So – if using only resistors, once the circuit is fully powered on, the resistors have “gone into effect” and thus … it is safer to run it constantly with a resistor constrained circuit than to turn it off and on with a resistor constrained circuit? Each new power cycle brings with it the over-voltage risk that, if passed safely, is fine until the next power cycle? Or am I mis-understanding what you wrote?C) So diodes always… immediately … constrain voltage… which includes LEDs that are related to diodes? No “warm-up” time?The idea of adding in 3 extra red LEDs is _by far_ the best choice for kid’s costume. Nice call. I’ll add some LEDs and presumably a resistor still needs to hang out in front of the first LED in the circuit..?Thank you Evil Mad Scientist!NickOctober 29, 2016 at 11:32 pm #22643Windell OskayKeymaster
(A) The IC is capable of sourcing enough current (40 mA) to destroy the resistor (typically rated for 30 mA maximum). Also, LEDs blow instantly, whereas the IC has some protection circuitry, and usually withstands being out of spec for a little while. So… no surprise. However, it’s not unheard of for an LED to be out for more mundane reasons, such as an incomplete solder joint that opened up.(B) That would be correct only if the current draw were constant.(C) Diodes are very different from resistors. They drop an essentially _constant_ amount of voltage, over a wide range of currents, whereas resistors have vanishingly small voltage drop at small current. If using diodes to drop the extra voltage, you won’t need a resistor in there.
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