Help on choosing the right resester - Page 1
Forum section: electronics / beginners

author: robertlopez925
author_rank: Contributor
timestamp: April 08, 2014, 06:42:11 pm
content: i have a grain of wheat bulb and i need 5v of power to the bulb my supply of power is putting out is around 18-23v ac ad a max of 12 amps i am still learning electronics so this will be a big help

author: Holograph
author_rank: Regular Contributor
timestamp: April 08, 2014, 06:49:58 pm
content: Just one bulb? How much power or current does it take? Does it specify whether it can run on AC, DC, or both (both is fairly likely)? And what is that power supply from?From the information you have given, assuming you only have 1 of these bulbs, you can either use a small transformer to step down that voltage and run the bulb with AC, or you can rectify it, add some capacitors to smooth it out, and use a cheap LM2596 buck converter module for a few bucks from Ebay.Your power supply has too much voltage to run one of these alone, so you must use either use multiple bulbs in series to spread the voltage out, or pad it down. You could also just use a resistor with a resistance about 3x (or a little more) as much as the bulb itself. But that would be inefficient and the resistor would need to be rated to 3x as many watts as the bulb. Which is why I recommend a transformer or buck converter (or a power supply with a lower voltage). These sorts of lights can get pretty hot, too. Might want to look into LEDs instead if possible.To calculate the resistance of the bulb, use Ohm's Law on its rated voltage and current.

author: robertlopez925
author_rank: Contributor
timestamp: April 08, 2014, 07:01:07 pm
content: yes just 1 bulb...i do not have any idea on the amount of current it takes the gran of wheats take i know they like 5v ac... our power supply is ac 650W 12Amp power supply it runs toy trains so the current can't be much to run a little of grain of wheat... and a step down transformer is hard to hind under a shell of a locomotive but a rectifier might work....

author: robertlopez925
author_rank: Contributor
timestamp: April 08, 2014, 07:48:46 pm
content: all right did a little experiment i had a national semiconductor LM2937 500MA Low dropout regulator exact model number is LM2937ET5.0so i am feeding it 19v Ac and on the output side is 2.69v out and the light was almost there i need maybe 1.5 to 2 v will the next model up the LM2937ET8.0 Give me more voltage or less if so how much more?or how much less ?

author: Holograph
author_rank: Regular Contributor
timestamp: April 08, 2014, 07:52:48 pm
content: Quote from: robertlopez925 on April 08, 2014, 07:01:07 pmWell, let me ask you again, a little more directly: Is it an option to use an LED? If you do use an LED, all you'd need is a rectifier, possibly a capacitor (smaller than if you were going to drive the incandescent with DC) to stop it from flickering, and a resistor. A resistor will be far more efficient with an LED because while it will still be dropping approximately the same voltage, there will be much less current, meaning less power wasted in the resistor. That makes your overall system more efficient, plus allows you to choose a low-power resistor (a normal 1/4 watt resistor would most likely be fine in this case). If you can accept some flicker in the LED, you can even leave the rectifier and capacitor out.If you're looking to use the bulb you have simply because you already have it (but if there's no other reason to specifically use it)... I honestly would say it's still not worth using.

author: Holograph
author_rank: Regular Contributor
timestamp: April 08, 2014, 08:01:43 pm
content: Quote from: robertlopez925 on April 08, 2014, 07:48:46 pmWhile a linear (LDO included) regulator is a valid option to lower the voltage, just so you know, it's still really, really inefficient. It's got the same problem as the resistor that I mentioned in my first reply - the only advantage is that you won't have to do any work to figure out the resistance of the lamp (and technically you'd be able to put in any lamp as long as it requires the same voltage). By the way, the low dropout has zero benefits here over a non-LDO linear regulator (though it's valid to use).a 500mA regulator at 5V is going to give you a max of 2.5W (P=IV). What wattage is the lamp?There are lots of regulators that can handle more current. Even some 7805s can handle 1-1.5A. If you use a regulator, I'd recommend it be able to handle at least 25% more power than the lamp draws. Also, depending on the power draw, it will probably need a heatsink. Like the resistor method, the regulator is still going to have to dissipate about 3-4x as much power as the load. Also keep in mind to use a regulator you'd be, again, rectifying and filtering your AC voltage first.I do not recommend using a regulator for this, at least not a linear one, and not with that power supply and bulb. Excluding the poor efficiency of the bulb itself (i.e. how much of the electrical power is converted to light - LEDs will probably be 3-5x as good at this than the lamp in question), your circuit is still going to be something like 20% efficient with that power supply, that lamp, and a non-switchmode regulator. That's terrible! That's why I mentioned a switchmode regulator (the LM2596 module, which again, you can get cheap on Ebay) as an option before. A linear regulator is basically a variable resistor (just one that adjusts itself via feedback) and is subject to the same sort of power sinking requirements.If you don't care at ALL about efficiency and want to do this with as few parts as possible, just say screw it and order a 25 or 50 watt power resistor from Ebay (shouldn't be more than a few bucks, but you'll have to wait for shipping from China) and use that. Sizing would be done via Ohm's Law - if you don't know the current or power draw of the lamp, stick it on a real 5V power supply and measure the current. That way you'll know the voltage of the lamp (5V) and the current (whatever you measure) and the resistor value you'd need to buy would be that, proportional to the voltage drop (i.e. 5V for the lamp, your supply voltage - 5V for the resistor - so multiply your lamp's calculated resistance by (Vsupply-5V)/5V). Resistors of that power level usually come in metal cases that should be capable of dissipating anything your mini lamp would require.

author: robertlopez925
author_rank: Contributor
timestamp: April 09, 2014, 07:47:56 am
content: Quote from: Holograph on April 08, 2014, 07:52:48 pmi am replacing most of the grain of wheats with these LED'S* Emitted Color : Warm White* Size: 3mm* Lens Color : Water Clear* Forward Voltage (V) : 3.2 ~ 3.4* Reverse Current (uA) : <=30* Luminous Intensity Typ Iv (mcd) : 12000* Life Rating : 100,000 Hours* Viewing Angle : 20 ~ 25 DegreeAbsolute Maximum Ratings (Ta=25°C)* Max Power Dissipation : 70mw* Max Continuous Forward Current : 20mA* Max Peak Forward Current : 75mA* Reverse Voltage : 5V* Lead Soldering Temperature : 240°C (<5Sec)* Operating Temperature Range : -25°C ~ +85°C* Preservative Temperature Range : -30°C ~ +100°Ci am using 1k resister in line to the led to regulate it down to a safe voltage for the led'sthe two remaining grain of wheats are hard to get to i was hoping to not change them out but if i haft to i willas for my LM2937 it has a huge heat sync matter in fact it has 5" x 32" for 1 LM2937which is huge i do agree it is not efficient but it does work...now something i just thought about is a 5v board i have seen 5v power boards some where all they do if i remember is regulate power down to 5v is this a option for what i wantFinally i found the grain of wheat resistance after hours of reaching it is a "5V 80MA" grain of wheat

author: David Hess
author_rank: Super Contributor
timestamp: April 09, 2014, 11:27:10 pm
content: A 5 volt grain of wheat bulb will be either 60 milliamps or 115 milliamps. For 23 volts worst case, the needed voltage drop will be 18 volts:18 volts / 0.06 amps = 300 ohms 1 watt resistor minimum18 volts / 0.115 amps = 157 ohms 2 watt resistor minimum