In DC systems, what could be an advantage of lower-voltage (higher-current) light bulb over a higher-voltage one, assuming the input power is the same? What could be a higher-voltage (lower-current) light bulb's disadvantage? Would the low volt high current lamp have a longer life because its filament has thicker wire that would take longer to evaporate ? Would the low volt bulb be less likely to fail when vibration is an issue ? Would the low volt lamp be able to run a higher filament temp leading to more lumens per watt?
Electronic – What are the advantages of low-voltage light bulbs
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Related Solutions
Basically, this won't really work - or work well. An alternative is presented at the end of this answer.
Before that, the calculations, and an explanation of why it won't work.
From Wikipedia:
The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating.
- Assuming the resistance of the filament was measured when the bulb was not lit, this gives a rough estimate of
77 x 15 = 1155 Ohms. - This implies a ~45 Watt bulb, let's go with 50 Watts for convenience. This back-calculates to a hot filament current of around
I = P / V = 217 mA = ~220 mA. - To obtain ~5 Volts at 220 mA,
R = V / I = 22.727 Ohms = ~ 22 Ohms, the nearest standard resistor value. - Power dissipated by this resistor:
P = V x I = 1.1 Watts, so let's go with a 5 Watt resistor to be safe. - LED choice: Let's say red LED, 1.8 Volt forward voltage, 20 mA nominal current. Such an LED will light up quite well at 6 mA (230 Volt situation) and will also stay within current limits at 12 mA (460 Volts situation).
- Therefore a suitable LED current limiting resistor is 560 Ohms
Using two LEDs wired in anti-parallel would be recommended, because LEDs typically have low reverse voltage ratings (often as low as 5 Volts). With two LEDs facing opposite ways, the maximum reverse voltage appearing across the non-conducting LED will be the forward voltage of the conducting one, at any point in the AC cycle. It'll work fine in DC as well, so all is good.
Now, the problems:
What happens when the bulb is just switched on, and the filament is cold?
- Resistance = 77 Ohms, therefore current = ~ 3 Amperes! That's before adding the bleed resistor of 22 Ohms
- Current with 22 Ohms added in series: 2.323 Amperes. Power across resistor: 119 Watts!
- Result: Fireworks.
What happens when the bulb is supplied with 460 Volts?
- Assuming it is a 230 Volts rated bulb: The filament burns out.
- If it is a 460 Volts rated bulb: Not much goes wrong other than the sense resistor blowing up anyway.
Solutions:
- Use the light bulb to illuminate the LDR directly.
- Use a high wattage bulb in series with the 50 Watt bulb, for the voltage divider arrangement.
- Use an alternative, such as a current sensing transformer, then use that CT's output in place of the LDR for sensing. A CT can be made by winding several turns of thin enamel wire around the wire supplying current to the light-bulb.
- For AC and DC current sensing, use a prebuilt module such as this one ($4.35 from eBay including international shipping) that incorporates a nice "lil black IC", the Allegro ACS712, and will provide safe, nicely isolated current sensing:
My first reaction is to put the two panels in series, then use a off the shelf buck regulator chip. If the two panels can't even put out 6 V or so in series (3 V per panel), then there is so little power available that it doesn't matter if the output is just shut off.
Best Answer
Low-Voltage incandescent lamps can withstand significantly-more vibration than high-voltage lamps. This is because the filament is thicker, which makes it more sturdy.
This property of low-voltage incandescent lamps is one of the many reasons that 42 Vdc vehicle electrical systems never became popular. There are other reasons - this was just one.