Hi, It depends on the lamp's rating. If you lower the voltage to half you have to double the current to get same Wattage. Can the battery do that with the lamp you have? Automotive type bulb draws quite bit of curent.
I have done that before. They usually do not burn out immediately, but glow brighter (and whiter) for awhile.
Of course that is not what I was talking about before. Maybe you missed that what I said was an ALTERNATIVE to using 12V.
?
I was responding to your suggestion, visible above, that "if the power supply was big enough" he could connect two 6V lamps in parallel to a 12V supply.
That won't work for very long.
And what differene does it make? High voltage or low voltage circuit. Basically it is all govenred by simple Ohm's law.
Where did you get the formulas for these? As you say, an incandescent lamp filament is far from a fixed resistor, but how does resistance and temperature change with voltage?
Dave
Put a speed controller on the motor. A simple rheostat (variable resistor) demonstrates how motor speeds can be changed.
Get various resistors. Demonstrate how the different color bands correspond to ohm numbers - and then measure them with the meter. Radio Shack once sold little color wheels. Dial in the color and read te number.
Reading the values and then measuring with the meter should teach something about how the real world 'varies' from what is should be.
Meanwhile some numbers for that light bulb. A 12 volt light operating at 6 volts will output about 1/10 the lumens and consume 33% of watts as compared to 12 volt power consumption. Obviously the bulb is less efficient at those lower voltages. At 6 volts, the bulb will last about 8200 times longer.
A 12 volt bulb at 9 volts will output 40% the light at 12 volts and consume about 60% of the wattage. Unlike Pop', numbers can be provided with replies.
Put a speed controller on the motor. A simple rheostat (variable resistor) demonstrates how motor speeds can be changed.
Get various resistors. Demonstrate how the different color bands correspond to ohm numbers - and then measure them with the meter. Radio Shack once sold little color wheels. Dial in the color and read te number.
Reading the values and then measuring with the meter should teach something about how the real world 'varies' from what is should be.
Meanwhile some numbers for that light bulb. A 12 volt light operating at 6 volts will output about 1/10 the lumens and consume 33% of watts as compared to 12 volt power consumption. Obviously the bulb is less efficient at those lower voltages. At 6 volts, the bulb will last about 8200 times longer.
A 12 volt bulb at 9 volts will output 40% the light at 12 volts and consume about 60% of the wattage. Unlike Pop', numbers can be provided with replies.
I hooked them up in parallel and they seem to work just fine. Incredibly bright too.
A decent approximation is that the resistance is proportional to the absolute temperature. A more accurate relationship is given at the end of this post.
For my calculation, I used an iterative approach which goes as follows (somewhat lengthy description, skip if not interested. NOT FOR MATHOPHOBES!):
Step "0". the design values:
First, assume three conditions at the filament's designed operating point, eg. power, voltage, and filament temperature (3000 Kelvin is a reasonable number). Current and resistance are then determined from power and voltage.
Step 1. 1st iteration for new operating point: Next, change the voltage but do the calculation for the same resistance, as if it were fixed. This gives a new power and current, which will be wrong, but our iterative approach will converge to the correct values. We also need to calculate a new temperature using the blackbody radiation law:
Power is proportional to T^4 or in other words T is proportional to P^0.25
The way we REALLY use this relation is:
T_1 = T_0 * (P_1 / P_0)^0.25 where "0" refers to the original operating point, and "1" refers to the updated temperature and power.
Step 2. 2nd iteration of values: Recalculate the resistance, R_2, using:
R_2 = R_0 * (T_1 / T_0)
Since voltage is fixed, we can recalculate power and current from
I_2 = V / R_2 P_2 = V * I_2
And then a new temperature from T_2 = T_0 * (P_2 / P_0)^0.25
Step 3. Recalculate the resistance: R_3 = R_0 * (T_2 / T_0)
Then recalculate I_3, P_3, and T_3 as in Step 2.
Steps 4, 5, etc. Keep repeating the process until the power, temperature, and resistance converge to steady values that don't change significantly.
That's it, really. A somewhat more accurate description (for tungsten, anyway) is that resistance is proportional to T raised to the power of 1.2. Accounting for this did not change the results by very much.
Regards,
Mark
That seems unlikely. There's probably a resistor somewhere. You may have a LED module with built-in resistor (it's still a resistor, where ever it is). Those modules are often being sold for use in flashlights now (just replace the bulb). It could be the battery. Little batteries (button cells) have significant internal resistance.
I actually tested that on a new LED (they're not THAT expensive). That LED isn't working any more.
LEDs in parallel with each other? It's unlikely that all would light, since the threshold voltages would be slightly different, and the one that's lowest would prevent the others from lighting.
As someone said earlier, your LEDs probably are actually LED modules, and come with built-in resistors. You have a separate resistor in series with each LED.
Modern LEDs can appear very bright. I noticed that with the holiday lights I had this year (yes, I know that's "last year", but it is still less than 2 months ago). Those LEDs look brighter than the miniature incandescent's.
That just might have worked IF I had forgotten what I said, and you had edited my quote to say that. "12V" does not appear in that quote at all.
OK, I left something out (saying to use 6V, or clarifying what "big enough" means). That's no excuse to stick in something that doesn't belong.
Apparently you *did* forget what you said, or at least you forgot what you were responding to -- which was the girl's belief that, to light two 6v lamps, she should connect them to a 12v supply.
*I* said that would work fine if they were connected in series -- which is true, as each lamp would see 6v. *You* said "or connect them in parallel if the power supply is big enough".I didn't stick *anything* in there, and you know it.
If they're all from the same batch, they're probably pretty well matched in forward voltage, and it only takes a little bit of series resistance within each LED to approximately balance the current between LEDs. This isn't a *good* way of connecting multiple LEDs, but it's not automatically doomed to failure.
The little 1 W and 3 W LEDs appearing in flashlights are now brighter than anything I ever saw from any incandescent bulb in the same size of flashlight (2 AA battery).
Dave
He said 3V LEDs in one post - that implies the internal resister.
Bob
Larry inspired greatness with:
They may have a resistor packaged with it, but a LED is a *DIODE*, and it requires a resistor.
Well, yes that is true. However they do not require an external resistor in order to make a complete lighting circuit. Whatever resistance is needed in built into the unit itself.
For the sake of my education let me ask. Doesn't an LED have a predictable voltage drop across the junction? I seem to recall 1.5 volts is that correct? If it is I only need as many LEDs in series as will equal the voltage and they will need no external resistance. Is my memory wrong?
-- Tom Horne
The voltage-current characteristic of an LED has a pretty sharp "knee" characteristic of most diodes. Very little current flows as voltage increases up to the turnon voltage, and then current rises very rapidly with very little change in voltage. The actual voltage depends on the type of LED (different colours use different chemistry and have different voltage) and temperature.
LEDs are easy to drive from a constant-current power source. You set the current you want, and the power supply adjusts its voltage to whatever is needed to make the LED draw that amount of current. Put a bunch in series, and all get the same current.
But most power sources are closer to constant-voltage instead of constant-current. It's nearly impossible to set the voltage needed for a particular current, since a small voltage change results in a large current change, and the voltage needed depends on temperature (which depends on current among other things!).
So you generally use a resistor in series with the LED, or string of LEDs. The power supply voltage is equal to the sum of the voltages of the LEDs in the string, plus a few extra volts. The resistor is sized to give the desired LED current with the "extra" voltage applied across it. This gives a system where small changes in LED turnon voltage produce small changes in the voltage across the resistor, which produce small changes in current through the LEDs. Without the resistor, you'd get large and unwanted changes in current.
As others pointed out, it is possible to build a system using only a LED and a battery *if* the battery's own internal resistance is enough to stabilize the system.
Dave
I considered that. "Built-in" is not equivalent to "nonexistent".
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