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Newbie:Current and voltage draw through light bulb

M

Michael

Jan 1, 1970
0
I am fairly new to this and have several (simple) question:

I have a krypton flashlight bulb (markings on bulb say 4.8v .5A). The
original flashlight that the bulb was removed from used 4 AA batteries.

1. Wouldn't the 6v from the batteries be 'pushing' the bulb a bit and
reducing its life?

2. If I want to wall-wart power this bulb, what voltage wall wart would be
recommended?

3. Since most wall warts (unregulated) supply higher voltage than the
sticker says, what method
should I use to bring down the voltage to the recommended voltage
(regulator, resistor, etc)
(I remember once trying to do something similar to this project and using a
resistor. Even
though the resistor was a high wattage resistor it got REALLY hot)

4. I think this question probably will have the same answer as #3, but if I
want to dim the bulb what
method would be recommended?

Thanks for you assistance
 
J

John Fields

Jan 1, 1970
0
I am fairly new to this and have several (simple) question:

I have a krypton flashlight bulb (markings on bulb say 4.8v .5A). The
original flashlight that the bulb was removed from used 4 AA batteries.

1. Wouldn't the 6v from the batteries be 'pushing' the bulb a bit and
reducing its life?

---
For alkalines, yes, but at 1/2 an amp the internal resistance of the
cells would drop the voltage somewhat, and as the batteries were
drained the voltage would also drop. For NiCd and NiMH
rechargeables I believe their "flat" voltage region is around
1.2V/cell, so that would work perfectly.
---
2. If I want to wall-wart power this bulb, what voltage wall wart would be
recommended?

---
4.8V @ 0.5A, but since that's not a readily available value I'd get
one rated at 5V @ 0.5A and drop the voltage to the lamp with a
resistor:


Vin - Vlamp 5.0V - 3.8V
R = ------------- = ------------- = 2.4 ohms
Ilamp 0.5A

The resistor would need to dissipate:


P = IE = Ilamp * Vin - Vlamp = 0.5A * 1.2V = 0.6 watts


Which means you should use a resistor capable of dissipating at
least 1 watt; a higher wattage allowing the resistor to run cooler.
---
3. Since most wall warts (unregulated) supply higher voltage than the
sticker says, what method
should I use to bring down the voltage to the recommended voltage
(regulator, resistor, etc)
(I remember once trying to do something similar to this project and using a
resistor. Even
though the resistor was a high wattage resistor it got REALLY hot)

---
Most wall-warts put out their rated voltage when they're fully
loaded, so in your case, since your load has an odd voltage
requirement you'd want to pick a wall-wart with a slightly higher
voltage but which is rated for the same output current as your load,
then drop the extra voltage with a series resistor, as illustrated
above.
 
H

Homer J Simpson

Jan 1, 1970
0
I have a krypton flashlight bulb (markings on bulb say 4.8v .5A). The
original flashlight that the bulb was removed from used 4 AA batteries.

1. Wouldn't the 6v from the batteries be 'pushing' the bulb a bit and
reducing its life?

When you draw .5 A from the batteries you wind up with 4.8 V (not strictly
true but it works).
2. If I want to wall-wart power this bulb, what voltage wall wart would be
recommended?

4.5 V.
4. I think this question probably will have the same answer as #3, but if
I
want to dim the bulb what
method would be recommended?

Use a multi voltage unit, run at 3 V.
 
J

jasen

Jan 1, 1970
0
I am fairly new to this and have several (simple) question:

I have a krypton flashlight bulb (markings on bulb say 4.8v .5A). The
original flashlight that the bulb was removed from used 4 AA batteries.

1. Wouldn't the 6v from the batteries be 'pushing' the bulb a bit and
reducing its life?

2. If I want to wall-wart power this bulb, what voltage wall wart would be
recommended?

5V ones are fairly common, use one of them.
3. Since most wall warts (unregulated) supply higher voltage than the
sticker says, what method
should I use to bring down the voltage to the recommended voltage
(regulator, resistor, etc)

5V ones are usually regulated :), or you could use a 4,5V unregulated one
that'd probably be close enough.
(I remember once trying to do something similar to this project and using a
resistor. Even
though the resistor was a high wattage resistor it got REALLY hot)

to drop 0.2V at 0.5A you need a ideally a 0.4 ohm resistor, 0.39 will
probably be will close enough, actually 5V would be close enough.

usually it's just easier to buy a nightlight,

Bye.
Jasen
 
M

Michael

Jan 1, 1970
0
John Fields said:
---
For alkalines, yes, but at 1/2 an amp the internal resistance of the
cells would drop the voltage somewhat, and as the batteries were
drained the voltage would also drop. For NiCd and NiMH
rechargeables I believe their "flat" voltage region is around
1.2V/cell, so that would work perfectly.
---


---
4.8V @ 0.5A, but since that's not a readily available value I'd get
one rated at 5V @ 0.5A and drop the voltage to the lamp with a
resistor:


Vin - Vlamp 5.0V - 3.8V
R = ------------- = ------------- = 2.4 ohms
Ilamp 0.5A

The resistor would need to dissipate:


P = IE = Ilamp * Vin - Vlamp = 0.5A * 1.2V = 0.6 watts


Which means you should use a resistor capable of dissipating at
least 1 watt; a higher wattage allowing the resistor to run cooler.
---


---
Most wall-warts put out their rated voltage when they're fully
loaded, so in your case, since your load has an odd voltage
requirement you'd want to pick a wall-wart with a slightly higher
voltage but which is rated for the same output current as your load,
then drop the extra voltage with a series resistor, as illustrated
above.


Thanks
 
M

Michael

Jan 1, 1970
0
Homer J Simpson said:
When you draw .5 A from the batteries you wind up with 4.8 V (not strictly
true but it works).


4.5 V.


Use a multi voltage unit, run at 3 V.

Thanks
 
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