Guest
is it possible to make an LED lamp with a 35v / 25mA supply? if so can
anyone teach me?
anyone teach me?
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D
David L. Jones
Guest
<giteco@gmail.com> wrote in message
news:211f3ccf-e1c7-4bae-a990-10072742353b@s38g2000prg.googlegroups.com...
with the LED connected to your 35V supply.
If you don't like the resistor getting too hot then try a 5W resistor or a
couple of 1W resistors in series or parallel to give you the same value.
Find the brightest LED you can that runs at 25mA.
You can also puts extra LEDs in series, dropping your resistor value
accordingly.
Your supply is capable of running better LEDs like say a 1W luxeon, but that
would need a suitable DC-DC converter, not nearly as easy unless you buy a
pre-built module.
Dave.
news:211f3ccf-e1c7-4bae-a990-10072742353b@s38g2000prg.googlegroups.com...
Yes, easy (assuming your supply is DC). Use a 1.5K 1W resistor in seriesis it possible to make an LED lamp with a 35v / 25mA supply? if so can
anyone teach me?
with the LED connected to your 35V supply.
If you don't like the resistor getting too hot then try a 5W resistor or a
couple of 1W resistors in series or parallel to give you the same value.
Find the brightest LED you can that runs at 25mA.
You can also puts extra LEDs in series, dropping your resistor value
accordingly.
Your supply is capable of running better LEDs like say a 1W luxeon, but that
would need a suitable DC-DC converter, not nearly as easy unless you buy a
pre-built module.
Dave.
E
Eeyore
Guest
giteco@gmail.com wrote:
Graham
Do you mean a single LED or multiple LEDs ?is it possible to make an LED lamp with a 35v / 25mA supply? if so can
anyone teach me?
Graham
Guest
On Mar 12, 9:08 pm, git...@gmail.com wrote:
accept, supply ALL details then we have something to go on.
Depends on the LED you want to use and what % of rated output you canis it possible to make an LED lamp with a 35v / 25mA supply? if so can
anyone teach me?
accept, supply ALL details then we have something to go on.
N
Nobody
Guest
On Thu, 12 Mar 2009 18:08:37 -0700, giteco wrote:
2. Divide 26V (75% of 35V) by their forward voltage to determine how many
LEDs to use. E.g. if the forward voltage is 3V, 26/3 = 8.66, so use 9 LEDs
in series.
3. That will drop 9*3 = 27V, so you need a resistor to drop the remaining
35-27 = 8V. At 20mA, you would need 8V/0.02A = 400 Ohms, for which the
closest E24 (or E12) value is 390 Ohms.
4. The power dissipation in the resistor is I*V = 0.02A * 8V = 0.16W, so a
1/4 Watt resistor will suffice.
5. Connect all of the LEDs and the resistor in series, i.e.:
[+] o---/\/\/\---|>|---|>|---|>|---|>|-- ... --|>|---|>|---o [-]
where /\/\/\ is the resistor and |>| is an LED.
Where does the 75% figure in #2 come from? So that typical variations in
the forward voltage of the actual LEDs and in the supply voltage don't
substantially alter the current drawn. A lower figure will be more robust
(i.e. less variation in current for a given variation in forward voltage)
but less efficient.
If you can find actual min/max forward voltage values for the LEDs,
and the actual min/max output voltage of the power supply, then choose the
number of LEDs such that the difference voltage to be dropped by the
resistor doesn't vary too much, as the current through the LEDs will vary
in direct proportion.
1. Decide which LEDs you want to use, and look up their forward voltage.is it possible to make an LED lamp with a 35v / 25mA supply? if so can
anyone teach me?
2. Divide 26V (75% of 35V) by their forward voltage to determine how many
LEDs to use. E.g. if the forward voltage is 3V, 26/3 = 8.66, so use 9 LEDs
in series.
3. That will drop 9*3 = 27V, so you need a resistor to drop the remaining
35-27 = 8V. At 20mA, you would need 8V/0.02A = 400 Ohms, for which the
closest E24 (or E12) value is 390 Ohms.
4. The power dissipation in the resistor is I*V = 0.02A * 8V = 0.16W, so a
1/4 Watt resistor will suffice.
5. Connect all of the LEDs and the resistor in series, i.e.:
[+] o---/\/\/\---|>|---|>|---|>|---|>|-- ... --|>|---|>|---o [-]
where /\/\/\ is the resistor and |>| is an LED.
Where does the 75% figure in #2 come from? So that typical variations in
the forward voltage of the actual LEDs and in the supply voltage don't
substantially alter the current drawn. A lower figure will be more robust
(i.e. less variation in current for a given variation in forward voltage)
but less efficient.
If you can find actual min/max forward voltage values for the LEDs,
and the actual min/max output voltage of the power supply, then choose the
number of LEDs such that the difference voltage to be dropped by the
resistor doesn't vary too much, as the current through the LEDs will vary
in direct proportion.