DIY LED Canister Lamp
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fire5man
Contributor
These are some great ideas. I am hoping to find a time and the way to make a light for me. I am not going a lot of low light diving but I plan on it in the future. As anyone done like a dyi video with explanations about everything how it works and goes together.
rescue15
Contributor
I Wish!!
Steve
lucca brassi
Contributor
Staring with prototype of buck ( step down ) constant current regulator . I 've chosen
LM22677 from National Semiconductors. Simulator on their page looks too much promising
- 95%steady state efficiency , if I'm on 90% Im on the top ; 2,8 A constant current , 11,8V V out for 3 SSC P7 C bin. Have bought delrin -POM for 3 tanks , waiting for SS latches Nielesen and AGRO glands.
At moment I have one doubt - switch. The question it is classic syle or proximity with reed magnetic switch.
Classic it is qite OK , but normaly fitting is a little confused for me ( almost all screw coil go in to the acryl and rubber cap go far over of top of switch and there become a possibility to penetrate rubber and water flod switch if you can imagine wht I think)
Magnetic can maybe disturb compass , speciall with those small powerfull magnets , but it is in water leak problem total safe.
LM22677 from National Semiconductors. Simulator on their page looks too much promising
- 95%steady state efficiency , if I'm on 90% Im on the top ; 2,8 A constant current , 11,8V V out for 3 SSC P7 C bin. Have bought delrin -POM for 3 tanks , waiting for SS latches Nielesen and AGRO glands. At moment I have one doubt - switch. The question it is classic syle or proximity with reed magnetic switch.
Classic it is qite OK , but normaly fitting is a little confused for me ( almost all screw coil go in to the acryl and rubber cap go far over of top of switch and there become a possibility to penetrate rubber and water flod switch if you can imagine wht I think)
Magnetic can maybe disturb compass , speciall with those small powerfull magnets , but it is in water leak problem total safe.
Packhorse
Contributor
What I know aboth the LM22677 you could write on the back of a postage stamp with a very large marker.
Can it be set up for current regulation at 2.8a with 95% efficiency with large differences in Vin and V out?
I have been wondering why there isnt a all in one adjustable current regulator IC with high efficiency.
Can it be set up for current regulation at 2.8a with 95% efficiency with large differences in Vin and V out?
I have been wondering why there isnt a all in one adjustable current regulator IC with high efficiency.
martzak
Contributor
for the purpose of a buck regulator for 'constant' current operation to power an LED try the Supertex hv9910b. It's considerably cheaper than LM22677 and controls an external nMOSFET that you can choose so you can run any current you want, limited by the MOSFET you select, not the IC. I recently used this IC to design a circuit to power a 350mA LED. I can easily redesign it to run 2.5A or more. This is what i do for a living and enjoy it, so let me know if want me to cook up a circuit design for you...what you want to run it off of, output, space constraints etc.
Martzak, that would be really awesome!!!
Let's say:
Vout: 3.7V * N of LED
Vin (min): as close to Vout as possible
Vin (max): the higher the better, at least Vout + 5V
Size: As small as possible, should fit in the handle of the MAG Light (C or D).
Would be cool to have an output current jumper for 1A, 1.4A, 2.8A to be able to run different LEDs
Not sure how to handle heat though. MOSFETS should become pretty hot running at 2.8A...
Additional feature request: integrated MOSFET circuit for reed switch.
Let's say:
Vout: 3.7V * N of LED
Vin (min): as close to Vout as possible
Vin (max): the higher the better, at least Vout + 5V
Size: As small as possible, should fit in the handle of the MAG Light (C or D).
Would be cool to have an output current jumper for 1A, 1.4A, 2.8A to be able to run different LEDs
Not sure how to handle heat though. MOSFETS should become pretty hot running at 2.8A...
Additional feature request: integrated MOSFET circuit for reed switch.
Last edited:
martzak
Contributor
MrSpock, give me a few days and I'll have a circuit schematic plus BOM for you...don't thank me yet. Let's see what I can come up with.
. Let's see what I can come up with.
lucca brassi
Contributor
Nice circuit martzak - love to be MOSFET located separatly - but it make circuit a little larger .
Previous year I done some lamp head just for test for MTB ; look aveable space for it look at
3w.elektronik.si/phpBB2/album_personal.php
I'd like to made this lamp purly D.I.R. just to switch between LED and HID.
Previous year I done some lamp head just for test for MTB ; look aveable space for it look at
3w.elektronik.si/phpBB2/album_personal.php
I'd like to made this lamp purly D.I.R. just to switch between LED and HID.
OP
bronk
Registered
Here is an excerpt from a longer article I am working on. Parts are should be here tomorrow for my wife's light. I will be using a new regulator design in it. This weekend I had the light down to 140' at Lake Wazee. 5 dives on one charge and still going strong, even in 41 degree water!.
Participating in the newsgroups shows there is a lot of confusion and mis-information on using LEDs. LEDs are current devices and as such require a constant current source supply or at a minimum current limiting resistor. If you drive LEDs from a voltage source with no (intentional) current limiting you will end up destroying them and letting the magic smoke out. This may take some time as they overheat and if Murphy's law holds this will be just as you reach max depth on your dive. So here is a brief tutorial on how and why to do the current limiting on the LEDs. The calculations are based around the the LEDs used, but by substituting the values of the devices you want to use, you can rework the calculations for your own devices.
Ohms law V = IR where V = applied voltage, I = current, R = resistance.
Reorganizing this to find the current: I = V/R So dividing the voltage by the resitance will give you the current in your circuit.
LEDs also have a characteristic called "forward drop". Think of it as either the minimum voltage to "turn on the device" or more accurately the voltage drop in the device that is developed regardless of the current through it. The forward drop does in fact vary with current, temperature and the manufacturing process (some part to part variation). However, for our purposes we will ignore this and just use some simple worst case calculations.. The CREE LEDs have a forward drop of 3.4 to 3.7 volts according to the data sheet. They also have a rated absolute max current of 1A. This is a "do not exceed" value.
The setup I used has the 3 LEDs placed in series. This is to match relatively closely with a 12v battery pack. With 3 LEDs in series the total forward drop will be somewhere between 3x3.4 = 10.2 and 3x3.7 = 11.1 volts. So for starters, if the battery is not at least at that voltage, the LEDs will not turn on at all. So you are thinking, "I have a 12 v battery pack so that is not a problem!" Not so. With a 12v NIMH battery pack the output voltage may be as high as 13.6 volts fully charged and you should not discharge it below 11.1v to avoid permanent damage to the battery cells. So while we have "enough voltage" to turn on the LEDs we are dealing with a voltage that varies as the batteries discharge and also without a lot of "headroom" in that as the battery approaches full discharge, there is little extra voltage available above that which is needed to barely turn on the LEDs..
So no for "why do I need a current regulator". There are 2 parts to the answer:
1. prevent destruction of the LEDs
2. maximize brightness of the LEDs during discharge of the battery pack
The light output of a LED varies in direct proportion to the current. So if you halve the current, you get half the light output.
The simplest way to provide current limiting (not regulation) is to have a resistor in series between the battery and LEDs. The circuit wiring itself inherently has some resistance which we want to keep as low as reasonably possible to get the battery energy delivered to the LEDs and not lost in the wiring. So for purposes of discussion, lets assume that you have 0.1 ohms of total wiring resistacne (battery connections, switch, cord). Modifying our ohms law equation to take into account the battery voltage and forward drop of the LEDs yields:
I = (Vbatt-Vf)/R where VBatt is the battery voltage and Vf is the forward drop of the string of LEDs.
Worst case max voltage and min Vf
I = (13.6-10.2)/.1 = 3.4 Amps - not conducive to long (any?) LED life as this is over three times as great as the absolute max rating . So this is the justification for some sort of current limiting
Worst case Min battery voltage and max Vf
I = (11.1-11.1)/.1 = 0. So the light would not even be lit. .
So now you know that current limiting must be used. Now lets look at what happens with a simple resistor as the current limiter. (simple and cheap)
Rearranging Ohms law again
R = V/I = (13.6-10.2)/1000mA = 3.4 Ohms
Now at 12v (12-10.2) / 3.4 = 530mA So you can see that the current at 12 v is down by nearly half from a fully charged battery.
You can measure the forward drop of the LEDs by simply hooking up a resistor in series (start with something a little larger, maybe 4 ohms) and then measuring the drop. Mine came out to 10.5 v Be sure you have the LEDs on a heat sink when doing this. They heat up quickly! I simply taped them down to a big chunk of aluminum for the first brief tests.
Doing the same calculations with the value of 10.5V forward drop gives:
R = 3.1 ohms, 1000mA at 13.6v, 480mA at 12v and 230ma AT 11.2 V. .
So if you are satisfied with a 4:1 brightness change during discharge you can be set with just a simple resistor . However, I wanted something with a more constant light output.
So that is the reason I went with an active current regulator (the LED dimmer circuit mentioned earlier).
Participating in the newsgroups shows there is a lot of confusion and mis-information on using LEDs. LEDs are current devices and as such require a constant current source supply or at a minimum current limiting resistor. If you drive LEDs from a voltage source with no (intentional) current limiting you will end up destroying them and letting the magic smoke out. This may take some time as they overheat and if Murphy's law holds this will be just as you reach max depth on your dive. So here is a brief tutorial on how and why to do the current limiting on the LEDs. The calculations are based around the the LEDs used, but by substituting the values of the devices you want to use, you can rework the calculations for your own devices.
Ohms law V = IR where V = applied voltage, I = current, R = resistance.
Reorganizing this to find the current: I = V/R So dividing the voltage by the resitance will give you the current in your circuit.
LEDs also have a characteristic called "forward drop". Think of it as either the minimum voltage to "turn on the device" or more accurately the voltage drop in the device that is developed regardless of the current through it. The forward drop does in fact vary with current, temperature and the manufacturing process (some part to part variation). However, for our purposes we will ignore this and just use some simple worst case calculations.. The CREE LEDs have a forward drop of 3.4 to 3.7 volts according to the data sheet. They also have a rated absolute max current of 1A. This is a "do not exceed" value.
The setup I used has the 3 LEDs placed in series. This is to match relatively closely with a 12v battery pack. With 3 LEDs in series the total forward drop will be somewhere between 3x3.4 = 10.2 and 3x3.7 = 11.1 volts. So for starters, if the battery is not at least at that voltage, the LEDs will not turn on at all. So you are thinking, "I have a 12 v battery pack so that is not a problem!" Not so. With a 12v NIMH battery pack the output voltage may be as high as 13.6 volts fully charged and you should not discharge it below 11.1v to avoid permanent damage to the battery cells. So while we have "enough voltage" to turn on the LEDs we are dealing with a voltage that varies as the batteries discharge and also without a lot of "headroom" in that as the battery approaches full discharge, there is little extra voltage available above that which is needed to barely turn on the LEDs..
So no for "why do I need a current regulator". There are 2 parts to the answer:
1. prevent destruction of the LEDs
2. maximize brightness of the LEDs during discharge of the battery pack
The light output of a LED varies in direct proportion to the current. So if you halve the current, you get half the light output.
The simplest way to provide current limiting (not regulation) is to have a resistor in series between the battery and LEDs. The circuit wiring itself inherently has some resistance which we want to keep as low as reasonably possible to get the battery energy delivered to the LEDs and not lost in the wiring. So for purposes of discussion, lets assume that you have 0.1 ohms of total wiring resistacne (battery connections, switch, cord). Modifying our ohms law equation to take into account the battery voltage and forward drop of the LEDs yields:
I = (Vbatt-Vf)/R where VBatt is the battery voltage and Vf is the forward drop of the string of LEDs.
Worst case max voltage and min Vf
I = (13.6-10.2)/.1 = 3.4 Amps - not conducive to long (any?) LED life as this is over three times as great as the absolute max rating . So this is the justification for some sort of current limiting
Worst case Min battery voltage and max Vf
I = (11.1-11.1)/.1 = 0. So the light would not even be lit. .
So now you know that current limiting must be used. Now lets look at what happens with a simple resistor as the current limiter. (simple and cheap)
Rearranging Ohms law again
R = V/I = (13.6-10.2)/1000mA = 3.4 Ohms
Now at 12v (12-10.2) / 3.4 = 530mA So you can see that the current at 12 v is down by nearly half from a fully charged battery.
You can measure the forward drop of the LEDs by simply hooking up a resistor in series (start with something a little larger, maybe 4 ohms) and then measuring the drop. Mine came out to 10.5 v Be sure you have the LEDs on a heat sink when doing this. They heat up quickly! I simply taped them down to a big chunk of aluminum for the first brief tests.
Doing the same calculations with the value of 10.5V forward drop gives:
R = 3.1 ohms, 1000mA at 13.6v, 480mA at 12v and 230ma AT 11.2 V. .
So if you are satisfied with a 4:1 brightness change during discharge you can be set with just a simple resistor . However, I wanted something with a more constant light output.
So that is the reason I went with an active current regulator (the LED dimmer circuit mentioned earlier).
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