TINY 4.5v 1w driver
- Thread starter da_guy2
- Start date
HarryN
Flashlight Enthusiast
Look at Doug S posts related to the new mini Li flashlight project. If you only need 4 - 6 V in, and 1 watt out, he just used a TI LDO, which is not that bad in this power / voltage range.
You can buy the part for < $ 2 - saw it at either Newark or Digikey - one of them. Shipping is pretty cheap as well.
You can buy the part for < $ 2 - saw it at either Newark or Digikey - one of them. Shipping is pretty cheap as well.
[ QUOTE ]
da_guy2 said:
Well two problems with that circuit. Fist of all its not small enough especialy if I try to make it. secondly once the battery voltage drops below vf it stops. That ALOT of wasted battery. Any other sugestions?
David
[/ QUOTE ]
That is incorrect. Once the battery voltage drops below the Vf of the LED at its rated current, the current begins to graudally decline as the input voltage declines. This is a misconception that people need to get over - LEDs will still work if they are fed voltage below their Vf.
For instance, take a particular luxeon that has a Vf of 3.3V at 350mA. If the luxeon is fed 3V, then around 100mA will still flow through the luxeon, giving about 1/3 the light, but the luxeon will still be producing light given that input voltage.
da_guy2 said:
Well two problems with that circuit. Fist of all its not small enough especialy if I try to make it. secondly once the battery voltage drops below vf it stops. That ALOT of wasted battery. Any other sugestions?
David
[/ QUOTE ]
That is incorrect. Once the battery voltage drops below the Vf of the LED at its rated current, the current begins to graudally decline as the input voltage declines. This is a misconception that people need to get over - LEDs will still work if they are fed voltage below their Vf.
For instance, take a particular luxeon that has a Vf of 3.3V at 350mA. If the luxeon is fed 3V, then around 100mA will still flow through the luxeon, giving about 1/3 the light, but the luxeon will still be producing light given that input voltage.
HarryN
Flashlight Enthusiast
Hi
I am not an electronics expert, but here is my very approximate interpretation of what I read on the forums.
1) Start with the assumptions
- Since you talk about 4.5 V in (starting), I am assuming that this is a 3 x 1.5 V setup. (such as 3 D, C, or AA cells)
- I also assume that since you are here on the forum, you are using high end batteries / cells, such as quality alkalines or even better, Li enhanced versions.
- Since efficiency is important to you, you have selected a 1 watt Luxeon with a good bin for electrical efficiency, such as a J bin (for low Vf) - nominal 3.1 - 3.3 V at 350 ma.
2) Converter Circuits
- Since LEDs need some kind of managed power flow, somehow, you need to either raise or lower the voltage coming from the batteries, and throttle the current flow.
- The further you are (Vin) (in either direction) from the Vf of the LED, the circuit becomes more complex, and the conversion efficiency TENDS to be lower.
- If you are using the batteries in a way which actually crosses from Vin that is initially higher than the Vf to one that is lower, there are very few (not 0) circuits that can do this without the current changing dramatically. I have not yet been able to purchase one, esp. as small as you seem to need.
- The circuit you found does look interesting, but I could not find a data sheet on it, so I could not really tell how it actually works (buck or boost) nor how efficient it really is. (maybe it is an LDO as well - no idea)
- Certainly, there is a very good case for switching circuits when you can buy them in the price / size / voltage range you need.
3) Batteries
- Staying with the assumption that you are indeed using 3 x 1.5 Alkalines or Li enhanced cells.
- Look at the data sheets from the mfg.
- In all of the ones I have seen, once the V of cell drops much below 1 V, most of the usable power is gone, and certainly, pulling 400 ma out of a cell with only 0.7 V on the terminals is going to be tough.
- What is even worse, if the V terminal is this low, to maintain a real 400ma to the LED, you really need to pull even more current (closer to 600ma) since it now has to convert current to voltage to make it all work.
4) LDO
- Again, assuming your starting 4.5 V in, and a J bin Vf Luxeon (or close)
- The difference between 4.5 V and 3.3 V is not very large, but this V difference does convert to heat.
- Since the battery voltage is dropping throughout its use, the LDO will essentially vary its resistance (and the effective loss) to maintain a relatively constant current flow until Vin is almost V out of the LDO.
- The LDO which Doug S picked requires only the LDO, no other peripherals, so there really is no "circuit" and these are really tweezer size parts.
- There are a few people on the forum that can build a switcher circuit more efficient than the LDO, but very few would actually achieve greater than 20 % longer useful run time than this very simple, very inexpensive, very small LDO in the battery voltage range you have picked.
5) Lengthening Run Time
- If you really want to be able to drain the last life out of your batteries, consider switching colors.
- RED LEDs have a Vf more like 2.5 V. Once the battery voltage is too low to run white LEDs, there is plenty left to run a red one.
Sorry if this became a long ramble.
HarryN
I am not an electronics expert, but here is my very approximate interpretation of what I read on the forums.
1) Start with the assumptions
- Since you talk about 4.5 V in (starting), I am assuming that this is a 3 x 1.5 V setup. (such as 3 D, C, or AA cells)
- I also assume that since you are here on the forum, you are using high end batteries / cells, such as quality alkalines or even better, Li enhanced versions.
- Since efficiency is important to you, you have selected a 1 watt Luxeon with a good bin for electrical efficiency, such as a J bin (for low Vf) - nominal 3.1 - 3.3 V at 350 ma.
2) Converter Circuits
- Since LEDs need some kind of managed power flow, somehow, you need to either raise or lower the voltage coming from the batteries, and throttle the current flow.
- The further you are (Vin) (in either direction) from the Vf of the LED, the circuit becomes more complex, and the conversion efficiency TENDS to be lower.
- If you are using the batteries in a way which actually crosses from Vin that is initially higher than the Vf to one that is lower, there are very few (not 0) circuits that can do this without the current changing dramatically. I have not yet been able to purchase one, esp. as small as you seem to need.
- The circuit you found does look interesting, but I could not find a data sheet on it, so I could not really tell how it actually works (buck or boost) nor how efficient it really is. (maybe it is an LDO as well - no idea)
- Certainly, there is a very good case for switching circuits when you can buy them in the price / size / voltage range you need.
3) Batteries
- Staying with the assumption that you are indeed using 3 x 1.5 Alkalines or Li enhanced cells.
- Look at the data sheets from the mfg.
- In all of the ones I have seen, once the V of cell drops much below 1 V, most of the usable power is gone, and certainly, pulling 400 ma out of a cell with only 0.7 V on the terminals is going to be tough.
- What is even worse, if the V terminal is this low, to maintain a real 400ma to the LED, you really need to pull even more current (closer to 600ma) since it now has to convert current to voltage to make it all work.
4) LDO
- Again, assuming your starting 4.5 V in, and a J bin Vf Luxeon (or close)
- The difference between 4.5 V and 3.3 V is not very large, but this V difference does convert to heat.
- Since the battery voltage is dropping throughout its use, the LDO will essentially vary its resistance (and the effective loss) to maintain a relatively constant current flow until Vin is almost V out of the LDO.
- The LDO which Doug S picked requires only the LDO, no other peripherals, so there really is no "circuit" and these are really tweezer size parts.
- There are a few people on the forum that can build a switcher circuit more efficient than the LDO, but very few would actually achieve greater than 20 % longer useful run time than this very simple, very inexpensive, very small LDO in the battery voltage range you have picked.
5) Lengthening Run Time
- If you really want to be able to drain the last life out of your batteries, consider switching colors.
- RED LEDs have a Vf more like 2.5 V. Once the battery voltage is too low to run white LEDs, there is plenty left to run a red one.
Sorry if this became a long ramble.
HarryN
pbarrette
Enlightened
Hi Da_Guy,
The circuit you have listed is most likely the ZXSC300 reference circuit. These are fairly easy to build with a little practice and I calculated the cost to be about $4-5 if you buy the parts for 1 from digikey. Of course, the costs go down somewhat as you buy more parts, which is why I bought enough to build ~20 of them.
The problem with this circuit, IMO, is that it is only a constant current at a specific, pre-set voltage. So, if you are using batteries, you will need to use batteries that are able to hold their voltage through most of the length of their charge. Otherwise, the current to the Luxeon will start to drop as the battery voltage starts to decrease.
Also, this circuit is a boost circuit, so it's really not going to be that useful for a 4.5v application since you are already exceeding the Vf of the LED. I don't know much about how this circuit operates in that situation, but you would likely end up wasting a lot of power in heat as the boost switcher tries to bring the V down to the Vf of the LED.
If you like, I could customize one of mine for your specified current/voltage/Lux-Vf and ship it out via USPS. But as it is completely handbuilt, and I'm not really in the business of making and selling circuits/lights I would have to charge ~$15-20 just to cover my time. So, in the end, you may be better off buying the one you found and changing the sense resistor to suit your needs.
Either way if you are using, say, 3x1.5v NiMH's, the minimum voltage would probably be (~3x0.9) = 2.7v before you have to worry about reverse loading the batts. That's still not as low as the 0.7v that the circuit will take, but remember that in this circuit, current drops as the battery voltage drops.
If anything, this circuit would be good for draining the last bits of life out of 3 alkalines, and not really that good with 3 fresh batts.
pb
The circuit you have listed is most likely the ZXSC300 reference circuit. These are fairly easy to build with a little practice and I calculated the cost to be about $4-5 if you buy the parts for 1 from digikey. Of course, the costs go down somewhat as you buy more parts, which is why I bought enough to build ~20 of them.
The problem with this circuit, IMO, is that it is only a constant current at a specific, pre-set voltage. So, if you are using batteries, you will need to use batteries that are able to hold their voltage through most of the length of their charge. Otherwise, the current to the Luxeon will start to drop as the battery voltage starts to decrease.
Also, this circuit is a boost circuit, so it's really not going to be that useful for a 4.5v application since you are already exceeding the Vf of the LED. I don't know much about how this circuit operates in that situation, but you would likely end up wasting a lot of power in heat as the boost switcher tries to bring the V down to the Vf of the LED.
If you like, I could customize one of mine for your specified current/voltage/Lux-Vf and ship it out via USPS. But as it is completely handbuilt, and I'm not really in the business of making and selling circuits/lights I would have to charge ~$15-20 just to cover my time. So, in the end, you may be better off buying the one you found and changing the sense resistor to suit your needs.
Either way if you are using, say, 3x1.5v NiMH's, the minimum voltage would probably be (~3x0.9) = 2.7v before you have to worry about reverse loading the batts. That's still not as low as the 0.7v that the circuit will take, but remember that in this circuit, current drops as the battery voltage drops.
If anything, this circuit would be good for draining the last bits of life out of 3 alkalines, and not really that good with 3 fresh batts.
pb
