Here is an entry for the DIY circuit.

The TPS61042 part I referenced in my Post #3 has an operating current of about 45μA, which is quite low but still significant in this application.
Depending upon the LED current needed, its overall efficiency may not be that high, perhaps not much better than a resistor in series or a linear constant-current source.

I think it is a boost regulator for only 6 volts. So not so sure.
Here is an entry for a DIY one. About 45% effecient on low and 60% on high. I think they may make a chip with a lower reference voltage which would help. @goingalong Let me know if you want to try one and I'll look.

 

Here is an entry for the DIY circuit.

I think it is a boost regulator for only 6 volts. So not so sure.
Here is an entry for a DIY one. About 45% effecient on low and 60% on high. I think they may make a chip with a lower reference voltage which would help. @goingalong Let me know if you want to try one and I'll look.

Yes, I missed the obvious, that the TPS61042 a boost, not a buck circuit.
Don't know if it can how difficult it would be to convert it to a buck circuit.
Perhaps just the addition of a P-MOSFET switch and flyback diode in series with the inductor input would work.

 

So I am concluding that there is a simple mcu in there along with a buck type regulator.

Most of those torch-type led flash lights are mcu based.

Your battery sounds unlike a typical 9v battery.

and I suspect that it doesn't have a down converter.

I'm thinking "could it be as simple as limiting the current to the LED by means of a humble (yet proportionally large) resistor?"

Inefficiency would kill it here.

 

There is a good article in the link below for a dimmer made up from basic components. The author claims 98% efficiency down to 4v input. Lots of bits and pieces in the circuit and just two ic's but it also has a good description.

http://www.opensourcepartners.nl/~costar/leddimmer/

• Not sure why dannyf thinks the pp3 9v is not typical, it looks pretty ordinary to me.
• If there is no down-converter the circuit is not going to be very efficient as there are lots of volts to lose going down from ~9 to ~3?

Anyway, to get myself back onto track, what I want to do is to create a highly efficient night light. It does not have to be a functional copy of the PALights Workman, that was just an example, nor does it need to last two years. Just for a start any practical lifetime can be doubled by adding a light sensor. On top of that a night light is better with a red led which disrupts night vision less and gives a lower threshold voltage.

So do we think it better to boost up from say a 1.5v C or D cell, or is it better to down-convert from an equivalent volume but higher voltage stack of smaller cells? Personally I am inclined to the former and one way to do that might be to use a pump circuit driven by a mcu - I have done that before to provide bias voltages for displays.

 

A 9 volt battery would be a good choice for the always on part of it. I have some rather old white LEDs that, in a darkened room, can be seen to glow at a few tens of microamps, and I imagine modern LEDs could do even better.

The always on glow could be fed with a simple resistor. A simple buck converter (preferably current fed) would power the LED in flashlight mode.

 

The always on glow could be fed with a simple resistor.

It is not very efficient, as the bulk of the voltage drop (=energy consumed) would be on that resistor.

The ideal I proposed earlier is better: it uses the mcu essentially as a variable resistor, and the led / leds as a step down converter: you can pick the led or leds so that its voltage drop from idle (=always on) to high beam doesn't vary much, say from 2v @ 10ua to 2.5v @ 10ma. Put two such leds between the battery and the mcu's Vdd, and make sure that the mcu drains about 10ua in always on and 10ma at high beam. The 10ma figure can be achieved via the number of output pins tied to ground, or pwm if you want it to be fancier.

 

A single resistor does not require the power dissipation overhead associated with a switching circuit and if only supplying a few tens of microamps, how much more efficient could a power supply possibly be? On top of that, a single resistor is likely to be the solution with the highest MTBF.

 

A single resistor does not require the power dissipation overhead associated with a switching circuit and if only supplying a few tens of microamps, how much more efficient could a power supply possibly be?

It depends. There are low idle current converters - some of them for example wouldn't turn on until the output voltage goes below a threshold. The issue with that approach is 1) complexity; 2) efficiency and 3) management of mcu supply voltage in "off state".

Plus, it is not that difficult to incorporate the necessary logic into the mcu to function as a dc/dc converter.

On top of that, a single resistor is likely to be the solution with the highest MTBF.

What's the practical difference between 1,000,000 year MTBF and 1,000 year MTBF?

 

There are low idle current converters - some of them for example wouldn't turn on until the output voltage goes below a threshold.

So how would they provide the required 'always on' feature?

 

So how would they provide the required 'always on' feature?

You can implement it in a variaty of ways. Putting the diodes on the supply line, as discussed earlier, is one approach. Another would be to put the diodes on the output, through a current sense resistor that controls the on/off of the smps chip: if the voltage across the current sense resistor is too low, the oscillator is turned on; until the voltage across the current sense resistor reaches the upper-end of the threshold.

It is called a "gated oscillator". Linear makes a lot of such chips. 1073 is probably the most famous of them all.

 

There is a good article in the link below for a dimmer made up from basic components. The author claims 98% efficiency down to 4v input. Lots of bits and pieces in the circuit and just two ic's but it also has a good description.

http://www.opensourcepartners.nl/~costar/leddimmer/
.............................

It may have 98% efficiency at higher LED currents, but at the low LED currents you want, the operating current of that circuit will likely give an overall efficiency much lower than just a resistor in series with the LED.
At such very low LED currents you need a circuit that takes only tens of microamps of operating current and that's very difficult to achieve with any switching design.

 

There is a good article in the link below for a dimmer made up from basic components. The author claims 98% efficiency down to 4v input. Lots of bits and pieces in the circuit and just two ic's but it also has a good description.

http://www.opensourcepartners.nl/~costar/leddimmer/

• Not sure why dannyf thinks the pp3 9v is not typical, it looks pretty ordinary to me.
• If there is no down-converter the circuit is not going to be very efficient as there are lots of volts to lose going down from ~9 to ~3?

Anyway, to get myself back onto track, what I want to do is to create a highly efficient night light. It does not have to be a functional copy of the PALights Workman, that was just an example, nor does it need to last two years. Just for a start any practical lifetime can be doubled by adding a light sensor. On top of that a night light is better with a red led which disrupts night vision less and gives a lower threshold voltage.

So do we think it better to boost up from say a 1.5v C or D cell, or is it better to down-convert from an equivalent volume but higher voltage stack of smaller cells? Personally I am inclined to the former and one way to do that might be to use a pump circuit driven by a mcu - I have done that before to provide bias voltages for displays.

You need to be careful just looking at efficiency claims. Most drivers supply much more current so the IC current doesn't make up a large portion of the total. But in your case the diode current is only about 600 ua with 200ua from the battery due to the down conversion.
The 358 shown in the schematic draw 1 to 2 ma - so more than the LED.
The circuit I posted in http://forum.allaboutcircuits.com/t...ficient-night-light.116336/page-2#post-909480 draws only a few ua.
I think you will need a DIY circuit be it hardware or micro as the very low current is also a problem. It's kind of the opposite of what chip makers aim for.
@dannyf , what micro would you use?

 

Many good comments above - I am having trouble keeping up.

Just testing the resistor idea first I find that using a warm white smd with 0.1mA and Vled at 2.5v I get around the right level of light output. This is with a 6.76k resistor in series. I have 0.647V across the resistor so it is taking V**2/R =0.062mW. The LED itself is taking V*I = 0.25mW. If I have that right the resistor is adding 25% extra demand.

It looks like ~100uA is the sort of current I need to be looking at so putting the led in series with a sleeping mcu doesn't seem like the right approach.

I have a high efficiency red led but not a smd so it is harder for me to judge equivalent light output (by eye of course) but I will try to see what I get tomorrow. I can only do this in the dark.

I will also turn on a picaxe and use the pwm and a constant current source to get similar output to the 6.76k resistor. That will give me some idea of the power that would be going into a mcu based approach. Crude but an indicator and excludes any converter losses.

 

0.1mA and Vled at 2.5v

In a linear system, the voltage has to be dropped somewhere. So in this case, assuming a 9v power supply, your efficiency would be 2.5v/9v=30%.

One way to improve efficiency would be to run a few leds in serial. If you put 3 of them in serial, your efficiency would be 2.5x3/9=80%.

Another way is to run the led at very low duty cycle: when the led is on, large amount of current goes through it, but very briefly. The rest of the time, the led is off -> 100% efficiency.

The bulk of the discussion was around how to set up the circuit so that it minimizes inefficiency.

 

Many good comments above - I am having trouble keeping up.

Just testing the resistor idea first I find that using a warm white smd with 0.1mA and Vled at 2.5v I get around the right level of light output. This is with a 6.76k resistor in series. I have 0.647V across the resistor so it is taking V**2/R =0.062mW. The LED itself is taking V*I = 0.25mW. If I have that right the resistor is adding 25% extra demand.

It looks like ~100uA is the sort of current I need to be looking at so putting the led in series with a sleeping mcu doesn't seem like the right approach.

I have a high efficiency red led but not a smd so it is harder for me to judge equivalent light output (by eye of course) but I will try to see what I get tomorrow. I can only do this in the dark.

I will also turn on a picaxe and use the pwm and a constant current source to get similar output to the 6.76k resistor. That will give me some idea of the power that would be going into a mcu based approach. Crude but an indicator and excludes any converter losses.

Your math is a little to fuzzy for me.
If you are still using the 9 volt battery and there is 2.5 volts across the LED there must be 6.47 volts not 0.647 volts across the resistor. That makes it 1 ma not 100 ua. I did the same experment with a cheap red LED and it didn't light with a 68K resistor, but had a nice glow with 1 ma. This all pushes me back to the buck regulator. You could get about 0.6 ma to the LED with 0.2 ma from the battery. This might give you the life you want from the LIPO 9 volt.

I will also turn on a picaxe and use the pwm and a constant current source to get similar output to the 6.76k resistor. That will give me some idea of the power that would be going into a mcu based approach. Crude but an indicator and excludes any converter losses.

You will still need something to limit the current thru the diode. Just pwm won't do it.

You haven't said anything about the circuit in post 21. Is there something you don't like about it?

 

Your math is a little to fuzzy for me.
If you are still using the 9 volt battery and there is 2.5 volts across the LED there must be 6.47 volts not 0.647 volts across the resistor. That makes it 1 ma not 100 ua. I did the same experment with a cheap red LED and it didn't light with a 68K resistor, but had a nice glow with 1 ma. This all pushes me back to the buck regulator. You could get about 0.6 ma to the LED with 0.2 ma from the battery. This might give you the life you want from the LIPO 9 volt.

You will still need something to limit the current thru the diode. Just pwm won't do it.

You haven't said anything about the circuit in post 21. Is there something you don't like about it?

The supply in this case was 2.5v across the led plus 0.647 across the resistor making 3.1v total (two aa cells in series). I did not then have a 9v pp3 to hand and what I was interested in was the current and voltage on the led for low light. Use 9v and the resistance would have to go up [(9-2.5)/0.1 kohms] and the power wasted by the led would climb of course.

Sorry I did not comment on your post there were quite a few coming in at the time and I briefly looked at the waveforms rather than the circuit, intending to get back to it once I had some handle on relative efficiencies. I have not used a comparator like that so it is new to me. I don't have anything like that in my bits and pieces box right now. As I see it D2 is going to take a significant amount of energy out of the circuit so it's forward voltage needs to be as low as possible. Would it be an option to put a led in it's place or is that too slow.

 

The supply in this case was 2.5v across the led plus 0.647 across the resistor making 3.1v total (two aa cells in series). I did not then have a 9v pp3 to hand and what I was interested in was the current and voltage on the led for low light. Use 9v and the resistance would have to go up [(9-2.5)/0.1 kohms] and the power wasted by the led would climb of course.

Sorry I did not comment on your post there were quite a few coming in at the time and I briefly looked at the waveforms rather than the circuit, intending to get back to it once I had some handle on relative efficiencies. I have not used a comparator like that so it is new to me. I don't have anything like that in my bits and pieces box right now. As I see it D2 is going to take a significant amount of energy out of the circuit so it's forward voltage needs to be as low as possible. Would it be an option to put a led in it's place or is that too slow.

Ahh, I see, a trick question.
In that case I think I would just use the circuit in post 8. It's only purpose is to keep the current constant from new to dead battery. That should give you at least 1.4 years (probably more) with the 1200 ma hr. battery.
You will probably need to buy some parts with any low current buck converter.
But if you decide to build it let us know and we can pick some more common parts than what I used in the simulation.

 

I'd go with Ronv's post #8 circuit, but replace Q1 with a FET. R2 could then be increased to 1meg or so, thus reducing overall current consumption.

 

I'd go with Ronv's post #8 circuit, but replace Q1 with a FET. R2 could then be increased to 1meg or so, thus reducing overall current consumption.

Thanks. I usually do use a FET in that position.

 

I'd go with Ronv's post #8 circuit, but replace Q1 with a FET. R2 could then be increased to 1meg or so, thus reducing overall current consumption.

Good idea Alec.