Best Chip/Step Up Circuit Design for the new 5W Luxeon Star?

MrAl · Jun 15, 2002

Hello Rene,

Wow, some pretty nice numbers!
Looks like your circuit is working
out pretty good 🙂

I just have one question, when you say
Rsense is equal to 1 ohm, where is
this resistor and what is it's function?

Do you have a schematic i could look at?

Thanks, and good luck with it,
Al

remuen · Jun 15, 2002

Hi Al

Rsense is the Rfb which is in series with the load and goes to ground. On his high end the feedback voltage to sense the current goes to the feedback pin of the IC.

I used the schematic on page 6 of the datasheet (typical operating circuit) but didn't use the shutdown and the adjust function of the chip. So SHDN is connected with Vcc and ADJ with REF.

The MAX1698 datasheet can be downloaded here: MAX1698

The same applies to the MAX1722 where I've also used the basic operating circuit shown in figure 7 on page 9 of the datasheet.

I someone is interested I can email you my first pcb layout (sqare board) as jpg picture.

vicbin · Jun 15, 2002

Hi Rene,

Great result !

So what is your plan for next phase to nex target for 700 ma output current ? (LS 5 W requirement)

Let say we ignore the space, complexity and cost consideration, I was wondering if it is possible to increase the output current by making twin circuit supplying single LS 5 W in parallel.

Back in my overclocking activities, I used to run two PC switching power supplies by running in parallel to increase output current for 12 V or in series to double output voltage for powering 12 or 24 volt peltier element.

Again, it is beyond my skill and knowledge level, just some wild idea !

PM sent for the circuit PCB.

Vic

MrAl · Jun 15, 2002

Hello there again Rene,

<BLOCKQUOTE>quote:<HR>Originally posted by remuen:
Hi Al
Rsense is the Rfb which is in series with the load and goes to ground. On his high end the feedback voltage to sense the current goes to the feedback pin of the IC.
<HR></BLOCKQUOTE>

I was afraid you were going to say that 🙂
With a resistor that large for the sense resistor, you cant measure efficiency by
measuring the output voltage from Vout to ground because the 1 ohm resistor eats
up power that doesnt get to the LED's. With a 6.0 volt output at 350ma,
the 1 ohm resistor eats 16 percent of the efficiency. This puts the
actual real eff at around 65 percent! Instead of measuring the output
voltage from the output to ground, measure from the output to the top
of the sense resistor, then use this quantity to calculate the eff and
see the difference. The real numbers will look much lower.

To improve the eff quite a bit, you can use a resistor voltage divider (or pot)
to divide Vref (REF) down to around 100mv to supply the Vadj (ADJ) pin. Doing
this you can then use a 0.1 ohm resistor for the Rsense which will boost the
actual efficiency quite a bit. The 0.1 ohm resistor will eat only less then
1 percent of the efficiency, so you could measure the output voltage from
the output to ground and get reasonable accuracy.

I wouldnt doubt it if your output current at low voltage input gets
much better also.

So far, all of the internal MOSFET chips i have seen have limiting problems
of one kind or another that arent indicated on the data sheets.
By going with a chip that allows you to use an external MOSFET, i think
you have selected the best choice for driving the 5w LS.

Good luck with your LED circuits,
Al

dat2zip · Jun 15, 2002

Rene,

Great results. I think you have a nice solution. I really like the external FET approach. For the 5W version you shouldn't need the other IC. In fact having the IC stop working is benificial when you start with higher battery voltages.

I can duplicate your circuit here to help with the inductor selection or help with making measurements of the actual waveforms.

I'm not liking the LM2621 solution. I can get it to work but, the mode it operates in is not the inductor current ramp waveform that would be ideal. My LM2621 is one micro-soic and one SOT23. I think your design is similar.

I like the possibility of adding the pot to vary the adj to vary the load current. Nice!

If I read the datasheet correctly 300mV sense (default) is ~10% loss in the sense resistor. The min adj is 50mV. I would think setting it down to 100mV and then allowing the trim pot to turn from 100mV to 50mV would give sufficient power control for the 5W LS.

I'm going to continue on the LM2621 for another couple of days. I could not release the current version even though it posts great numbers is the fact that I cannot gurantee that stablility. It can go into different modes of operations and most modes tend to shutdown the IC for thermal reasons.

I would be afraid that most of them would end up shutting down and never ever starting up. It would just sit there and blink the LED ON/OFF/ON...

What FET did you use?

-Wayne

remuen · Jun 15, 2002

<BLOCKQUOTE>quote:<HR>Originally posted by vicbin:

So what is your plan for next phase to nex target for 700 ma output current ? (LS 5 W requirement)
<HR></BLOCKQUOTE>

I do not have real plans. As said it looks like my inductor already has been run into saturation. So I will try to use a inductor with a lower inductivity (means higher current and less resistance) but I don't know whether and how this will work. And I think I will also try what dat2zip did with the LM2621 means connecting two inductors in parallel. But as I'm not the real specialist I need some help to make the necessary measurements and improvements.

Btw, I've got your PM and will send you the layout.

<BLOCKQUOTE>quote:<HR>Originally posted by MrAl:

I was afraid you were going to say that 🙂
With a resistor that large for the sense resistor, you cant measure efficiency by
measuring the output voltage from Vout to ground because the 1 ohm resistor eats
up power that doesnt get to the LED's.
<HR></BLOCKQUOTE>

Don't be afraid, Al

- I didn't measure the output voltage to ground. I measured it across the load means from Vout to the high end of the Rsense as you suggested. So the results are the real efficiency!

<BLOCKQUOTE>quote:<HR>Originally posted by MrAl:

To improve the eff quite a bit, you can use a resistor voltage divider (or pot)
to divide Vref (REF) down to around 100mv to supply the Vadj (ADJ) pin. Doing
this you can then use a 0.1 ohm resistor for the Rsense which will boost the
actual efficiency quite a bit. The 0.1 ohm resistor will eat only less then
1 percent of the efficiency, so you could measure the output voltage from
the output to ground and get reasonable accuracy.
<HR></BLOCKQUOTE>

I need some more explanation to understand this. As far as I see I have the maximum output when connecting the ADJ pin to the REF pin. Maybe I didn't understand something correctly?

The Vsense is 300mV so I needed this 1 ohm resistor only for my first test run. For the second test run I reduced the value of the Rsense to 0.75 ohms. To get the wanted output current of 700 - 800 mA Rsense has to be reduced to something about 0.3 to 0.4 ohms. Of course: The lower the better!

<BLOCKQUOTE>quote:<HR>Originally posted by dat2zip:

I can duplicate your circuit here to help with the inductor selection or help with making measurements of the actual waveforms.
<HR></BLOCKQUOTE>

That would be great! You know I'm not a specialist and don't have the necessary equipment. Having an idea and bring it to work is quite a different thing. So I need every help. You also know that I do have problems to get some parts as eg. a better inductor. So eventually we all want the same: A 5W boost converter for the new 5W LS!

<BLOCKQUOTE>quote:<HR>Originally posted by dat2zip:

My LM2621 is one micro-soic and one SOT23. I think your design is similar
<HR></BLOCKQUOTE>

The MAX1698 is a 10uMax and then I used the MOSFET (SOT23) and the external Schottky diode (a ZHCS1000 also in SOT23). And of course for my special combination MAX1698/MAX1722 also a MAX1722 in a SOT23 housing. I'll send you my pcb layout so you will see how I put them all on the same little board ...

<BLOCKQUOTE>quote:<HR>Originally posted by dat2zip:

What FET did you use?
<HR></BLOCKQUOTE>

It's a Fairchild FDN337N - the one I could buy here without any problems ...

I think if all of you specialists are going to work on this circuit the MAX1698 has the potential to deliver these 5W output we want to have ....- and together with the MAX1722 we also could use it also for a 2cell application

. So please help me! I can't do it on my own!

MrAl · Jun 15, 2002

Hi Rene,

The 1 ohm resistor eats up efficiency no matter how you measure the
output voltage, so if you could get away with using a 0.1 ohm resistor
instead that would be much better.

If you look at the data sheet, you will see that if you connect a
100mv dc voltage to the ADJ pin, that means you only require something like
1/4 of that at the FB pin, meaning you can use a series resistor with
less the resistance then when you connect FB to the REF pin.
That's the reason for using a voltage divider at the ADJ pin.

It's great that you measured efficiency directly across the LED's the first
time around, but if you also do this circuit change you will still see
an increase in efficiency.

You simply choose the two divider resistors such that the voltage at
the ADJ pin is about 146mv. This could be accomplished with two resistors
8.56k and 1k. Then, replace the sense resistor with a 0.1 ohm resistor.

Check efficiency again.

Good luck with your LED circuits,
Al

Xcandescent · Jun 15, 2002

<BLOCKQUOTE>quote:<HR>Originally posted by MrAl:
If you look at the data sheet, you will see that if you connect a
100mv dc voltage to the ADJ pin, that means you only require something like
1/4 of that at the FB pin, meaning you can use a series resistor with
less the resistance then when you connect FB to the REF pin.
That's the reason for using a voltage divider at the ADJ pin.<HR></BLOCKQUOTE>
I had a look at that data sheet (since I'm considering using this in my GBA light circuit). Am I correct in assuming that the only purpose to having a higher voltage at the ADJ pin is to allow brightness adjustment by using a pot to reduce Vadj -- and that, if you just want fixed output, it's more efficient to decrease Vadj and recalculate Rfb with the new value?

-XCN-

Xcandescent · Jun 15, 2002

<BLOCKQUOTE>quote:<HR>Originally posted by dat2zip:
I'm not liking the LM2621 solution. I can get it to work but, the mode it operates in is not the inductor current ramp waveform that would be ideal. My LM2621 is one micro-soic and one SOT23. I think your design is similar.<HR></BLOCKQUOTE>

What about the LTC1871? I posted a link to this a while ago because the sample circuits were fairly high power (5V @ 7A with 3.3V input) compared to what we're trying to do here. I find that unusual considering that most of the IC's out there seem to either not be able to handle, or just barely handle the 5-6W required for a 5W LS circuit.

Of course, I'll be damned if I actually understand what that part *does*, but I figure someone here can sort it out.

-XCN-

MrAl · Jun 16, 2002

Hi there Xcand,

<BLOCKQUOTE>quote:<HR>Originally posted by Xcandescent:
Am I correct in assuming that the only purpose to having a higher voltage at the ADJ pin is to allow brightness adjustment by using a pot to reduce Vadj -- and that, if you just want fixed output, it's more efficient to decrease Vadj and recalculate Rfb with the new value?
<HR></BLOCKQUOTE>

Yes that's the exact idea.

LTC1871 chip:
The LTC1871 chip isnt really a current mode
converter like the MAX chip. It's only
current mode in that it measures current
though the inductor, not the output.
The MAX chip does sense output current,
and so it will suit our purpose of driving
an LED much better.
The chips look almost the same otherwise.
Renes' choice of chips was the best one,
as it has true current feedback, plus
uses an external MOSFET. These two
features make it the best choice for going
to 700ma output current, or even higher,
unless someone comes up with a simpler
circuit that has both these features of
course 🙂

The writeup in the LTC1971 data sheet is
very good though, so i think it's worth
a read. It's very detailed and in depth.
One point they make about saving efficiency
isnt really true for our app because they
assume a very high output current (7A or so).
With our max final current at 10 times less
then this (0.700 amps) we wont have this
problem if the sense resistor is made
equal to 0.05 ohm or even a little less.

To explain the difference between using a
1 ohm sense resistor and an 0.1 ohm sense
resistor for 350ma output a little:

Lets say we have 6.00 volts output
with 350ma flowing, using a 1 ohm sense
resistor.
Since there is 350ma flowing through 1 ohm,
we have 0.350 volts across the resistor.
Across the load we have 5.65 volts.
The input power to the load is
6*0.350=2.1 watts.
The true output power is only the power that
reaches the load, which is
5.65*0.350=1.9775 watts.
Calculating Po/Pi we get:
1.9775/2.1 = 0.94, or 94 percent.
This means we lost 6% in the sense resistor.

Now lets calculate the eff lost in the
0.1 ohm sense resistor:
0.1 ohm at 0.350 amps means there is
0.035 volts across the 0.1 ohm resistor.
The output voltage is thus
5.965 volts.
The true output power delivered to the load
is
5.965 * 0.350 = 2.08775 watts.
The input power is still 2.1 watts.
Again calculating eff:
eff=Po/Pi=
2.08775/2.1=0.994, which means
now only 0.6 percent of the eff is lost
in the sense resistor.

This means the lower the sense resistor the
better. For 700ma output, 0.086 ohms
would be even better yet.
Since the min
spec on the ADJ pin is 50mv, it's best
to keep the ADJ pin at 100mv or so, so
going lower then 0.086 ohms for 350ma output
and 0.043 ohms for 700ma output isnt good
either.

11.5k and 1k for the resistor divider
would provide 100mv at the ADJ input,
in which case you could use about
0.086 ohms for 350ma output and
0.043 ohms for 700ma output.
This resistor combo keep the loss
in the sense resistor down to around one
half of one percent in efficiency.
This of course means that if you had
a circuit that was 100% eff and you
added this sense resistor, you would
still get 99.5% efficiency.

Good luck with your LED circuits,
Al

remuen · Jun 16, 2002

Hi Al

Thank you very much for your explanation.

I must have been stupid that I completly have ignored this ADJ function.

I didn't want to use it in this first test circuit and didn't look what it does.

I just tried to solder a small SMD pot to this ADJ pin (with wires) but I'm sorry I'm not able to solder wires to this da... little pins without making a shortcut....

So I have to make a new pc board where I can put the pot on the board. Next time I am going study the datasheet more carefully before etching and assambling a board! But as a beginner I still am allowed to make some mistakes ...

Xcandescent · Jun 16, 2002

MrAl:

Thanks for the explanation! I'm still not up on what everything means (like running in "current mode"), but I had assumed that, even if the LTC1871 regulated voltage, it could be made to regulate current using a variety of techniques discussed in other threads (not that I would know how to do it -- just that it could probably be done).

Obviously 7A is overkill, but it's the only IC I've seen so far that seems capable of handling the 5W without breaking a sweat, unlike the other IC's which seem to be having issues (though it seems a lot of those hinge on the inductor being saturated).

re: Really small resistances. Getting a fractional resistance is something I'm not familiar with. I guess one could put a bunch of resistors in parallel to do it, but more often than not people seem to use thin wires for the job, which goes completely over my head. (That bit about folding a piece of wire in half and soldering it at some point to a board completely lost me in one of your other circuits.) I guess I would assume there's a formula for obtaining a desired resistance given a length of wire and its gauge? That doesn't even cover the idea of using traces on a PCB to simulate a resistive wire (again ... how would one determine the necessary resistance from the traces?)

-XCN-

MrAl · Jun 17, 2002

Hello again,

<BLOCKQUOTE>quote:<HR>Originally posted by remuen:

Thank you very much for your explanation.

I must have been stupid that I completly have ignored this ADJ function.

I didn't want to use it in this first test circuit and didn't look what it does.

<HR></BLOCKQUOTE>

Stupid? I dont think so!
The fact that you found the perfect chip
has to account for something doesnt it?
The chip you found has the best features
built in:
1. External mosfet, ie you can choose a modern very low Rds part
2. Easy to add current feedback, ie no additional active circuits needed

What more could you ask for 🙂

Xcandescent:
It's ok that the LTC1871 can handle 7A and
we might only use it for 0.7A, that's not
the point. The point is, we are looking at
two chips:
both have external mosfets,
both can handle 0.7 amps,
but only one can handle a current feedback
topology without adding another active
device (such as an op amp or active
current sensor).
Which do we choose:
the one that has everything on the same chip,
or the one that requires us to buy another
part?
Add to that the higher cost
of the LTC1871 (2 to 3 times higher).
The choice probably boils down to
whether or not we want to use copper
wire as our sense resistor or not.
If we want to use copper wire, maybe using
one of the Zetex high side current monitors
would be better, as they offer some degree
of temperature compensation when using copper,
but we would have to add that to the
parts list.
With the Max chip which Rene found,
you can use a standard resistor value for
the sense resistor (like 0.1 ohm or 0.050
ohm) and then adjust the final output
current by either swapping a resistor in
the ADJ voltage divider or just using a
pot.

If you are interested in using copper wires
as sense resistors, you have to keep in mind
there is a positive temperature coefficiency
associated with copper. The resistance gets
higher and higher as temperature rises.
Some devices are made to take advantage of
this (Zetex) but some arnt.

I could post some formulae for copper traces
if your interested. You will have to know
the width, thickness, and total length of
the copper trace in order to calculate the
resistance.

Good luck with your LED circuits,
Al

Xcandescent · Jun 17, 2002

<BLOCKQUOTE>quote:<HR>Originally posted by MrAl:
It's ok that the LTC1871 can handle 7A and
we might only use it for 0.7A, that's not
the point. The point is, we are looking at
two chips:
both have external mosfets,
both can handle 0.7 amps,
but only one can handle a current feedback
topology without adding another active
device (such as an op amp or active
current sensor).
Which do we choose:
the one that has everything on the same chip,
or the one that requires us to buy another
part?
Add to that the higher cost
of the LTC1871 (2 to 3 times higher).<HR></BLOCKQUOTE>

Like I said -- I've no clue what I've stepped into, so if there's an IC for cheaper that can do what we want, it's all good.

I only brought it up on the off-chance that the other IC's can't handle the power requirements (which is something that'll have to be borne out in practice).

<BLOCKQUOTE>quote:<HR>If you are interested in using copper wires
as sense resistors, you have to keep in mind
there is a positive temperature coefficiency
associated with copper. The resistance gets
higher and higher as temperature rises.
Some devices are made to take advantage of
this (Zetex) but some arnt.

I could post some formulae for copper traces
if your interested. You will have to know
the width, thickness, and total length of
the copper trace in order to calculate the
resistance.<HR></BLOCKQUOTE>

I only brought up the usage of copper wiring for resistance because it seems to be common in quite a few circuits -- and isn't something I'm familiar with. At my barely experienced level, if I have the choice between working with standard value resistors and a piece of wire whose resistance changes with temperature, I'm much more inclined to go with the former.

That said, the more knowledge, the better. Formulae for copper traces and copper wire would be appreciated, as would an explanation of how to compensate for the temperature coefficient.

-XCN-

dat2zip · Jun 17, 2002

Here is my data for the Current mode converter I did a while back called badboy. This regulator is a true current mode regulator which regulates the current to the load (LS).

I setup this run one for 666mA of regulation. Two 0.15 ohm SMT resistors in parallel.

Code:

This is a single IC solution. Fits on a 14mm size board (Mini Maglight style). Output current mode regulation.

Output current set by low value sense resistor (0.06 to 0.1 ohms).

Two 0.15 ohm resistors in parallel were used for this run. A 0.15 and a 0.12 would set the current regulation to around 0.75A.

There were reasons I had not evaluated this converter for usage in the 5W configuration were:

The testing I did for the 2AA battery setup got poor efficiency. Thinking back I realized the lower voltage supply might not be turning on/off the transistor very fast leading to poor efficiency. Raising Vin for the 5W does improve the turn on/off of the internal switch element.

2ndly: My initial target of badboy was to compete for the king of the hill in the Mini Maglight arena and drive the LS to 1A. And this IC does do it but is extremely tempermental and overheats rapidly. This setup was pushing the part well outside of useable parameters.

Because of the King of the hill testing I did I kinda came to the conclusion efficiency sucked soo bad it would not be useful for other configurations.

Sincy I'm still struggling with the LM2621 and the ZXCT current sense amplifier I've stepped back to re-consider the options.

This part came back up for re-evaluation and the run above shows good promise.

I could in a couple days clean up the layout, finalize the components and send out for some small qty of boards.

So, if this is OK with everyone here, I'm going to proceed with this design.

MrAl · Jun 18, 2002

Hello again,

datzip:
That circuit looks very promising, yes.
What chip does that use?

Xcand:
I didnt mean to sound like i was putting that chip down or anything🙂
Actually, the more chips we have in the pool the better, in case we
decide later to try something else, so anything you find might be of use
at some point anyway.

As far as the trace resistance goes, the formulae are pretty simple.
You just have to make some measurements and decide how wide and long
to make the trace.
These are in convenient units of inches, ohms, and degrees C.

r = resistivity of copper = 6.7878e-7(1+.00393(T-20)) {ohm inches, T in deg C}
where
T is the temperature of the copper.

R= total resistance = r*L/A
where
r is resistivity in ohm inches
L is length in inches
A is cross sectional area in square inches

It's also a good idea to establish a rule of thumb for the
maximum current allowed for a given trace (or wire) area.

The approximate 10 minute extrapolated fusing current is:
i(600)=800*A*(48-Ta*17/400)/pi {for 0<=Ta<=400}
where
Ta is the ambient temperature, zero to 400 degrees C.
A is cross sectional area in square inches.

We now end up with two conditions that our trace (or wire)
must meet:
1. R must be equal to the correct value for the circuit requirements.
2. i(600) must be less then i(average) through the resistor.

It would be easy enough to set the minimum cross sectional area
by solving the eq for i(600) for A, then divide by the trace
thickness to get the width, then calculate the length L to satisfy the
total resistance R.

Temperature compensation using copper wire resistors for the sense resistor:

When using the copper wire instead of the standard resistor the value will
change as the temperature changes which may or may not upset the output current
setting, depending on what the chips' sense voltage level does.

Taking the ratio of the resistivity r at T=21 to the
resistivity r at T=20:

r(20)= 6.7878e-7(1+.00393(20-20))
r(21)= 6.7878e-7(1+.00393(21-20))

ratio=r(21)/r(20)=1.00393
or
approximately +0.4 percent per degree C.

Now many of the Zetex chips sense voltages are sensitive to a package temperature
change of this same amount, +0.4%/degC. This means the sense voltage level
changes with temperature at the rate of 0.4 percent per degree C.
Looking back at a resistor whos' resistance changes by 0.4%/degC and also with a
constant current flowing through it we find that the voltage across it changes by
the same rate with temperature: +0.4 percent per degree C.
With both the chips' sense voltage changing and the resistors' "across" voltage
changing the current though the resistor stays the same.

Of course if you are using a chip whos' sense voltage doesnt change
by +0.4 percent per degree C then using a copper wire resistor will cause
a -0.4 percent per degree C error in output current regulation. In other words,
the current will decrease by -0.4 percent per degree C. This may or may not
be desireable, as you may wish to protect another element from burning up
due to an ambient temperature rise.

The opposite is also true:
If you use a chip whos' sense voltage does change with temperature and you
use a sense resistor whos' value doesnt change very much over a wide temp
range (such as a 1% film resistor), you end up with about a +0.4 percent per
degree error, or a current increase of +0.4 percent per degree C.

This is the basic idea behind temperature compensation using a
copper sense resistor.

Good luck with your LED circuits,
Al

vicbin · Jun 18, 2002

Hi Rene,

Got your email for the layout, thanks.

<BLOCKQUOTE>quote:<HR>Originally posted by remuen:

I must have been stupid that I completly have ignored this ADJ function.

I didn't want to use it in this first test circuit and didn't look what it does.
<HR></BLOCKQUOTE>

Don't think so, currently your finding so far is the best solution IMHO, with the 1722 combination , two cells is now possible, MAG 2C or 2D is feasible. This 172x combination is really a great finding for powering the MAX1698. It is a proof that you're not what you claimed !

Looking at your PCB layout, I was wondering have ever consider replacing the 1722 with the fixed voltage version ? e.g. 1724EZK30 for 3 Volt , more space can be saved since it doesn't need those 2 feed back resistors. Again just my 2 cents.

I'm so exciting to hear for your next run result with a bigger inductor and with the new Vadj (smaller Rfb).

Vic

remuen · Jun 18, 2002

<BLOCKQUOTE>quote:<HR>Originally posted by dat2zip:

Here is my data for the Current mode converter I did a while back called badboy. This regulator is a true current mode regulator which regulates the current to the load (LS).
<HR></BLOCKQUOTE>

Hi Wayne
Heja, pretty nice results with the badboy!

IMO it's a good thing that you brought up this chip again as the MAX1698 causes me some troubles with the rev1 board (that's the new design with only a 0.1 ohms sense resistor and a pot for the ADJ). I'll come back on this as soon as I have checked it more closely.

<BLOCKQUOTE>quote:<HR>Originally posted by vicbin:

Looking at your PCB layout, I was wondering have ever consider replacing the 1722 with the fixed voltage version ? e.g. 1724EZK30 for 3 Volt , more space can be saved since it doesn't need those 2 feed back resistors. Again just my 2 cents.
<HR></BLOCKQUOTE>

Hi Vic

IMHO you're right - the MAX1724 would be the better choice concerning the number of parts. In this case I just took the chip I had in stock and this was the MAX1722 as I already used this little chip to drive one or two Nichias in parallel. There is only one question: How behaves the MAX1698 with only 3 volts input (I adjusted the MAX1722 to about 4.1 volts).

Jonathan · Jun 18, 2002

<BLOCKQUOTE>quote:<HR>Originally posted by vicbin:
Don't think so, currently your finding so far is the best solution IMHO, with the 1722 combination , two cells is now possible, MAG 2C or 2D is feasible. This 172x combination is really a great finding for powering the MAX1698.<HR></BLOCKQUOTE>

Yes, bravo for figuring out a combination of components which might actually run a 5W lamp with a couple of volts input.

I'm still in favor of using more cells and a buck converter rather than a boost converter. 8AA cells in a 2D mag body will be able to provide more total energy than the 2D cells that they replace.

However 8AA cells is yet another level of modification...in theory, using the boost converter circuit, one could have a circuit board which perfectly fits in the head of the maglight, connects where the bulb connects, and leaves the rest of the flashlight entirely unchanged.

If you do this, I would _strongly_ suggest running the light with D sized NiCd cells, batteries or D sized NiMH cells, rather than alkaline cells.

I spent some time looking at 'Energizer' datasheets. Their D cells are rated at 18,000 mAh, but only at a 25mA draw. Try to pull a couple of amps out of them, and the voltage falls through the floor. (Note that this internal resistance effect is a great advantage with the direct drive LS devices.)

The NiMH D cells are rated at 8,000 or 9,000 mAh (if you get good ones), or about half what the Alkaline cells are rated at. But they will be able to deliver a much greater fraction of this at high current. Alkaline cells have a nominal voltage of 1.5V, which at first looks better for this application than the 1.2V of a NiMH...but under load the NiMH will actually have a higher output voltage.

If your application demands alkaline cells (say because it has to sit on a shelf for a while), then go with the AA cells. You could use 4AA cells in parallel to replace 1 D cell, which will give substantially better results at the current levels that we are talking about. (According to the datasheet, if you compare a D cell at 2A to an AA cell at 0.5A, you will get about twice the run time from the AA cell. Put four AA cells together, and you are back at your full 2A of current.)

However, once you've gone to the 8AA cells, it may pay to wire them in series, get the 12V, and then step down. The benefit is that you have to jump through fewer hoops to get the buck converter to work, the input and output capacitors are smaller, the inductor and switch currents are lower, and if you come up with a simple circuit, it will be useable for the 'Light Up the World' project.

-Jon

remuen · Jun 20, 2002

Hi again

In the meantime I've build my rev1 board which now uses only a 0.1 ohms current sense resistor (as suggested by MrAl) and a 250kohms pot at the ADJ pin.

This rev1 board caused me quite some troubles as I produced the first time some smoke and flashs with one of my booster circuits

and had to replace the chips ...

Ok, now here the results with this MAX1698 rev1 in a comparison with the former results of the rev0 version:

Code:

I could not increase the input voltage to more than these 4.6 volts as after a few seconds the MAX1698 reduced its output to about 150mA. After reducing the input voltage I again could go up to 4.5/4.6 volts input. Seems there is some kind of a thermal switch off function as soon as we reach a certain output power (or better a certain loss = heat in the chip).

Interesting is that the input current stays always at about the same level! This may be caused by an internal peak current limit of 1.5 amps (see page 7 of the data sheet in the 'Inductor Selection' section).

Added later:
------------
A strong indication that there must be some kind of current limiting is:
Reducing the input voltage fast from eg. 4 volts to 2 volts the input current increases to over 2 amps. This is not anly visible on my DMM but also the current limiter of my regulated power supply which is a 2 amps version gets into action. But then within 1 to 2 seconds the input current of the MAX1698 boost converter sinks to these ~1.3 amps I measured as input current.

So my question to the experts is: Can we eliminate this peak current limiting function eg by connecting the MOSFET directly to ground instead to the CS pin? IMO it is only needed to get a better regulation with the MAX1698. Or is there another way to outwit this function?

A curious thing: I could not adjust the output current with that pot at the ADJ pin!

So I think these results prove that the MAX1698 has the potential as a boost converter for the new 5W LS

. Who is going to improve it? IMO I reached my personal limits!

Best Chip/Step Up Circuit Design for the new 5W Luxeon Star?

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