Simpler Constant Current Zetex Circuit (ZLT-CCR-2)

Hello again to those interested in
constant current boost regulators.

This is another new circuit for constant current output regulation
for the Zetex 300 chip (ZLT).
This one is even simpler then the last one, but it doesnt
compensate for the variations in LED voltage, so each
LS will have to be set up individually as with the
original ZLT circuit.
It of course does compensate for input
voltage variation as the data below shows.


The new circuit is at: http://hometown.aol.com/xaxo/page3.html

As shown, this one only requires the addition of two SM resistors.


For nominal 2.4 volt operation:

Start with the original ZLT circuit.
Disconnect pin 4 from the sense resistor.
Change sense resistor to half it's original value.
Insert a 5 ohm resistor between pin 4 and the sense resistor.
Connect a 1150 ohm resistor between pin 4 and Vcc (+Battery).

Both resistors can be SM. Sense resistor is a lenght of copper wire,
about half it's original length (half of that which was in the ZLT).


Test circuit by turning on and checking for overheating in the
transistor, and adjust current output by adjusting sense resistor.
Use an LS simulator for the first run.

DATA

ZLT-CCR-2:
2.80v 314ma
2.60v 331ma
2.40v 340ma
2.20v 342ma
2.00v 339ma
1.80v 329ma
1.60v 312ma
1.40v 289ma
1.20v 258ma
1.00v 221ma
0.80v 177ma

Pretty good for the price of two additional SM resistors if you ask me


Good luck with your LED circuits,
Al

Hi Al,

Amazing, it much more simple and good regulation too !!

At the data it seems top current at -> 2.20v 342ma , is it possible to make it at 2.40v ? Yes, it is related to flat NiMH cell discharge property at 1.2 V.

Regards,
Vic

Damn, Al, if you don't stop coming up with better ideas, I'm never gonna get this built...

/frank

Okay, Al, where do you buy your resistors?

4.4 ohms in the CCR, 5 and 1150 ohms in the ccr-2. Okay, I guess Digi-Key sells 1150's, and I guess I can stack two 10's to get a 5, but you must have access to a stash of parts that I don't have.

You have a better feel for the space available inside a Mag than I do (I haven't tried to fit one of these in yet), but how do you feel about the reasonableness of trying to layout the circuit with pads for pins that would directly connect the board with the bulb socket inside the maglite (rather than the two wires shown on the 5thcolumn site)?

I have a layout for the original ZLT, and now the CCR-2 that does this, but before I start etching PCB's I was wondering if you knew of anything that would keep this from working.

Thanks,

/frank

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR> You have a better feel for the space available inside a Mag than I do (I haven't tried to fit one of these in yet), but how do you feel about the reasonableness of trying to layout the circuit with pads for pins that would directly connect the board with the bulb socket inside the maglite (rather than the two wires shown on the 5thcolumn site)? <HR></BLOCKQUOTE>

If you intend to go with the design from the 5thcolumn site, I assume you will end up using a tail cap switch. There is barely enough space in the minimag head to install both the circuit and the optics while still having the light screwed together securely. You may have to remove the lens cover of the minimag as well as the top half of the switch assembly to get everything to fit inside. On top of that, you will need to unscrew the head so much that the O-ring on the battery barrel is practically showing. A better approach would be to remove the entire switch assembly, insert the circuit into the battery barrel, and drill holes through the board to route the necessary leads to the other side. The LS can be placed on the outside of the battery barrel for better heat sinking.

Hello again,

Vic:
Actually 342ma isnt all that much different then 340ma, is it?
Setting the max at 2.2 volts means 2.4 and 2.0 volts are both the same.
Also, adjusting R1 to 0.014 ohms makes it 350ma:

R1=0.014 ohms
R3=5 ohms
R5=1150 ohms

Vin __ Iout
2.40v 350ma
2.00v 350ma
1.60v 322ma
1.20v 266ma
0.80v 182ma

With a real life circuit it might be necessary to play around with the
values of R1 and R5, but R3 should be able to stay at 5 ohms.
If you want to try to double R3 to 10 ohms, then also double R5.
R1 sets the max current out, while R5 sets the regulation percent over
a range of input voltages, and some values work better then others.
1150 ohms looked the best.

Frank:
Hee hee, i just thought i would try and come up with something a little
simpler then the original network. Remember though, this new one doesnt
compensate for changes in LED voltage, only input voltage (which isnt that bad
after all).
See above for other possible resistor values.
I dont see any problem with adding two SM resistors to the board, as this shouldnt
take up that much more space really. Maybe on the other side might work also.
The ZLT worked very well, so i dont see why this new circuit shouldnt work either.
The new one (ZLT-CCR-2) doesnt have the input surge problem either, which is
a nice side benefit.
Before you etch more then one pcb you should always prepare a prototype anyway,
with this or any other circuit. Once you test one, you know it works the way
you want it too and you can go on to make more boards if you so wish too.

Minimag LS conversion:
Personally, i originally wanted to convert a minimag too, but then realized a
few things:
1. it was always going to mean using two AA batteries, which dont last that long
unless you use NiMH cells, which have a nasty self discharge.
2. the LS/o has to be ground down to fit
3. a special switch has to be purchased
4. other problems that plage the minimag still exist such as:
a. The paint scratches too easy:
I dont know anyone who uses their minimag, converted or not, that
doesnt have at least several scratches on it after a few months.
I still try to take care of my Nichia converted minimag, but even though
i never use it anymore, i got several scratches on it just from carrying
it around in it's belt holster.


Im moving toward using the ABS plastic cases now. They are rugged,
lightweight, and even if scratched it doesnt show up very much.
Some sizes fit a Li-ion cell nicely, and with a special circuit
to take advantage of the unique Li-ion/LS compatability, it
works out really great.

I see Vegeta found some other problems too with converting the minimag.
I suspect that future LED's might be better suited to this light case.


The basic circuit explanation is about the same as the ZLT-CCR, except it
doesnt measure output current.
In the ZLT, as the battery voltage runs down, although the inductor charges up to the
same current on each cycle, it takes longer and longer to do so, hence the
duty cycle gets lower and lower. If the inductor could be charged to a
somewhat higher and higher current as the voltage dropped lower and lower,
some compensation would be noticed in the average current delivered to the
output. Since the input voltage is dropping, using this voltage as a bias
on the inductor current measurement would mean the inductor current would
increase as the input voltage dropped in order to keep the sense voltage
at a constant 17mv or so. It would then only be a matter of determining the
relationship between the percent input drop and the required inductor
current increase to determine the values. More simply, bias the sense
voltage at half the nominal value and set the inductors' contribution
to the sense voltage at half of what it was before and see what happens
This means the input voltage partly determines the inductors' current peak,
rather then just keep charging the inductor to the same peak regardless of input.
This means there is more energy stored in the inductor to help compensate for the lowering duty cycle.
Since Vin goes down as the battery drains, the inductor current has to
go higher and higher in order to maintain the 17mv sense voltage.
Makes a lot of sense, and it happens to work nicely too.
I havent bothered to try to determine the exact relationships mentioned above,
because i think in the real world the nonlinearity of the inductor will be
too hard to determine, and if it has a profound affect on this relationship,
no equations for this will be worth the effort. It will still boil down
to a little bit of trial and error on the exact resistor selections for
R1 and R5. Adding to this complication is the possible variation in
the sense voltage threshold, which may vary quite a lot; the difference
between 17mv and 20mv will have quite an impact on the value of R1.
Besides this, the simulations seem to
indicate the values good enough to start with.
Also, R1 should be a copper wire to help compensate for the temperature
variation of the chip itself. This is nice in a way, because thin copper
wire is cheap and it's easy to adjust the length.

Oh one more little thing:
Increasing the inductors' current peak
decreases the duty cycle even more, but
because the energy stored is proportional
to iL squared and the energy transferred to
the output is proportional to the duty
cycle (a linear relationship) eventually
a favorable iL squared and duty cycle is
reached where more energy is transferred
because the iL squared energy increase
increases faster then the linear duty cycle
decreases the energy transfer.


Good luck with this and other LED circuits,
Al

So how do you deal with heat dissipation in the plastic cases?

/frank

Hi Frank,

That's a very good question, as the plastic
case doesnt conduct heat from the LS to the
outside of the case as well as say the
aluminum of the minimag (as you obviously
already realized)
This means you need air holes or slots.
With my particular case, the board of
the LS is slightly larger then the height
when fully closed, so rather then grind it
down i left one end of the narrow case open.
This creates just enough of a slot along
both sides where heat can escape via
convection, plus gives the flashlight an
aerodynamic look and feel


Good luck with your LED circuits,
Al

Hello again,

Here is a better explanation of the circuit
and the resistor selection procedure.


Referring to the schematic at: http://hometown.aol.com/xaxo/page3.html

Solving the resistor network R1, R2, R3 knowing the inductor current I and the
input voltage V leads to the equation:
Vs=V*R13/R135+I*R1/R135*R5
where
Vs is the sense voltage presented to the chip, and
R13=R1+R3, and
R135=R1+R3+R5

For a linear inductor the required peak current increase is proportional to the
decrease in input voltage. To describe this we first define
x=Va/Vb, where
Va is nominal input (or high battery voltage input set point)
Vb is drained voltage input (or low battery input set point)

Now since we have two set points at which we want perfect regulation,
we calculate the sense voltages for each condition:
Vs1=Va*R13/R135+Ia*R1/R135*R5
Vs2=Vb*R13/R135+Ib*R1/R135*R5

where
Ia is peak current at nominal input, and
Ib is peak current at drained input that is required to keep Iout
the same as when full voltage was applied to the input.
Vs1 is sense voltage at high input
Vs2 is sense voltage at low input

Using
Vb=Va/x,
Ib then becomes:
Ib=Ia*x

Now since Vs1 has to equal Vs2 when the input drains down to the second set point,
we set the two equations (for Vs1 and Vs2) equal to each other.
Solving in terms of Va, Ia, and x, we end up with:
Va*R13+Ia*(1-x)/(1-1/x)*R1*R5=0

To simplify things a little, we also define:
xx=(1-x)/(1-1/x)=-x

In terms of xx we now get:
Va*R13+Ia*xx*R1*R5=0

Rearranging the first equation a little we get:
Vs*R135=Va*R13+Ia*R1*R5

Taking these last two equations simultaneously, we have a system of
two equations in two unknowns (R1 and R5, we will set R3=5 ohms later).
[1] Va*R13+Ia*xx*R1*R5=0
[2] Vs*R135=Va*R13+Ia*R1*R5

Solving [1] for R5 and substituting the result into [2] we get a quadratic
equation in the unknown variable R1:
R1*R1*(Va/xx+vt-Va)+R1*(R3*Va/xx+vt*R3-Va*R3-Va*vt/Ia/xx)-R3*Va*vt/Ia/xx=0

Note also that the threshold voltage vt was substituted for the sense voltage, Vs.
This is normally about 0.017 or 0.018, but can be as high as 0.025 volts.

Looking at the quadratic, we see the coefficients:

a=Va/xx+vt-Va
b=R3*Va/xx+vt*R3-Va*R3-Va*vt/Ia/xx
c=-R3*Va*vt/Ia/xx

which represents the coeff's of:
R1*R1*a+R1*b+c=0

Using a decent value of R3 set equal to 5 ohms (to keep network bias current low),
we can then solve for R1.
Solving for R1 we get two solutions. The positive solution will be the correct one.
If by chance you get two positive solutions, try each one in the two equations
[1] and [2] above to see which one is correct.

Now knowing R1 and R3, we can solve for R5:
R5=-Va*(R1+R3)/[Ia*xx*R1]

Thus, we obtain the values of R1 and R5 that will provide for perfect regulation
at the two selected input voltage points. For points in between, they should be
fairly close too, although they wont be perfect. Somewhere in between there
will be a high point, where output current is maximum.


NUMERICAL EXAMPLE

Lets say we want constant current output at the two input voltage points:
2.4v and 1.666667.
(The second point was chosen at 2.4/1.44, and 1.44 is the square of 1.2, which
makes the highest input 20% higher than the midpoint, no big deal really though).
Lets also say that we measure 0.5 amps peak inductor current at 2.4 volts input
in the uncompensated circuit when we have our nominal output current present(usually
350ma for the LS).
Lets also say that our chip has a voltage threshold of 0.018 volts, so
we set vt=0.018 .


First, we calculate x:
x=Va/Vb=2.4/1.6666667=1.44
then xx:
xx=-x=-1.44

Now set R3=5 and calculate a,b, and c for the quadratic:
a=Va*Ia/Ia/xx+vt-Va=-4.04866667
b=R3*Va*Ia/Ia/xx+vt*R3-Va*R3-Va*vt/Ia/xx=-20.18333333
c=-R3*Va*vt/Ia/xx=0.30000000

Now we solve the quadratic
R1*R1*a+R1*b+c=0
using a numerical solver for R1 and get two solutions:
R1=-5, and
R1=0.0148196
Since the first is negative, and R1 wont be negative, we use
the second solution, R1=0.0148196.

Knowing R1 and R3, we solve for R5:
R5=-Va*(R1+R3)/[Ia*xx*R1]=1127.963

So here we have found the three required resistors:
R1=0.0148196
R3=5
R5=1127.963

With a linear inductor, the circuit with these values
would provide the same current at 1.666667 volts input as
with 2.40 volts input, and there will be a slightly higher
point somewhere in between.

To check the midpoint high current, we go back to:
Vs=V*R13/R135+I*R1/R135*R5
and solve for I:
I=(vt-V*R13/R135)/(R1/R135*R5)
Now with vt=0.018
and V=Vm=(Va+Vb)/2 which is the midpoint input voltage, we get
I=0.61 amps.
To see how much higher this is than it would be if it were also perfectly
regulated, we first calculate the
inductor current that should be present at the midpoint voltage Vm:
Ix=Va/Vm*Ia=0.59 amps.
then the ratio of this inductor current Ix to I:
I/Ix=1.034 times higher, or
3.4 percent higher output current at the midpoint.
This is still a safe operating point for the LS, and would result in
a current of about 362ma if the output was originally set at 350ma.
If you want to be perfectly safe, set the nominal output at 340ma instead
of 350ma.

Since the input voltage is present as soon as the switch is closed,
there wont be any current surge when the circuit is first switched on.


CONCLUSION

Using this circuit with a linear inductor makes it possible to select
two points on the input voltage curve in which to provide perfect
current regulation. The regulation is very good in between these two
points also, so the circuit should regulate pretty good.
The inductor only need be linear in that it has to store energy
proportional to it's current squared over the required range, not necessarily
over the full possible range of the inductor. For inductors that dont
meet this criterion some deviation in the regulation will be noticed,
and just how much depends on the actual physical device.
A possible solution to a highly nonlinear device would be to experiment
with R5 to see if other values work better. R1 can then be changed to
set nominal output current.

Whelp, im off to try to do some small chip soldering again, those
0.0256 pitch parts sure are small and the leads are so close
together

Good luck with your LED circuits,
Al

mr.al, you're a madman. it's constantly amazing to me how much effort and thought you put into your projects. very well done.

now if i could stick all of your knowledge directly into my head...

Hi there papasan,

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by papasan:
mr.al, you're a madman. it's constantly amazing to me how much effort and thought you put into your projects. very well done.
<HR></BLOCKQUOTE>

Hee hee papasan, i guess ive always been
interested in the understanding of these
kinds of circuits.
I wish i could say that all that was
perfectly 100% accurate, but i noticed that
my assumption of the relationship between
the required increase in peak inductor current
with the decrease in input voltage (as the
batteries drain down) isnt exactly linear,
so i might come up with a second method
where you measure the inductor peak at
the two points you want to regulate and
then calculate the two resistors R3 and R5
after doing that. This will also take into
account some normal nonlinearity in the
inductor too, which is almost always present.
In the mean time, i noticed that that
original procedure got me pretty close to
the final values as is, so it works pretty
decently anyway i guess.
With the new method of making two measurements
first instead of one, it should get even more
accurate then before.

Good luck with your LED circuits,
Al

Hello again,

Here is the second method for finding
R1 and R5. This is almost the same
but requires two measurements of the
inductor current peak and so should be
much more accurate.


Again starting with the network (R1, R3, R5) equations for Vs:

Vs1=Va*R13/R135+Ia*R1/R135*R5
Vs2=Vb*R13/R135+Ib*R1/R135*R5

where
R13=R1+R3
R135=R1+R3+R5
Va=high battery voltage (2.4v)
Vb=low battery voltage (1.6667v)
Ia=peak inductor current with high battery voltage in and 350ma out
Ib=peak inductor current with low battery voltage in and 350ma out
Vs1=sense voltage with high battery in
Vs2=sense voltage with low battery in

This requires measurements of the peak inductor current at the two
input voltages Va and Vb after readjusting the sense resistor
(in the original ZLT circuit) to get 350ma output.

Again, since Vs1 must equal Vs2 we set their two equations equal:

Va*R13/R135+Ia*R1/R135*R5=Vb*R13/R135+Ib*R1/R135*R5

reducing, we get:

(R1+R3)*(Va-Vb)/(Ib-Ia) = R1*R5

Now defining
VI=(Va-Vb)/(Ib-Ia)

we can write this as:
(R1+R3)*VI = R1*R5

Taking either Vs equation again and simplifying a little:
Vs*R135=Va*R13+Ia*R1*R5

and now taking these last two equations simultaneously again we get:
[1] (R1+R3)*VI = R1*R5
[2] Vs*R135=Va*R13+Ia*R1*R5
which is again a system of two equations in two unknowns (we set R3=5 ohms and Vs=0.018).

Solving [1] for R5 and substituting into [2] we again get a quadratic
equation in the unknown variable R1:
R1*R1*a + R1*b + c = 0
where
a=Vs - Va - Ia*VI + Vs
b=Vs*R3 - Va*R3 - Ia*VI*R3
c=Vs*R3*VI

Solving this for R1 and then subsituting R1 into [1] and solving for R5 we
get the two required values of R1 and R5.

The difference between this method and the first method of finding
R1 and R5 is that this method doesnt presume any special relationship
between the input voltage and the required increase in peak inductor
current to get a constant current output as the battery drains down.
This means it should work better even with quite a bit of nonlinearity
in the inductor.

Yes there are a few equations to solve, but the resulting cost increase
for two SM resistors will be minimal.


Good luck with your LED circuits,
Al

Hello again,

Procedure and Numerical Example
using the second method.


With this more accurate method we need to make two measurements of
the peak inductor current I and the input voltage V.

For this example we will assume that we wish to set up the
circuit to output 350ma at 2.4 volts and also at 1.667 volts
as before.

The general procedure:
We start with the original Zetex circuit with a single current
sense resistor, adjust the input voltage to high input (2.4v)
and adjust the sense resistor to provide 350ma output and
take an inductor peak current reading, then, adjust the input
voltage to low input (1.667v) and then re-adjust the current sense
resistor to provide 350ma output again and take another inductor
peak current measurement.
Then, we calculate the values of R1 and R5 from this data.
Finally, we test the midpoint voltage input to make sure
it doesnt result in an output current that is too high.


Setting the input voltage at Va=2.4 volts and adjusting the
current sense resistor so that we obtain an output of
350ma through the LED, we measure the inductor current
peak and call it Ia.

Setting the input voltage at Vb=1.667 volts and adjusting the
current sense resistor so that we again obtain an output of
350ma through the LED, we measure the inductor current
peak and call it Ib.


Numerical Example

Lets say we measure the peak inductor current to be 0.6 amps
at 2.4 volts with 350ma output,
and after adjusting the voltage down to 1.667 volts and also
adjusting the sense resistor to again give 350ma output we
measure 1.06 amps peak inductor current.

This gives us:
Va=2.4v
Vb=1.667v
Ia=0.6 amps
Ib=1.06 amps

Now that we have Va, Vb, Ia, and Ib we have all the information
needed to calculate the two resistor values R1 and R5.

First we calculate a, b, and c for the quadratic using a value
of 5 ohms for R3 and a value of 0.018 for Vs
and recalling that VI=(Va-Vb)/(Ib-Ia):

a = 2*Vs - Va - Ia*VI
b = Vs*R3 - Va*R3 - Ia*VI*R3
c = Vs*R3*VI

so from that we find:

a = -3.3205217
b = -16.692609
c = 0.14347826

which are the coefficients of the quadratic:
R1*R1*a + R1*b + c = 0

Solving this numerically or using the quadratic formula:
x1 = [-b+sqrt(b*b-4*a*c)]/(2*a) {sqrt = square root}
x2 = [-b-sqrt(b*b-4*a*c)]/(2*a)

we find that the correct value for R1 is x2, which is:
R1=0.00858067 ohms.

Substituting this value in equation [1] from the last post:
[1] (R1+R3)*VI = R1*R5

and solving for R5 we get:

R5=(R1+R3)*VI/R1
or
R5=930.544 ohms.

Now that we have the best choice of resistors, we calculate the
approximate mid point error to check for high current at mid
voltage input (2.0 volts):

First, calculate the midpoint current from Ia and Ib:
Im = (Ia+Ib)/2 = 0.83 amps

Then, calc the current resulting from using the network R1, R3, and R5 from
Vs=V*R13/R135+I*R1/R135*R5

at the midpoint voltage by setting:

V=2.0 volts (the midpoint input voltage)
and
Vs=0.018 (the sense voltage level for the Zetex 300 chip)
and recalling that
R13=R1+R3 and
R135=R1+R3+R5

and then solve for I:

I = R5*(Vs - V*R13/R135)/(R1*R135)
or
I=0.845 amps.

Now we calculate the ratio I/Im and we get:
I/Im = 1.018
which is about two precent high, and would result in a mid point
output current of about two percent higher, or:
0.350*1.02 = 357ma
instead of 350ma.

This seems like a safe level, so this would work out fine.
In practice however, the midpoint current could be a little
higher than this estimate, so the output current at
the midpoint input voltage of 2.0 volts should be checked also.
If it turns out to be higher then we want (such as 370ma) we
may wish to adjust R1 slightly or go back and repeat the whole
procedure only this time setting up the circuit to output
a nominal 340ma. Even at 6% high, the circuit would then output
between 340ma and 360ma for input voltages between 1.667v and
2.400v.

Good luck with your LED circuits,
Al

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by MrAl:
Procedure and Numerical Example using the second method.
<HR></BLOCKQUOTE>

Hi Al

Astonishing! Unbelievable! I should be ashamed looking on my 'try and error-methods'! So all this is far beyond my level. I only can try to put all these formulas in an Excel sheet and use them as soon as I have time to again work with the ZLT ....

There is only one question: Is there any possibility to measure the inductor peak current without a scope?? And if yes is the result good enough to make these calculations?

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by remuen:
Hi Al

There is only one question: Is there any possibility to measure the inductor peak current without a scope?? And if yes is the result good enough to make these calculations?<HR></BLOCKQUOTE>

That's a very good question Rene,
as most people dont own scopes
and dont want to buy one.
Not only that, the sense voltage level
might vary from chip to chip too.

The only other way is to try to measure
the two sense resistor values instead
of the two peak current levels.
Then one might imply what the peaks
are for both input voltages.
Since this is somewhat impractical
too (it can be done though)
maybe the best way is to just install
a 2k pot for R5 and starting with around
1.9k, adjust R1 (wire length) for
350ma output at 2 volts input. Then,
check output current for inputs of
2.4 volts and 1.667 volts. If it's not
to your liking, readjust the pot for
1.8k and try again. If this too doesnt
work good enough, decrease to 1.7k
and try again, etc, etc, untill you
get a range that you like.
The range of values R5 might take on
goes from about 500 ohms min to about
1.9k max.

The benefits are pretty good, as the circuit
no longer will burn up the LS if the input
goes a little higher then expected.
Only two more very small SM resistors seems
like a fair price to pay in real estate.

Using two in parallel might be able to drive
the 5w LS, although i havent looked at this
in depth yet.

Lastly, please remember when working with
this circuit and adjusting values that
whenever you turn back on to check the
transistor for overheating. This is the
most important test you could ever do.

Good luck with your LED circuits,
Al

********************************************************************************
VERY IMPORTANT! MUST BE READ! VERY IMPORTANT! VERY IMPORTANT! MUST BE READ! VERY IMPORTANT!
********************************************************************************

Hi Al

Seems nobody has ever tried your CCR2 circuit. I now did it.

Hey, all I can say: Great - no, it's fantastically , phaenomenally ...

First of all: The ZLT in its basic circuit works great. I wanted to drive two LS from 4.8 volts input voltage with it but was not sure whether the ZLT can do this or not. It can!

And as I was interested in this CCR2 addition from the beginning I now have tried to build it. Ok, I first had some troubles as I took some standard resistors with long wires just because they are much easier to solder than SMD parts. The ZLT didn't really like this as the output current jumped down from eg 250 to 200mA and again up to 250mA and down and up without any recognizable reason. Changing to SMD resistors these stupid jumps vanished and that's what I eventually could measure:

Uin | Iin | Uout | Iout | Eff
-----------------------------
6.0 | 375 | 6.42 | 310 | 88.5
5.5 | 408 | 6.40 | 305 | 87.0
5.0 | 451 | 6.37 | 303 | 85.6
4.8 | 481 | 6.38 | 310 | 85.7
4.5 | 520 | 6.37 | 310 | 84.4
4.0 | 579 | 6.31 | 299 | 81.5

This is exactly what I wanted to get. These LS usually will be driven with 4 NiCdO or NiMH cells (4.8 volts) but it could well be that someone will put 4 alcalines (6 volts) in it so this ZLT-CCR2 won't mind!

Al, thank you for this circuit!
Duggg, Al, Mercator, thank you for introducing the ZLT to the CPF!

René

Thats great news

I'm assuming from your eff figures that the two LS were in series...

Maybe there is hope that we can use the ZLT circuits to drive the 5Ws, say two in parallel as MrAl suggested.

Rumuen, you deserve a hug! I've always been a fan of the ZLT. Any chance you could test it with voltages down as low as 2.7v?

Originally posted by Lucien:

I'm assuming from your eff figures that the two LS were in series...

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<font size="2" face="Verdana, Arial">Obviously yes, as the output voltage is > 6volts. I wanted both LS to be driven with the same current.

Originally posted by Lucien:

Maybe there is hope that we can use the ZLT circuits to drive the 5Ws, say two in parallel as MrAl suggested.

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<font size="2" face="Verdana, Arial">The ZLT in its basic form cannot be used for a 5W LS even though I drove my two LS in series with more than 400mA from 4 to 4.8 volts input (!) before I added the additional resistors for the CCR2 circuit. But it would be interesting to connect two FMMT617 in parallel (and not two ZLTs) and test it with a 5W LS. IMHO this should work as well - at least with input voltages =/> 3.6 volts (3 or more NiMHs).

Originally posted by TripleDouble:
Any chance you could test it with voltages down as low as 2.7v?

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<font size="2" face="Verdana, Arial">As one has to optimize the CCR2 network for the voltages/currents one want to use it with (see MrAl's posts above) this ZLT-CCR2 will not regulate as good with lower voltages as it now does between 4 and 6 volts input. And as I don't want to damage my board with desoldering these resistors more than once you have to wait till I build one for a lower input voltage. But I have no doubts the CCR2 circuit will work with lower voltages as good as it does with higher input voltages.

Originally posted by remuen:
The ZLT in its basic form cannot be used for a 5W LS even though I drove my two LS in series with more than 400mA from 4 to 4.8 volts input (!) before I added the additional resistors for the CCR2 circuit. But it would be interesting to connect two FMMT617 in parallel (and not two ZLTs) and test it with a 5W LS. IMHO this should work as well - at least with input voltages =/> 3.6 volts (3 or more NiMHs).

Click to expand...

<font size="2" face="Verdana, Arial">Understood, but I meant to use 2 ZLT circuits in parallel, driving the same load. Alternatively, I wonder is 400 mA+ is enough for those wanting to underdrive their luxeons.

If you connect two FMMT617s in parallel, isin't there the possibility that the output of the zetex chip would not be enough to drive both transistors at the required base current. I say this because I think I remember that the zetex output drive x transistor gain was an issue in the original ZLT+ thread.

Battery configs are a given, powering the circuit with anything less than 3 cells imho would not be worth it, even if possible.