How does this circuit provide constant current?

[jeff1500]

Here's a link to a very well made LS flashlight conversion using a Linear Technology LT1308B:

http://nedkonz.dhs.org:8080/Ned/27

Looks like it has some extra circuitry that changes a step up voltage regulator into a constant current provider.

He says:

The secret here is to use the low-battery comparator (LBI and LBO pins) to sense the current. When the voltage across R5 is lower than 200mV, the LBO pin is pulled low, causing the output voltage to increase. So the output is a sawtooth of a few mA around the current setpoint.

<font size="2" face="Verdana, Arial">How does this work?

 

[Jonathan]

Ned Konz is a member here...
http://www.candlepowerforums.com/cgi-bin/ultimatebb.cgi?ubb=get_topic&f=3&t=001584

Most chips have a single comparitor, used to detect and adjust the output voltage. These chips have two comparitors; the LBI/LBO pins are normally used to shut down the chip when the input battery voltage falls below a minimum.

What Ned does is use this comparitor to regulate the output, by running the output current throug a low value sense resistor. The circuit turns on and off to maintain about 200mV across this sense resistor.

-Jon

 

[MrAl]

Hello there,

To add to what Jonathan was saying:

When the sense voltage rises above 200mv the
chip shuts the output off, and this means the
output current starts to ramp down to a lower
level. Since the 200mv depends on the current
THROUGH the led, when the output current gets
too high it shuts the output off, so the current
starts to ramp down to a lower level.
When the current gets down to a low enough level,
the sense voltage falls to a value lower
then 200mv and the chip senses this and allows
the output to start pulsing again which causes
the output current to again rise. This cycle
repeats over and over and usually at a frequency
lower then the normal oscillation frequency of
the chip. This also causes sub harmonics to
appear at the output across the LED.

The eff noted on that web site cant
really be 90%, and that was determined through
doing a simulation anyway.
The sense resistor eats up 6% of the efficiency
by itself, so my guess is that you will see
84% tops on efficiency for that circuit with
a good inductor. Using a high speed op amp
amplifier circuit you can get most of that 6%
back, but it does add another chip to the
pc board.

Good luck with your LED circuits,
Al

 

[jeff1500]

I think I'm starting to understand. Do you agree with the following: ?

Vout = 1.22(1 + (R1/R2))

These are the R1 and R2 values from the datasheet schematic not the constant current schematic.

The output voltage goes up when R1 gets large and R2, which connects to ground, gets small. So connecting FB to ground makes the output voltage spike up.
------------------------
Then on the constant current circuit:

The sense resistor is in series with the output led so it's plus side voltage, Vsense, increases with output current.

At startup, Vsense, which equals V-LBI, is low, so that makes LBO ground, which makes FB ground, which makes Vout jump up.

High Vout charges C5 through R2. D1 prevents backflow from C3 when it discharges.

Then higher Vout increases current through the output led, raising Vsense = V-LBI, so LBO goes to open circuit (high impedence).

V-FB is then whatever FB sees at the plus side of C5 through R3. This makes Vout drop to a low value.

So when you tune the values just right, the pulsing of the chip from Vout = high to Vout = low, provides a stream of current that's about constant.

In mass production, led current will be set by the value of Rsense. Rsense can be manufactured with close tolerance. This allows assembly of lights with leds that have different voltage drops for the same current flow.

What's the purpose of D2?

 

[MrAl]

Hello Jeff,

You are right about the voltage calculation.

The current regulation with temperature
is only about as good as +/- 10 percent.

What schematic were you referring to when you
mentioned D2 ?

Good luck with your LED circuits,
and Merry Christmas,
Al

 

[jeff1500]

Originally posted by MrAl:
What schematic were you referring to when you mentioned D2 ?

<font size="2" face="Verdana, Arial">This one:

http://nedkonz.dhs.org:8080/Ned/27

 

[MrAl]

Hi again Jeff,

That site seems to be down or something.
Do you have another site where the schematic
is located?

If you want you can put the schematic on
your own web site and i'll take another
look.

Take care,
Al

 

[MrAl]

Hello again Jeff,

That site was working so i got to look at
the schematic again.

Now that you mention it, there are a few strange
things about that circuit.

First off, the LBO pin is connected to the
FB pin. Doesnt make any sense.

Secondly, the SHDN pin should be tied high.

Thirdly, the resistor FB network isnt correct at all.
This is probably what brought in that zener diode
"D2".

One thing nice about using this 'B' version
chip is that the LBI fluctuation with temperature
is much better then older versions. This means
it might be possible to regulate the output
current to within about three percent over
the full temperature range !
This makes this circuit a very very good
candidate for driving any LS, with a 'few'
small changes 🙂

First off, i'd get the circuit more up to the
data sheet standard and calculate the correct
resistor values for the FB network. I'd also
lower the impedance of this network lower then
the data sheet recommends while im at it.

Second, i'd tie the SHDN pin high.

Third, i'd use the Zetex 2000 Schottky diode.

Fourth, since adding the 0.68 ohm resistor in
series with the LS reduces efficiency, i'd try
building a network to fool the circuit into
working with a smaller series resistor value.
Target value would be 0.1 ohm resistor, but
the tradeoff would be decreased temperature
stability. Since to begin with it's about 3%,
i wouldnt want it to get any higher then about
8% over the full range (that's about +/- 4% ).

Target eff would be 85% or better with the
network, 80% without it.

Good luck with your LED circuits,
Al

 

[jeff1500]

Sounds like there are lots of ways to do things.

I got a max1797 voltage regulator working on a SurfBoard adapter the other day. The Evaluation kit for that chip runs up to 500 mA.

I wonder if it's worth the effort to use this method to make it run constant current?

 

[bikeNomad]

Originally posted by MrAl:

The eff noted on that web site cant
really be 90%, and that was determined through
doing a simulation anyway.
The sense resistor eats up 6% of the efficiency
by itself, so my guess is that you will see
84% tops on efficiency for that circuit with
a good inductor. Using a high speed op amp
amplifier circuit you can get most of that 6%
back, but it does add another chip to the
pc board.

<font size="2" face="Verdana, Arial">I got about 80% efficiency on a single LED at 310mA, as I noted. The 90% estimate was for 3 LEDs in series; the 200mV is only a 2% loss in that case (rather than 6%). I suspect that in practice (with the small inductor I used) it would be nearer to 86% or so.

 

[bikeNomad]

Originally posted by MrAl:

Now that you mention it, there are a few strange
things about that circuit.

First off, the LBO pin is connected to the
FB pin. Doesnt make any sense.

Secondly, the SHDN pin should be tied high.

Thirdly, the resistor FB network isnt correct at all.
This is probably what brought in that zener diode
"D2".

One thing nice about using this 'B' version
chip is that the LBI fluctuation with temperature
is much better then older versions. This means
it might be possible to regulate the output
current to within about three percent over
the full temperature range !
This makes this circuit a very very good
candidate for driving any LS, with a 'few'
small changes 🙂

First off, i'd get the circuit more up to the
data sheet standard and calculate the correct
resistor values for the FB network. I'd also
lower the impedance of this network lower then
the data sheet recommends while im at it.

Second, i'd tie the SHDN pin high.

Third, i'd use the Zetex 2000 Schottky diode.

Fourth, since adding the 0.68 ohm resistor in
series with the LS reduces efficiency, i'd try
building a network to fool the circuit into
working with a smaller series resistor value.
Target value would be 0.1 ohm resistor, but
the tradeoff would be decreased temperature
stability. Since to begin with it's about 3%,
i wouldnt want it to get any higher then about
8% over the full range (that's about +/- 4% ).

Target eff would be 85% or better with the
network, 80% without it.

Good luck with your LED circuits,
Al

<font size="2" face="Verdana, Arial">This is a bit strange. However, you've missed a couple of things.

First, the FB circuit's "correct" configuration would assume that you're running constant-voltage, not constant current. If you look at the schematic, you'll see that both the FB and LBI/LBO amplifiers are part of the same feedback loop. I strapped them together (and added the divider/zener circuit) to get the loop response that I wanted. So there is nothing "incorrect" about my use of FB; it's just that this isn't a constant-voltage circuit, so it's not used that way.

Second, without adding an op amp or other gain it's going to be impossible to "fool" the LT1308B into working constant-current with less than 200mV drop across the sense resistor. This is due to its on-chip reference.

Third, the SHDN pin was in fact strapped high (this wasn't on the circuit diagram, but was on the assembled circuit; you can see the jumper going from pin 3 to pin 6 in the photo). I've made a note to that effect. Thanks for reminding me.