Flash Light Project (Feb 08 N&V)

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gatoruss
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Flash Light Project (Feb 08 N&V)

I have been reading the Flash Light Project article in N&V Feb08, and am wondering if someone else who might have read same could field a question? I am a total newbie, so this may be a dumb question, but here goes.

The author states that if 5-10 LEDs are used that R2 is not longer needed. I am having trouble following this? Assuming that the output voltage is 30V (as he later suggests), I can understand why a current limiting resistor would not be necessary with 9 LEDs were uses, as the aggregate forward voltage for the LEDs would exceed 30V. But if 5 or 6 LEDs were used, this would not seem to be the case?

Please note, I am by no means suggestion that the author's analysis is incorrect. I realize I am missing something and hope some can point me in the right direction.

Thanks.

MrAl
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Hi there,

If you could post the circuit that would help others that dont have
LEDs vs Bulbs, LEDs are winning.

Bigglez
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Re: Flash Light Project (Feb 08 N&V)

Greetings (No Name Supplied),
gatoruss wrote:I am a total newbie, so this may be a dumb question, but here goes.

The author states that if 5-10 LEDs are used that R2 is not longer needed. I am having trouble following this?
There ar no "dumb" questions!

The uC controlled charge pump circuit presented in the
article is energy limited, meaning that it can't produce
enough energy at its output to harm the long string of LEDs.
Normally an LED is protected by a series "ballast" resistor.
So now the ballast resistor can be removed, thus saving the
power loss in that component.

The charge pump built around the uC initially used the
internal analog comparator for feedback control. With the
second version, that the author used with nine LEDs,
the the charge pump runs at max and the feedback
loop is removed saving a few more components.

An analogy would be if you had a not-so-good car,
and it never went fast enough to exceed the speed
limit you could drive it all day and not look at the
speedometer!

gatoruss
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Re: Flash Light Project (Feb 08 N&V)

Bigglez wrote:Greetings (No Name Supplied)
Sorry...the name is Russ
Bigglez wrote:The uC controlled charge pump circuit presented in the
article is energy limited, meaning that it can't produce
enough energy at its output to harm the long string of LEDs.
Normally an LED is protected by a series "ballast" resistor.
So now the ballast resistor can be removed, thus saving the
power loss in that component.

The charge pump built around the uC initially used the
internal analog comparator for feedback control. With the
second version, that the author used with nine LEDs,
the the charge pump runs at max and the feedback
loop is removed saving a few more components.
Is this true for any number of LEDs? I was under the impression that the feedback control limited the pump when the number of LED <5. I can understand why there isn't enough energy when 9 LEDs are used, but what about when 4<# LEDs<9? What discriminates the amount of energy in that configuration? Isn't a resistor needed?

What aspect of the circuit design limits the energy? Or is the limitation inherent in the nature of a DC boost converter? The article has prompted me to find more information on how a DC boost converter works. But I am still struggling to understand.

The literature I have read suggests that the following relationship is applicable:

Iout = IL(avg)* (1-Duty Cycle)

Does this explain the limitation? An inductor and duty cycle is chosen that limits the output current to an amount that is safe for the LEDs? But that wouldn't explain the need for the feedback control for situations where the number of LEDs < 5 (unless I misunderstand the role of the feed back circuit)?
MrAl wrote:If you could post the circuit that would help others that dont have access to it understand what you are asking.
I apologize...I do not have access to an electronic copy of the circuit. Also, the article is copyrighted, may not be appropriate to post? The circuit is essentially a DC Boost converter (see http://en.wikipedia.org/wiki/Boost_converter for circuit diagram) where "Load" is a string of LEDs and inductor is L=100micr-H with Ipeak of 300 mA. Control circuit is a transistor controlled by a uC.

Thanks for the help.

Russ

Edited to fix current formula

ericnlyvone
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Russ wrote:
Is this true for any number of LEDs? I was under the impression that the feedback control limited the pump when the number of LED <5. I can understand why there isn't enough energy when 9 LEDs are used, but what about when 4<# LEDs<9? What discriminates the amount of energy in that configuration? Isn't a resistor needed?

What aspect of the circuit design limits the energy? Or is the limitation inherent in the nature of a DC boost converter? The article has prompted me to find more information on how a DC boost converter works. But I am still struggling to understand.

Russ

It has been a little while now since I read the article. As I recall, the charge pump circuit took a battery voltage of 4.5V and "stepped" it upto 12V(if I remember correctly). This was accomplished by uC, the transistor it "pulsed" on and off, and the diode connecting between the inductor and the cap. During the time the uC turned on the transistor to "Charge" the inductor, the diode "disconnected" the cap from the inductor circuit to keep the capacitor voltage from leaking back into the uC. Once the inductor was charged the uC turned off the transistor which in turn "reconnected" inductor to the capacitor. This enabled the inductor to transfer its charge into the capacitor. The end result is that approx. 12V ends up accros the capacitor and is the "new" voltage source that powers the LED's.

Now, remember that an LED is a device that,once turned on, has a "constant" voltage drop and will adjust its current to accomdate the voltage. In the case of the flash light circuit, the LED's were 1.2 volt units that wanted to opperate at 20mA. There is a rule in electronics that says the sum of the "voltage drops" of all circuit components must equal the voltage source. In this case the source voltage (from the charge pump) is approx. 12V. If each LED "drops" 1.2V the "extra" voltage has to be "used up" by something else. This is the purpose of the "Ballast" resistor. It's accually "using"up the extra voltage from the source to protect the LED's from and overvoltage condition. This is why the author is able to eliminate the resistor when using 9 to 10 LED's. If you have 10 LED's that use 1.2V each connected in series (as they are in the circuit) the sum of the individual "voltage drops" equals 1.2V * 10 = 12V which is equal to the source voltage. Therefore, since all of the sorce voltage is acconted for, there is no need for a resistor. In the case where you have 9 LED's, each LED would have to drop approx. 1.33Volts instead of 1.2V. This is apparently "close enough" to work without the resistor. But when you have any less than 9 LED's, you begin to run the risk of actually damaging the LED's unless you have the resistor.

I hope this helps.

dyarker
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The author of the article started out with the common misconception that the controlling parameter for LEDs is voltage like it is for incandescent lamps, it isn't, it is current. By the end of the article he had enough LEDs in series that the voltage booster sort-of became the ballast.

On a voltage (X) verses current (y) graph of an LED, the line is almost vertical in the region where light is emited. Very small changes in voltage cause enormous changes in current. Trying to control the voltage to light an LED is a doomed effort. No matter how careful you are in adjusting the voltage, even a change in temperature can allow enough current to flow to destroy the LED.

Incandescents on the other hand are self limiting for current. Apply a voltage (up to a little more than it's rating) and current flows. As current flows the bulb heats up, this increases the resistance which prevents further increase in current.

For incandescents control the voltage, for LEDs control the current.

I don't want to slam the author of article too hard. At least he had the initiative to sit down and write about his experiments. (does make me wonder about the technical editor at N & V though)

Cheers,
Dale Y

Bigglez
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Greetings Mrai,
MrAl wrote:If you could post the circuit that would help others that dont have access to it understand what you are asking.
Here you go!

philba
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Not having seen the article, I could be out in left field. It seems to me that there is a simpler way to go - most of the power IC companies (National, Analog, ...) have specialized boost switch mode ICs designed to drive LEDs. These typically support curren tlimitation and a PWM/dimming control input. I'd use that long before I'd build a uC based boost circuit.

Bigglez
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Re: Flash Light Project (Feb 08 N&V)

Greetings Russ,
Welcome to the forum!
gatoruss wrote:Is this true for any number of LEDs?
No.
I was under the impression that the feedback control limited the pump when the number of LED <5. I can understand why there isn't enough energy when 9 LEDs are used, but what about when 4<# LEDs<9?
I don't think the author tried with different numbers of
LEDs, or, that data was not published.
What discriminates the amount of energy in that configuration? Isn't a resistor needed?
In Fig 5 the uC uses 'burp mode'
to skip cycles when the LED current through R2 (Vs) is greater
than the reference voltage from D1 (Vr).
What aspect of the circuit design limits the energy? Or is the limitation inherent in the nature of a DC boost converter? The article has prompted me to find more information on how a DC boost converter works. But I am still struggling to understand.
The energy available is limited by several factors,
firstly, the time allowed for current to rise in the inductor
is limited by the switch ON period. Secondly, the inductor
is not perfect, and will saturate if too much current is
allowed to flow. Thirdly, the voltage applied to the inductor,
which is limited by the inductor's DC resistance (plus the
switch's resistance).
The literature I have read suggests that the following relationship is applicable:

Iout = IL(avg)* (1-Duty Cycle)

Does this explain the limitation?
Not the max current, but that is correct for the energy
transfered in one cycle.
An inductor and duty cycle is chosen that limits the output current to an amount that is safe for the LEDs?
I can't speak for the author of that article, but likely he is
reporting on his experiments, and at nine LEDs he
discovered a limit in the energy from his booster.
But that wouldn't explain the need for the feedback control for situations where the number of LEDs < 5 (unless I misunderstand the role of the feed back circuit)?
Fig.5 does have an energy limit using feedback (as noted above).

Bigglez
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Greetings (No Name Supplied),
ericnlyvone wrote:Russ wrote:
I see this is your first post, welcome to the forum!
Could you use the quote tools in future posts?

Bigglez
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Greetings (No Name Supplied),
ericnlyvone wrote: As I recall, the charge pump circuit took a battery voltage of 4.5V and "stepped" it upto 12V.(if I remember correctly).
The boost circuit has unlimited output voltage in theory,
the author reports that it generates 30V for the nine
LED version (without feedback to the uC).
ericnlyvone wrote:In the case of the flash light circuit, the LED's were 1.2 volt units that wanted to opperate at 20mA.
The LEDs used in the article are 3.5V at 20mA. The author
attempted to obtain 3V @ 20mA LEDs as they would be
compatible with a two cell flashlight. Most LED flashlights
have either a voltage booster or use three cells (4.5V).
ericnlyvone wrote: In this case the source voltage (from the charge pump) is approx. 12V. If each LED "drops" 1.2V the "extra" voltage has to be "used up" by something else. This is the purpose of the "Ballast" resistor.
There is no "ballast" resistor in the charge pump design.
For the two LED circuit the current in the LEDs is monitored
by R2, converted to a voltage, and fedback to the uC.
When the current in the LEDs exceeds 20mA the voltage
across R2 is greater than the voltage developed by D1,
and the uC skips cycles ("Burps").
It's accually "using"up the extra voltage from the source to protect the LED's from and overvoltage condition. This is why the author is able to eliminate the resistor when using 9 to 10 LED's. If you have 10 LED's that use 1.2V each connected in series (as they are in the circuit) the sum of the individual "voltage drops" equals 1.2V * 10 = 12V which is equal to the source voltage.
The energy from the booster provides approximately
30V to operate nine LEDs that each require 3.5V as
9 x 3.5 = 31.5V, which will vary slightly as the LEDs
are randomly selected.

Because the energy is limited to be safe for the LEDs
there is no need for a "ballast" resistor.

This is a subtle but clever design, and it can be better
understood by thinking of the booster as a "battery
with a resistor hidden inside". The internal resistance
of the booster/battery limits the current and protects
the LEDs. Some commercial flashlights apply the
battery (without an internal resitor) to the LEDs, and
rely on the cell's internal resistance to limit the
current. This was mentioned by the author for his
first LED flashlight experiment.

Bigglez
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Greetings Dale,
dyarker wrote: The author of the article started out with the common misconception that the controlling parameter for LEDs is voltage like it is for incandescent lamps, it isn't, it is current.
Actually, the author stated:
G.Y.Xu wrote:I started to think about and work on how to transform all of my traditional bulb flashlights into LED flashlights.
dyarker wrote: By the end of the article he had enough LEDs in series that the voltage booster sort-of became the ballast.
After the first one, which was a simple LED for bulb swap,
the author turned to a booster design built around a very
cheap uC (\$0.52). The author attempted to source higher
current white LEDs for direct two-cell battery connection,
but was limited to finding only 3.0V 10mA or 3.5V 20mA devices.
dyarker wrote: Trying to control the voltage to light an LED is a doomed effort. No matter how careful you are in adjusting the voltage, even a change in temperature can allow enough current to flow to destroy the LED.
As a practical matter it would seem that today's LEDs are
tough enough to survive direct connection to a battery.
Said another way, the battery's internal resistance acts
to limit the current and protect the LED. On another forum
the direct connection of LEDs to ICs and batteries is a popular
beginner error. Not to mention the fans of LED throwies that tie
a battery and LED together.
dyarker wrote:(does make me wonder about the technical editor at N & V though)
Send Bryan ([email protected]) an email, reader feedback is always welcome!

gatoruss
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Thanks to everyone. This has been very helpful.

I gather from the responses that if less than 9 LEDs are used with the circuit in question (and with no feedback control), a "ballast resistor" would likely be required.

What had kind of confused me was the following observation by the author:
G.Y.Xu wrote:In other words, when the number of LEDs equals five or more, we can safely remove the MCUâ€™s analog comparator circuit components R1, D1, and R2, without having to worry about the LEDs being overdriven. This is an advantage. Remember, R2 also wastes some energy (heat), although it is needed as a voltage sensing component.
I suppose that he could have been suggesting that the LEDs were "tough enough" to handle the voltage and current?

In any event, it is not my purpose to be critical of the author. I thought the article we very helpful and interesting. It really peaked my curiosity and got me thinking.

Thanks again.

Russ

Bob Scott
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This is an iteresting thread, so I searched for the ATtiny11 datasheet:

http://www.atmel.com/dyn/resources/prod ... oc1006.pdf

It's now obsolete - "not recommended for new designs". Too bad.

Bob

Bigglez
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Greetings Bob,
Bob Scott wrote:ATtiny11 datasheet:
It's now obsolete - "not recommended for new designs".
According to the manufacturer, in this Ap Note, the Attiny13
should be used.

This "upgrade" has a negative impact on the BOM cost, though.