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'[EE] Alternate Method of Increasing Brightness?'
2011\01\22@142257 by Josh Koffman

face picon face
Hi all,

This is a sort of rethink to my questions of last night about a boost
regulator. I'm wondering if I'm approaching the problem incorrectly.

My goal is to be able to flash an LED string for a short duration at a
higher brightness than its constant brightness. The idea is similar to
a camera photo flash. The actual burst of light needs to be 1/60s
(0.0167s), and the repeat rate won't be faster than 1/8s (0.125s).

The LED string is sealed, and contains a current limiting resistor. I
can't change that. It runs at 12V. I did some testing with my bench
supply and a light meter and I found that if I increased the voltage
to 24V, did a quick pulse on/off (to avoid LED overheating), I was
getting an acceptably bright pulse.

My initial thought for how to do this in circuit was to use a boost
regulator, and then control the output via FET (controlled by a PIC).
I'm starting to wonder if I'm not going to be able to cram that
circuit into my allocated space. So now I'm wondering if there's a
better way to do this, perhaps more along the lines of an RC network
that's discharged into the LED string. I haven't ever done anything
like that, so I'm not sure what I'm thinking is possible. If I charge
up a cap and then discharge it through the LEDs, I would still need to
be charging it at the higher voltage due to that pesky limiting
resistor in the LED string, correct?

If that is the case, it would seem I still need a boost regulator. I
think one of the things that's hampering me in that search is my
current requirement. If I was charging a large cap, I should be able
to get away with a smaller current rating on the regulator, correct?

Here I start entering unfamiliar territory. Suppose I want a recharge
time of 1/8s (.125). Doing some calculation, if I use a cap with a
value of 1000uF, that would mean a resistor of 125 Ohms. What I'm
unsure of is the current draw as its charging. Based on some graphs I
saw, the current draw should start out high, but I'm not sure how
high. I'm also not sure what would happen to the boost regulator with
such and uneven load.

So...is my thinking correct? Or am I way off base here? I've been
through a number of websites, but I'm having difficulty figuring out
the current draw of charging circuit. I also haven't dealt with
controlling and routing the discharge into the LEDs yet.

Thanks,

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\22@152818 by Christopher Head

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On Sat, 22 Jan 2011 14:22:56 -0500
Josh Koffman <spam_OUTjoshybearTakeThisOuTspamgmail.com> wrote:
[snip]
{Quote hidden}

I'm assuming you don't have a 24V supply available, hence the boost
converter. There are other alternatives to boost converters, especially
if all you need is double the voltage. You could consider a charge
pump, which might be smaller if the voltage ratio isn't too big (as it
seems not to be here).

Chris
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2011\01\22@153032 by Carey Fisher

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On Sat, Jan 22, 2011 at 2:22 PM, Josh Koffman <.....joshybearKILLspamspam@spam@gmail.com> wrote:

{Quote hidden}

Set up a circuit that charges two capacitors in parallel, then switches them
to series to discharge into the LED string.

Carey Fishe

2011\01\22@153535 by RussellMc

face picon face
> My goal is to be able to flash an LED string for a short duration at a
> higher brightness than its constant brightness. The idea is similar to
> a camera photo flash. The actual burst of light needs to be 1/60s
> (0.0167s), and the repeat rate won't be faster than 1/8s (0.125s).

If LEDs are being operated at about their rate value then what you
propose will significantly decrease their lifetime BUT this is likely
to be acceptable in many cases.

What brand LEDs (and part number?),
 What diameter. What current rating. ?

Time constant of R & C is t = RC.
Cap will charge to about 60% in that time then  to about 80% in
another time constant etc.
So 2 or 3 tcs is enough for "full" charge.

Energy in Cap is 0.5 x C x V^2.
Usual energy = V x I x t Joule or VI Joule per second.
So cap energy and steady energy into whole string can be compared.

Using cap on top of supply will double supply in a burst. Current peak
will rise to >> usual current so adding a new series limiting resistor
of > Rexisting to >> R existing is wise.

"Ideal is to increase Vin in a steady step for as little as acceptable
for as little time as is acceptable for result.
Minimising Vin minimises I peak. Modern white LEDs do not like high I
peak and are much affected by short high I compared to lower longer
Ipulse.
As above re lifetime probably being OK.

Camera flash LEDS have very very very short lifetimes by most standards.

Sufficient unto a boost converters specifications are the performance thereof.
this varies widely by design and can be designed as required.



        Russell

2011\01\22@155234 by Dwayne Reid

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face
At 12:22 PM 1/22/2011, Josh Koffman wrote:
>Hi all,
>
>My goal is to be able to flash an LED string for a short duration at a
>higher brightness than its constant brightness. The idea is similar to
>a camera photo flash. The actual burst of light needs to be 1/60s
>(0.0167s), and the repeat rate won't be faster than 1/8s (0.125s).
>
>The LED string is sealed, and contains a current limiting resistor. I
>can't change that. It runs at 12V. I did some testing with my bench
>supply and a light meter and I found that if I increased the voltage
>to 24V, did a quick pulse on/off (to avoid LED overheating), I was
>getting an acceptably bright pulse.

How stiff is your 12v supply?

Can you charge a capacitor to 12V, then put that capacitor in series with the existing 12V supply so as to get your 24v pulse?

Alternatively, charge 2 capacitors to 12v each, then switch them in series.

The first method is easier and requires fewer switches: feed the (+) terminal of the cap with a suitable diode / current limit resistor (permanent connection) from your 12v rail; the (-) end of the cap has a spdt switch (or suitable power devices) that connects it to either 0V or to the +12v rail; and a final switch that connects the load to the (+) end of the cap (or switches the bottom end of the load if you want to use a N-channel device).

You should choose an appropriate amount of capacitance so that the capacitor is not fully discharged at the end of your flash period.  Otherwise, connect another diode across the capacitor so that it can't become reverse biased.

dwayne

-- Dwayne Reid   <dwaynerspamKILLspamplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax
http://www.trinity-electronics.com
Custom Electronics Design and Manufacturing

2011\01\22@172409 by IVP

face picon face
> My goal is to be able to flash an LED string for a short duration at
> a higher brightness than its constant brightness. The idea is similar to
> a camera photo flash. The actual burst of light needs to be 1/60s
> (0.0167s), and the repeat rate won't be faster than 1/8s (0.125s)

Josh, this may or not be (very) relevant

I'm currently reading up on LED lamps, and this is one of the Cree
articles I looked at yesterday, dealing with failure mechanisms

http://www.cree.com/products/pdf/XLamp-Pulsed-Current.pdf

from this page

http://www.cree.com/products/xlamp7090_xre.asp

Jo

2011\01\22@224619 by RussellMc

face picon face
> I'm currently reading up on LED lamps, and this is one of the Cree
> articles I looked at yesterday, dealing with failure mechanisms

> http://www.cree.com/products/pdf/XLamp-Pulsed-Current.pdf

That's an extremely good document to have available, not because it
says anything (that I noticed) stunningly new or different that can't
be understood from available data sheets and common sense BUT because
it puts a reputable manufacturers stamp of approval on the common
sense conclusions that any number of people seem to want to ignore or
criticise.

A very significant point is that modern white phosphor LEDs specify
the ratio of Ipeak to Ioptg max as not much above 1. In some cases
Imax and Iptgmax are the same, ratios of only about 1.1:1 are common
enough and as high as 2;1 would be rare. Add thermal data and a little
more and you come up with the sort of results that have been given in
this thread and in the paper Joe referenced.

Disturbing is the fact that the overall mix of factors is that Cree
actually DERATE the mean power able to be put into a pulsed LED.
eg they say

Based on the 1 KHz pulse testing we have reviewed in this application
note, Cree suggests the following guidelines for
pulsed current operations:

1. For duty cycles between 51-100%, do not exceed 100% of the maximum
rated current;
2. For duty cycles between 10-50%, do not exceed more than 200% of the
maximum rated current;
3. For duty cyles less than 10%, do not exceed more than 300% of the
maximum rated current

It seems to me that there is an element of arbitrariness in there and
also that the results are slightly conservative in some areas, but
only slightly.

A rule of thumb which falls on the optimistic side of what Cree says
is "Total power in should never be more than Pmax_DC and maximum
current ever is <= 3 x Imaxoptg.

That's going to probably reduce total lifetimes at the upper end but
probably not vasyly.

NB LED life time is decreased *independently* by either of increased
current or increased temperature.



        Russell





>
> from this page
>
> http://www.cree.com/products/xlamp7090_xre.as

2011\01\22@225843 by RussellMc

face picon face
2:1 ish capacitor voltage pump

Vin   D1A
D1K   LED+
Vin   Q1+
Q1-   Q2+
Q2-  gnd
Vin  D2A
D1K R1A
R1B  D2K
D2K  C1A
C1B Q1- (=Q2+)

E&OE - drew from above - looks OK.



Q1 is a N-Channel MOSFET
Q2 is a N-Channel MOSFET
D1, D2 = most any diode
R1 = cap discharge resistor for flashing.
C1 = flashing energy cap

Q1 off Q2 on - Cap charges via D2.
Q1 on, Q2 off - Cap discharges via R1 into Led.

Q1 and Q2 on - magic smoke.




             Russel

2011\01\23@004711 by Harold Hallikainen

face
flavicon
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Interesting ideas proposed so far. Besides the dissipation in the LEDs,
you have to consider the dissipation in the internal current limit
resistor. Hopefully, if your duty cycle is under 50%, stuff will keep
working.

I did a strobe light system for motorcycles a few years ago. This ran off
the 12V battery. I had a single LM317 based current regulator driving the
tops of all the LED strings. Each LED module had a series string of
photoflash LEDs in a module. The bottom of each module was pulled to
ground for a short pulse by an FET driven by a PIC. The pulse sequence and
rate could be varied.

In your application, I think I'd just try to come up with a constant +24V
supply. You may be able to do this with a charge pump converter. I've used
one from TI that could output something like 300mA, but I think it will be
difficult finding a charge pump chip that will do much power at 24V. So,
you're probably back to a boost converter. Linear Technology has some nice
simple boost converter chips, and maybe even some modules that have
everything in one assembly. I think such a single chip solution will be
simpler than trying to build your own charge pump.

Harold



-- FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available

2011\01\23@021255 by RussellMc

face picon face
> In your application, I think I'd just try to come up with a constant +24V
> supply. You may be able to do this with a charge pump converter. I've used
> one from TI that could output something like 300mA, but I think it will be
> difficult finding a charge pump chip that will do much power at 24V. So,
> you're probably back to a boost converter. Linear Technology has some nice
> simple boost converter chips, and maybe even some modules that have
> everything in one assembly. I think such a single chip solution will be
> simpler than trying to build your own charge pump.

If this a one off / some off / high volume application?

The circuit that I described using semi-pseudo-SPICE-code is in fact a
voltage doubling (almost) charge pump.

As I described it it drives the LED string via an external current
setting resistor. But, short R1, add a reservoir cap C2 at the output
and one more diode (location left as an exercise for the student but
clue: it stops the pump capacitor C1 from back charging from C2 when
Q2 is turned on.

You now have an ~= 24V supply and can current regulate the LED to
control brightness and "flashing.
The above stats allow resistor value and

The converter needs a square wave of probably asymmetric  mark space,
but a square wave would do.
Having no overlap on the two FET gates is a very good idea (tm) but a
complete driver could be achieved with ed CD40106 or equivalent, glue
parts as above plus a few more glue bits for the oscillator.

HOWEVER - we have allowed the discussion to mislead us away from what
is actually wanted (or some of us have :-) ).

Aim is to provide extra brightness and the 24V figure provided is a
"probably good enough" voltage but limits the result  achievable. Josh
should ideally tell us more. At least, what is I_LED at 12V and 24V
and how many LEDs of what colour (white?). LED part numbers would
help. Is the resistor value known.

If 12->24 doubles current then with the above information we can have
a reasonable idea of the resistor value (if not already known or
knowable).

Using a "normal" SMPS/Boost converter with current feedback will allow
the LEDs to be flashed or held at any desired brightness with ease.

___________

I don't think you say how many LEDs are in the string.

I_LEDs = (Vin = (N x V_LED))/R

VLED ~~~~ constant.
4 LEDs is marginal to dead on 12V.
3 LEDS allows about 3V max drop across R at 12V so about 15 at 24 V so
I_LEDs increases by about 5:1 !!!

@ LESDs allows about 6V drop at 12V and 18V at 24V or about 3:1 I_LED ratio..

IF the LEDs are operated at << Imax at 12V then all may be well.

If I_LED at 12V ~> Imax_optg then the above suggests "there will be trouble".

3:1 on bright is at Cree max limit and 5:1 is above it.
Duty cycle is about 7:1 so Cree suggestion is well violated if LEDs
are run at max voltage and there are 3+ LEDs in string.

More data will help tie down the above guesstimates.


Russel

2011\01\23@022119 by RussellMc

face picon face
> > I'm currently reading up on LED lamps, and this is one of the Cree
> > articles I looked at yesterday, dealing with failure mechanisms
>
> > http://www.cree.com/products/pdf/XLamp-Pulsed-Current.pdf
>
> That's an extremely good document to have available, not because it
> says anything (that I noticed) stunningly new or different that can't
> be understood from available data sheets and common sense BUT because
> it puts a reputable manufacturers stamp of approval on the common
> sense conclusions that any number of people seem to want to ignore or
> criticise.

And finally ... :-)

One common method of driving LEDs from smaller smps boost converters
is to "ring" an inductor into the LED, allowing the LD to act as
rectiifer with no series diode and no smoothing capacitor.The LED is
driven by pulsed DC at the smps frequency - really essentially AC
pulses. The current will tend to be in very narrow high magnitude
peaks.

Based on what has been said in this thread, this is an extremely poor
practice. The LED will be pulsed by unknown by high  magnitude
short duration pulses. If the LED is driven to about design brightness
then it will be vastly overdriven current wise.

Circuits like this will shorten LED life, possibly very substantially,
and you'd hope that application note circuits would not show this
system. Alas, many do.

Perhaps a year ago wrote to Zetex about a typical and bad example of
this in one of their app notes. I received no response.
A bit sad given Zetex's competence in product performance.  Mayhaps
their designers and app note writers don't talk to each other.




   Russell



            Russel

2011\01\23@053732 by cdb

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face
part 1 392 bytes content-type:text/plain; charset="us-ascii" (decoded quoted-printable)


:: 2:1 ish capacitor voltage pump
::
:: Vin   D1A
:: D1K   LED+
:: Vin   Q1+
:: Q1-   Q2+
:: Q2-  gnd
:: Vin  D2A
:: D1K R1A
:: R1B  D2K
:: D2K  C1A
:: C1B Q1- (=Q2+)

Quick schematic of above using AutotraxEDA

Colin
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part 2 20587 bytes content-type:image/png; name="russells_flash.PNG" (decode)


part 3 181 bytes content-type:text/plain; name="ATT00001.txt"
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2011\01\23@074741 by Dave Tweed

face
flavicon
face
Colin wrote:
> Russell wrote:
> :: 2:1 ish capacitor voltage pump
> ::
> :: Vin   D1A
> :: D1K   LED+
> :: Vin   Q1+
> :: Q1-   Q2+
> :: Q2-  gnd
> :: Vin  D2A
> :: D1K R1A
> :: R1B  D2K
> :: D2K  C1A
> :: C1B Q1- (=3DQ2+)
>
> Quick schematic of above using AutotraxEDA

Obviously not. What keeps Vin from destroying the LED via D1?

Russell never said where the cathode of the LED goes, and what's the purpose
of D1 anyway? If you just leave it out, then the LED is simply on at less
than full brightness when not flashing.

-- Dave Twee

2011\01\23@091518 by Olin Lathrop

face picon face
Harold Hallikainen wrote:
> In your application, I think I'd just try to come up with a constant
> +24V supply.

Why?  That sounds like extra complexity.

If you only have 12V and need to boost up to 24V to occasionally drive the
LED string, then it would be simpler to control the boost switcher as needed
instead of first making 24V, then controlling that down to drive the LEDs.

Since the OP apparently wants short light pulses somewhat like a strobe,
there may be no need for regulation at all.  If the pulse energy is low
enough, a single boost cycle might do it.  Feed the LED string from the 12V
supply thru a Schottky diode.  Also feed it thru a separate inductor
followed by another Schottky diode.  To do a pulse, you pull the point
between inductor and diode low a fixed time to charge up the inductor from
12V, then turn the switch off.  The energy in the inductor will be dumped
into the LED string.

If you can't reasonably get a inductor to hold enough energy for the pulse,
then you'll have to use multiple pulses.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\23@092029 by Olin Lathrop

face picon face
cdb wrote:
> Quick schematic of above using AutotraxEDA

That's obviously a bad idea since the current thru D1 and D3 from voltage
source V1 is not limited in any way.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\23@121547 by Dwayne Reid

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face
This does not allow the LED string to ever turn off.

Note that Josh said that the brightness was not adequate with 12V in but was adequate with 245V in.  I read from that statement that the LEDs are indeed ON while 12V is applied to them.

Regarding the current limit resistor: Josh said that the LED strings had internal current limit resistors that were not accessible.  I don't think that you need another current limit resistor (but I could be wrong).

dwayne


At 08:58 PM 1/22/2011, RussellMc wrote:
{Quote hidden}

-- Dwayne Reid   <EraseMEdwaynerspam_OUTspamTakeThisOuTplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax
http://www.trinity-electronics.com
Custom Electronics Design and Manufacturing

2011\01\23@141809 by cdb

flavicon
face


:: Obviously not. What keeps Vin from destroying the LED via D1?

I realised late last night, the LED cathode wasn't supposed to connect to gnd, but being lax I went to bed rather than redraw and repost it.

Sorry for the confusion I caused.

Colin
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2011\01\23@182139 by RussellMc

face picon face
part 1 2148 bytes content-type:text/plain; charset="iso-8859-1" (decoded quoted-printable)

Obviously my pseudoSPICEcode drawing and checking skills were wanting.
Soory
Try attached ugly cut and paste.
Not at PC with access to schematic software etc etc

LED and load swapped and slightly altered.
Cap is charged by D2 and pumped by FETS.
LED load is driven by D1 or by Cap via R1.

*** NB NB NB ***
Electrons can jump over small breaks in circuit in this tolopogy :-).
eg like one immediately under LED string.

Cap charges via D1 when Q2 is on
Cap -ve pulled to V+ when Q1 is on and Q2 off.
Cap+ now discharges via R1.
R1 limits flash brightness, duration and phase of Moon.

An (erk) '555 could drive FET gates with suspect crossover.

Not that FETs as shown need to be N Channel so that body diodes behave.
This means Q1 gate needs to be above Vin but can be biased from Vout.
Gets messay albeit cheap and easy enough.

A simple inductive SMPS gets very attractive.
My often mentioned olde MC34063 & friends IC will do what you want with 1 IC
and a few glue parts.
eg take fig6 here and add one extra diode, add an extra diode from Vin to
Vout and adjust R2 to suit.
Inductor also to suit.
Enable high voltage operation as required by enabling/disabling smps by any
of several means.
Switch is good for 1.5A AFAIR BUT do design for package dissipation
eg  41/  85 / 97 degrees C/Watt   DIP / SOIC / QFP packages
Efficiency is lower than modern ICs and opn at not much over 100 kHz means
inductor is larger than some BUT quite likely OK in this application.


fig 6
http://focus.ti.com/lit/ds/symlink/mc33063a.pdf


Russell


On 23 January 2011 23:37, cdb <@spam@colinKILLspamspambtech-online.co.uk> wrote:

{Quote hidden}

> -

2011\01\23@224829 by Richard Prosser

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On 24 January 2011 12:20, RussellMc <RemoveMEapptechnzTakeThisOuTspamgmail.com> wrote:
{Quote hidden}

Here's an alternative - similar idea to Russells
Don't pay too much attention to device types or component values -
thay will need tuning to your application but could be used as a very
rough starting point . Possibly.
(Your device is the LED / Resistor at the top. It can equally well be
moved to the earthy end of the battery etc.)
RP


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2011\01\23@234111 by RussellMc

face picon face
part 1 1297 bytes content-type:text/plain; charset="iso-8859-1" (decoded quoted-printable)

On 24 January 2011 16:48, Richard Prosser <RemoveMErhprosserspamTakeThisOuTgmail.com> wrote:

> Here's an alternative - similar idea to Russells
> Don't pay too much attention to device types or component values -
> thay will need tuning to your application but could be used as a very
> rough starting point . Possibly.
>  (Your device is the LED / Resistor at the top. It can equally well be
> moved to the earthy end of the battery etc.)
> RP

That bends the brain just the wee-est amount until a few clever points
are noted.
Plus it is a flasher only with the LED flashing bright, then tailing off.
You may need a second Q to get it back to a DC or flash mode as mine
was. Maybe not.

C! charges via path D3 R1 D1 C1 D2 - so starts bright and tails off.
Pulse on Q1 base turns Q1 on, clamps C1 top to gnd so C1 bottom goes to  -Vin

Q2e is drive to -Vin so Q2_b is at ground so Vbe is applied to Q2 to turn it on.
LED now conducts Via Q2 to bootm on C1 so sees Vin + Vcap.
As C1 discharges brightness drops.

More but that will do.

Min is cleaner IMHO and does more like wyat is wanted.
But I lke the brain bending aspects of this one.

Mine, (which is standard voltage doubler) can use bipolars instead of
FETS with due care.

  Russell


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2011\01\24@015305 by Richard Prosser

picon face
On 24 January 2011 16:40, RussellMc <apptechnzEraseMEspam.....gmail.com> wrote:
{Quote hidden}

>

2011\01\24@141124 by Josh Koffman

face picon face
On Sat, Jan 22, 2011 at 3:52 PM, Dwayne Reid <RemoveMEdwaynerEraseMEspamEraseMEplanet.eon.net> wrote:
> How stiff is your 12v supply?

I'm not 100% sure. I think it will end up being a switching power
supply, so I'm guessing putting as large an input cap as I can fit
might not be a bad idea. My first thought was to change the input PSU,
but that turns out to not be an issue unfortunately.

{Quote hidden}

Interesting!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@141445 by Josh Koffman

face picon face
First off, an appology. I started to plan a reply to everyone's
responses in a single email and it quickly got too complicated. Thank
you to everyone who replied, and I will be replying with further
information where clarification is needed.

On Sat, Jan 22, 2011 at 3:34 PM, RussellMc <RemoveMEapptechnzspam_OUTspamKILLspamgmail.com> wrote:
> What brand LEDs (and part number?),
>  What diameter. What current rating. ?

I don't know about the LEDs themselves. I can't access them as they
are embedded in a plastic material. They are spec'ed to operate at
12V, and when I measured the current draw at 12V I saw that they were
drawing 600mA. There are longer and shorter strips in this product
line, so I believe they are running a series/parallel configuration,
and adding more "blocks" in parallel when needed.

> Time constant of R & C is t = RC.
> Cap will charge to about 60% in that time then  to about 80% in
> another time constant etc.
> So 2 or 3 tcs is enough for "full" charge.
>
> Energy in Cap is 0.5 x C x V^2.
> Usual energy = V x I x t Joule or VI Joule per second.
> So cap energy and steady energy into whole string can be compared.

Ok, so assuming I need 24V at 3A for 1/60s, that gives me roughly 1.2J
(24x3x0.1667), correct?
Then I can rearrange the equation above (E=0.5xCxV^2) to solve for C
(C=E/(0.5xV^2). Then I solve and get 0.0041667F, or roughly 4200uF, if
my units are correct (I may be way off). How does that sound?

> Using cap on top of supply will double supply in a burst. Current peak
> will rise to >> usual current so adding a new series limiting resistor
> of > Rexisting to >> R existing is wise.

Without knowing the value of the resistor in place already, how would
I calculate the new one? This might become self evident as I work
through this.

> Sufficient unto a boost converters specifications are the performance thereof.

:)

Thanks!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@141527 by Josh Koffman

face picon face
On Sat, Jan 22, 2011 at 5:24 PM, IVP <RemoveMEjoecolquittTakeThisOuTspamspamclear.net.nz> wrote:
> Josh, this may or not be (very) relevant
>
> I'm currently reading up on LED lamps, and this is one of the Cree
> articles I looked at yesterday, dealing with failure mechanisms
>
> http://www.cree.com/products/pdf/XLamp-Pulsed-Current.pdf
>
> from this page
>
> http://www.cree.com/products/xlamp7090_xre.asp

Interesting reading. Thanks Joe!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@142941 by Richard Prosser

picon face
On 25 January 2011 08:15, Josh Koffman <EraseMEjoshybearspamspamspamBeGonegmail.com> wrote:
{Quote hidden}

Josh,
You should be able to work out the value of the series resistor by
measuring the current at two (or more) different voltages.

Assume constant voltage drop on the LEDs. Then R = (Vin-Vled) / I
Vled will change slightly with current but this should get you
reasonably close. Graphing over a voltage range would let you get
closer by extroplation.

RP

2011\01\24@143746 by Josh Koffman

face picon face
On Sun, Jan 23, 2011 at 12:47 AM, Harold Hallikainen
<haroldSTOPspamspamspam_OUThallikainen.org> wrote:
> In your application, I think I'd just try to come up with a constant +24V
> supply. You may be able to do this with a charge pump converter. I've used
> one from TI that could output something like 300mA, but I think it will be
> difficult finding a charge pump chip that will do much power at 24V. So,
> you're probably back to a boost converter. Linear Technology has some nice
> simple boost converter chips, and maybe even some modules that have
> everything in one assembly. I think such a single chip solution will be
> simpler than trying to build your own charge pump.

That was my initial thought too. My problem is that it's proving
harder to find a boost converter that has a high enough current rating
that will fit in the allotted space. I wonder what would happen if I
used a lower current boost converter and a large output cap. Then used
a switch (MOSFET, etc) to pulse the LED for a short duration.

I imagine it's somewhat similar to the other ideas proposed, but I
need to keep studying the various ideas so I understand them a bit
better.

Thanks!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@150714 by Josh Koffman

face picon face
On Sun, Jan 23, 2011 at 2:12 AM, RussellMc <spamBeGoneapptechnzSTOPspamspamEraseMEgmail.com> wrote:
> If this a one off / some off / high volume application?

I'd call it a some off, budget/size conscious project. How's that!

> The circuit that I described using semi-pseudo-SPICE-code is in fact a
> voltage doubling (almost) charge pump.

More questions about that in a further email.

> Aim is to provide extra brightness and the 24V figure provided is a
> "probably good enough" voltage but limits the result  achievable. Josh
> should ideally tell us more. At least, what is I_LED at 12V and 24V
> and how many LEDs of what colour (white?). LED part numbers would
> help. Is the resistor value known.

At 12V, current is 600mA. At 24V I read 2.8A. No information is
available on the LEDs or the internal resistor.

> If 12->24 doubles current then with the above information we can have
> a reasonable idea of the resistor value (if not already known or
> knowable).
>
> Using a "normal" SMPS/Boost converter with current feedback will allow
> the LEDs to be flashed or held at any desired brightness with ease.

I don't think I'll need to hold the LEDs at brighter than normal other
than for a flash. I will likely need a mode where I can PWM the LEDs
up to their regular brightness at 12V, but brighter than that would
just be a flash.

> I don't think you say how many LEDs are in the string.
>
> I_LEDs = (Vin = (N x V_LED))/R
>
> VLED ~~~~ constant.
> 4 LEDs is marginal to dead on 12V.
> 3 LEDS allows about 3V max drop across R at 12V so about 15 at 24 V so
> I_LEDs increases by about 5:1 !!!

I think there is a series/parallel arrangement going on here. I know
that there are more than 4 LEDs in total, but I would guess that
there's a 4 LED building block, and there are a few blocks.

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@151026 by Josh Koffman

face picon face
On Sun, Jan 23, 2011 at 12:15 PM, Dwayne Reid <KILLspamdwaynerspamBeGonespamplanet.eon.net> wrote:
> Note that Josh said that the brightness was not adequate with 12V in
> but was adequate with 245V in.  I read from that statement that the
> LEDs are indeed ON while 12V is applied to them.

Correct, they are on with 12V.

> Regarding the current limit resistor: Josh said that the LED strings
> had internal current limit resistors that were not accessible.  I
> don't think that you need another current limit resistor (but I could
> be wrong).

Yup, they are internal and I can't get them out.

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@153534 by Josh Koffman

face picon face
On Sun, Jan 23, 2011 at 6:20 PM, RussellMc <EraseMEapptechnzspamEraseMEgmail.com> wrote:
> LED and load swapped and slightly altered.
> Cap is charged by D2 and pumped by FETS.
> LED load is driven by D1 or by Cap via R1.
>
> *** NB NB NB ***
> Electrons can jump over small breaks in circuit in this tolopogy :-).
> eg like one immediately under LED string.

Ah, wireless!

> Cap charges via D1 when Q2 is on
> Cap -ve pulled to V+ when Q1 is on and Q2 off.
> Cap+ now discharges via R1.
> R1 limits flash brightness, duration and phase of Moon.

Ok, I understand the basic concept of how the cap is being switched. I
think I'm a bit confused in how D1, D2, and R1 interact. When Q1 is on
and Q2 off (flash), what stops C1 from discharging via D1 right back
into itself?

> Not that FETs as shown need to be N Channel so that body diodes behave.
> This means Q1 gate needs to be above Vin but can be biased from Vout.
> Gets messay albeit cheap and easy enough.

I'm not sure I understand this. Q1's gate needs to be above Vin so the
body diodes conduct in the correct direction, right? I'm not sure how
to achieve that by using Vout. This circuit won't neccesarily be
operated in a continuous flashing cycle. In fact I'd guess it's more
likely that it will sit charging the cap (Q1 off, Q2on), wait some
arbitrary amount, then a flash.

One thing I like about this setup is that I think I'd be able to add
another MOSFET and PWM the LED directly off the 12V line for times
when constant light is required. These would be two very different
modes, so I could ensure there would be no crossover with Q1/Q2 and
the charge pump circuitry.

{Quote hidden}

Is this a suggestion as an alternative or adjunct to your charge pump?

Thanks!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@153800 by Josh Koffman

face picon face
On Mon, Jan 24, 2011 at 1:52 AM, Richard Prosser <@spam@rhprosser@spam@spamspam_OUTgmail.com> wrote:
> Yep, that pretty much describes the operation. It was originally
> intended as a relay drive circuit to allow 24V relays to operate from
> a 12-15V supply. Once pulled in the standing current was enough to
> hold the relay in. Not recommended for critical applications however!
>
> As only a quick pulse of energy was required in this case I thought it
> may be applicable.  Q1 can be just a mechanical switch if required and
> the "off" current is effectively zero once C1 is charged.

Cool idea, thanks Richard!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@153809 by Josh Koffman

face picon face
On Sun, Jan 23, 2011 at 11:40 PM, RussellMc <spamBeGoneapptechnzspamKILLspamgmail.com> wrote:
> That bends the brain just the wee-est amount until a few clever points
> are noted.
> Plus it is a flasher only with the LED flashing bright, then tailing off.
> You may need a second Q to get it back to a DC or flash mode as mine
> was. Maybe not.
>
> C! charges via path D3 R1 D1 C1 D2 - so starts bright and tails off.
> Pulse on Q1 base turns Q1 on, clamps C1 top to gnd so C1 bottom goes to  -Vin
>
> Q2e is drive to -Vin so Q2_b is at ground so Vbe is applied to Q2 to turn it on.
> LED now conducts Via Q2 to bootm on C1 so sees Vin + Vcap.
> As C1 discharges brightness drops.

Thank you for that explanation!

> Mine, (which is standard voltage doubler) can use bipolars instead of
> FETS with due care.

What would be an advantage of going that way? I am under the (possibly
wrong) impression that a FET would be more suitable as it will have a
lower on Rds on.

Thanks!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\24@154952 by Olin Lathrop

face picon face
Josh Koffman wrote:
> At 12V, current is 600mA. At 24V I read 2.8A. No information is
> available on the LEDs or the internal resistor.

Actually there is, from the measurements you cite.

At low voltages, the LEDs will be off and the string will draw very little
current.  This means all the voltage is accross the LEDs and practically
none accross the resistor.  As you increase the string voltage, the voltage
on each LED will rise.

Eventually the voltage on each LED will rise so that it starts conducting
for real.  At that point the LEDs are acting like conducting diodes, so
their voltage only goes up a little with increased current.  Another way to
say this is that their dynamic impedence goes way down.

Now as the string voltage is increased, most of the increase appears accross
the resistor.  This allows finding the resistor value.  In your case the
dynamic resistance from 600mA to 2.8A is

 Ohms = Volts / Amps

 = (24V - 12V) / (2.8A - 600mA)

 = 12V / 2.2A = 5.5 Ohms

That is the series sum of the fixed resistor plus the dynamic impedence of
the LEDs over that current range.  The fixed resistor is probably in the 4.5
to 5 Ohm range.  However, what you really want to know is the dynamic
impedence of the whole string, and it doesn't matter much what contribution
comes from the LEDs and what contribution from the resistor for most
purposes of driving the string.

Now that you can see the current rises by 2.8A / 600mA = 4.7 times, do you
really want to do the flash at 24V?  That's not going to be good for the
LEDs at all.

How long is a flash supposed to last?


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\24@160121 by Olin Lathrop

face picon face
Josh Koffman wrote:
> What would be an advantage of going that way? I am under the (possibly
> wrong) impression that a FET would be more suitable as it will have a
> lower on Rds on.

What you ultimately want is the lowest possible voltage drop accross the
switch when on.  FETs look roughly like low value resistors when on (Rdson),
and bipolars roughly like voltage sources (saturation voltage).  As the
voltages being switched get higher, the roughly fixed voltage drop of a
bipolar becomes smaller as a fraction of the total and therefore less of a
problem.

Let's say a saturated bipolar in your situation does 500mV.  That's 1/24 of
the overall voltage, or about 4%.  Whether that's below the don't care level
depends on things in your application you haven't told us, but unless you're
doing something unusual 4% is probably not a big deal.  For a FET to drop
the same 500mV it would need to have a Rdson of 500mV / 600mA = 830mOhms.
That's quite high for even cheap FETs for that current and voltage range, so
no problem.

Bipolars tend to be cheaper, but in your small quantities that's not too
relevant.  Both will be cheap enough.  So what it then comes down to is
whether it's more convenient for the circuit to provide the gate voltage for
a FET or the base current for the transistor.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\25@175311 by Josh Koffman

face picon face
part 1 2576 bytes content-type:text/plain; charset="iso-8859-1" (decoded quoted-printable)

On Mon, Jan 24, 2011 at 4:02 PM, Olin Lathrop <.....olin_piclistspam_OUTspamembedinc.com> wrote:
> Let's say a saturated bipolar in your situation does 500mV.  That's 1/24 of
> the overall voltage, or about 4%.  Whether that's below the don't care level
> depends on things in your application you haven't told us, but unless you're
> doing something unusual 4% is probably not a big deal.  For a FET to drop
> the same 500mV it would need to have a Rdson of 500mV / 600mA = 830mOhms.
> That's quite high for even cheap FETs for that current and voltage range, so
> no problem.
>
> Bipolars tend to be cheaper, but in your small quantities that's not too
> relevant.  Both will be cheap enough.  So what it then comes down to is
> whether it's more convenient for the circuit to provide the gate voltage for
> a FET or the base current for the transistor.

First off, thank you for that explanation, that makes a lot more sense
now. I think that it might be easier to use a bipolar transistor
rather than the FET, unless maybe I try to find a logic level FET. It
seems that current should be a bit easier for me to deal with, rather
than another voltage.

I've attached a schematic of what I'm thinking. I've picked components
based on what I think were appropriate values and what Digikey has
available. I'm quite sure I've gotten at least some of it wrong, so
please feel free to suggest alternatives. I don't have values for the
resistors yet, partially because I'm not 100% sure how to calculate
them all, and partially because I'd like to know if my design is even
viable. I tied the bases of T3 and T4 together as an idea, I could
send them to dedicated pins, but this would be a way to hopefully make
sure T1 and T2 can't turn on together.

For those who are curious, here are the links to more information
about my specific choices:
T1 and T2
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=ZXTN25060BZCT-ND

T3
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=MMBT4401-FDICT-ND

T4
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=MMBT4403-FDICT-ND

D1 and D2
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=S1PB-M3/84AGICT-ND

C1
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=493-1326-ND

All thoughts appreciated!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams


part 2 13476 bytes content-type:image/png; name="ChargePumpIdea.png" (decode)


part 3 181 bytes content-type:text/plain; name="ATT00001.txt"
(decoded base64)

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2011\01\25@175957 by Josh Koffman

face picon face
On Tue, Jan 25, 2011 at 5:52 PM, Josh Koffman <TakeThisOuTjoshybear.....spamTakeThisOuTgmail.com> wrote:
> I've attached a schematic of what I'm thinking. I've picked components
> based on what I think were appropriate values and what Digikey has
> available. I'm quite sure I've gotten at least some of it wrong, so
> please feel free to suggest alternatives. I don't have values for the
> resistors yet, partially because I'm not 100% sure how to calculate
> them all, and partially because I'd like to know if my design is even
> viable. I tied the bases of T3 and T4 together as an idea, I could
> send them to dedicated pins, but this would be a way to hopefully make
> sure T1 and T2 can't turn on together.

Uh, clearly I wasn't thinking with my small transistors. New circuit
coming soon.

-j
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\25@183018 by Josh Koffman

face picon face
part 1 634 bytes content-type:text/plain; charset="iso-8859-1" (decoded quoted-printable)

On Tue, Jan 25, 2011 at 5:59 PM, Josh Koffman <TakeThisOuTjoshybearKILLspamspamspamgmail.com> wrote:
> Uh, clearly I wasn't thinking with my small transistors. New circuit
> coming soon.

Here's the latest!

T3 switches the "Charge" transistor on, and keeps the cap charging
while the PIC output is low. T5 acts as an inverter for T4, which
keeps the "Flash" transistor off as long as the PIC output is low.

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams


part 2 13278 bytes content-type:image/png; name="ChargePumpIdea_v3.png" (decode)


part 3 181 bytes content-type:text/plain; name="ATT00001.txt"
(decoded base64)

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2011\01\25@200119 by Dwayne Reid

flavicon
face
Couple of things:

1) the LEDs never turn off.  They always see 12V applied to them via D2.

2) T3 not necessary.  Instead, connect the left side of R2 to T5c.

3) I really RP's circuit.  Why not give it a try?

dwayne


At 04:29 PM 1/25/2011, Josh Koffman wrote:
{Quote hidden}

-- Dwayne Reid   <RemoveMEdwaynerspamspamBeGoneplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax
http://www.trinity-electronics.com
Custom Electronics Design and Manufacturing

2011\01\25@200411 by RussellMc

face picon face
General idea is right but tere are a number of "things needing attention" [tm].

T3/T4 driving sense is wrong.
Magic smoke will happen - or would if T1 had drive which it doesn'.
when you fix T1 drive below MS will occur if you haven;t fixed T3 T4
drive.

The / a problem with bipolar transistors is that they are 1 quadrant
devices (in normal operation).
This circuit tends to have polarities "jumping around" and both
base/gate drives and current fows must be kept track of.

With T1 NPN as shown it needs base drive to above C1- to turn on fully.
When it is on fully C1- is about at +12 so T1b needs to be above 12V.
A resistor from T1b to +12V will give you most of 12V.

To drive C1- all tjhe way to 12V a resistor to D1/R1 junction will
give you +24V ish bias or ~ 12V more than V+.
As this point tracks up as T1 turns on it will give you positive
feedback when T1 is driven on which may actually help switching.

T2 may need a high value resistor to ground to deal with leakage currents.

In fact all transistors need a base to emitter leaj\kage resistor
except when the driver provides  lower resistance clamp when off.
eg T3 and t4 are driven explicitly high and low by the PIC so need no
b-e resistor.
T1, T2 may. 100k usually OK. even 1M.





On 26 January 2011 12:29, Josh Koffman <spamBeGonejoshybear@spam@spamspam_OUTgmail.com> wrote:
{Quote hidden}

>

2011\01\25@204539 by Josh Koffman

face picon face
On Tue, Jan 25, 2011 at 8:03 PM, RussellMc <apptechnzEraseMEspamgmail.com> wrote:
> General idea is right but tere are a number of "things needing attention" [tm].

Well, I'm fairly impressed I got the general idea right.

> T3/T4 driving sense is wrong.
> Magic smoke will happen - or would if T1 had drive which it doesn'.
> when you fix T1 drive below MS will occur if you haven;t fixed T3 T4
> drive.

I'm not 100% sure I understand what you mean by the drive. Let's see
if I puzzle it out below.

> The / a problem with bipolar transistors is that they are 1 quadrant
> devices (in normal operation).
> This circuit tends to have polarities "jumping around" and both
> base/gate drives and current fows must be kept track of.

I am definitely willing to go with FETs, I had moved to BJTs based on
idea that it would be easier to get higher current rather than higher
voltage to help with the switching. Evidently I was a bit wrong in my
thoughts. If I did switch back, would I leave out the resistors talked
about below? I'd be happy to lose the extra components.

> With T1 NPN as shown it needs base drive to above C1- to turn on fully.
> When it is on fully C1- is about at +12 so T1b needs to be above 12V.
> A resistor from T1b to +12V will give you most of 12V.
>
> To drive C1- all tjhe way to 12V a resistor to D1/R1 junction will
> give you +24V ish bias or ~ 12V more than V+.
> As this point tracks up as T1 turns on it will give you positive
> feedback when T1 is driven on which may actually help switching.

Just to clarify, are you saying to add two resistors?

> T2 may need a high value resistor to ground to deal with leakage currents..
>
> In fact all transistors need a base to emitter leaj\kage resistor
> except when the driver provides  lower resistance clamp when off.
> eg T3 and t4 are driven explicitly high and low by the PIC so need no
> b-e resistor.
>  T1, T2 may. 100k usually OK. even 1M.

I thought about this, and clearly then forgot about it. I'm going to
guess you meant T3 and T5 might not need one as they are the two
driven directly from the PIC. So, I should look at adding a leakage
resistor to ground on T4. For T2, I add one from base to ground.  T1
gets a ground resistor as well as the one or two talked about above.

I'm not sure if these changes fix the T3/T4 driving sense problem. I'm
also unsure if tying the bases together to avoid T1/T2 both on
simultaneously as I did was a good idea.

I feel like I'm getting closer to something I can safely try out. I
know there is some upcoming calculations for those resistor values
though.

Thanks!

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\25@205158 by Josh Koffman

face picon face
On Tue, Jan 25, 2011 at 8:01 PM, Dwayne Reid <RemoveMEdwaynerEraseMEspamspam_OUTplanet.eon.net> wrote:
> 1) the LEDs never turn off.  They always see 12V applied to them via D2..

I wondered about that. I'm not sure how to fix that. Aren't they also
being fed by D1, through R1?

> 2) T3 not necessary.  Instead, connect the left side of R2 to T5c.

Interesting. So in that case T4 functions as the inverter and prevents
the two transistors from both being on?

> 3) I really RP's circuit.  Why not give it a try?

A few reasons. First off, I don't fully understand it, even with the
excellent explanation someone posted. I'm working on it, but I'm not
there yet. Second, I like this idea because if I thought it might be
easier to also add a 12V PWM mode to this one. That said, on looking
at RP's circuit again, I think if I added a transistor from the R1/D1
junction to ground, I should be able to PWM on the low side. I'm
unsure what this would do to the flash circuit though.

Thoughts?

Josh
-- A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
        -Douglas Adams

2011\01\26@075246 by Olin Lathrop

face picon face
Josh Koffman wrote:
> I've attached a schematic of what I'm thinking.

The way T3 and T4 are driven will cause both to be on during transitions.
Think what would happen if the PIC output were held at 2.5V.  Bipolars
generally take longer to turn off than to turn on, so both transistors will
be on for a while after each transition.

T1 is never going to come on as you have it.  It looks like maybe you
intended T1 to be a PNP?

If you could drive T2 from a separate PIC pin, then you could lose T4 and R2
completely, and also guarantee break before make timing.

It looks like your intent is to have T2 on during normal operation.  That
charges C1 to 11V.  Then to do a extra brightness pulse, T2 is first turned
off then T1 turned on.  This raises the bottom of C1 to the power voltage,
with the top side twice (minus the diode and transistor drops) the power
voltage.  The energy in C1 produces the extra brightness.  C1 decays to
empty, and you start the cycle over again.

However, D1 and R1 make no sense.  It looks like R1 is meant to make the C1
charge current finite.  That's a good idea, but it's placement isn't.  Note
that D1 defeats that purpose.  As you have it now, R1 does nothing to limit
the C1 charge current.  It only gets in the way during the extra bright
pulse.

You could lose D1 and put R1 in series with the collector of T2.  That
limits the charge current, but won't get in the way during discharge.


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2011\01\26@080106 by Olin Lathrop

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Josh Koffman wrote:
> Here's the latest!

This is going from the just broken to downright silly.

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2011\01\26@113637 by Dwayne Reid

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Couple of more things to add to this:

4) The +5V connection to R7 and T4 should be to +12V instead.  Note that the emitter of T1 can never be higher than whatever voltage is on the emitter of T4 less one diode drop.  There is NO downside to having those two points at +12V rather than +5V.

5) This could be simplified significantly: change T1 to a PNP and swap the E-C connections.  Tie the bases together and to the collector of T5.  Note that R7 is tied to +12V, not +5v.  Eliminate T4 & T3 and all associated passives.

This turns your charge pump into an emitter-follower circuit - its not as efficient as having saturated switches but is much simpler and safer - you can never have the situation of both T1 & T2 being saturated at the same time.

dwayne


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-- Dwayne Reid   <EraseMEdwaynerspam@spam@planet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax
http://www.trinity-electronics.com
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