Chip with simple program for Toy

[Michael A. Terrell]

Fred Abse wrote:

On Sat, 09 Aug 2003 05:13:16 +0100, Michael A. Terrell wrote:

She had a degree

Probably in geography

Its August 5, 2003, so I'm 51 today!
Michael A. Terrell
Central Florida

Happy birthday to you
Happy birthday to you
Happy birthday, dear Michael
Happy birthday to you!

--

--

Thank you. I forgot to remove that from my sig file.


BTW: The idiot wanted a bigger office so she threw out several
thousand data books, reference books and all the original disks and
manuals for all the design software, just so she could move her desk
into the library.


Michael A. Terrell
Central Florida

 

[bob cannetti]

Does the "Voltage Drop" of an LED vary directly with the amount of
current going thru it? If not, what are the common voltage drops for
different LED's?
thanks, bob.

"Kevin Aylward" <kevin@anasoft.co.uk> wrote in message news:<Ap1Za.12776$R6.1078266@newsfep2-win.server.ntli.net>...

John G wrote:
I don't think you can calculate the voltage drop of a LED.

You can, but its a bit tricky. There is a formular for the Vsup.D.R.
circuit here
http://www.anasoft.co.uk/EE/widlarlambert/widlarlambert.html

It depends on the construction and comes from the LED manufacturer's
data sheet and is determined by the construction theniques. Super
bright LEDS have a higher forward voltage than others.

Yes. But its one generic formula.

Id = Is.exp(Vd/Vt.N)

Vd is the diode voltage, N a constant (~1-5 for leds), Vt is the thermal
voltage.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Costas Vlachos]

"Frank Pickens" <frankpickens@verizon.net> wrote in message
news:3F34C2E0.A115ED0C@verizon.net...

Lacking a low voltage xformer with a center tap, will this scheme work
to get dual
voltages from a single secondary ??
http://www.micro-parts.com/frank/dual.jpg
Thanks
Frank

I'd use a 9V xformer, and larger than normal filter caps, due to the
half-wave rectifiers. 6.3V is too low (remember, you must take into account
diode drops, mains line variations and load). 6.3VAC will give you just over
8VDC after the filter caps, which is barely enough for the 7x05. Apart from
this, the circuit should work OK.

Regards,
Costas
_________________________________________________
Costas Vlachos Email: c-X-vlachos@hot-X-mail.com
SPAM-TRAPPED: Please remove "-X-" before replying

 

[Joshua K Drumeller]

I would use a bridge rectifier. after the transformer and cut the other 2
diodes out. Look on the data sheets of the lm337 and lm317 to get a much
better dual voltage power supply.

Josh

"Frank Pickens" <frankpickens@verizon.net> wrote in message
news:3F34C2E0.A115ED0C@verizon.net...

Lacking a low voltage xformer with a center tap, will this scheme work
to get dual
voltages from a single secondary ??
http://www.micro-parts.com/frank/dual.jpg
Thanks
Frank

 

[Joshua K Drumeller]

The diode idea that tempus Fugit suggest is a good way to do it.
The capacitor may be there just to protect the laser from AC Voltages. Even
though you are running from a DC battery, when you first turn on the power
to the laser there would be a sharp voltage spike from 0 to 4.5V. This
causes some very high frequencies to run through your circuit. The capacitor
can protect against this.

Josh

"Ewan Sinclair" <ewan_sinclair@hotmail.com> wrote in message
news:6C9Za.34840$jL2.2493099@ursa-nb00s0.nbnet.nb.ca...

I've got a bunch of laser pens that I'd like to run from a fixed power
supply rather than the batteries that they came with. The thing is, their
batteries produce 4.5v, but I can only seem to find 5 volt regulators.
While
they seem to be happy at 5, I'd like to pull the voltage down by 0.5v if
there is a simple way, just to be on the safe side. Are there any simple
ways for this?

Also, I'm still not too sure why you have to put a capacitor across the
output and ground terminals of the regulator, what does it do? I gather
that
its voltage rating must be at least double the voltage that will pass
through it, and that the capacitance needs to be proportional to the
current
you will draw.

I'm not sure how much current I wll be wanting (I want several supplies,
with various numbers of pens runing off them), it could range from
35-1000mA, so what cap value should I use?

Sorry for all the idiotic questions!

Ewan

 

[Kevin Aylward]

"Kevin Aylward" <kevin@anasoft.co.uk> wrote in message
news:<Ap1Za.12776$R6.1078266@newsfep2-win.server.ntli.net>...
John G wrote:
I don't think you can calculate the voltage drop of a LED.

You can, but its a bit tricky. There is a formular for the Vsup.D.R.
circuit here
http://www.anasoft.co.uk/EE/widlarlambert/widlarlambert.html

It depends on the construction and comes from the LED manufacturer's
data sheet and is determined by the construction theniques. Super
bright LEDS have a higher forward voltage than others.

Yes. But its one generic formula.

Id = Is.exp(Vd/Vt.N)

Vd is the diode voltage, N a constant (~1-5 for leds), Vt is the
thermal voltage.

bob cannetti wrote:
Does the "Voltage Drop" of an LED vary directly with the amount of
current going thru it?

Well, yes, thats what the above equation means.

Vd = vt.N.ln(Id/Is)

If not, what are the common voltage drops for

different LED's?

However, because of the log variation, it doesn't vary very much. If the
current changes by a factor of 10:1, the voltage only changes by 60mv.
Typical led voltages are 2-3 volts at their specified operating current.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[dan williams]

after having worked things out a little more and trying things, I want
to go over this again and confirm some things...

John Popelish wrote:

dan williams wrote:
(snip)
Please tell me where I go wrong, there are an uncomfortable number of
arbitrary limits made, I dont know if their right. you may want to read
it trough before making comments.
(snip)
Lets take the case at hand:

Input: 10-15VDC
Ouput: 5VDC (30V Isolation) 3A

{snip}

(Ignoring the turns ration for the moment Since the output voltage
will be pretty constant, you won't need its time to vary much. You
input varies over a 2 to 3 ratio, so it will take this range of on
times to produce the same volt seconds area in your timing diagram.
If we allow the full duty cycle (low input voltage, highest output
current) case to reach 50% (because equal charge and discharge times
are pretty efficient and you want efficiency at maximum output). So
the worst case will be 50% duty cycle on the verge of continuous
current operation. 25 us to charge and 25 us to discharge. In order
to have 15 watts to be the average input power at 10 volts in, the
average current (neglecting losses) will be (draw that triangle
current wave and figure out how you get a peak current from the
average) from the 15/10*4=6 amps. Lets guess 20% more for losses. So
the switch must handle 7.2 amps, peak under this case.

You seem to be aiming at 80% efficiency, my trails have topped out at

5W out, 5V
12.47V in .67A = 8.35W in,
60%

the primary problem seems to be the output rectifier, I'm using a
MUR820, from trial and error it seems to be the best of the various
diodes I have.

(Its an ultrafast, 8A 200V)

But its dropping (from the scope) about 1V forward, and appropriotly
generating heat. As I understand synchronous rectification isn't an
option if I want isolated outputs. (Though I have an idea using a
secondary winding on the transformer to trigger a mosfet that is charge
reversed when "on" and forward changed when off (as so it dosn't
conduct) I doubt it will work as I have never seen it done before.

how would one kill this major inefficiency?

I would choose N to give a worst case load duty cycle of 50% and see
how badly that treats the switch at turn off. With 9.9 volts in and
5.5 volts out (approximately) it will be very close to a 2:1
transformer.

you rework this to 1.8:1 later I take it this one is an approximation?

I usually use a rule of thumb to not use any power transistor at a
higher peak current than one which has half of the peak current gain
on the bets versus current graph.

"Current gain" = bipolar
I'm in the habbit of working out how much power loss one would get using
a bipolar and finding out what RDSon would be required by a mosfet to
match it. In all the low voltage converters I'm working with (trying to
get some working (well)) I find any mosfet with less than about .2Ohms
beats a bipolar. these are easily found. I can see that in higher input
voltage applications it becomes cheaper to use a bipolar.

For the 20 kHz case with 50% duty cycle at worst case (and throwing in
another .1 volt loss for primary resistance), I would get about:
Lprimary = 9.8 volts*(25 us / 7.2 amps) = 34 uh

If N is the primary / secondary turns, then I need N to be 9.8/5.5=1.8
to allow the 50% duty cycle.

2:1 still applicable at that point?

Before you design the snubber, think about how you minimize this
uncoupled energy by optimizing the transformer design. For such a low
isolation voltage requirement, the ultimate in low leakage inductance
would be to wind three wires (trifilar) and use two of them connected
in series as the primary and one as the secondary. This causes other
problems with high coupling capacitance, but it just about eliminates
the snubber problem. Putting half of the primary turns on each side
of the secondary is a more practical possibility.

2:1...

I think you are doing this wrong.

Turns = sqrt(47.42uh/2540mh)*1000=4.3 turns

here is an interesting problem I have run up against latley.

There are two ways of sizing a given inductor, (from what I can tell)
Faradays law B = (E * 1000000000)/(4.44 * A * N * f)
[ B = flux density (gauss)
E = inductor charge voltage,
A = inductor area cm^2
N = turns
f = frequincy (Hz)]

Ampers law H = (0.4 * Pi * N * I) / l
[ H = "magnetizing force" oersteds (divide by 14 for gauss?)
Pi = 3.1415926535....
N = turns
I = Inductor current
l = magnetic path length (cm)]

http://www.bytemark.com/products/ipmatsat.htm
<a href = "http://www.bytemark.com/products/ipmatsat.htm">graphs&lt;\a&gt;

The primary goal is to make sure the inductor dosn't saturate. In
swithcing power supplies there are two types of magnetizing actaions at
work, depending on the topology. These are AC magnetizing forces and DC
magnetizing forces

both the equations work on different principals. one effectivly says to
put more turns on to avoid saturating and the other says to use less
turns. AAARRRGGG! this makes sense when its realized that one is voltage
based and the other current based.

In something like a flyback converter, running at, or less than 50%
duty, all the magnetization forces are in AC form, so only faradays's
formula is used. In any supply that can enter continous mode, ampers's
law also has to be brought into effect, using the sum of the equations
to find the peak magnetic flux.

this all seems to come into effect when I'm trying to figure out how
many turns to use and what my minimum sized core should be.

the formula I stated later:

V = ((Ipeak)^2 L u .4 Pi) / (B^2 * 10^-8)

it seems to be based on ampers's law, and commes up with the most
outrageus numbers! I can understand that the amount of power a core can
handle is partly a factor of its volume, and would like a descent
equation as such. I also understand that core selection is primarily
based on being able to fit the number of required turns on a core as so
it donsn't saturate. e.g. tying to fit 3 coils of 67 turns of number 22
gauge wire on a .375" core is not going to work nomatter how good at
winding you are.

Flyback transformers (or inductors) normally use cores with gaps in
the magnetic path to allow the core to store more energy without
saturation, Of course, this lowers their Al value, requiring more
turns, which requires more air gap to handle the higher amp turns
without saturating, etc. But energy goes as amp turns squared, while
Al drops as amp turns, so it settles down to a solution. Basically,
you can guess a gap, and check how many turns that gap will require to
get the inductance, and then check the peak amp turns and see how high
the core flux will go. For flyback designs, I like to stay below
about half of the saturation flux rating.

Kevin mentioed staying just under the max rating, the problem it seems
with this, is that you get a lot of core lossses.
I picked up somewhere that 2000 Gauss in material 77 will loose about
1.5W!?
Atop of this, Amidon charts show that maximum application flux is a
function of frequincy and maximum core flux.(as I understand them)

http://www.amidoncorp.com/images/aai_drawings/2-39.gif
&lt;a
href="http://www.amidoncorp.com/images/aai_drawings/2-39.gif"&gt;graph&lt;\a&gt;

I gather from design documents in general, that 2000 Gauss is a
proctical maximum for material 77 cores.
I have dabbled with a few other (band new, mystery E cores.) cores
around the shop, and found that at even 900Gauss (which seems to be well
within tolerance of about any material) some cores still saturate...

I'm not sure what the AL vlaue is based on...

its all so much more complicated than way back when I was told for 60Hz
iron transformers
"keep the turns to about .5 volts/turn"...

Summary of questions that reamin:

How do I get rid of about 2W of losss on my output doide?
What method do I use for (good approximiation) figuring out core size?
Should I use the Al value, or faradays law (at, say 2000 gauss) for
determining number of turns?

dan

--
Dan Williams, Owner
Electronic Device Services
(604) 741 8431
RR8 855 Oshea rd
Gibsons BC Canada
V0N 1V8

 

[dan williams]

I dont think I caught this all on the first pass(es)...

for the application at hand:

Input: 10-15 VDC
Ouput: 5 VDC 3A
f = 20Khz
L = 34uH

Kevin Aylward wrote:

{snip}

The *key* formula for a transformer is

ET=NBA

To determine the minimum area of the core.

as in

NA = ET/B

Where:
N isn't known untill a core is selected,
A isn't known untill a core is selected,
E is known (or within .7 volts)
T is known (1/f)
B is known from a presumed core material (e.g. 77)

I'm going to say I'm using material 77 at 2000 gauss
10 volts, 20Khz

NA = (10)/(2000)(20000) = .00000025 This number seems quite wrong...

if a person used
N = ET/BA

they could look up number of turns as per core area's available, at some
point one would make sense (er go trying to fit 67 turns of number 22
wire onto a .375" toroid wont work"

for b = 2000, E = 10, T = 1/20000
assuming #22 wire (3 amps) 1 primary and 2 secondaries (1 for feedback)

window area N
-----------------------------------
0.312 0.047 5.31914893617021e-06
0.312 0.094 2.65957446808511e-06
0.312 0.094 2.65957446808511e-06
0.52 0.07625 3.27868852459016e-06
0.54 0.0825 3.03030303030303e-06
0.748 0.11623 2.15090768304224e-06
0.748 0.21473 1.16425278256415e-06
0.9 0.25 1e-06
0.906 0.30149 8.29214899333311e-07
0.75 0.1875 1.33333333333333e-06
1.252 0.425 5.88235294117647e-07

if frequincy is suppoed to be in Mhz....
window area N
-----------------------------------
0.312 0.047 5.31914893617021
0.312 0.094 2.65957446808511
0.312 0.094 2.65957446808511
0.52 0.07625 3.27868852459016
0.54 0.0825 3.03030303030303
0.748 0.11623 2.15090768304224
0.748 0.21473 1.16425278256415
0.9 0.25 1
0.906 0.30149 0.829214899333311
0.75 0.1875 1.33333333333333
1.252 0.425 0.588235294117647

its better, but dosn't prove the point.



presuming the numbers made sense, the turns would go from impractically
large for the core, to possable, to rediciously small for the core.

Then one chooses the primary inductance such that its magnetising
current in not too large a fraction of the load current, for obvious
reasons. i.e.

Imag =ET/L

hmmm, this was already worked out to be 7.2A, which is definitly NOT in
the catagory of "such that its magnetising
current is not too large a fraction of the load current" should, .. can
this be reworked, how far?

This allows the minium length to be determined from Le=Al.N^2/L

I would think one is best to use ampers law if you want to customize the
magnetizing force, but were supposed to use faradays law when involved
with discontinious situations....


another way I thought to look at this, for a discontinous supply, is to
look at the power going in per cycle, seeing as the applied voltage will
be constant and the current will effectivly be a linear ramp. this would
give a person a maximum? continous inductance, any more inductance and
you would not be able to pump enough power into the inductor per
cycle...
Watching a supply on the scope didn't help things much, teh currrent on
my test supply was not a linear ramp from zero, but had an offset of
about 1 amp to it, from what I understand this means that it (a flyback
converter) was running in continous mode... which it shouldn't have been
as its duty was under 50% can continious operation occur at any duty?

The above will give the minium volume.

*After* this is done, one can have a look to see if this minium volume
generates too much loss.

I take it your reffering to the losses that occur when your magnetizing
flux is too high, e.g. using 4500 Gauss with material 77 as opposed to
2000 gauss?


Please do tell more about how to make this method work.

dan

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

--
Dan Williams, Owner
Electronic Device Services
(604) 741 8431
RR8 855 Oshea rd
Gibsons BC Canada
V0N 1V8

 

[Frank Pickens]

Thanks Josh.
Frank

Joshua K Drumeller wrote:

I would use a bridge rectifier. after the transformer and cut the other 2
diodes out. Look on the data sheets of the lm337 and lm317 to get a much
better dual voltage power supply.

Josh

"Frank Pickens" &lt;frankpickens@verizon.net&gt; wrote in message
news:3F34C2E0.A115ED0C@verizon.net...
Lacking a low voltage xformer with a center tap, will this scheme work
to get dual
voltages from a single secondary ??
http://www.micro-parts.com/frank/dual.jpg
Thanks
Frank

 

[Fred Abse]

On Sun, 10 Aug 2003 01:59:40 +0100, Michael A. Terrell wrote:

BTW: The idiot wanted a bigger office so she threw out several thousand
data books, reference books and all the original disks and manuals for
all the design software, just so she could move her desk into the
library.

Definitely geography.

Probablt summa cum laude, too.

--
Then there's duct tape ...
(Garrison Keillor)
nofr@sbhevre.pbzchyvax.pb.hx

 

[John Popelish]

dan williams wrote:

John Popelish wrote:

{snip}

(Ignoring the turns ration for the moment Since the output voltage
will be pretty constant, you won't need its time to vary much. You
input varies over a 2 to 3 ratio, so it will take this range of on
times to produce the same volt seconds area in your timing diagram.
If we allow the full duty cycle (low input voltage, highest output
current) case to reach 50% (because equal charge and discharge times
are pretty efficient and you want efficiency at maximum output). So
the worst case will be 50% duty cycle on the verge of continuous
current operation. 25 us to charge and 25 us to discharge. In order
to have 15 watts to be the average input power at 10 volts in, the
average current (neglecting losses) will be (draw that triangle
current wave and figure out how you get a peak current from the
average) from the 15/10*4=6 amps. Lets guess 20% more for losses. So
the switch must handle 7.2 amps, peak under this case.


You seem to be aiming at 80% efficiency, my trails have topped out at

I am guessing that 80% is about the achievable efficiency to simplify
the calculations. Including each of the actual losses is how typical
methods get so complicated.

5W out, 5V
12.47V in .67A = 8.35W in,
60%

the primary problem seems to be the output rectifier, I'm using a
MUR820, from trial and error it seems to be the best of the various
diodes I have.

(Its an ultrafast, 8A 200V)

But its dropping (from the scope) about 1V forward, and appropriotly
generating heat. As I understand synchronous rectification isn't an
option if I want isolated outputs. (Though I have an idea using a
secondary winding on the transformer to trigger a mosfet that is charge
reversed when "on" and forward changed when off (as so it dosn't
conduct) I doubt it will work as I have never seen it done before.

how would one kill this major inefficiency?

Start out by switching to a 20 volt 3 amp schottky diode. A 1N5820
should work fine with .475 volts maximum forward drop at 3 amps. Part
of the problem with the diode is that 4 to 1 peak to average current
waveform.

I would choose N to give a worst case load duty cycle of 50% and see
how badly that treats the switch at turn off. With 9.9 volts in and
5.5 volts out (approximately) it will be very close to a 2:1
transformer.


you rework this to 1.8:1 later I take it this one is an approximation?

Yes. It is a simplifying approximation. Since you are using a
separate winding to produce the output voltage, the duty cycle is a
degree of freedom.

I usually use a rule of thumb to not use any power transistor at a
higher peak current than one which has half of the peak current gain
on the bets versus current graph.


"Current gain" = bipolar
I'm in the habbit of working out how much power loss one would get using
a bipolar and finding out what RDSon would be required by a mosfet to
match it. In all the low voltage converters I'm working with (trying to
get some working (well)) I find any mosfet with less than about .2Ohms
beats a bipolar. these are easily found. I can see that in higher input
voltage applications it becomes cheaper to use a bipolar.

This all sounds reasonable.

For the 20 kHz case with 50% duty cycle at worst case (and throwing in
another .1 volt loss for primary resistance), I would get about:
Lprimary = 9.8 volts*(25 us / 7.2 amps) = 34 uh

If N is the primary / secondary turns, then I need N to be 9.8/5.5=1.8
to allow the 50% duty cycle.


2:1 still applicable at that point?

50% duty cycle takes precedence, in my mind. But if I was going to
wind the transformer multi filar, I would switch horses and go with a
low integer ratio of primary to secondary turns and shift the duty
cycle a bit.

Before you design the snubber, think about how you minimize this
uncoupled energy by optimizing the transformer design. For such a low
isolation voltage requirement, the ultimate in low leakage inductance
would be to wind three wires (trifilar) and use two of them connected
in series as the primary and one as the secondary. This causes other
problems with high coupling capacitance, but it just about eliminates
the snubber problem. Putting half of the primary turns on each side
of the secondary is a more practical possibility.


2:1...

See? I told you so. ;-)

I think you are doing this wrong.

Turns = sqrt(47.42uh/2540mh)*1000=4.3 turns


here is an interesting problem I have run up against latley.

There are two ways of sizing a given inductor, (from what I can tell)
Faradays law B = (E * 1000000000)/(4.44 * A * N * f)
[ B = flux density (gauss)
E = inductor charge voltage,
A = inductor area cm^2
N = turns
f = frequincy (Hz)]

This is most applicable to voltage transformers, where you are
calculation the zero load magnetic field of the core, to make sure the
core does not saturate, and the magnetizing current is small, compared
to the load current. The load current has little effect on the
magnetic field. The only purpose of the core is to generate winding
voltage by having rate of change of flux happening in the coils. The
actual value of the flux is totally unimportant to the coils and equal
peak flux in each direction is assumed.

Ampers law H = (0.4 * Pi * N * I) / l
[ H = "magnetizing force" oersteds (divide by 14 for gauss?)
Pi = 3.1415926535....
N = turns
I = Inductor current
l = magnetic path length (cm)]

I think this is most applicable when the transformer is being used as
an energy storage mechanism, and this is certainly what you are using
it for in the flyback regulator. The transformer properties are
optional. In this case, the load current translates directly into
peak ampere turns the core must withstand without saturating.
1/2 * Lpri * Ipri_peak^2 represents the peak energy that is put in the
flux, and that equals (except for losses)
1/2 * Lsec-peak * Isec^2, the energy per cycle exiting the inductor.
In this case flux in only one direction is assumed, so the core flux
swing capability is cut in half when using it this way.

http://www.bytemark.com/products/ipmatsat.htm
a href = "http://www.bytemark.com/products/ipmatsat.htm"&gt;graphs&lt;\a

The primary goal is to make sure the inductor dosn't saturate. In
swithcing power supplies there are two types of magnetizing actaions at
work, depending on the topology. These are AC magnetizing forces and DC
magnetizing forces

both the equations work on different principals. one effectivly says to
put more turns on to avoid saturating and the other says to use less
turns. AAARRRGGG! this makes sense when its realized that one is voltage
based and the other current based.

In something like a flyback converter, running at, or less than 50%
duty, all the magnetization forces are in AC form, so only faradays's
formula is used. In any supply that can enter continous mode, ampers's
law also has to be brought into effect, using the sum of the equations
to find the peak magnetic flux.

this all seems to come into effect when I'm trying to figure out how
many turns to use and what my minimum sized core should be.

the formula I stated later:

V = ((Ipeak)^2 L u .4 Pi) / (B^2 * 10^-8)

it seems to be based on ampers's law, and commes up with the most
outrageus numbers! I can understand that the amount of power a core can
handle is partly a factor of its volume, and would like a descent
equation as such. I also understand that core selection is primarily
based on being able to fit the number of required turns on a core as so
it donsn't saturate. e.g. tying to fit 3 coils of 67 turns of number 22
gauge wire on a .375" core is not going to work nomatter how good at
winding you are.



Flyback transformers (or inductors) normally use cores with gaps in
the magnetic path to allow the core to store more energy without
saturation, Of course, this lowers their Al value, requiring more
turns, which requires more air gap to handle the higher amp turns
without saturating, etc. But energy goes as amp turns squared, while
Al drops as amp turns, so it settles down to a solution. Basically,
you can guess a gap, and check how many turns that gap will require to
get the inductance, and then check the peak amp turns and see how high
the core flux will go. For flyback designs, I like to stay below
about half of the saturation flux rating.


Kevin mentioed staying just under the max rating, the problem it seems
with this, is that you get a lot of core lossses.
I picked up somewhere that 2000 Gauss in material 77 will loose about
1.5W!?

I'll take your word for it, except that this is usually speced as
watts per cubic centimeter, or something equivalent. It is usually a
bit lower if driving flux in only one direction compared to passing
symmetrically through zero. The BH loop is usually a bit flatter for
unidirectional flux loop that are only driven up and left to fall back
toward zero, undriven. The area of the loop represents the total loss
per cycle per core volume. In addition to this fairly fixed loss per
cycle, you have to add the eddy current conduction losses and this is
a function of the volts per turn you are applying and the resistivity
of the material.

Atop of this, Amidon charts show that maximum application flux is a
function of frequincy and maximum core flux.(as I understand them)

I think this is advice based on some total loss per volume not
exceeding some maximum value. It summarizes the effect of both
hysterisis losses and conduction losses.

http://www.amidoncorp.com/images/aai_drawings/2-39.gif
a
href="http://www.amidoncorp.com/images/aai_drawings/2-39.gif"&gt;graph&lt;\a

I gather from design documents in general, that 2000 Gauss is a
proctical maximum for material 77 cores.

It looks like that is what this chart shows. But is the chart for
bydirectional flux or unidirectional flux. The first, I think. So I
think this is a bit conservative for unidirectional flux applications
(only half the peak to peak flux excursions, so much less area in the
BH loop). I would have to think about the effect on the conduction
losses.

I have dabbled with a few other (band new, mystery E cores.) cores
around the shop, and found that at even 900Gauss (which seems to be well
within tolerance of about any material) some cores still saturate...

High frequency materials often have lower saturation flux levels, but
any power material should handle well over 2000 gauss.

I'm not sure what the AL vlaue is based on...

Core permeability, flux path length, and core cross sectional area.

its all so much more complicated than way back when I was told for 60Hz
iron transformers
"keep the turns to about .5 volts/turn"...

That "law" assumes a narrow range of power rating.

Summary of questions that reamin:

How do I get rid of about 2W of losss on my output doide?
What method do I use for (good approximiation) figuring out core size?
Should I use the Al value, or faradays law (at, say 2000 gauss) for
determining number of turns?

Forget faraday's law. It is for AC voltage transformers where the
core just couples two windings so their voltages are instantaneously
proportional.

--
John Popelish

 

[cpemma]

bob cannetti wrote:

Does the "Voltage Drop" of an LED vary directly with the amount of
current going thru it? If not, what are the common voltage drops for
different LED's?
thanks, bob.

You'll find Vf : If curves in the Kingbright datasheets

http://www.kingbright.com/product/html/detailcategory.php?version=english&amp;file=detailcategory&amp;PriCategory=13,
or at any other led maker's.

A typical blue will drop about 3.4v @ 2.5mA, 3.7v @ 10mA and 4.2v @ 30mA.

White and some pure green are in the same ballpark, red, orange, yellow,
amber and other greens are around 1.8v @ 5mA up to 2.2v @30mA.

It does vary with the exact chemistry, so check the datasheet for accuracy.

 

[cpemma]

A E wrote:

Ewan Sinclair wrote:

I've got a bunch of laser pens that I'd like to run from a fixed
power supply rather than the batteries that they came with. The
thing is, their batteries produce 4.5v, but I can only seem to find
5 volt regulators.
Keep in mind that if you are using those cheap dollar store laser
pointer thingies, they are terrible designs to follow. The 4.5V
figure might be too high for the diode. I did a test once. I noticed
that these pointers where pretty bright when new and rapidly
decreased. This reminded me of how regular LEDs behave when
overdriven. So I took two new pointers. One I put on a constant 20mA
supply, and the other I used as is. After 1 minute, the 20mA laser
was as bright as new, but the other one was now dimmer. These
pointers are cheap crap. They save cost by not putting a resistor in
series with the batteries. So make sure that 4.5V is correct by
testing the diodes first with a constant current, you might be
surprised to see how bright the laser diodes can be if properly
driven. Same goes for the 'fake money' detector near-UV led pens.
They also just short the battery pack on the LED, and the brightness
goes way down, but the LED itself when driven correctly stays very
bright.

Good point, backed up by a comment I saw recently that running from a
low-impedance 5v supply blew one of these right sharpish. Wouldn't the OP be
better running them on 5v (or any other voltage his source will give) as a
regular led, with a series resistor? Any idea of the forward voltage @ 20mA?

 

Well, I had a look at those datasheets and there was no circuit
for a supply with both +ve and -ve supplies. We must use the
two diode technique I believe, if not, can you explain how ?

 

[Dr. Anton Squeegee]

In article &lt;3F3672D3.CE839141@nf.sympatico.ca&gt;, Terry says...

Preparing to install a Home Alarm System (Tandy 49-8450). Pretty
straightforward for anyone with electrical background.

Tandy? Bleah... my opinion, anyway. Then again, I will freely
admit to being very picky where security hardware is concerned. I would
have looked at DSC.

Anyway, that's neither here nor there. On to your question...

One item, relative to how to mount the magnet portion of the
magnetic door switches, caused surprise, it reads, exactly;
"Do not mount the magnet directly on any metal surface (even
aluminum). If you do so the magnet weakens after a period of less
than one year. Mount the magnet on at least a 1/4 inch (6mm)
standoff." It happens to be on page 13 of the 'instructions'!

Can anyone comment on this; I can sort of see the presence of

&lt;snip&gt;

I would -really- question it. My own security system uses a
magnetic switch on our sectional garage door. Said switch is made by
Sentrol, a very well-known maker of security products, and its magnet is
actually encased in an aluminum housing.

I could understand why a steel door might be a problem (then
again, I never had any trouble with the magnets for our system on our
old steel front door), but I don't see how aluminum could affect things,
being that it's non-ferrous.

HOWEVER -- the possibility exists that the composition of the
magnet itself may have something to do with it. Consider that
Tandy/Radio Shack will always select the cheapest available parts for
stuff like that, whereas a manufacturer such as Sentrol would not.

If you're worried about it, I would advise locating a commercial
security hardware wholesaler (Tri-Ed, maybe?) in your area, or some
place that sells such to DIY folk. Purchase a commercial-grade
switch/magnet combo, and you should have no problems.

--
Dr. Anton Squeegee, Director, Dutch Surrealist Plumbing Institute
(Known to some as Bruce Lane, KC7GR)
kyrrin a/t bluefeathertech d-o=t c&amp;o&amp;m
"Quando Omni Flunkus Moritati" (Red Green)

 

[Terry]

Terry had wrote:

Preparing to install a Home Alarm System (Tandy 49-8450). Pretty
straightforward for anyone with electrical background.

An item, relative to how to mount the magnet portion of the
magnetic door switches reads;
"Do not mount the magnet directly on any metal surface (even
aluminum). If you do so the magnet weakens after a period of less
than one year. ......... "

Many thanks to Dr. Anton (Our Tandy unit was gift btw!) and Frank
P. who replied with advice which seems to indicate that mounting
magnet of magnetic door switches on aluminum OK.
That may be a good point about very poor quality magnet materials
though. Terry.

 

[Fred Abse]

On Mon, 11 Aug 2003 03:15:18 +0100, Jacobe Hazzard wrote:

1 - Would it be worth it to put in a switching output stage to save
power? I have designed PWM circuits before, but don't know if I could
make one linear enough for audio- are there commonly available chips
that will do this?

This used to be called "Class D"

Do a search on that.


--
Then there's duct tape ...
(Garrison Keillor)
nofr@sbhevre.pbzchyvax.pb.hx

 

[ddwyer]

In article &lt;5778ec55.0308070950.1c0d8add@posting.google.com&gt;, eromlignod
&lt;eromlignod@aol.com&gt; writes

Mark Fergerson &lt;mfergerson1@cox.net&gt; wrote in message news:&lt;3F31416C.5080002@cox
.net&gt;...
eromlignod wrote:

The part I orignally had asked the group about was the reading of the
pickups. You can use a similar "shift register" method that I do for
the strings to address the pickups (just shift a single "1" through
the register for polling). The problem is that I need to
simultaneously disconnect the sustaining signal from the coil and
connect the signal line to the counter so that I can get a reading
(like a double-throw switch). Rather than use a lot of fancy analog
switch chips, I was hoping that there was an easier solution involving
back-to-back NPN/PNP transistors or something. I need something
cheap.

John Fields asked the critical question to determine if
4066-type analog switch chips (likely cheaper than
discretes) can handle the signals involved and have adequate
"off" resistance for isolation. If so, you're good to go.

Mark L. Fergerson

In looking up the 4066 analog switch, I accidentally turned up a
CD4053 analog MUX which looks like it does what I want. It is a
3PDT-type switch, which gives me the double-throw capability. I just
wish there were more than three poles. Oh well, they are only about
$0.40 ea. and I would need about 73 of them, but that's only about
$30. I don't know if I'll ever do any better. Thanks for the help.

Don
Check Maxim-IC.com switches ;really wide range.

--
ddwyer

 

[Genejocky]

Thank you John G &amp; Anand for your help.

I'll try connecting the unused cores to the cable shielding. Perhaps
the additional shielding will sort things out.
I can't see any other reason for the cable's failure to function.

I'm trying this because I'd like to aoid buying a "usb over cat5"
setup as they are a bit excessively expensive here.
The theory behind what I'm trying sees ok though so I'm persisting
with the concept.

Thanks Gang

Gene

chrisjhodges@yahoo.com (Chris Hodges) wrote in message news:&lt;d36a713e.0308080228.6bb01224@posting.google.com&gt;...

genejocky2000@yahoo.com (Genejocky) wrote in message news:&lt;5b42948.0308070128.27cb9c7a@posting.google.com&gt;...
Hey Brett,

I'm trying to keep the USB Specs in mind (what I know of them anyhow).
If there are any specifics you could throw at me I would be very happy
to run with them.
The current process has progressed this far.
- taken functional USB a to USB b cable &amp; cut it in half.
- crimped shielded RJ45 clips/connectors to the ends of the halves.
- used a RJ45 coupler to check the cable
- checked the wiring with a continuity tester &amp; all seem ok.

I must admit that I have tried to send data to a printer through this
cable &amp; found that it failed.
I'm very interested to know what is reason for this failure.

Does anyone have any suggestions??

Lack of a screen on the USB may well cause this. It's possible that
commoning-up the unused cores and connecting them to the screen at
both ends would do it. I'm still not sure why you're doing it this
way though. You can buy USB A-Male to A-female extension cables for
about the same cost as a USB cable, or buy a 5m USB cable and splice
that on the your device.

If you're trying to do more than 5m, as (I think) has been stated
earlier you _will_definately_ need an active device.

I've beeen using a USB-over-CAT5 kit (Ł225 ~ $340 I guess) for 40m +
hub and interconnects.

btw I've found that routing USB through a 4PDT switch works, although
it seems to take longer to register the device when switched in (USB
plug make power before comms, my switch doesn't).

Chris

 

[Bob Masta]

On 12 Aug 2003 02:58:12 -0700, kvlb@techie.com (K Baker) wrote:

Hi - I'm pronbably going to open a can of worms here, but am looking
for some advice.

I have a bizzare set-up - which I won't explain in any more detail
than necessary. I have at the moment a continuous sinewave (in time
this will be a decaying sinewave), that needs to be read using a DAQ
card in a PC. I can get the DAQ card reading, but not at a fine enough
level. (I am well into noise range, but this isn't a long term
problems as I can match the signal and de-noise it - if I can capture
it in the first place). I have a two amp set-up - we know the first
amp is doing it's job, but the latest amp, which includes a band-pass
filter to remove some of the noise, seems to be making matters worse
rather than better (although some of the noise is gone I've lost
signal strength - and the gain is set to 100!). Now, the most
important thing as far as I'm concerned is the information in the
signal, which in time will be how quickly the voltage decays, so I
need the voltage passed on primarily. Now, from what I can understand,
the current set up where the impedance input to my card is 1MOhm with
a 10pF capacitor in parallel should mean I don't need to worry about
the impedance of anything else for voltage transfer - or have I got
this backwards.

Also, if anyone can give me an idiot's guide to including a line
driver (I have ICs coming out my ears!) and how to set it up to match
the impedances so that I can at least get maximum power excahnge I
would be grateful. Or if you need more info, let me know and I'll give
you what I can.

First of all, you need not be concerned about "matching" impedances;
you are interested only in voltage transfer, to you simply want your
amp's input impedance to be much higher than your signal's source
impedance. But you say that your first stage seems to be OK, so
this is probably not the problem there.

The fact that your signal is reduced by the second stage might be
indicating that you've got your filter mismatched to your signal.
However, I'd be very leery of putting any filtering in here at all,
unless you are really sure you need it. The reason is that you
are interested in the time-decay of the waveform, and that will
be seriously changed by any filter that can remove significant
noise.

Since you are already talking about (I think!) coherent (synchronous)
averaging to remove the noise, why not just go with that? Your
preamps should boost the signal until the peaks (noise or signal)
are as high as feasible to still assure you never clip them, then
deal with everything digitally. If your averaging is synchronous
with the response (by triggering from the same stimulus that
produces the response, for example) then you will not change the
true waveshape at all, and noise will be reduced by the square
root of the number of presentations averaged.

Once you get this working, then you might want to look at
input filters. That way you can see their effect on the
waveshape.



Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
Shareware from Interstellar Research
www.daqarta.com