Chip with simple program for Toy

[Peter Bennett]

On Mon, 22 Dec 2008 11:28:28 -0800 (PST), panfilero
<panfilero@gmail.com> wrote:

Hello,

ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:

http://www.national.com/ds/LM/LM340.pdf

On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.

Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...

would my way be ok? or is there a beinifit to folowing the example on
the datasheet?

much thanks!
J.

The adjustable regulator circuit raises the "ground" pin of the
regulator above ground, but the regulator still tries to keep 12 volts
between its output pin and its ground pin. The circuit shown, when
used with a 12 volt regulator, will allow you to adjust the output
voltage from 12 volts to about 22 volts, with your 24 volt input.
Using the LM340-5 5 volt regulator in this circuit will let you adjust
the voltage from 5 to 22 volts, with your 24 volt input.

A better regulator choice would be an LM317, which is designed to be
used as an adjustable regulator. The same circuit would allow you to
adjust the output voltage from 1.2 volts to 22 volts.

As another poster mentioned, you _will_ have to put the regulator on a
good heatsink as it will have to dissipate
(24V - output voltage) * output current watts.

For the 300 mA at 10 volts you want, that's 4.2 watts.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[IanM]

panfilero wrote:

IanM wrote:
panfilero wrote:
Hello,
ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:
http://www.national.com/ds/LM/LM340.pdf
On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.
Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...
would my way be ok?
*NO*
or is there a beinifit to folowing the example on> the datasheet?
*YES*
snip
You can start learning by answering the questions (in the group
sci.electronics.basics *ONLY*):

What is your load and what is its minimum and maximum current requirements?

What is your supply voltage and what is it from?

N.B LM340-12 is absolutely the *wrong* choice unless you only want to
adjust from 12V to lets say 15V.

That's my good deed for the day done, Merry Christmas all.

My supply voltage is around 24V, and I want a 10V output... so I was
thinking about putting a voltage divider at the output of the
regulator and using a pot to dial in 10V out... the load will have a
draw of around 300mA.

Could you elaborate a little more on please on what they are trying to
do on that datasheet, and on what would be wrong with my approach?

To get acceptable regulation the current in the pot should be around ten
times the load current. You need 300mA at 10V. The pot you are using as
the voltage divider would need to be a 4 ohm one rated at over 36 watts
and the regulator wouldn't be able to supply enough current and would
either shut down or burn up.

If you try to use a higher value lower wattage pot, you will have crap
regulation. Pot's wattage ratings actually define the maximum current
the track can take. You need 300 ma out at 10V so if one reduces that
10:1 divider current to load current ratio to 1:1 and tolerate crap
regulation, you will have 600mA flowing in the top part of the track
which by definition *must* be dropping 2V. The remainder of the track
has 10V across it and is passing 300mA. This would need a 36 ohm pot, a
value that is not going to be easy to find.

OK, try a 50 ohm pot. The regulation will now be truly awful. The top
part of the track is carrying 516mA, the lower part, 216mA and all
should work. One small problem, the pot needs to be rated at over 13.5
watts or it will overheat and catch fire. A 50 ohm loudspeaker control
pot is good for about 3.5 watts, you need 5 times better.

I hope you can now see why a resistive dropper, especially one based on
a pot is impracticable at more than a few tens of milliamps output current.

On the data sheet they are showing you how to adjust the output voltage
of the regulator upwards by 'lifting' the voltage at the common terminal
as it *always* acts to keep its designed output voltage 12V above its
common. They can do this because the current in the common leg is about
6mA, though it would be advisable to design the potential divider to
pass around 10 times that. You need to start with a LM340-05 regulator
to be able to get 10V out though.

Personally, I'd put a small 6.2V Zener diode in series with the common
leg, cathode towards the regulator which would give 10.2V (+- 0.1V)
output and forget the resistors which would work for *most* 10V loads.

N.B. the capacitors shown in the data sheet are *essential* or it can
start oscillating at a frequency too high to hear, get very hot and burn
up, possibly shorting and taking your load with it.

PS - I thought a 12V regulator outputs 12V... but you said you can use
it to adjust between 12-15V... is this true?
You can go a lot higher than 15, but it would be preferable to start

with a higher voltage regulator. You *cant* go below 12 (without some
trickery with a negative supply to the common terminal). See data sheet
for examples.

"> N.B LM340-12 is absolutely the *wrong* choice unless you only want
to
adjust from 12V to lets say 15V."

You've had another suggestion in S.E.D to use a LM317T. Its fully
adjustable with a minimum output of 1.2V, and the data sheet is much
more understandable.
<http://www.national.com/ds/LM/LM117.pdf>

Basically, its designed to keep its output 1.2V above its Adj terminal
(the equivalent of common on the 78xx or LM340-xx series regulators).
It also works with 100 times less current through that pin. This means
that it works the same way by lifting the voltage on the Adj terminal
but you can use pots of normal (well under 1 watt) ratings and values of
several kiloohms. Much easier. It still needs the caps for the same
reason.

If you are going to need various voltages in the future, get a few
LM317T regulators (you can get any voltage if you have a selection of
resistors handy), but the 5V and 12V fixed regulators are useful too.
If its a one off need for a fixed 10V, the 5V regulator with the 6.2V
Zener is the simplest circuit and is reliable.

What is your 24V supply and what is your 10V 300mA load? There may be
other problems.

N.B. If the supply is from a vehicle, BEWARE that whenever the starter
motor or other high current devices operate there are often transients
up to several hundred volts which *will* blow the regulator and your
load. If its only plugged in with the engine off and cold it will be
fine, but sooner or later you know it will get plugged in with the
engine running :-(

 

[IanM]

IanM wrote:
<snip>

pass around 10 times that. You need to start with a LM340-05 regulator
to be able to get 10V out though.

Personally, I'd put a small 6.2V Zener diode in series with the common
leg, cathode towards the regulator which would give 10.2V (+- 0.1V)
output and forget the resistors which would work for *most* 10V loads.
snip

OOPS, I made a mistake. That would give you 11.2V (+- 0.1V).

It should be a 5.1V Zener with a LM340-05 regulator for 10.1V (+-0.1V) out.

 

[panfilero]

On Dec 22, 6:55 pm, IanM <Inva...@totally.invalid> wrote:

panfilero wrote:
IanM  wrote:
panfilero wrote:
Hello,
ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:
http://www.national.com/ds/LM/LM340.pdf
On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.
Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...
would my way be ok?
*NO*
or is there a beinifit to folowing the example on> the datasheet?
*YES*
snip
You can start learning by answering the questions (in the group
sci.electronics.basics *ONLY*):

What is your load and what is its minimum and maximum current requirements?

What is your supply voltage and what is it from?

N.B LM340-12 is absolutely the *wrong* choice unless you only want to
adjust from 12V to lets say 15V.

That's my good deed for the day done,  Merry Christmas all.

My supply voltage is around 24V, and I want a 10V output... so I was
thinking about putting a voltage divider at the output of the
regulator and using a pot to dial in 10V out... the load will have a
draw of around 300mA.

Could you elaborate a little more on please on what they are trying to
do on that datasheet, and on what would be wrong with my approach?

To get acceptable regulation the current in the pot should be around ten
times the load current.  You need 300mA at 10V. The pot you are using as
the voltage divider would need to be a 4 ohm one rated at over 36 watts
and the regulator wouldn't be able to supply enough current and would
either shut down or burn up.

If you try to use a higher value lower wattage pot, you will have crap
regulation.  Pot's wattage ratings actually define the maximum current
the track can take. You need 300 ma out at 10V so if one reduces that
10:1 divider current to load current ratio to 1:1 and tolerate crap
regulation, you will have 600mA flowing in the top part of the track
which by definition *must* be dropping 2V.  The remainder of the track
has 10V across it and is passing 300mA. This would need a 36 ohm pot, a
value that is not going to be easy to find.

OK, try a 50 ohm pot. The regulation will now be truly awful. The top
part of the track is carrying 516mA, the lower part, 216mA and all
should work.  One small problem, the pot needs to be rated at over 13.5
watts or it will overheat and catch fire.  A 50 ohm loudspeaker control
pot is good for about 3.5 watts, you need 5 times better.

I hope you can now see why a resistive dropper, especially one based on
a pot is impracticable at more than a few tens of milliamps output current.

On the data sheet they are showing you how to adjust the output voltage
of the regulator upwards by 'lifting' the voltage at the common terminal
as it *always* acts to keep its designed output voltage 12V above its
common.  They can do this because the current in the common leg is about
6mA, though it would be advisable to design the potential divider to
pass around 10 times that.   You need to start with a LM340-05 regulator
to be able to get 10V out though.

Personally, I'd put a small 6.2V Zener diode in series with the common
leg, cathode towards the regulator which would give 10.2V (+- 0.1V)
output and forget the resistors which would work for *most* 10V loads.

N.B. the capacitors shown in the data sheet are *essential* or it can
start oscillating at a frequency too high to hear, get very hot and burn
up, possibly shorting and taking your load with it.

PS - I thought a 12V regulator outputs 12V... but you said you can use
it to adjust between 12-15V... is this true?

You can go a lot higher than 15, but it would be preferable to start
with a higher voltage regulator. You *cant* go below 12 (without some
trickery with a negative supply to the common terminal).  See data sheet
for examples.



"> N.B LM340-12 is absolutely the *wrong* choice unless you only want
to
adjust from 12V to lets say 15V."

You've had another suggestion in S.E.D to use a LM317T. Its fully
adjustable with a minimum output of 1.2V, and the data sheet is much
more understandable.
http://www.national.com/ds/LM/LM117.pdf

Basically, its designed to keep its output 1.2V above its Adj terminal
(the equivalent of common on the 78xx or LM340-xx series regulators).
It also works with 100 times less current through that pin.  This means
that it works the same way by lifting the voltage on the Adj terminal
but you can use pots of normal (well under 1 watt) ratings and values of
several kiloohms. Much easier.  It still needs the caps for the same
reason.

If you are going to need various voltages in the future, get a few
LM317T regulators (you can get any voltage if you have a selection of
resistors handy), but the 5V and 12V fixed regulators are useful too.
If its a one off need for a fixed 10V, the 5V regulator with the 6.2V
Zener is the simplest circuit and is reliable.

What is your 24V supply and what is your 10V 300mA load?  There may be
other problems.

N.B. If the supply is from a vehicle, BEWARE that whenever the starter
motor or other high current devices operate there are often transients
up to several hundred volts which *will* blow the regulator and your
load.  If its only plugged in with the engine off and cold it will be
fine, but sooner or later you know it will get plugged in with the
engine running :-(

IanM, I appreciate you taking the time to give me such a detailed
answer, and I am really trying to understand it:

1. When you say "To get acceptable regulation the current in the pot
should be around ten
times the load current."

ok, I believe you are referring to my voltage divider solution with
the reference of the voltage regulator hooked up to ground.
So, I'm curious where you got the '10 times the load current' rule...
why is that?

comparing what you are saying to john field's answer above he says:
PD = (Vs - Vl) Il = (12V - 10V) * 0.3A = 0.6 watts

which doesn't sound bad... but he's not following the same conventions
your pointing out... his way seems like the voltage divider method
aint bad....

help?

Thanks!
J.

 

[John Fields]

On Tue, 23 Dec 2008 08:43:07 -0800 (PST), panfilero
<panfilero@gmail.com> wrote:

On Dec 22, 6:55 pm, IanM <Inva...@totally.invalid> wrote:
panfilero wrote:
IanM  wrote:
panfilero wrote:
Hello,
ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:
http://www.national.com/ds/LM/LM340.pdf
On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.
Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...
would my way be ok?
*NO*
or is there a beinifit to folowing the example on> the datasheet?
*YES*
snip
You can start learning by answering the questions (in the group
sci.electronics.basics *ONLY*):

What is your load and what is its minimum and maximum current requirements?

What is your supply voltage and what is it from?

N.B LM340-12 is absolutely the *wrong* choice unless you only want to
adjust from 12V to lets say 15V.

That's my good deed for the day done,  Merry Christmas all.

My supply voltage is around 24V, and I want a 10V output... so I was
thinking about putting a voltage divider at the output of the
regulator and using a pot to dial in 10V out... the load will have a
draw of around 300mA.

Could you elaborate a little more on please on what they are trying to
do on that datasheet, and on what would be wrong with my approach?

To get acceptable regulation the current in the pot should be around ten
times the load current.  You need 300mA at 10V. The pot you are using as
the voltage divider would need to be a 4 ohm one rated at over 36 watts
and the regulator wouldn't be able to supply enough current and would
either shut down or burn up.

If you try to use a higher value lower wattage pot, you will have crap
regulation.  Pot's wattage ratings actually define the maximum current
the track can take. You need 300 ma out at 10V so if one reduces that
10:1 divider current to load current ratio to 1:1 and tolerate crap
regulation, you will have 600mA flowing in the top part of the track
which by definition *must* be dropping 2V.  The remainder of the track
has 10V across it and is passing 300mA. This would need a 36 ohm pot, a
value that is not going to be easy to find.

OK, try a 50 ohm pot. The regulation will now be truly awful. The top
part of the track is carrying 516mA, the lower part, 216mA and all
should work.  One small problem, the pot needs to be rated at over 13.5
watts or it will overheat and catch fire.  A 50 ohm loudspeaker control
pot is good for about 3.5 watts, you need 5 times better.

I hope you can now see why a resistive dropper, especially one based on
a pot is impracticable at more than a few tens of milliamps output current.

On the data sheet they are showing you how to adjust the output voltage
of the regulator upwards by 'lifting' the voltage at the common terminal
as it *always* acts to keep its designed output voltage 12V above its
common.  They can do this because the current in the common leg is about
6mA, though it would be advisable to design the potential divider to
pass around 10 times that.   You need to start with a LM340-05 regulator
to be able to get 10V out though.

Personally, I'd put a small 6.2V Zener diode in series with the common
leg, cathode towards the regulator which would give 10.2V (+- 0.1V)
output and forget the resistors which would work for *most* 10V loads.

N.B. the capacitors shown in the data sheet are *essential* or it can
start oscillating at a frequency too high to hear, get very hot and burn
up, possibly shorting and taking your load with it.

PS - I thought a 12V regulator outputs 12V... but you said you can use
it to adjust between 12-15V... is this true?

You can go a lot higher than 15, but it would be preferable to start
with a higher voltage regulator. You *cant* go below 12 (without some
trickery with a negative supply to the common terminal).  See data sheet
for examples.



"> N.B LM340-12 is absolutely the *wrong* choice unless you only want
to
adjust from 12V to lets say 15V."

You've had another suggestion in S.E.D to use a LM317T. Its fully
adjustable with a minimum output of 1.2V, and the data sheet is much
more understandable.
http://www.national.com/ds/LM/LM117.pdf

Basically, its designed to keep its output 1.2V above its Adj terminal
(the equivalent of common on the 78xx or LM340-xx series regulators).
It also works with 100 times less current through that pin.  This means
that it works the same way by lifting the voltage on the Adj terminal
but you can use pots of normal (well under 1 watt) ratings and values of
several kiloohms. Much easier.  It still needs the caps for the same
reason.

If you are going to need various voltages in the future, get a few
LM317T regulators (you can get any voltage if you have a selection of
resistors handy), but the 5V and 12V fixed regulators are useful too.
If its a one off need for a fixed 10V, the 5V regulator with the 6.2V
Zener is the simplest circuit and is reliable.

What is your 24V supply and what is your 10V 300mA load?  There may be
other problems.

N.B. If the supply is from a vehicle, BEWARE that whenever the starter
motor or other high current devices operate there are often transients
up to several hundred volts which *will* blow the regulator and your
load.  If its only plugged in with the engine off and cold it will be
fine, but sooner or later you know it will get plugged in with the
engine running :-(

IanM, I appreciate you taking the time to give me such a detailed
answer, and I am really trying to understand it:

1. When you say "To get acceptable regulation the current in the pot
should be around ten
times the load current."

ok, I believe you are referring to my voltage divider solution with
the reference of the voltage regulator hooked up to ground.
So, I'm curious where you got the '10 times the load current' rule...
why is that?

comparing what you are saying to john field's answer above he says:
PD = (Vs - Vl) Il = (12V - 10V) * 0.3A = 0.6 watts

---
That's the power being dissipated by the 680 ohm resistor, not the
regulator.

The power being dissipated by the regulator will be the difference
between the supply voltage and the output voltage multiplied by the load
current:


Pd = (24V - 12V) * 0.3A = 3.6 watts.

which doesn't sound bad... but he's not following the same conventions
your pointing out... his way seems like the voltage divider method
aint bad....

---
It's very bad.

What I'm talking about is:

+24V>----[LM340-12]---+
| |
| [RS]
| |
| +--10V
| |
| [LOAD]
| |
GND>---------+--------+

Which is only good if you have a constant-current load, as I pointed out
earlier.


While what you seem to be talking about is:

+24V>----[LM340-12]---+
| |
| [POT]<--+
| | |
| | [LOAD]
| | |
GND>---------+--------+-----+

Yes?

JF

 

[Hammy]

On Mon, 22 Dec 2008 11:28:28 -0800 (PST), panfilero
<panfilero@gmail.com> wrote:

Hello,

ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:

http://www.national.com/ds/LM/LM340.pdf

On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.

Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...

would my way be ok? or is there a beinifit to folowing the example on
the datasheet?

much thanks!
J.
Just use something like the ncp3063

http://canada.newark.com/on-semiconductor/ncp3063bpg/dc-dc-converter-ic/dp/27M2538

You wont need a heatsink and it's cheap even with the inductor it
shouldnt cost much more then two bucks if that.

Onsemi even has a spread sheet with schematic that will calculate all
external components for you. I would give you a link but their site is
down now.

http://www.onsemi.com/

When the site is back up search for NCP3063 you'll find app notes and
a excel spreadsheet.

This is similiar to the old mc34063 but faster switching freq.150kHz
with internal switch.

 

[Pat Cheney]

On Fri, 26 Dec 2008 14:20:44 -0500, tnom@mucks.net wrote:

That's not all that's proprietary on that camera.
Both the battery and the users manual are proprietary.

We did worry a lot about the proprietary battery.

But at least that proprietary battery lasts (much) longer than AA
batteries, which, as you noted, we prefer.

I'm not sure what your point is about the manual because typically, while
traveling abroad, in a pinch, one can download the PDF easily.

As for that proprietary USB interface that Casio camera reviews missed,
does anyone know WHY Casio put a proprietary USB interface in their EXILIM
camera? Was it price? Was it competition?

Is there any advantage to the proprietary Casio USB data interface?

 

[dj_nme]

Pat Cheney wrote:

On Fri, 26 Dec 2008 14:20:44 -0500, tnom@mucks.net wrote:

That's not all that's proprietary on that camera.
Both the battery and the users manual are proprietary.

We did worry a lot about the proprietary battery.

But at least that proprietary battery lasts (much) longer than AA
batteries, which, as you noted, we prefer.

I'm not sure what your point is about the manual because typically, while
traveling abroad, in a pinch, one can download the PDF easily.

The above two items seemed to have been included in the list of
proprietary items because they are usually only useful for that
particular camera or family of cameras.
I would be surprised if you actually needed the manual when travelling
and if you haven't got the battery in the camera, it's rather silly
taking the camera in the first place! ;-)

As for that proprietary USB interface that Casio camera reviews missed,
does anyone know WHY Casio put a proprietary USB interface in their EXILIM
camera? Was it price? Was it competition?

Is there any advantage to the proprietary Casio USB data interface?

The obvious advantage to Casio is that if you misplace the lead, you
must buy a new one from them and assures them of a small revenue stream
from their camera users.
It's a disadvantage (as you've pointed out) to the consumer who has
bought the camera, but was "clever" enough to put the cable somewhere
"too safe" to be able to ever find it again.

 

[Pat Cheney]

On Sat, 27 Dec 2008 11:36:52 +1100, dj_nme wrote:

It's a disadvantage (as you've pointed out) to the consumer who has
bought the camera, but was "clever" enough to put the cable somewhere
"too safe" to be able to ever find it

I agree with you. Worse than misplacing the cable, it looks like all your
other mini-usb cables (except upon close inspection), so, it gets lost
among them since all our equipment comes with mini-usb cables (or we don't
buy them).

All I was asking was for the dpreview and steve's digicam web sites to care
enough about testing the camera to put the word "proprietary" in front of
the description of the usb connector.

John Navas pointed out that all combined USB/AV connectors use proprietary
connectors ... which had we had this secret decoder ring with us when we
read the reviews, we would have realized the connector was proprietary (and
therefore useless to us).

After the fact, if we search for "Casio proprietary usb" we do find some of
the better reviews which do put the word "proprietary" in front of "USB";
but we don't find dpreview nor steve's digicam in this list of better
review sites.

Unfortunately, these are all for other cameras than the one we bought so
we'd also have to be omnipotent to know that it applied to our camera also.

We really wonder why dpreview and steves' digicam are so out of tune with
the user but that is now water under the bridge.

See ...

http://reviews.digitaltrends.com/review/4701-3/casio-exilim-ex-z77-review-full-review
"Sadly, Casio felt it necessary to equip the Exilim EX-Z77 with a
proprietary mini-USB jack, forcing users to keep yet another USB cable at
hand. With so many other camera manufacturers using standard mini-USB
ports, it's a wonder Casio doesn't make it easier on consumers by hopping
on the universal mini-USB bandwagon."

http://www.amazon.com/Casio-Exilim-EX-S10SR-2-7-Inch-Digital/dp/B0012XTILG/ref=cm_cr-mr-title/177-0142153-6788251
"The USB cable is proprietary. While the camera input sort of looks like a
mini-USB, it is NOT! Come on Casio! I am glad you dumped the dock but where
is USB!!!"

http://www.laptopmag.com/review/cameras/casio-exilim-zoom-ex-z77.aspx
"The SD card slot and proprietary mini-USB port sit on the underside of the
camera."

 

[John Navas]

On Fri, 26 Dec 2008 23:10:13 -0800, Pat Cheney <pcheney@ymail.com> wrote
in <nzk5l.8901$W06.3697@flpi148.ffdc.sbc.com>:

On Sat, 27 Dec 2008 11:36:52 +1100, dj_nme wrote:

It's a disadvantage (as you've pointed out) to the consumer who has
bought the camera, but was "clever" enough to put the cable somewhere
"too safe" to be able to ever find it

I agree with you. Worse than misplacing the cable, it looks like all your
other mini-usb cables (except upon close inspection), so, it gets lost
among them since all our equipment comes with mini-usb cables (or we don't
buy them).

All I was asking was for the dpreview and steve's digicam web sites to care
enough about testing the camera to put the word "proprietary" in front of
the description of the usb connector.

There's no way they could get everything -- you're being unrealistic,
and not taking responsibility for your own mistaken assumption.

John Navas pointed out that all combined USB/AV connectors use proprietary
connectors ... which had we had this secret decoder ring with us when we
read the reviews, we would have realized the connector was proprietary (and
therefore useless to us).

It's no secret. Even though you didn't know it, it is well-known. Your
problem came from not properly informing yourself. You made an
assumption on your own initiative that turned out to be wrong. That's
your responsibility and only your responsibility. If that assumption
was so important, then you should have checked it in advance by
contacting the manufacturer, and you didn't do so.

--
Very best wishes for the holiday season and for the coming new year,
John

 

[Pat Cheney]

On Fri, 26 Dec 2008 23:23:24 -0800, John Navas wrote:

It's no secret. Even though you didn't know it, it is well-known.
If that assumption was so important, then you should have checked
it in advance by contacting the manufacturer, and you didn't do so.

Hi John,

You are correct.

I did not realize that everyone knew a combined AV/USB connector couldn't
possibly use a mini-USB or micro-USB port even when it's described as a
"USB" port and even though it sort of looks like a mini USB connector.

I'm surprised that the reviewers didn't mention this but if it is, as you
said, a common fact that combined AV & USB ports can't possibly be standard
USB even if they say they are, then I must agree with you that it's our
fault for not being informed.

But wasn't that the reason we read the reviews in the first place?
To be informed?

Pat

 

[John Navas]

On Fri, 26 Dec 2008 23:50:36 -0800, Pat Cheney <pcheney@ymail.com> wrote
in <i9l5l.8904$W06.8809@flpi148.ffdc.sbc.com>:

On Fri, 26 Dec 2008 23:23:24 -0800, John Navas wrote:

It's no secret. Even though you didn't know it, it is well-known.
If that assumption was so important, then you should have checked
it in advance by contacting the manufacturer, and you didn't do so.

Hi John,

You are correct.

I am indeed.

I did not realize that everyone knew a combined AV/USB connector couldn't
possibly use a mini-USB or micro-USB port even when it's described as a
"USB" port and even though it sort of looks like a mini USB connector.

Actually "USB/AV Connector, USB2.0 (Full-Speed) compatible", which is
exactly what it is.

I'm surprised that the reviewers didn't mention this but if it is, as you
said, a common fact that combined AV & USB ports can't possibly be standard
USB even if they say they are, then I must agree with you that it's our
fault for not being informed.

They don't say that.

But wasn't that the reason we read the reviews in the first place?
To be informed?

Again, if that assumption was so important, then you should have checked
it in advance by contacting the manufacturer, and you didn't do so.

You are trying to blame others for your own mistaken assumption and lack
of care. Try taking responsibility for a change.

--
Very best wishes for the holiday season and for the coming new year,
John

 

[John Navas]

On Tue, 30 Dec 2008 07:25:19 +0800, "Norman Webb" <tekrec@iinet.net.au>
wrote in <49594c69$0$22123$5a62ac22@per-qv1-newsreader-01.iinet.net.au>:

You are missing the point. A review should point out technical issues.
We read a review to evaluate whether the gear is suitable for our purpose.

We can't ask the intelligent question if we don't know what to ask.

There is no technical issue.

(Top posting deliberately)

Please don't.

--
Very best wishes for the holiday season and for the coming new year,
John

 

[Norman Webb]

You are missing the point. A review should point out technical issues.
We read a review to evaluate whether the gear is suitable for our purpose.

We can't ask the intelligent question if we don't know what to ask.

(Top posting deliberately)

John Navas wrote in message ...

On Fri, 26 Dec 2008 23:50:36 -0800, Pat Cheney <pcheney@ymail.com> wrote
in <i9l5l.8904$W06.8809@flpi148.ffdc.sbc.com>:

On Fri, 26 Dec 2008 23:23:24 -0800, John Navas wrote:

It's no secret. Even though you didn't know it, it is well-known.
If that assumption was so important, then you should have checked
it in advance by contacting the manufacturer, and you didn't do so.

Hi John,

You are correct.

I am indeed.

I did not realize that everyone knew a combined AV/USB connector couldn't
possibly use a mini-USB or micro-USB port even when it's described as a
"USB" port and even though it sort of looks like a mini USB connector.

Actually "USB/AV Connector, USB2.0 (Full-Speed) compatible", which is
exactly what it is.

I'm surprised that the reviewers didn't mention this but if it is, as you
said, a common fact that combined AV & USB ports can't possibly be
standard
USB even if they say they are, then I must agree with you that it's our
fault for not being informed.

They don't say that.

But wasn't that the reason we read the reviews in the first place?
To be informed?

Again, if that assumption was so important, then you should have checked
it in advance by contacting the manufacturer, and you didn't do so.

You are trying to blame others for your own mistaken assumption and lack
of care. Try taking responsibility for a change.

--
Very best wishes for the holiday season and for the coming new year,
John

 

[Henrik Sandberg]

MYRTYNEITÄ ydinherraskaisia.

YLE 8.5.2008. Haastateltavan energiaherran suusta: "Olemme joutuneet
huomaamaan, ettemme kovista yrityksistämme huolimatta kyenneet vakuuttamaan
Suomen kansalle ydinvoiman tarpeellisuutta lainkaan!" .. ..

Ja melkein itkua vääntävää ydinlobbariherraa yksi toisensa perään
kuvaruutuun enemmän kuin harminsa kiukkua nieleskellen. Eikä mairitteluun
todella ole syytä. Suomen kansa oli pannut karut faktat esiin. Gallupin
mukaan vain 20 % oli ylipäätään edes yhtä ydinvoimalaa halukas rakentamaan.
Teollisuuspiirien himoitseman triplan ydinhankke- emieliin peilaili 4 %
kannatus! Eikä edes ydinalan OL-3 plutoniumpoltolle ja
isotooppiuraanikaivoshankkeille uraanirikastamoineen välttämättömän
NATO-jäsenyyden suosio ole ylittänyt 20 % pöyristyttävän pientä
kannatusnyanssiaan kovasta, megalomaanisesta painostuksesta piittaamatta.
Toki jo Niinistön pitämän nuorisoparlamentin 80 % ydinvastaisuuden kanssa
kaikki kivasti stemmaa.

Ehkä silti kaikkein tärkein on yhä sanomatta. Nimittäin tuollaisilla
ydinprojektiensa tahkoamilla kannatusluvuilla on päivän selvä viesti niin
ydinintoisille puolueille yleensä, kun erityisesti ydinryvettyneelle
Kokoomukselle, KTM:lle kuin ihan koko tulevan hallituskuvion varmennukseksi.
Jokainen joka tällaisiin lukuihin luo ydinhankkeitaan tietää tarkoin, että
kansa viskaa vaaleissa oppositioon takuuvarmasti JOKAISEN
ydinkannatuspuolueen ministeriöineen!..

Ei siis mikään ihme, että ydinherrojen rasvaiset hikivanat kasvavat vaalien
lähetessä. Ei lupaa muuten jatkossa ruusutarhoja yhdelle sun muille
ydinyhtiömonopoleillemme. Ja HYVÄ NIIN!)

 

[MikeWhy]

I realize it's an old conversation, but...

What's involved in building a simple 50 MHz oscilloscope? I have a spare
Xilinx Spartan DSP board, and an unused LTC 105 MSps A/D chip in the part
box. The board's 125 MHz clock is different enough from the sampling rate, I
think, to jitter the trigger point for meaningful averaging. It seems almost
too simple. Aside from a low noise amp, what am I missing? The board has a
VGA port, 128 MB 5ns DDR2, and tons of I/O.


"David L. Jones" <altzone@gmail.com> wrote in message
news:485dfbd6$1@dnews.tpgi.com.au...

"Talal Itani" <titani@verizon.net> wrote in message
news:6zk7k.871$9J.437@trnddc06...

One model that I have used is this one:
http://www.dynoninstruments.com/products_elab080.php
It's a DSO, logic analyzer, and waveform generator all in one.
It's only 80 MS/s but for $500 you can't expect much more. I've actually
found
myself using the digital waveform generators on this thing quite often.

Thanks, this is nice, yet I wished it had a higher sampling rate. Are
you aware of any others?

Most low to medium end PC based DSO's are all a similar sample rate, i.e.
a few hundred MHz.
Because they all use off-the-shelf FPGA's and memories in their design,
and that's about as high as you can go cheaply.
When you start talking 1GS/s+ you are into the high end domain of the big
manufacturers of professional oscolloscopes.

Agilent make a PC based DSO that might suit you if you *really* want a PC
based scope:
http://www.home.agilent.com/agilent/product.jspx?nid=-536902447.774929.00&cc=US&lc=eng
200MHz, 1GS/s, 32Mpoint memory, $1600

Stop being cheap, you *need* at least one real bench scope for your lab,
even if it's a lower end mixed signal scope like a Rigol:
http://cgi.ebay.com.au/RIGOL-DIGITAL-Oscilloscope-DS1102CD-100M-COLOUR-LCD_W0QQitemZ150261816918QQihZ005QQcategoryZ45008QQrdZ1QQssPageNameZWD2VQQcmdZViewItem

Dave.

 

[Jim Granville]

MikeWhy wrote:

I realize it's an old conversation, but...

What's involved in building a simple 50 MHz oscilloscope? I have a spare
Xilinx Spartan DSP board, and an unused LTC 105 MSps A/D chip in the
part box. The board's 125 MHz clock is different enough from the
sampling rate, I think, to jitter the trigger point for meaningful
averaging. It seems almost too simple. Aside from a low noise amp, what
am I missing? The board has a VGA port, 128 MB 5ns DDR2, and tons of I/O.

Just code
And a variable gain pre-amp/attenuator, if you want a useful Volts/Div
dynamic range.
-jg

 

[David L. Jones]

On Jul 1, 5:28 pm, "MikeWhy" <boat042-nos...@yahoo.com> wrote:

I realize it's an old conversation, but...

What's involved in building a simple 50 MHz oscilloscope? I have a spare
Xilinx Spartan DSP board, and an unused LTC 105 MSps A/D chip in the part
box. The board's 125 MHz clock is different enough from the sampling rate, I
think, to jitter the trigger point for meaningful averaging. It seems almost
too simple. Aside from a low noise amp, what am I missing? The board has a
VGA port, 128 MB 5ns DDR2, and tons of I/O.

A "simple" oscilloscope is easy, real easy, as you suspect. Buffer/amp
+ ADC + memory + basic sample control + PC interface + a few lines of
code to display samples.

Making it more useful on the other hand is much more work.
You need wide range input attenuators and amps plus triggering and
protection stuff. Lots of switching control stuff involved too. The
control side gets more complicated with pre/post triggering, buffering
and other niceties.

A decent DSO project involves a lot of software, and this is where
many of the cheap Chinese PC based DSO's fall down.

BTW, 50MHz analog bandwidth is going to need a lot more than 105MS/s
to be useful in single shot mode.

Been there, done that quit a few times.

Dave.

 

[Roger_the_Dodger]

I would have just bought my own software.

There is plenty of free stuff and paid for packages.

There is often something on ebay for a few pounds.

 

[JeffM]

"Roger_the_Dodger" <cresswellavenue@ talktalk.net> wrote:
There is often something on ebay for a few pounds.

Joel Koltner wrote:
Nothing that's at the same level of quality/utility as, e.g., Altium,
Pulsonix, PADS, or even Eagle is "a few pounds."

I thought I recognized that rap. Yup, I called it.
http://groups.google.com/groups/search?enc_author=_U-3sxwAAACljgN-V29TdGsdWVTrgV-GXiYp9-orUyGBN00G0sVMjw&scoring=d

Heee's baaack. (Marra, aka cadman, aka Nigel Wright, aka...)
http://groups.google.com/group/sci.electronics.cad/browse_frm/thread/7eb8dfaa19ed5709/a74114ef2d864b4a?q=his-ethics+zzz+Nigel.Wright+ugly.schematics+KNOW-YOUR-VENDOR+god.awful