Chip with simple program for Toy

[Jonathan Kirwan]

On 20 Oct 2004 06:23:23 -0700, kianmeng.tey@gmail.com (KM) wrote:

When the digital line is logic low at 0 volt the transistor turn on
and current flow via the resistor R2. (choose R1 & R2 to be loage
value eg. 47k not to stress the IC sink cabability ) and when the the
line turn to high the transistor turn off.

Thanks! After my first post on this (where I'd entirely forgotten to mention
the method), I decided to dance around your posted circuit in my second post but
to hold back from actually posting it until a third addition I wanted to make
today, wondering if someone would mention this exact alternative! Now, no need.

Some hobbyist thoughts:

A fair design might use a 220k in the base, planning a current gain of only 25
for reasonable saturation, ((5-.2)/47k)/((1.5-.6)/220k), and not much increase
in drive current over what's already needed. However, it's speed tops out at
about 50kHz or so, while still pulling 100uA drive current when low. My first
posted example design (using 1/2 sized R1 and R2 of what I posted) is also good
to about 50kHz, but by comparison pulls only tens of nanoamps of drive current.
Also, my second post which includes the circuit quite similar to yours, but with
the base resistor removed and an emitter resistor added, and designing it for
about the same 100uA drive current and similar output impedance yields a top
speed of about 500kHz -- 10 times better. The output swing is a little less, 4V
instead of 5V, but that's not enough to account for the difference. Why then?

Jon

 

[Lord Garth]

"Kelvin@!!!" <kelvin604@shaw.ca.ca> wrote in message
news:IWydd.780060$M95.106748@pd7tw1no...

Hi:
how dose bypass capacitor filter out the noise on the signal?
say i have a 74HC74 chip... with a bypass capacitor connecting the Vcc and
Gnd, i get a really nice wave form. but w/o one, i got noice all over the
place! can't even get a waveform on the scope...
it's like magic!
i know how a cap can filter out high freq. signal. but just can't figure
out
how a bypass cap work....

thank you for any answers...
--
{ Kelvin@!!! }

The noise spikes are high frequency. Does it help you to think of the cap
as a small
local power supply?

 

[Mortel]

"karan" <karan@iitk.ac.in> wrote in message
news:547d92f1.0410172312.54d6b724@posting.google.com...

Does there exist a software which outputs a circuit design for a given
input function and output function?

Thanks
Karan

Wow, this is a big question.

There's not a program but a language, called VHDL for Very High speed
Hardware Definition Language. Try googling on that you will find a lot of
information.

But VHDL is for logic systems wich are defined, synthetized and either
burned on an FPGA or sent to the foundry to make thousands of ASICs of it.

Have fun.

For analog design, you might find someting but I dont know it.

cheers

Mortel

 

[Stefan Arentz]

"Kelvin@!!!" <kelvin604@shaw.ca.ca> writes:

Hi:
how dose bypass capacitor filter out the noise on the signal?
say i have a 74HC74 chip... with a bypass capacitor connecting the Vcc and
Gnd, i get a really nice wave form. but w/o one, i got noice all over the
place! can't even get a waveform on the scope...
it's like magic!
i know how a cap can filter out high freq. signal. but just can't figure out
how a bypass cap work....

thank you for any answers...

I just found this with Google:

http://www.seattlerobotics.org/encoder/jun97/basics.html

Excellent piece of info!

S.

 

[rayjking]

Rene,

You are correct. If you measure watts to be equal then you are correct
except for the power factor which is ( Power = E x I x Cos of the angle
between the voltage and current ) ( theta ) which is different for loads
other than resistive. Just measuring the voltage and just measuring the
current ( E x I ) gives you apparent power which is always more than that
which is creating heat-watts.

Ray


..
"peterken" <peter273@hotmail.com> wrote in message
news:j3ydd.284682$FW5.14407816@phobos.telenet-ops.be...

yes, if they all converted exactly the same amount of energy into
something
else (movement, light, sound)
no, if they all convert different amounts of energy into something else

the bulb will convert an amount into light, the non-light-energy will be
turned mostly into heat
the heater will have losses in light-radiation (usually IR), the rest will
be dissipated as heat
the radio will convert energy into sound, the rest will be dissipated as
heat

so it all depends on the efficiency of every aparatus that will be
compared


"Rene" <nospam@nospam.nospam> wrote in message
news:UYidd.394645$mD.336927@attbi_s02...

Suppose that I have a light bulb consuming 100 watts, a radio also
consuming
100 watts and an electric heater also consuming 100 watts.

If I were to measure the heat generated by all these 3 devices, would I
get
the same amount of heat? In other words, would all devices consuming the
same wattage (solid state, no moving parts) generate the same amount of
heat?

Thank you.

 

[CFoley1064]

Subject: Re: LDR / photo resistor - increasing resistance with increasing
light???
From: myothermailishotmail@gmail.com (mike)
Date: 10/20/2004 1:50 AM Central Daylight Time
Message-id: <f6382c10.0410192250.7f7a9a34@posting.google.com

wow, thanks for the response. here is exactly what i want to do:

i live in a basement w/o windows, and i want to rig inside light so it
coincides with the outside light. i was figuring i could run an LDR
placed outside, to a dimmable ballast in my basement apt.

i have an electronic dimming ballast, made by the now defunct JRS
Technology (#C232120RS501). it has a dimming control by way of a 0 to
10v class 2 circuit, 0.5mA output. it is 0-10v out, and by connecting
a 0-100k ohm potentiometer to this circuit i can dim the connected
flourescent bulbs no problem-- the greater the resistance, the greater
the V, the greater the resulting brightness.

however, what i want to do is connect an LDR in such a way that a
greater amount light hitting the LDR makes the bulbs brighter. this
means that in order to get brighter bulbs, i need a greater voltage
accross this circuit, which means i need greater resistance connected
to this circuit (i think, right?)

i have an LDR (from a radioshack multi-pack) connected in series (with
a 0-100k trimmer/potentiometer) to this 0-10v circuit, and it does
what i want it to do, only in reverse. that is, as lighting
conditions on the LDR get brighter, the dimmer makes the bulbs get
dimmer.

this makes sense since an LDR decreases resistance as more light hits
it. what i need is increasing resistance with more light. or a
circuit that simulates this.

if this looks familiar, i posted 6 weeks ago before i had any idea
what i was talking about (thanks to those who offered suggestions):

http://groups.google.com/groups?selm=f6382c10.0409011320.6725a530%40posti
ng.google.com

Hi, Mike. I'm hearing you say you tried the RS photoresistor, but it operated
in a bass-ackwards fashion. You want resistance to increase as light
increases.

You might want to try something like this (view in fixed font or M$ notepad):

VCC
+
|
| VCC
.-. 1/2 LM358 +
| | VCC |
1K| | |\| |
'-' .---|-\ ___ |/
| | | >-|___|--|2N3904
.----o-----|---|+/ 100 |>
| | | |/| |
| | | GND | ___ 1 H11F1 6
~~.-. .-. '-----------------o-|___|---o--. .---o--->
~~| | | | 150 ohms | |
LDR| |1K| | V ~~ ||-+
'-' '-' - ~~ || To Ballast
| | 2 | ||-+
| | .---------o--' | 4
| | | '---o--->
| | |
V | |
- | ===
| | GND
=== ===
GND GND .------------.
| |
1N4001 | |\ |
'---|-\ |
| >-----'
.---|+/
| |/
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

There are a couple of components here you'll have to purchase, and I'm afraid
they're special order at RS. You can try mouser.com for better prices and good
service for the hobbyist. You also might want to pick up a 9VDC wall wart for
the Vcc in the diagram.

The two components you need are an LM358 (dual op amp) and a H11F1 (an opto
analog output FET). The distinguishing characteristic of the H11F1 is that, as
current through the LED increases, the resistance of the FET decreases. You
can use that to advantage with your CdS photoresistor by seting up a voltage
divider, and using the CdS to shunt the lower resistor. That voltage is picked
up by the op amp, and used to drive the LED.

Notice that, as ambient light increases, the voltage decreses, so the current
througfh the LED decreases, causing the resistance the ballast sees as
increasing. Also note that this assumes your CdS LDR has less than a couple
hundred ohms "ON" resistance. Since there's such a wide variety in the RS CdS
parts, you'll probably have to tweak this circuit for your part to make it
work.

I hope this is of value, even though it's a little on the complex side. If you
need more help, feel free to email. Please put "LDR" in the heading to avoid
having the spam filter dump you.

Good luck
Chris

 

[rayjking]

Peter,
With the rising oil/energy prices I may convert back to head phones in the
less than 5 watt range.

Ray





"peterken" <peter273@hotmail.com> wrote in message
news:RNBdd.284950$KB7.14624822@phobos.telenet-ops.be...

Ray

Correct, but as I read the question
- a light bulb and a heater are both resistive loads
- a radio is a more complex load

As I see the question I read "all these apparatus consuming the same
wattage..."
so I in fact read "do all these apparatus turn the applied (identical)
power
into heat"

Answer is no, given the fact all of them are designed for different
purposes
so some of the applied power is turned into something else as intended per
design, and the rest into heat

Proof for this can be established empirically

Or just take next situation :
I'd like to stand next to my 1kW heater on a cold day, it gets me warm
I wouldn't like to stand next to my 1kW amplifier on a cold day, it mostly
creates noise but gives me ampel heat




"rayjking" <rayjking@bellsouth.net> wrote in message
news:d5Bdd.38181$pi7.10280@bignews4.bellsouth.net...
Rene,

You are correct. If you measure watts to be equal then you are correct
except for the power factor which is ( Power = E x I x Cos of the angle
between the voltage and current ) ( theta ) which is different for loads
other than resistive. Just measuring the voltage and just measuring the
current ( E x I ) gives you apparent power which is always more than that
which is creating heat-watts.

Ray


.
"peterken" <peter273@hotmail.com> wrote in message
news:j3ydd.284682$FW5.14407816@phobos.telenet-ops.be...

yes, if they all converted exactly the same amount of energy into
something else (movement, light, sound)
no, if they all convert different amounts of energy into something else

the bulb will convert an amount into light, the non-light-energy will be
turned mostly into heat
the heater will have losses in light-radiation (usually IR), the rest
will
be dissipated as heat
the radio will convert energy into sound, the rest will be dissipated as
heat

so it all depends on the efficiency of every aparatus that will be
compared


"Rene" <nospam@nospam.nospam> wrote in message
news:UYidd.394645$mD.336927@attbi_s02...

Suppose that I have a light bulb consuming 100 watts, a radio also
consuming
100 watts and an electric heater also consuming 100 watts.

If I were to measure the heat generated by all these 3 devices, would I
get the same amount of heat? In other words, would all devices consuming
the same wattage (solid state, no moving parts) generate the same amount
of heat?

Thank you.

 

[Adam Aglionby]

Trimmed crosspost abit because server will think its spam...
Hi from sci.engr.lighting

"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in message
news:cl5mu4$g2e$1@uns-a.ucl.ac.uk...

Hello,

I recently saw an advert for some LED lights for filmmaking. They looked
perfect - very efficient, flicker-free, dimable from 0-100% etc. The
problem is that they're extortionately priced. So now I want to make my
own
LED lights for use on film...

Efficent ,er, um, see rhose MR16s your using at the moment, theyre probably
abit more efficient..

Has anyone tried this? What should I be careful of? Can I vary the
colour
temperature of the lights by pushing more or less current through the
LEDs?

Not really, whites because theri blue with a phosphor go a bit of angry blue
when you overdrive them though....

Even better - does anyone know of any LED lights suitable for film that I
could buy off the shelf here in the UK?

Sure I saw Kino Flo with some protype Luxeon LED film lights a while back,
but then they make fluro compact film lights which fro general light are
probably still a better bet.

Here's my dream LED light:

- dimmable from 0-100% with no change in colour temp (ultimately I'd like
to
build in a remote control so I can change dim the light whilst I'm looking
through the viewfinder on my camera)

Thats kind of doable , but current white LEDs are basic phosphor wise,with a
high colour temperature and not great Colour Rendering Index.
RGB colour mixing does not give a good white , some units have started
using RGB and Amber to warm it up a bit.

- cheap!

cheap or bright?
cannae have both

- highly efficient

If you don`t need alot of light, say an LED macro ring it is efficient for
the purpose, as replacement Redhead, not yet.

- stable and predictable colour temperature (it would be very cool if I
could change the colour temp with a switch... my research into LEDs so far
has hinted at the possibility of changing the colour temperature by
increased in the current).

Going cooler CT wise in white,
but cooking LED in practice,
lowering efficiency, LEDs hit sweet spot at exceedingly low currents
Negative temp co-efficient means output goes down as heat goes up
Lowering lumen maintenace , cooked phosphors and LED dice put out less light
as they age, hard life will age them faster

Please do let me know your thoughts - any leads you can give me will be
very
well received

http://www.ledmuseum.org/

http://members.misty.com/don/

HTH
Adam

Thanks,
Jack

 

[tempus fugit]

Basically, the cap at the Vcc of the chip diverts any noise spikes to
ground. Think of it as water flowing downhill instead of up. The cap
provides a low impedance path for any noise spikes to ground, so they go to
ground rather than into the chip.


"Kelvin@!!!" <kelvin604@shaw.ca.ca> wrote in message
news:IWydd.780060$M95.106748@pd7tw1no...

Hi:
how dose bypass capacitor filter out the noise on the signal?
say i have a 74HC74 chip... with a bypass capacitor connecting the Vcc and
Gnd, i get a really nice wave form. but w/o one, i got noice all over the
place! can't even get a waveform on the scope...
it's like magic!
i know how a cap can filter out high freq. signal. but just can't figure
out
how a bypass cap work....

thank you for any answers...
--
{ Kelvin@!!! }

 

[James Varga]

Okay - I've started trying to make some d/s boards and have run into
another
funny. It seem that the image after developing is 'smudged' - its not
actually smudged but it looks like that. It looks like the image is
missing
an area like it has been rubbed off or something.

This is as good as I've gotten it so far - as you can see there is still a
problem with the second side, although the one side is perfect. Each step is
done vertically so I'm really struggling to figure out whats going wrong.

http://www.jamesvarga.com/projects/beblu/pcbtest3.jpg

Any ideas?

 

[Michael Buchholz]

"James Varga" <james@jamesvarga.com> schrieb:

This is as good as I've gotten it so far - as you can see there is still a
problem with the second side, although the one side is perfect. Each step is
done vertically so I'm really struggling to figure out whats going wrong.

http://www.jamesvarga.com/projects/beblu/pcbtest3.jpg

Any ideas?

Not long enough exposed or Lamp not centered.
Maybe (if You use a 'face tanner') one of the Tubes emits fewer UV?

just my 2 cent.

Michael.

 

[Michael Buchholz]

"Nicholas O. Lindan" <see@sig.com> schrieb:

Side 2 looks like it was one (or more) of:

o under etched
o under developed resist
o bad artwork
o under exposed resist

My guess: under etched.

Sorry to disagree:
Underetched woud show blank copper where the Traces are
too big, but as You can see there ist violet resist all
over the traces.
If the resist would change colour at the faulty Traces,
i woud suggest underdevelopment, but i still think
there's something wrong with the exposing process..

Michael.

P.S.: If this resist don't change colour when being exposed,
it can also be underdevelopment (Board not lifted while
developing, so there was no NaOH at he bottom side?)

 

[Jonathan Kirwan]

On 21 Oct 2004 02:04:54 -0700, "mark.mcgee@csfb.com" <mark.mcgee@csfb.com>
wrote:

Jon your first non-inverting circuit with the pair of transistors
NPN&PNP - I get the same results as you, using a DC sweep on the input
from 0v to 1.5v in 0.1v steps. Although, the IQ2 (IIRC) was negative -
is that what you'd expect because of the PNP transistor?

Spice uses an entirely consistent point of view for currents. But it isn't
always obvious from looking at a schematic. In the case of one I occasionally
use, flipping a resistor's orientation inverts the reading when I click it. Yet
it looks the same on the schematic. I suspect this has nothing to do with the
way spice analyzes it, but everything to do with the way the resistor's pin
numbers are seen by the GUI interface and mapped back to the currents. I live
with it. To be honest, I don't usually worry about the sign of the current
because I usually know what it's supposed to be doing.

However, I can't get anything to happen with the later suggested
single transistor cct with the input at the emitter, and base held
permanently at 1.5v - I get the collector permanently sitting at 5v
across the whole input sweep. With or without a resistor on the
emitter and collector, both 47k as suggested.

This one?

5V
|
|
\
/ R2
\
|
1.5V +----> out
| |
| |/c
'----| Q1
|>e
|
|
\
/ R1
\
|
in >------'

If so, just plug in R2=10k and R1=2k, for one example. But I'm not sure what
you are doing, so it's hard to say. If you set both R1 and R2 at 47k, I think
the behavior should be that since you get only about 1V across R1 that you'd
only get about 1V across R2, so I wouldn't expect the collector to go lower than
4V. But I would expect something when the input is near 0V in your .dc
analysis. I assume you are specifying the input source for your .dc analysis,
right?

Your fast switching, but inverting with three transistors, feed back(?)
and capacitors totally lost me I'm afraid. I'm interested to learn
what's going on here if you have the time to explain.

It should wait until you understand the above circuit better, since that is one
third of the transistor sections in it. And keep in mind I'm only a hobbyist
and do not have all the better intuitions here. But I do think I can explain
the basic aspects and let others correct me where they feel the need, when you
feel you can follow the above circuit.

Speed wise, I don't think I need anything too fast - the clock pulse
duration is 1.25ms, I make that 800Hz, is that right? So nothing too
complicated/fast is required anyway.

That's VERY easy, then. In this case, KM's suggestion is excellent. Very few
parts and it works well. I'd try a 10k in the collector and a 56k in the base,
for example. I don't know if you have a scope, but if for some reason the low
voltage on the collector isn't quite as low as you'd like, you can reduce the
56k somewhat. But I think a beta of 30 should be fine for reducing the V(CE)
near enough to zero. I don't think going much below 30 will buy you much in
getting the lower voltage any closer to 0V.

I did find a couple of IC's yesterday for level shifting, but they are
14-pin and do additional stuff, which I don't really need. The IC's I
found were Texas Instruments CD40109B - CMOS Quad Low-to-high voltage
level shifter, and ON Semiconductor MC14504B Hex Level Shifter for TTL
to CMOS or CMOS to CMOS. The TI chip has enable pins in addition -
which I don't care about and can just fix enabled, but the ON looks
somehow simpler, but has this TTL/CMOS level switch configuration - I
think I need to configure it for TTL to CMOS for 1.5v to 5v level
shifting? Which do you prefer? Or can you suggest something simpler?

I've not used any of these (I'm a hobbyist, remember?), but I'm sure that folks
like Spehro or others will be able to tell you some useful information.

'simpler' to me would be using the BJT in KM's arrangement. Of course, with
lots of LEDs, maybe one of the ICs would be 'simpler.' There are also some very
fancy ones designed exactly for serial loading and driving up to some 36 LEDs
independently in a single (big) package. But I also hate TQFPs and SC70s and
their tiny little pins for surface mounting, when all I'm trying to do is just
make something fun in a one-off. I'd much prefer to just wire-wrap or solder.
So that would push me back to the BJT unless I could find a DIP IC that fits
nicely into a wire-wrap socket or else buy a cheap pc board adapter for
something I could solder well.

I'm a TO-18 kind of person, I guess.

Jon

 

[John Popelish]

Jack// ani wrote:

whoops, what an ugly mistake i did; feeling very shameful.
should be delta to star. really frustrating isn't it……should go and die somewhere.

If that is the biggest mistake you make this year, you are doing
better than most of us.

--
John Popelish

 

[Rich Grise]

On Wed, 20 Oct 2004 19:23:20 +0000, Kelvin@!!! wrote:

Hi:
how dose bypass capacitor filter out the noise on the signal?
say i have a 74HC74 chip... with a bypass capacitor connecting the Vcc and
Gnd, i get a really nice wave form. but w/o one, i got noice all over the
place! can't even get a waveform on the scope...
it's like magic!
i know how a cap can filter out high freq. signal. but just can't figure out
how a bypass cap work....

thank you for any answers...

Well, the gag straight answer is, a bypass wroks by filtering out the high
frequency signals from the power supply leads. ;-)

When a totem-pole-output chip switches, both output devices are on
simultaneously, briefly. During this short transition period (you'd
have to look up the spec - probably on the order of 1 ns), they
draw a heavy current pulse. The charge on the capacitor provides
charge to power that current pulse, and then recharges at its
leisure. So it filters out the high-frequency component of the
signal.

This is also known as a "decoupling" capacitor, which they generally
use instead of "bypass" when you're doing digital. Technically,
a "bypass" simply presents a low impedance to the signal at the
frequency in question - notably, to complete an RF circuit.

HTH!
Rich

 

[Rich Grise]

On Wed, 20 Oct 2004 08:25:31 -0700, Khalid KHALIFA wrote:

Hi all,
I'm new in electronics. i try to learn how to make simple dongles.
I need some documents to begin. I need especially documents about
progs to use and documents describing the life cycle of a
programme/circuit (creating driver, circuit and tools to use).

------------------------------------------------------------------------------
/\
/! \ I'm not looking for a cooked pie that you have made efforts to
learn how to
make it, but i need your help to reach the begining of the way.
------------------------------------------------------------------------------

best regards,
Khalid.

(Sorry for the bad english).

No problem, your English is excellent - better than many English-speakers!

See if any of these can help:

http://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&q=%2Bbeginning+%2Belectronics+%2Btutorial&btnG=Search

Have Fun!
Rich

 

[Rich Grise]

On Wed, 20 Oct 2004 07:43:31 -0700, Sean wrote:

Hi,

I've known this was 'taboo' for as long as I can remember. What I'd
like to know is why is it a bad idea to connect multiple power strips
(with surge suppression and/or line filters) in series; especially
when using computers.

I found several references which say not to do it, but no simple
practical explanation as to why.

The only way I can think to demonstrate it is to get a half a dozen of
them and connect them up in series, and stick meters at the wall,
between each, and at the end. I suspect that doing so will show some
form of degradation, but what I'm not sure what it will be or why it
will occur.

Any detailed explanation and or pointer to a web site where this is
explained would be appreciated.

The breaker in the first strip has to handle the whole load, so
when you add the 14th computer, the whole string goes down. ;-)

Cheers!
Rich

 

[Rich Grise]

On Wed, 20 Oct 2004 10:59:51 -0400, rayjking wrote:

Hi,

One reason is due to the fast rise of current in transients the inductance
of the added length of wire defeats the transprotection.
One inch of wire ( type used in the power strips ) is about 19nh. if you add
the 1000 amps possible and the length of wire at a high di/dt then many
volts can be generated and addition the phone leadin may have lower
impedance ( more current ) than the power leads due to the transmission line
effect of the telephone line.

Right. So be sure to pile up all of your electrical appliances next to
the wall outlets. ;-)

Cheers!
Rich

 

[Rich Grise]

On Wed, 20 Oct 2004 18:44:22 -0400, rayjking wrote:

Peter,
With the rising oil/energy prices I may convert back to head phones in the
less than 5 watt range.

With headphones, a few milliwatts are probably sufficient. It's amazing

how little actual sound power it takes to be really, really loud. One
watt of sound power will fill a room uncomfortably loud. Any amp. power
over that is nothing but "reserve." Well, and the inefficiencies, of
course, like speakers that need 100 watts just to get the cone up out
of bed. ;-)

Cheers!
Rich

 

[Rich Grise]

On Tue, 19 Oct 2004 23:50:20 -0700, mike wrote:

wow, thanks for the response. here is exactly what i want to do:

i live in a basement w/o windows, and i want to rig inside light so it
coincides with the outside light. i was figuring i could run an LDR
placed outside, to a dimmable ballast in my basement apt.

I seem to remember this from all those months ago, and it seems like
someone had suggested another LDR inside, with a comparator/difference
amp, to make the inside lights _really_ track the outside. Then, as
long as you get it scaled right, you won't have to worry about polarity,
or even nonlinearity.

Of course, on cloudy days, it could be a little dim. )-;

Have Fun!
Rich