Chip with simple program for Toy

[Andrew Holme]

Danny T wrote:

Ban wrote:

hey Danny,
there are many possibilities constructing H-bridges, but they
narrow down
with your low voltages. If you have a 4V supply and your motor is
drawing 1A
average, or 3A peak, you can best use a relay, maybe of the
bistable kind
like Nais TQ2-L2-5V. This would allow the full voltage swing and
you could
PWM the motor with an inexpensive NMOS. A bistable relay uses
energy only
when changing states and also stays latched on power off. I costs
4bucks
single quantity.

But then I won't understand the differences between bipolar,
darlington,
FETs etc.

Darlingtons are not applicable as they use at least 1V each, so for
your
motor there is only 2V left. :-(

In the circuit of Fig 3.14, there's also a diode drop across the
1N5408.

My power supply can be increased - though if it means they'd flatten
my
batteries quicker, I'm open to other methods!

But you can use bipolar transistors, like the FZT788 PNP for the
power stage
and a FCX688 for the NPN. Those can be driven by a small (20mA)
current from
a PIC or via another buffer transistor. These transistors have
around
2X250mV= 0.5V saturation voltage(at 2A), so this would leave 3.5V
across the
motor.

e.g.
http://www.armory.com/~rstevew/Public/Motors/H-Bridges/HBridge_NPN-PNP.gif

I recommend this circuit: it'll happily run at lower voltages; you can
drive it directly from PIC outputs; you could substitute different
(suitably beefy) back e.m.f diodes, if you can't get the ones shown.

With your low voltages I would not recommend FETs, they need higher
gate
voltages and do not perform well with 4V.

The gate voltage is 4-5V (from my PIC) and the N-channels I've got
seems
to work great. What's the difference (in basic terms!) between a
MOSFET
(& JFET for that matter), bipolar and darlington? I'm sure they're
covered in the book I'm reading (The Art of Electronics), but I want
to
order as many of my bits as I can asap, so I can fiddle with things
and
try things out!

To answer in context:

1. Darlingtons are 2 bipolar transistors. They require less base drive
current than one bipolar, but drop more voltage.
2. MOSFETs are not suitable for low-voltage H-bridges for reasons
already stated.
3. JFETs are small signal (not power) devices.
4. Complementary bipolar (NPN+PNP) is the way to go.

I can't find anything marked as bipolar on Rapid - can you see
anything
you think would be good from here:


http://www.rapidelectronics.co.uk/rkmain.asp?PAGEID=20671&CTL_CAT_CODE=30397

You want a "high power silicon NPN transistor" and a "high power
silicon PNP transistor." They have TIP31/32. They are bipolar
transistor.

Look up "darlington", "bipolar" (etc) in the AoE index.

 

[jws63]

Hello. I'm having the same problem. Anybody has any sugestions?

 

[Rich Grise]

On Fri, 14 Jan 2005 18:10:53 +0000, urjant wrote:

On Fri, 14 Jan 2005 17:52:02 GMT, Rich Grise <richgrise@example.net
wrote:


sorry i guess i should have stated that i bought is off EBAy a couple
months back so i'm not ezpecting returns as an option and would like
to develop a plan to use what i have if possible.

Then I guess it would do to figure out exactly what you _do_ have. Get
that done, and compare that to what you want done, and then we can work
from there.

Good Luck!
Rich

 

[Ban]

etaoin999@yahoo.com wrote:

I am not an Electrical Engineer. All I know about the topic is:
a) If a lamp socket says "60W Bulb Max", then a higher wattage bulb
would risk overheating.
b) Power(Watts)=Volts*Amps

Why don't packagers of computer peripheral devices list their power
requirements?? Isn't it in their interest to protect their customers
from premature hardware failure from using their product improperly??

I wanted to add USB2.0 capability to my (elderly) Dell notebook so I
could use a portable disk drive for backup and overflow storage.
I found two products from Apricorn that seemed to fill my capacity and
form-factor needs:
1) USB2.0 Cardbus adapter (EZUSB2CB) is self-powered by the TypeII
card socket, and provides two powered USB ports: 5V,100mA and
5V,500mA. 2) Disk Drive (EZBUS40) needs 5 Volts, but the only other
power specs are: "Bus powered or auxiliary power if needed";
auxiliary power cables are provided for USB port or PS2 port power.

Now I have six possible combinations for cabling the devices, and no
guidance on what is right or wrong!! I was forced to experiment, and
my results are shown below.

Dell PS2-port power specs: 5V * 100mA = .05W
Cardbus power specs: 5V * 100mA = .05W 5V * 500mA = .25W

Cardbus Socket the Disk is connected to:
100mA........500mA........Auxiliary Power source
===============================================
nothing......nothing......None
OK??.........OK...........100mA PS2 socket
.............Boot Failed..100mA Cardbus socket
Boot Failed...............500mA Cardbus socket

"nothing".....Drive did not power-up
"Boot Failed".Drive power light was flashing, followed by "Blue Screen
of Death" in a USB boot subroutine
"OK"..........Drive booted normally and functioned normally.
"OK??"........Drive booted normally and functioned normally, even
though it probably shouldn't??

WOULD SOMEONE PLEASE ADVISE ME ON THE CORRECTNESS OF THESE
CONCLUSIONS??
1. Power supply by USB2CB cardbus device is only .25W total, across
BOTH ports - not .30W as I had assumed.
2. Power draw by EZBUS40 disk drive is more than .25W, which caused
the USB boot failure.
3. I think I risk PS2 socket failure from a sustained "over-draw"
situation, if I continue with PS2 auxiliary power and the disk drive
connected to the 100mA USB port.

Best is if you go to the hardware store where you bought the items and have
them installed by the sales person. I suspect a driver problem or a problem
adressing USB2. You will eventually need WindowsXP and maybe an external
wall wart.
--
ciao Ban
Bordighera, Italy

 

[j.b. miller]

Google is your friend..

several companies have high price 'convertors', plug and go types...
or
you can do it yourself for about $5, all you need is a small
microcomputer( I use the now 'obsolete' PIC16C84 ) and some time. Easy to
program, TONS of info on the web about it, just need to ask Google.

Seems to be a 'classic' project for colleges,etc.

 

[CBarn24050]

Subject: LCD controlling with comparators
From: Danny T danny@nospam.oops

PIC for each digit, 8 input pins, and 7 output pins (7 segments per
digit).

Sounds like your driving your lcd with dc, it wont last long.

What's the best (least components etc.) to do this in "the real world"?

An lcd controller chip, find 1 to mach your display.

 

[Danny T]

CBarn24050 wrote:

PIC for each digit, 8 input pins, and 7 output pins (7 segments per
digit).

Sounds like your driving your lcd with dc, it wont last long.

Whoops?

What's the best (least components etc.) to do this in "the real world"?

An lcd controller chip, find 1 to mach your display.

heh. I meant without using something like that - let's say I'd invented
my own 13 segment display for a petrol gauge on my car or something!
--
Danny

 

[John Fields]

On Sat, 15 Jan 2005 18:52:37 +0000, Danny T <danny@nospam.oops> wrote:

Anthony Fremont wrote:

What kind of LCD do you have? You may need to be careful with how you
drive the digits.

57-0160 from rapidelec.co.uk

It seems to be available in a 4-pin option, but I ordered the "31/2"
assuming it meant 21 pins on each side, but it only has 40 (no idea what
the "4" flava is then!)

CBarn24050 just suggested I shouldn't be driving it from DC - is this
really a big problem? It seems to be working fine, my clock has two
digits! Other two can wait till after tea - I've already wired about 60
pins together across 4 PICs...

http://dantup.me.uk/tmp/wires.jpg


There's a bunch of ways this can be done, but yours wins the Rube
Goldberg award. ;-) You should multiplex the digits from just one pic.
You can use PORTB to connect to all the segments in parallel and then
turn on the anode (or cathode whichever it is) from PORTA pins. You put
the segments for the first digit on PORTB and then drive PORTA, 0 low
(or high) to illuminate the first digit. You then turn off the PORTA,0
pin and put the data values for the second digit onto PORTB, then turn
on PORTA, 1. Cycle thru PORTA, 2 and PORTA, 3 then back to PORTA, 0.

If you setup say a timer interrupt to occur every 10mS or so, you can
switch digits in the ISR and leave them till the next interrupt 10mS
later. That would update the entire display every 40mS or 25 times per
second. Your eyes will never know the difference and it will only need
11 i/o pins.

This sounds like a fantastic idea *but* of the 40 pins on my display,
the only ones marked on the datasheet are one for every segment/dp on
the screen, plus two "com" (one on each side). I assumed it meant
"common" and grounded it, while connecting the other pins to pic
outputs, and they drive fine. I assume your method isn't useful for this
particular LCD?

---
You have what's called a "static" display, and the pins marked "com"
are connected to the backplane, which is the transparent electrode
which forms one side of the capacitor comprising each segment and the
backplane. MOST static LCD displays aren't designed to be driven by
DC, and what will happen is that the ITO will plate out of either the
backplane or the segment and migrate to the opposite electrode, with
the eventual result that the capacitor will be destroyed and the digit
will become unreadable. The proper way to drive static LCDs is with
square wave AC; the backplane and the segment being driven in phase
when the segment isn't supposed to be displayed, and out of phase when
it is, like this for the segment to be off:

__ __ __ __ __ __
SEGOFF__| |__| |__| |__| |__| |__|

__ __ __ __ __ __
BP __| |__| |__| |__| |__| |__|


and like this for it to be on:

__ __ __ __ __ __
SEGON |__| |__| |__| |__| |__| |__

__ __ __ __ __ __
BP __| |__| |__| |__| |__| |__|


This task _can_ be accomplished with a ľC and a longish shift register
By EXORring the backplane and the segments at about 30Hz., but it's
often done with a device like a National MM5483 so all you have to do
is shift data into it and strobe the output registers without any
extra software overhead.

http://cache.national.com/ds/MM/MM5483.pdf

If you've got a 3-1/2 digit seven-segment display, that's 23 segments,
and the 5483 can drive 32 segments and a backplane, so that leaves you
with nine extra segment drivers for decimal points, annunciators, or
the colon if you're doing a clock. Best of all, Digi-Key's got them
in 40-pin DIP packages for about five bucks.

--
John Fields

 

[Jonathan Kirwan]

On 15 Jan 2005 20:18:23 -0800, "tuxtlequino" <tuxtlequino@yahoo.com> wrote:

If E=IR can we just then increase the resistance and get as many volts
as we want?

If you already know the I and you already know the R, then it allows you to
compute the voltage which must be present across the two terminals in order to
make that particular current through that particular resistance.

However, it is the more usual case in simple circuits that you have a given
voltage and you either want to calculate the expected current, given some
particular resistance... or else you want to calculate the desired resistor to
use in order to achieve some given current value.

Your next question gets to this...

The schematics in the book explains the number of volts (for example a
9 volt battery), but they never mention the amps in the battery, how
can we then figure out how much resistance do we need?

The battery supplies whatever current is required by the circuit. Most circuits
will specify the desired voltage (say, 9 volts) and will have some particular
effective resistance (not usually specified, though.) When you connect up the
circuit, the battery presents a particular voltage to the circuit and the
current simply is what it is, based on what the circuit requires.

If you hook up a 1k Ohm resistor to your 9V battery, 9 milliamps will flow. If
you hook up a 10k Ohm resistor to your 9V battery, 0.9 milliamps will flow (or
900 microamps, as you prefer to imagine it.) The amount of current flowing
depends on the "load" resistance. But the voltage at the terminals remains the
same, with a battery -- 9 volts, in this case.

Batteries actually prefer it if the circuits require less current. They last
longer and they provide the correct voltage for longer, as well. If the battery
were an "ideal battery" it would provide 9 volts no matter what current is, but
real batteries are limited.

A more useful specification for a battery, other than its voltage, is the total
amount of energy it holds. Batteries don't really "hold amps" in them, they
hold potential energy. As the energy is consumed, by way of the voltage and
amps required over time, they gradually expire. While that takes place, their
ability to maintain the specified voltage degrades somewhat and their ability to
supply higher levels of current suffers. But it is the circuit that determines
the "amps" that the battery needs to supply it. Not something inside the
battery.

Still, batteries *do* have limitations in current. A car battery has a very
high current rating and is able to supply very high currents into very tiny
resistances, while still maintaining their voltage. A small hearing aid battery
(a so-called button battery), on the other hand, may not be able to supply more
than a milliamp or two and still hold their proper voltage, too. Something will
suffer, if the circuit demands more than some tiny current limitation for those
button batteries.

The voltage rating may be the same, but bigger batteries are usually able to
handle more current for longer times, because they store more energy and because
they are designed for higher current requirements. For example, a D-cell has
the ability to supply much higher current than a AAA-cell does, while still
holding a proper 1.5 volts. If the circuit requires too much for the AAA, for
example, the AAA voltage itself may suffer *and* the battery may also warm up
and the total lifetime of use will also suffer. In the same situation, the D
cell may be perfectly fine supplying the higher current, not getting warm at
all, and holding its proper voltage the entire time.

If I get a LED that says in the package 2 Volts, and .005amp. I
understand that I need to divide 7/.005 to figure out how much
resistance I need. But why .005? Where did it came from, why, what is
the resistance of the LED?

The LED "requires" about 2V to light up. With .005 amps (5 milliamps), the LED
will be reasonably visible. Since your 9V battery has too much voltage for the
LED to operate properly, you need to "throw away" part of that. To do so, you
take the difference of what you have in the battery and what you need at the LED
to compute the part to throw away. This is the 7 volts, as you said. Now,
since you also want about 5 milliamps in the LED at the same time, and since the
resistor is to be in series with the LED, whatever current the LED requires must
come through the resistor. So the current through the resistor must be equal to
the desired current through the LED. (Otherwise, the charge would have to "pile
up" somewhere.) If you compute your resistor to be 7V/.005, or 1400 Ohms, and
use that value then it will be true that the voltage taken away by the resistor
(opposing the battery voltage) will be 7 volts *if* it turns out that 5
milliamps occurs. But if the LED takes a little less than 2 volts then the
actual current through the resistor will be a little more and if the LED takes a
little more than 2 volts, then the actual current will be a little less.

Now, if I have a system of 9 volts (I bought an electric kit with a
schematic explaning how it works!), and then there is a resistor for
4.7MOhms, what is my amperage

If the resistor is hooked directly across the two battery terminals, then the
current through the resistor will be 9V/4.7Meg or slightly less than 2
microamps.

Would I use the same formula(e=ir) to figure out?

Yes, after some algebra to solve for I.

Does the use of capacitors change my amperage (I have seen them in my circuit!)?

It makes your current time-dependent. That's too complex for you to worry
about, just yet.

Jon

 

[Dr Engelbert Buxbaum]

Bob Masta wrote:

Hi, everyone.

For years I've heard vague stories about people receiving radio
broadcasts through their dental work. The stories have all the
characteristics of a classic urban legend, and I'm pretty sure
that's exactly what they are.

Hmm, I could imagine the amalgam to act as a crystal detector, like in
the early radios in the 1920's. But what would act as ear phone, and
what as resonance circuit?

 

[Danny T]

Anthony Fremont wrote:

57-0160 from rapidelec.co.uk

I found it, but no datasheet.

Datasheet came with it (cos it's not on the web):

http://dantup.me.uk/tmp/datasheet.gif

The table in the bottom right is blown up here:

http://dantup.me.uk/tmp/pins.gif

Matrix type LCDs have to be scanned, the pixels can never be allowed to
see any DC voltage or they will cook. Your display does not appear to
be in this category. Your display is a very fixed purpose display.
Your display will only display 3 1/2 digits, that means the first column
can't show all the numbers 0 - 9, only 0-1 or 0-3 depending on the
display.

It'll only do 1 on the first character. I thought maybe it was good for
clocks (as I'm doing now), but there's no "decent" am/pm indicator, and
it can't do 24hr, so maybe it's even more useless!

http://dantup.me.uk/tmp/wires.jpg

You should see the BASIC-52 computer I made with scads of jumpers for
the external latches and ram for the 8052. Kinda neat about 3.5" square
and runs on a 9V battery, just hook up to the serial port, push the
reset button, hit the space bar (or is it return) and it's 1977 again.
;-)

heh, at least it does something - this just lights up a few segments,
and it's got more wire in it than my entire apartment!!

There's a bunch of ways this can be done, but yours wins the Rube
Goldberg award. ;-)

This appears to be the case. If you don't have access to the cathodes
of each individual segment, then you can't multiplex it in the way I
described. You should be using a Hitachi type LCD display anyway, or 7
segment LEDs. ;-)

So, given the LCD, and the available hardware, would you agree that my
"Rube Goldberg" clock, is indeed, the best way I could've done it? ;P

The 5 PICs I've used... 4 are just simple registers (cos I don't have
any) like this:

Start
btfsc PORTB, 5 ; If RB5 is high
movf PORTB, W ; Copy input!
movwf PORTA ; Move to output
GOTO Start ; Start again

END

And the "driver" I haven't done yet. I know in the code above, if RB5
goes low between the first two instructions, it might not work properly,
it should copy PORTB into a register, then work on it from there -
however, ther's *no way* I'm getting those chips out of those
breadboards with all those wires in (around 80 wires so far, and the
driver chip isn't there), so it can stay for now ;-)


--
Danny

 

[Danny T]

Robert Monsen wrote:

I would devote a shift register to each digit, just to simplify the
setup. Having a table mapping binary data to 7 segment codes simplifies
matters.

This is essentially what I did, I've made 4 PICs control the 4 digits (I
don't have any shift registers, and couldn't find any on maplin or Rapid
- am I searching for the wrong thing?!)

Finally, you can easily get your timing data off of the AC line, if your
clock is meant to be plugged in. It is a nice way to go, since it varies
only a small amount, so your clock is more accurate. If you use the
internal oscillator, it can easily vary by a few percent in either
direction, which will cause your clock to run fast or slow. You can
compensate for this a bit, but temperature also affects the oscillator
in the pic, so the only really good solution is an external oscillator
or crystal.

I considered this, but it was really just something to keep me busy
until wheels and op-amps come for my robot. It'll be taken apart soon -
I just wanted to check my understanding of assembly programming a little
more, and wanted something to do with the bits I have (Even if you can
buy bits to do this stuff already!)

That said... Only my driving chip would need the crystal, and I do have
2x4Mhz that I bought a while back (I didn't realise they had internal
ones). Since my driving chip has spare pins (the only one that does!!),
I might hook it up to that. What's the best way to time something like
this - the only delays I've done so far are done by decrementing a
register and skipping if 0 - I'm sure there's a nicer way to do it?

Ta


--
Danny

 

[Danny T]

John Fields wrote:

You have what's called a "static" display, and the pins marked "com"
are connected to the backplane, which is the transparent electrode
which forms one side of the capacitor comprising each segment and the
backplane. MOST static LCD displays aren't designed to be driven by
DC, and what will happen is that the ITO will plate out of either the
backplane or the segment and migrate to the opposite electrode, with
the eventual result that the capacitor will be destroyed and the digit
will become unreadable. The proper way to drive static LCDs is with
square wave AC; the backplane and the segment being driven in phase
when the segment isn't supposed to be displayed, and out of phase when
it is, like this for the segment to be off:

__ __ __ __ __ __
SEGOFF__| |__| |__| |__| |__| |__|

__ __ __ __ __ __
BP __| |__| |__| |__| |__| |__|


and like this for it to be on:

__ __ __ __ __ __
SEGON |__| |__| |__| |__| |__| |__

__ __ __ __ __ __
BP __| |__| |__| |__| |__| |__|


This task _can_ be accomplished with a ľC and a longish shift register
By EXORring the backplane and the segments at about 30Hz., but it's
often done with a device like a National MM5483 so all you have to do
is shift data into it and strobe the output registers without any
extra software overhead.

http://cache.national.com/ds/MM/MM5483.pdf

If you've got a 3-1/2 digit seven-segment display, that's 23 segments,
and the 5483 can drive 32 segments and a backplane, so that leaves you
with nine extra segment drivers for decimal points, annunciators, or
the colon if you're doing a clock. Best of all, Digi-Key's got them
in 40-pin DIP packages for about five bucks.

I see (oops)... What I don't understand though, if "com" is for the
backplane - what about ground? :-\

--
Danny

 

[Danny T]

K1G1 wrote:

I've often noticed on some circuit boards that there is a type of black
epoxy that seems to be covering ICs. Is this black epoxy put in place to
hide the IC being used to protect the design or is it a type of cheap
integrated circuit?

I saw the same inside a Dongle issued to stop software copy protection,
it consisted of just a chip with the legs connected to the parallel
pins.. I assumed it was to hide the type of chip, but I don't know for
sure

--
Danny

 

[Anthony Fremont]

"Danny T" <danny@nospam.oops> wrote in message
news:41ea4b9d$0$14606$ed2619ec@ptn-nntp-reader01.plus.net...

Anthony Fremont wrote:

Bummer, that's it huh? Unfortunately, that doesn't tell us anything
important about how to drive the display other than the pin-out.
Maybe
you shouldn't purchase any more surplus LCDs from them. Hitachi
based
displays are the common display for tinkering. You can get them in
anything from a 1x8 to a 4 * 80. They are usually as cheap as the
one
you have, but have much more utility.

Yeah - I didn't buy it for any purpose really, was just interested in
seeing how it worked - something extra for debugging output and was
fairly cheap


Well if you had to make it out of only the parts you used, then ok.
However, if you had a couple of shift registers on hand, you could
have
used only one PIC.

That's why I programmed 4 pics to work as I understood shift registers
would (I didn't consider serial-in), it's the main driving pic that'll
be the clever stuff
(if I get a clock working quickly, I might add time set / alarm
buttons,
like most alarm clocks )

I'm bad about not finishing my projects all the way. I breadboard them
up, tinker with them till they work and then get bored and move on to
something else. I'm going to solder up my clock listening project since
I actually have some use for that. ;-)

Out of curiosity, why did you chose PORTA to drive the segments and
PORTB to read data in?

For parallel data, I needed 7 outputs (7 segments) and 8 inputs (7
segments plus clock/latch (what's the right word?))

On your slave PICs you might consider it to be a "chip enable" pin, or
even a "read".

My programmer doesn't like turning MCLR off (internal), so that left
me
with 7 I/O pins on one port. I used the same pin on the other port as
the clock/latch, so I could just copy (other 7 bits would be in the
right place). Copying over MCLR doesn't matter since it's ignored.

I didn't have enough room on my breadboard, so I had to fudge it
slightly... I put two chips upside down, with the bottom left 4 pins
in
the same rows as the top right 4 of another pic. Then I wired those 4
inputs up the other way, since all intputs are driven from the same 7
outputs of the driving pic, with just the clock (RB5) being wired
seperately.

The reason I ask is that the RA4 pin is an open collector when used as
output. This means it needs a pullup resistor when used as an output
pin. Therefore, you'll see allot of it being used as an input pin
whenever possible. It's also useful for talking to Dallas 1-wire
devices. ;-)

And the "driver" I haven't done yet. I know in the code above, if
RB5
goes low between the first two instructions, it might not work
properly,

Why do you say that?

How about this:

Start
btfss PORTB, 5 ; If RB5 is low
goto Start ; then no data is available
movf PORTB, W ; Otherwise, copy input!
movwf PORTA ; Move to output
GOTO Start ; Start again
END

Thinking about it, if I clock with two instructions, eg:

; Set output pins for char 1
; Clock digit 1
; Clock digit 1 off
; Set output pins for char 2
; Clock digit 2
; Clock digit 2 off

But my worry was, if after your first insutrction (which was RB5 is
high), the driving pic sends RB5 low and then gears the outputs for
the
next character. By the time you get to the next instruction (moving
PORTB to W) I might've changed the data. It's what we like to call
"random failure"!

Since it may take two instructions to change the output, it's not a
problem, *BUT* since the 4 "shift registers" are running on internal
clocks, I don't know how accurate they are, and the driving pic could
potentially clock RB5 and change the 7 output pins in between - ?

You have to give time for the slave PIC to "see" the data. If your
slave PIC has a worst case timing of say 10 cycles to recognize its
"enable" pin and retreive the data, then the master PIC has to allow at
least that much time every time it presents a digit. Then you wont have
random failures.

In cases like this, it's ok to just check the port pin without
making a
shadow copy of the whole register. It's output pins that you need
to be
concerned about the read-modify-write issues.

What I don't like, is the reading a value, satisfying a condition,
then
using a "new" copy of the data, which might not satisfy the criteria.
It's assuming that the data doesn't change between instructions,
which,
given the data is provided externally, is entirely possible!

I'm not sure what you mean by "new" copy of the data.

 

[Michael Bohlender]

Hi there,

"K1G1" <none225@someone.com> wrote in
news:W1oGd.8806$IV5.4480@attbi_s54:

I've often noticed on some circuit boards that there is a type of
black epoxy that seems to be covering ICs. Is this black epoxy put in
place to hide the IC being used to protect the design or is it a type
of cheap integrated circuit?

Thanks

most of the time this epoxy is used to replace the IC case!

This technology is known as "Chip on Board" where the silicon chip is glued
directly to the board and then bonded to the PCB connection pads just as
it would be done inside the normal plastic case.

The black cover is put over the chip and its bonding wires after testing
the device for correct function.

Examples for that are:
- LCD displays with integrated controller
- dongles that consist only of one chip
- cheap clocks and watches ...

Bye
Michael

 

[Danny T]

Michael Bohlender wrote:

most of the time this epoxy is used to replace the IC case!

This technology is known as "Chip on Board" where the silicon chip is glued
directly to the board and then bonded to the PCB connection pads just as
it would be done inside the normal plastic case.

Is that done because they're cheapskates, or another reason?

--
Danny

 

[Danny T]

Danny T wrote:

(if I get a clock working quickly, I might add time set / alarm buttons,
like most alarm clocks )

And this:

http://www.totalrobots.com/access_files/speech.htm

A talking clock!! ;-)

--
Danny

 

[Danny T]

Anthony Fremont wrote:

When an interrupt occurs, the GIE flag is cleared. This prevents the
scenario you described. When the ISR returns, it uses a retfie
instruction which sets the GIE bit again so that another interrupt can
occur. If you forget to clear the flag for the interrupt that you just
processed, then you'll be processing it again as soon as you execute the
retfie. ;-) This is a common mistake for newbies.

Well, I've gotten stuck already... Posted in alt.microsomething.8bit (TB
usefully says a.m.8bit) if you'd care to look :-(

--
Danny

 

[peterken]

"Danny T" <danny@nospam.oops> wrote in message
news:41ea6e18$0$47684$ed2619ec@ptn-nntp-reader02.plus.net...

Michael Bohlender wrote:

most of the time this epoxy is used to replace the IC case!

This technology is known as "Chip on Board" where the silicon chip is
glued
directly to the board and then bonded to the PCB connection pads just as
it would be done inside the normal plastic case.

Is that done because they're cheapskates, or another reason?

--
Danny

yes, it's cheaper
it also takes less space
the epoxy is to protect the chip as well as the bonding wires