Chip with simple program for Toy

[BobGardner]

Computer architecture books..... 'How to master Autocad in 21days'..... '3D
Studio Max for Dummies'.... oh... you didnt mean archtecture using computers?
Never mind....

 

[joji john]

Current is a flow rate, so where's the current going? I'd expect the
current Ic to flow through the transistor, through the emitter and on
to gnd (conventional current flow)? I thought that Ic = Ib*Hfe
(roughly)? Shouldn't Ie then = Ib+Ic? Therefore if I put a resistor
between emitter and gnd, I should get varying (& amplified) voltage
across it because of ohms law?
What am I failing to understand?

To explain this you will have to dig deep into the construction of the
semiconductor transistor. Lets refer to the npn transistor.
You have connected something like this:
Vcc (+5V) to collector; input to base; and emitter
to a resistor and you are tapping the voltage across the resistor wrt
ground? Correct me if I'm wrong.
The transistor may be simplified and thought of as 2 junctions on n-p
junction (Collector-Emitter) and another p-n junction (Base-Collector).
Junctions here may be akin to diodes for sake oof simplicity without loss of
detail.

Now when the base is at 0 V and the collector tied to 5V whic makes
p-side=0V and n=5V which reverse biases the collector base junction(diode)
which results in just minimal (reverse) current injected into the emitter
side from collector. The base -collector has no potential dufference.

Now when the voltage at the base is pumped up to 1.5 V, still the CB
junction is reverse biased but the threshold voltage for the base-emmiter
voltage is exceeded (roughly 0.7-1V) and the considerable forward current
flows to the collector from the base. Remember CB junction is still reverse
biased and minimal (microamps) of current os sourced from collector side.
Now after the 0.7-1V drop across the BE junction the volatage at the
resistor is hardly a fraction of the input (no amplification) and shape of
the output is smilar t input ; thus "follwing the input".

THink of the transistor as two switches if you still find it difficult to
follow.
I hope you can follow .
Joji

 

[CFoley1064]

Subject: need help with quick and dirty triac triggering please.
From: "Kevin" Some1Not@microsoft.com
Date: 10/16/2004 3:20 PM Central Daylight Time
Message-id: <cks009$3m0$1@newsg3.svr.pol.co.uk


Hi, wondering if some kind sole can help me out.

problem is timer module on washing machine is none working with regard to
motor drive, I think the tda1085c chip is dead, machine is about 5 years old
and new timer is apparently about a 100GBP so ordered new machine, trouble
is one we want will take 10 days min so I figure I ought to be able to
manually control motor speed and get some use in the mean time.

It's quite some time since I did any meddling with electronics, I still have
quite a lot of caps and resistors though I seem to have forgotten most of
how to use them!

Anyway what I'm attempting is to put an RxC network into the old triac gate
feed instead of from the triac controller chip. I've managed to find a 0.1uf
capacitor of a type I beleive to be suitable for 250 vac,( I'm in uk so 250
vac mains), also I think I can get a 120vac feed from the large dropper
resistor that's on the board, I've got a 50k duel pot and lots of other
fixed resistors of 0.25 maybe some 0.5 watt.

The triac is marked btb15-600bce, I assume that means it's a 15 amp 600volt
but I havn't been able to get a datasheet and don't know what the trigger
values are, I also don't have any diacs so would like to drive it direct
from my RxC network, it's only a temporary measure for a few uses.

If someone can tell me whether it's feasable without the diac and perhaps
the resistance values to use with the 0.1uf cap, I'll be isolating the pot
body inside the machine out of the way with just the plastic shaft
pertruding. I also expect to have to adjust the wash speed for each
different load but it will be easier than trips to the launderette and
cheaper, so if anyone can help it would be great.

thanks in advance
kevin.

Hi, Kevin. It's a little more complicated than you envision. It isn't worth
the hassle, especially if you're not sure what you're doing. And you are
talking about potentially lethal voltage in the presence of a tub of sudsy
water and a big grounded chassis.

Try the laundromat. You may spend a few more pounds, but you don't want the
excitement you may be missing as a result.

Sorry
Chris

 

[John Popelish]

Jack// ani wrote:

John Popelish <jpopelish@rica.net> wrote in message news:<417299D8.2A952AC1@rica.net>...

If the load is primarily capacitive, all the rectifier current is
confined to the brief periods when the transformer voltage is higher
than what is stored in the capacitor, so the RMS current can get quite
a bit higher than the average current. This affects the current
rating required in the transformer.

Just two more questions, John

1. Reactance offered by a capacitive load is always higher that a
resistive load, then why should it extract more current?

Reactance is a linear concept that applies to sine waves. To get your
mind around a capacitive input rectifier filter, you have to think in
the more general differential description of capacitance. I=C*(dv/dt)
current equals the capacitance times the time rate of change of
voltage across the capacitor.

As long as the capacitor voltage is equal to or greater than the
transformer voltage, the rectifier isolates the two. But the moment
the transformer wave rises above the capacitor voltage, the rectifier
is essentially a short circuit, and the voltage on the capacitor must
rise as fast as the transformer wave is rising, regardless of how much
current that takes. So the current into a capacitive filter is narrow
sort of half sine wave pulses that occur on the part of the
transformer voltage wave just before the peak voltage. Since the
transformer windings are heated by the RMS current, a pulse waveform
like this has a much higher RMS value than the average of the current
in those pulses. This is what the 'squared' part of the RMS does. It
is not unusual to have to double (or more) the transformer RMS current
rating relative to the DC output average current when using a
capacitor input rectifier filter to take care of this higher winding
RMS current. The exact ratio depends on how much leakage reactance
there is between primary and secondary windings that tends to spread
out the charging pulses, by sagging the waveform a bit while the cap
is charging, lowering the slope a bit. If you look at the transformer
waveform with a scope, you can see the flattened spot on the wave
where the cap is charging just before peak voltage.

2. Why do capacitor charges to peak ac voltage if placed across a
bridge rectifier? Without a cap i measured 11v, and it increased to
16v, after hooking a cap?

The diodes act as check valves, pumping the cap up all the way to the
transformer waveform peak voltage and then turning off, leaving that
voltage trapped in the capacitor. A resistor load on the rectifier
keeps it on the whole waveform so the resistor voltage is the same as
the transformer waveform (except for the inversion of one half). your
meter reads the average of the rectified transformer waveform instead
of the peak value.

Thanks for your help

--
John Popelish

 

[petrus bitbyter]

<mark.mcgee@csfb.com> schreef in bericht
news:1098099096.177919.30680@z14g2000cwz.googlegroups.com...

Hi

I'll check out the quad comparators, thanks for the idea. Never used
them before.

Back to transistors though, because I really should get to understand
these things. I obviously need to understand the first principals. So
if you can't amplify voltage at the emitter, that infers that the
current also isn't amplified at that point?

Current is a flow rate, so where's the current going? I'd expect the
current Ic to flow through the transistor, through the emitter and on
to gnd (conventional current flow)? I thought that Ic = Ib*Hfe
(roughly)? Shouldn't Ie then = Ib+Ic? Therefore if I put a resistor
between emitter and gnd, I should get varying (& amplified) voltage
across it because of ohms law?
What am I failing to understand?

Cheers,
Mark

Guess you fail to know some facts about transistors. That Ic=Ib*Hfe is a
good one to remember but it is only valid under some conditions. One is that
the base-emitter diode should be forward biased. A forward biased silicium
diode requires some 0.6V across it. Another is that the transistor should
not be saturated. Even within this conditions Hfe is not contstant over the
whole Ic range.

If you build an emitterfollower, with Vcc =5V and an input voltage of 0V the
base-emitter diode is not forward biased. There will be neither Ib nor Ic
except from some leakage.

If you raise Vin to 1.5V some Ib will flow generating an Ic and both will
flow to the emitter. Out of the emitter it mets the resistor and will build
some voltage across it. This voltage can be calculated easily. A 1.5V input
minus a 0.6V forward biased diode gives 0.9V left over for the resistor.
This voltage is - within reasonable limits of course - independent of the
resistor. If you lower the resistor, Ib will increase and Ic will raise
accordingly until the forward biased base-emitter diode is about 0.6V and
again there is 0.9V left for the resistor.

You can also raise Vin to let's say 2.5V. Again. the currents increase,
keeping Vbe at about 0.6V and left 1.9 for the resistor. So the voltage on
the emitter follows the voltage on the base minus that 0.6V. Which explains
the name emitterfollower.

If you really want to do more with electronics I advise to get some book.
Horowitz for example wrote a good and wellknown one.

petrus bitbyter


---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.778 / Virus Database: 525 - Release Date: 15-10-2004

 

[Bob Masta]

On 16 Oct 2004 13:11:28 -0700, rcam001@hotmail.com (rob) wrote:

Hi...I need to determing how much electricity a solar water heater is
replacing. The solar heater preheats the water before it enters the
electric heater...I need a cheap solution to measuring the electrical
usage over a months time period....I am thinking an hours meter
somehow wired might work....Any suggestions?????

Thanks Rob

This should be pretty simple since in theory you could
just put an electric (not electronic) clock across the element.
But the element runs on 220 in most installations, so
you'd need a transformer or something. And in most
electric water heaters there are actually 2 separate
elements, one at the bottom that does most of the
heavy lifting, and one at the top that kicks in if the
cold-water line in the tank reaches that high under
heavy use. So you'd need two separate clocks and
transformers.

In my own case many years ago, I wanted to see the
effect of super-insulating the tank. I just hooked
little NE2 neon bulbs across the elements, each with a
CdS photocell next to it and wrapped in electrical
tape to block stray light. Used a chart recorder I
happened to have on hand to monitor activity.
If you have some sort of recording device, this
approach may work for you as well.

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

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

 

[Roger Johansson]

"JeffM" <jeffm_@email.com> wrote:

I Googled the site and it looked like a normal HTML site
http://www.google.com/search?&q=site:futurlec.com+printed-circuit
so I went to the site again.
Guess what--it's not a Flash site any more.

I have disabled flash in my browser so I didn't even try before.
But I tried it now and it is nice, low prices and low freight costs.
All kinds of components and instruments.

Quoting from the site:
..........
"SHIPPING METHODS
For Australian, Canada, UK, EU and USA customers,
we can offer domestic postage rates as follows,

USA, Canada, UK, European Union(EU) and Australia

Standard Post
For orders up to US$30, Delivery Charge - US$3.00.
For orders US$30 to US$49, Delivery Charge - US$5.00
For order US$50 to US$99, Delivery Charge - US$8.00
For order US$100 and above, Delivery Charge - US$12.00

Express Post
For orders up to US$100, Delivery Charge - US$15.00.
For order US$100 and above, Delivery Charge - US$30.00"
..........

Those freight costs are basically the same as when I order from the
companies in my own country, so it is not a problem.


--
Roger J.

 

[Terry Pinnell]

myothermailishotmail@gmail.com (mike) wrote:

Is there such a thing as a LDR that increases resistance with
increasing light?

Or, is there a simple circuit that will invert the resistance of a
common LDR?

Disclaimer: I have very little experience with electronic/circuits.
Please be gentile.

Any help is appreciated. Thanks!
m

What is the objective? If you want to control a circuit using an LDR,
you would normally arrange the LDR with a normal resistor (or pot or
preset) as a voltage divider pair. You'd then use the output voltage
from that as input to a subsequent (analog or digital) circuit
section. So, simply reversing the position of the LDR and the other
resistor would reverse the operation of the circuit.

As an example, see the first section of my garden lamp circuit here:
http://www.terrypin.dial.pipex.com/Images/GardenLamp.gif

In the unlikely event that I wanted my lights to go on at dawn rather
than dusk, I could achieve that by swapping the positions of the LDR
and the resistors.


--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[John Popelish]

mike wrote:

Is there such a thing as a LDR that increases resistance with
increasing light?

I don't know of any.

Or, is there a simple circuit that will invert the resistance of a
common LDR?

This is fairly weary in some cases.

Disclaimer: I have very little experience with electronic/circuits.
Please be gentile.

Any help is appreciated. Thanks!
m

Tell us all the details you can think of about what you are trying to
accomplish. This will save us a lot of useless questions.

--
John Popelish

 

[Rich Webb]

On 19 Oct 2004 03:26:02 -0700, "mark.mcgee@csfb.com"
<mark.mcgee@csfb.com> wrote:

Thanks for the suggestions. I was put off by the cost of the tools,
but the actual items are cheap enough to have a play with, thanks.

You can (and should) use a proper crimper in a production environment.
For hobby or "just learning" pretty much anything that works, works.
You'll probably "explode" a few IDC connectors -- at least I've HEARD
that can happen, yeah that's it, just heard about it... ;-)

--
Rich Webb Norfolk, VA

 

[Tim Auton]

"Rene" <nospam@nospam.nospam> wrote:

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?

The light bulb and the radio will put out a fraction of their energy
as light or sound respectively, so it's not all heat. However, it
doesn't take long before that light and sound will hit something, not
bounce off and be converted into heat.


Tim
--
Copyright, patents and trademarks are government-granted,
time-limited monopolies. Intellectual property does not exist.

 

[fengjianqing]

Why you can see so many websites have flash. Because the servicer tell to
the owner of the website-- the flash is beautiful and wonderful for
others' first eye to see you website. And from this feather the servicer
company can earn more.

For the beginning business with China, I also 100% agree you should take
care about the quality, the price, the time... But if you do not to try
you do not know, you will lose chance. More and more company purchase
directly from China. Why? It is not only because the lower price, also
because you pay the lower money you can get the same quality products or
the higher quality. This is true!
So at first give yourself chance to try. You will earn from China.

Regards,
Feng

 

[Jonathan Kirwan]

On Mon, 18 Oct 2004 18:28:48 GMT, I wrote:

To invert the inverter:

5V
|
|
\ 5V
/ R2 |
\ |
| |<e
+----| Q2
| |\c
|/c |
in >----| Q1 +-----> out
|>e |
| \
\ / R3
/ R1 \
\ |
| |
gnd gnd

There are actually several different arrangements.

I thought I'd expand by adding some of those possibilities and then asking the
experts for their analysis. I prefer the above arrangement because it appears
to do better on power and on speed. But here are some other arrangements (and
why not add more if you want to...):

One interesting one is something like this:

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

In this case, the signal remains non-inverted and it uses a single transistor.
However, the output doesn't quite make it to ground. More like about 1.1V or
so. I kind of look at this as 'yanking the chain' on Q1, as Q1 is nailed down
to the 1.5V and the input "pulls" a current, when low.

It's fairly fast. But what makes it faster than other choices below?

In any case, this was the basis for the faster circuit I'd mentioned on the
previous post -- with the right values, the circuit can produce just the right
range of pulses for driving a 'high-side' transistor. And if you look, you'll
see this little piece in the three transistor circuit in the previous post.

Two other circuits, poorer I think than the two transistor one I offered up
earlier, follow. This next one is the worse of the two:

5V
|
|
\ 5V
/ R2 |
\ |
| R4 |<e
+--/\/\--| Q2
| |\c
R1 |/c |
in >--/\/\--| Q1 +-----> out
|>e |
| \
| / R3
gnd \
|
|
gnd

Very slow.

Here's a faster one, almost as fast as the early one I mentioned:

5V
|
|
\ 5V
/ R2 |
\ |
| |<e
+--------| Q2
| |\c
\ |
/ R4 +-----> out
\ |
| \
R1 |/c / R3
in >--/\/\--| Q1 \
|>e |
| |
| gnd
gnd

Would any of the resident experts care to discuss these variations on a theme
(or the ones in my previous post) and offer others, perhaps? I think it could
be useful to see how one approaches thinking about these.

Jon

 

[burbeck]

hi wong,
kirchoffs law, try
http://math.fullerton.edu/mathews/n2003/KirchoffMod.html

regards
On 19 Oct 2004 22:54:39 -0700, tatto0_2000@yahoo.com (Wong) wrote:

What theorem or method can be applied to calculate the total
resistance at point A to B?

----- A -----
| |
| |
R1 R2
| |
| |
------R3------
| |
| |
R4 R5
| |
| |
----- B -----

Thanks.

 

[Steve]

<mark.mcgee@csfb.com> wrote in message
news:1098102991.141487.152470@f14g2000cwb.googlegroups.com...

Hi

Is there available a PCB header pin set, with matching socket which can
be assembled by an amateur without expensive specialist tools?
Regards,
Mark

For exactly that, programming PICs in-circuit, we put a simple IDC header on
the board and used a ready assembled cable with female connectors at both
ends to connect temporarily to the programmer (actually via a male-male
convertor, but you might do it another way). In the UK, for instance,
assembled cables go for about 1GBP in one offs from Farnell, e.g.:
http://uk.farnell.com/jsp/endecaSearch/partDetail.jsp?SKU=3525909&N=401

Regards,
Steve

 

[Tim Auton]

"Rene" <nospam@nospam.nospam> wrote:

The light bulb and the radio will put out a fraction of their energy
as light or sound respectively, so it's not all heat. However, it
doesn't take long before that light and sound will hit something, not
bounce off and be converted into heat.

So, does this mean that at the end, they will all be generating the same
amount of heat?

Yes. If you put the devices in light-proof, sound-proof boxes then you
would measure that all the boxes would be heated by the same 100W.
It'll generally all get converted to heat wherever the light and sound
goes, but it's tricky to measure if some of the light gets all the way
to Alpha Centauri!

Very strictly speaking some may not end up as heat in certain
circumstances - sound or light may end up using some energy breaking
chemical bonds for example, but for a radio and light bulb that's
insignificant. Even for devices designed to break bonds (like the
ultrasound thingies used to smash gallstones) it's a tiny fraction
that doesn't end up as heat very quickly.


Tim
--
Copyright, patents and trademarks are government-granted,
time-limited monopolies. Intellectual property does not exist.

 

[rayjking]

Mike,

If you add a cad photocell in series with a 10k resistor that is connected
to a low current ( 10ma or more ) 10/12 volt dc power supply the signal you
wish will appear across the 10k resistor. The better circuit would
substitute the 10k for a 100k pot and a series 470 ohm ( safety resistor )
to allow adjustments.

Ray



"mike" <myothermailishotmail@gmail.com> wrote in message
news:f6382c10.0410190935.5f72e77e@posting.google.com...

Is there such a thing as a LDR that increases resistance with
increasing light?

Or, is there a simple circuit that will invert the resistance of a
common LDR?

Disclaimer: I have very little experience with electronic/circuits.
Please be gentile.

Any help is appreciated. Thanks!
m

 

[rayjking]

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.

Ray



"Sean" <stuffduff@cox.net> wrote in message
news:3222fe00.0410200643.700cfe17@posting.google.com...

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.

Thanks a 10E6!

 

[rayjking]

You also need to add a resistor in series with the transistor base to
protect the halls and connect a clamp circuit across the motor coils to keep
the transistors from failing.
The motor rpm is limited by the rate of current build up and should have the
same rate of current decay or faster. What ever the battery voltage is the
clamp should be higher. one way is to add a clamp diode across each motor
coil with a zener diode in series with the clamp diode that is at least 30%
higher than the battery voltage. The transistor voltage will see the battery
voltage + 30% + the battery voltage.

Fets are best.

Ray






"smpaladin" <smpaladin@yahoo.com> wrote in message
news:6753235c.0410171702.de6d3c6@posting.google.com...

This is a depiction of a circuit that I am planning to make.
http://www.geocities.com/smpaladin/maglevcircuit.gif

It is a linear motor; Whenever a magnet is passed over the hall
sensor, it turns on the Power transistor, and therefore turns on the
Electromagnet coil. In the picture, there are only four sets of the
hall sensor/transistor/coil. In the real circuit, I am planning on
having 12 sets of them.

So will this circuit work properly? And how much power will I need?
You would think that the circuit would need a lot of power, but then
you realize that only one set of the hallsensor/transistor/coil will
be turned on at any one time.

Any advice is greatly appreciated.

 

[John Larkin]

On 20 Oct 2004 07:43:31 -0700, stuffduff@cox.net (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.

Thanks a 10E6!

There's really no reason not to. If they all have circuit breakers,
which most do, the upstream guys will trip if they're overloaded by
the total load. If you put the lighter loads towards the end of the
string, voltage drop along the total length of power cord will be
minimized. As far as length goes, it's not much different from a long
extension cord.

John