Chip with simple program for Toy

[Bob Eldred]

"smokie" <stillbreezy@gmail.com> wrote in message
news:1112337309.835937.306160@z14g2000cwz.googlegroups.com...

what happens if you short the output on an inverting amplifier to its
inverting input?

Gain from the input resistor becomes zero. But, it turns into a
non-inverting, unity gain amp relative to the plus input of the amp.
Inverting gain = Rf/Ri = 0/Ri = 0. Non-inverting gain = Rf/Ri + 1 = 0/Ri + 1
= +1.
Bob

 

[Andrew Holme]

Thomas wrote:

Is there any way you could assist in the selection of the correct
components or point me in the right direction as to the right RC
Integrator? Thanks for the input!

The time constant depends on the product R*C. Many different value pairs
are possible; however, you might start with (say) 100k and 10uF. Look at
the signal on an oscilloscope and experiment with the values. Large values
for R will minimise loading of the input signal. You want the output to be
a steady DC level with minimal ripple. Increasing R*C will reduce ripple.
By minimal ripple, I mean relatively small (say <5%) compared to the level
changes produced by moving the gear selector. This enables the comparator
to distinguish between ripple and gear shifts.

 

[Active8]

On Fri, 01 Apr 2005 08:27:58 GMT, Lord Garth wrote:

"smokie" <stillbreezy@gmail.com> wrote in message
news:1112337309.835937.306160@z14g2000cwz.googlegroups.com...
what happens if you short the output on an inverting amplifier to its
inverting input?


Here's a place that will explain it to you:

http://www.williamson-labs.com/480_opam.htm

It always amazes me... the little things you can learn from the most
basic books and articles. Like... no one ever told me god created
PNP transitors. Is this more persecution, Dr. Laura, and how should
I smite them?

Seriously, I have a real basic book which wouldn't have taught me
much except that it has a chapter on components. Different types of
Rs, Cs, etc. Or there's my physics book that's short on words and
big on equations contrasted with an EE series (9 volumes) which
covers 1st year physics with more explanatory words. It just adds a
bit.
--
Best Regards,
Mike

 

[Rich Grise]

On Thu, 31 Mar 2005 22:35:09 -0800, smokie wrote:

what happens if you short the output on an inverting amplifier to its
inverting input?

You get a zero-gain inverter. I saw one of these in use once, and
my engineer was baffled by it. "What the heck is a zero-gain inverter
for?" and you could almost see the question marks around his head
like in the cartoons.

It provides a virtual ground. I believe that the reason for doing
it that way has something to do with temperature compensation or
input offset current of the downstream devices.

Or, if you lift the non-inverting input, it's a voltage follower.

Cheers!
Rich

 

[John Fields]

On 1 Apr 2005 04:43:35 -0800, "Thomas" <cordless89@hotmail.com> wrote:

Is there any way you could assist in the selection of the correct
components or point me in the right direction as to the right RC
Integrator? Thanks for the input!

---
I've posted an easy two-chip solution for you on abse under the same
subject as this thread.

--
John Fields
Professional Circuit Designer

 

[John Popelish]

JeffM wrote:

http://my.execpc.com/~endlr/index.html

The last item on the page
is a comment by the site owner about his employment status.
It contains this link:
http://my.execpc.com/~endlr/will_design20_.jpg

Oh crap!

 

[Lord Garth]

"SklettTheNewb" <SteveKlett@gmail.com> wrote in message
news:1112116734.773746.194200@l41g2000cwc.googlegroups.com...

I would like to calculate how long an LED will stay lit given a
specific power source. I need to know what the smallest battery I can
use in a project and still get n hours of operation.

For example:
if I have a typical LED and a 1.4v battery 130mah (hearing aid
battery), how long will it stay lit to the point where you can't
visually determine it's dimming..

What is the formula?

Also, does a flashing LED consume a non-linear amount of power? In
other words, if I had an led that was on 50% of the time, could I
double the duration value from the above said formula?

As you can tell, I don't really know anything. I just have an idea for
something and am trying to determine how feasible it is...

The answer will depend upon how much current you pass through the LED
and the voltage drop associated with that particular LED.

Pulsing the LED will save the battery but don't forget that the flasher
circuit
consumes power as well so a 50% duty cycle won't double the battery life.
Battery life would be greatly extended if you had a very short on time and a
long interval between repetitions. There are advertising buttons packaged
into products as attention getters. I would guess that many get thrown into
the trash.

 

[Andrew Holme]

smokie wrote:

Ri = 1K Rf = 10K and a 910 ohm from the non-inverting input to
ground.

how would the output be afffected if i swapped the 910 ohm with a 10K
resistor?

Depends on the input bias current - a (very) small current which flows out
of op-amp inputs. For a JFET-input op-amp, with a bias current of 200pA, it
would shift the ground reference level by (10e3-910) * 200e-12 = 1.8
micro-volts ! I leave it for the student to calculate the effect of this on
output voltage (of a x10 amplifier) ...

 

[Bob Masta]

On 1 Apr 2005 13:50:45 -0800, "gtslabs" <gts@nb.net> wrote:

I want to measure the distance a glass hydrometer bulb drops over time
in a solution of suspended solids. This distance is only about 4
inches. Since I can not modify the hydrometer bulb dramatically I
thought I could add a drop of metallic paint on the top tip. Then I
want to measure the distance from a fixed point to the top of the
bulb/metallic paint. My resolution needs to be about 1 mm. I am
looking to collect the data with some type of acquisition system.

What is the best type of sensor that could measure the vertical
distance assuming the bulb might float form side to side about 1 inch
max?

Thanks

I wonder if you could constrain the hydrometer to stay on-axis
and use an LVDT (Linear Variable Differential Transformer).
This is a standard way of measuring linear displacements,
consisting of an iron core that moves in a tube with a winding
on each end. An AC excitation signal is coupled between
the windings, with the percent of coupling proportional to
the core displacement.

In your application, you'd have to allow for the weight of
the core affecting the hydrometer depth, but this should
be pretty straightforward. This isn't as elegant as the
massless optical approach, but has the advantage that
you might be able to do it with an off-the-shelf sensor.

Best regards,




Bob Masta
dqatechATdaqartaDOTcom

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

 

[Ban]

smokie wrote:

Ri = 1K Rf = 10K and a 910 ohm from the non-inverting input to
ground.

how would the output be afffected if i swapped the 910 ohm with a 10K
resistor?

There would be a change in offset voltage by offset current x 9090.
There would be also an increase in noise voltage by around 3-5dB, depending
on the current- and voltage noise figures of the opamp.
The lowest noise would be achieved with the +input grounded.
--
ciao Ban
Bordighera, Italy

 

[Colin Dawson]

"SklettTheNewb" <SteveKlett@gmail.com> wrote in message
news:1112281089.235621.104250@o13g2000cwo.googlegroups.com...

I have been messing with CircuiMaker and doing "spice simulations"
I have also been reading from these different sites:
http://www.kpsec.freeuk.com/ohmslaw.htm
http://www.play-hookey.com/digital/ripple_counter.html
http://www.allaboutcircuits.com/vol_1/chpt_2/index.html

I'm slowly getting my head around the current and voltage ideas. I
have encountered a strange situation w/ a spice simulation like this:


LED 680
.--|<-----|___|-----.
| |
| |9V
| ---
| -
| |
|-------------------'
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

if I place the probe before the 680 ohm resistor, I get 9v, if I place
it between the resistor and the LED I get 1.79V and 10.60mA

I'm trying to understand how I got that number, I know that;
V
I = -
R

if I use that formula with my values:
9
- = 0.013
680

- that makes no sense.
I'm missing something VERY basic and simple, but I can't figure it out.


Any help or guidance would be REALLY appreciated at this point.
Thanks,
Steve

Maybe I can help a little. I'm using Livewire, to simulate the circuit.
Let's add a few points to the circuit to help make things a little easier to
explain...


Probe B Probe A
| |
| |
LED | 680 |
.--|<-----|___|-----.
| |
| |9V
| ---
| -
| |
|-------------------'
|
|
Probe C


The current for this circuit is 10.44mA.
The Voltate at the power source is 9v exactly (as I'm using a circuit
simulation this won't change unlike real life, but let not complicate things
with that)

The measured voltages are as follows

Probe A is 9V
Probe B is 1.9V
Probe C is 0V

The difference in voltage between Probe A and Probe B is 7.1V Remember
there's still 1.9V left to drive the LED

So from that

9V/680R = 0.01044A

This is in Amps. My Simulation shows the current in Milliamps so I need to
multiply my figure by 1000

0.01044A*1000 = 10.44mA


So what's the resistance for the LED?

V/I=R

The measured voltage for the LED is 1.9V (Probe B) so

1.9V/0.01044A = 181.992R (182R will suffice)

Add that to the 680R and lets do the entire circuit.

9V/(680R+182R) = 9V/862R
= 0.01044A
= 10.44mA


I hope this helps a little.

Regards

Colin Dawson
www.cjdawson.com

 

[jsmith]

Possibly you may not need the electrolytic capacitor since your B+ is a
battery??

"kell" <kellrobinson@billburg.com> wrote in message
news:1111729615.820255.39480@g14g2000cwa.googlegroups.com...

I was experimenting driving various mosfets and igbts with a NE555 at a
low freq around one per sec turning a small bulb on and off just to
check how well the 555 works before I try to build the actual circuit I
have in mind, which will have an inductor operating at a few KHz in a
topology similar to a boost converter. (The inductor is intended to
reach a peak current of several amps.) I used no gate resistor, just
connected pin 3 directly to the gate.
Anyway, when I tried the IRF740 and a IGBT designed for car ignitions
(HGTP14N...), which both come in the TO-220 case, the bulb turned on
and off cleanly. But when I tried a couple of devices that come in the
big TO-247 case, the bulb didn't turn off cleanly. With the IRG4PC30F
IGBT, the bulb dimmed before it turned off. With the STW18NB40 mosfet,
the bulb didn't even turn off, it just flickered a little.
I used a sealed lead acid battery and connected the Vcc pin of the 555
through a 75 ohm resistor to B+. I had a 100uF electrolytic across the
power pins.
I was powering the 555 through a resistor because that is the way I
intend to use it in the final circuit, as a way of protecting it from
the inductive spiking.
What is it about these big mosfets and IGBTs, that a 555 can't turn
them off?

 

[Roger Johansson]

Andrew Howard wrote:

Could someone please explain how a signal can be transmitted on a
single wire? I have always been taught that a signal, or current,
needs a circuit to work.

Is the second "wire" a path between you and the negative via your
legs, the ground, the table, then the casing or something?

Yes. And there is a capacity in big bodies, a buffer of charges, both
positive and negative which usually are in balance, and it works as a
kind of fixed point voltage-wise.

If we think of a short piece of water pipe, not connected to anything
at the ends, just air outside. A small amonut of water inside the pipe,
and a pump in the middle of the pipe.

Then if we start the pump it will pump the water in one direction,
but after a very short time there is no more water to pump. If the
amount of water from beginning in the pipe is only a nanoliter and the
pump pumps 1 liter per second the water will be gone in one nanosecond.

If the water from the beginning is a million liters, if the pipe is
connected to a swimmingpool, the pump can pump water for a million
seconds in one direction before the water stops coming.

The capacity is very different in these cases, but there is a certain
capacity
for delivering current in both cases.
The capacity determines how strong AC currents we can create if there
is no DC connection which completes the circuit.

Even if you body does not touch the ground, and you touch the probe to
an oscilloscope you can see a strong signal on the scope. That is
because your body has a certain capacity, measured in nanoFarad, and
that is enough to deliver AC current, as the AC current goes forth and
back. And because magnetic noise induces a voltage in your body.

If you touch the probe with a much smaller object, like a 1mm piece of
copper wire, you will not see much difference on the scope screen.
Because that little object cannot deliver much as much AC current as
your body, or the earth.

Another example, which probably has a different explanation, is
antennas on radio transmitters. How do they transmit anything, when
there is only a single wire?

The higher the frequency the easier it is to use these small
capacitances of the neutral parts of the antenna as virtual ground
planes.

At lower radio frequencies the earth itself forms the neutral part of
the antenna.

The antenna converts electric signals into electromagnetic waves which
can travel through space.

Well, the signal is actually an electromagnetic wave already when it
travels through the antenna wire. The antenna is just a bridge between
electromagnetic waves in conductors and electromagnetic waves in space.



--
Roger J.

 

[Don Bruder]

In article <424ffb65$0$43995$14726298@news.sunsite.dk>,
"Roger Johansson" <no-email@no.invalid> wrote:

Well, the signal is actually an electromagnetic wave already when it
travels through the antenna wire. The antenna is just a bridge between
electromagnetic waves in conductors and electromagnetic waves in space.

Sounds like a variation on the explanation a Ham operator once gave me -
Boiled down from a conversation that lasted about half an hour, I came
away with the understanding that a transmitter and receiver operate in a
similar-in-result, although very different in actual mechanism, way to a
physically huge, ultra-tiny-rating capacitor, with each set's antenna
acting as one plate of the cap, and the "universe between them" as the
dielectric. It makes sense on one level, but on others, it kinda falls
apart. I've still never managed to wrap my head around the details of
"why" radioo works, even though I like to think I have a reasonable
grasp on the "how" part.

--
Don Bruder - dakidd@sonic.net - New Email policy in effect as of Feb. 21, 2004.
Short form: I'm trashing EVERY E-mail that doesn't contain a password in the
subject unless it comes from a "whitelisted" (pre-approved by me) address.
See <http://www.sonic.net/~dakidd/main/contact.html> for full details.

 

[Jim Gregory]

Andrew
You are invoking at least two phenomena, open-circuit AC conduction and
electromagnetic wave radiation or radio (meaning wireless) transmission and
reception.
The earth plane has a lot to do with it in some cases. Don't forget the
neutral wire in a power line is earthed and that there are stray
inductances, capacitances and resistances, deemed as some of the losses, in
transmission. You are a huge capacitor and humidity plays some part. You
cannot touch something and send DC, well, only a static discharge.

In portable radio (covers radio, bluetooth, mobile phone, Tvs, etc) Txs and
Rxs, the chassis plane is unwittingly the 2nd wire.
Antennae and tuned capacitive circuits are responsible for conveying to/from
the frequency band/s concerned through the ether, in air or not in air.

"Andrew Howard" <ask.me@somewhere.net> wrote in message
news:3UR3e.22854$C7.7787@news-server.bigpond.net.au...

Could someone please explain how a signal can be transmitted on a
single wire? I have always been taught that a signal, or current, needs a
circuit to work. But then I see cases that when only one wire is used to
carry a signal.

eg. You touch the input on an audio amplifier, and you get a buzzing
noise, even when you only touch the positive wire.(the power supply used
doesn't have a grounding pin)

Is the second "wire" a path between you and the negative via your legs,
the ground, the table, then the casing or something? Or is it that there
is a path between negative and positive within the amp with some
resistance, which has a voltage drop, creatinge a potential difference
between the positive an negative? Am I even on the right track?

Another example, which probably has a different explanation, is
antennas on radio transmitters. How do they transmit anything, when there
is only a single wire?


Thanks,
Andrew Howard

 

[Roger Johansson]

Don Bruder wrote:

Sounds like a variation on the explanation a Ham operator once gave
me - Boiled down from a conversation that lasted about half an hour,
I came away with the understanding that a transmitter and receiver
operate in a similar-in-result, although very different in actual
mechanism, way to a physically huge, ultra-tiny-rating capacitor,
with each set's antenna acting as one plate of the cap, and the
"universe between them" as the dielectric. It makes sense on one
level, but on others, it kinda falls apart. I've still never managed
to wrap my head around the details of "why" radioo works, even though
I like to think I have a reasonable grasp on the "how" part.

If we think about photons we usually think of visible light photons,
like extremely small particles or small bundles of electromagnetic
waves.

It is more difficult to image the photons created by the 1440 kHz MW
radio transmitter in Luxembourg, (it is still exists).

It creates photons which are in the kilometer size range.

It may not help but rather make it all even more strange

But we have to piece together all these aspects of electromagnetic
waves into one mental concept.

Electromagnetic waves as mainly magnetism, as mainly electrical forces,
as photons.

It is like we are blind people trying to find out what an elephant is.
We touch different parts of the elephant and see different pictures,
but to understand the whole elephant we need to put all the pictures
together. And try to make some sense of it.



--
Roger J.

 

[petrus bitbyter]

"Andrew Howard" <ask.me@somewhere.net> schreef in bericht
news:3UR3e.22854$C7.7787@news-server.bigpond.net.au...

Could someone please explain how a signal can be transmitted on a
single wire? I have always been taught that a signal, or current, needs a
circuit to work. But then I see cases that when only one wire is used to
carry a signal.

eg. You touch the input on an audio amplifier, and you get a buzzing
noise, even when you only touch the positive wire.(the power supply used
doesn't have a grounding pin)

Is the second "wire" a path between you and the negative via your legs,
the ground, the table, then the casing or something? Or is it that there
is a path between negative and positive within the amp with some
resistance, which has a voltage drop, creatinge a potential difference
between the positive an negative? Am I even on the right track?

Another example, which probably has a different explanation, is
antennas on radio transmitters. How do they transmit anything, when there
is only a single wire?


Thanks,
Andrew Howard

Andrew,

Guess you do not mean an antenna. The so called single wire divices use one
wire for power and signal *and* a common ground. Long signal lines,
telegraph lines for instance, used old mother earth as a common. An example
of the Dallas (Maxim) single wire thermometers can be found on:
http://pdfserv.maxim-ic.com/en/an/app162.pdf

petrus bitbyter

 

[JeB]

On 3 Apr 2005 14:12:23 -0700, ymustuknow@gmail.com wrote:

I am wondering if anybody knows of any power source(s) that produces or
contains LOW VOLTAGE and HIGH AMPERES?

you need to give more specifics.

a computer power supply might do the job.

 

[mike]

ymustuknow@gmail.com wrote:

I am wondering if anybody knows of any power source(s) that produces or
contains LOW VOLTAGE and HIGH AMPERES?

TYVM in advance!!!

ymustuknow@gmail.com

ARC WELDER
mike

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[News Me]

ymustuknow@gmail.com wrote:

I am wondering if anybody knows of any power source(s) that produces or
contains LOW VOLTAGE and HIGH AMPERES?

Auto battery..?

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