Chip with simple program for Toy

Hi,

Why does my 9v D.C. transformer only give a max. amps. of 0.22A?.
I've looked inside it and it doesnt seen to have any resistor in
series to restrict the current.
A 6v. battery charger can give out 4 amps.,which proves its nothing
to do with the voltage of input or output.
But everything to do with losses its iron core and the copper
wire that it is wound with.


Cheers - Joe
 
hartly wrote:
Why does my 9v D.C. transformer only give a max. amps. of 0.22A?.
I've looked inside it and it doesnt seen to have any resistor in
series to restrict the current.
A 6v. battery charger can give out 4 amps.,which proves its nothing
to do with the voltage of input or output.
Thanks.
The output current has to pass through the resistance of the secondary
winding. This uses up voltage and produces heat. The output current
(transformed by the turns ratio), in addition to the magnetization
current has to pass through the primary winding. This also drops
voltage and produces heat. A transformer is rated for output current
based on acceptable total voltage drop (called regulation, expressed
as %) and temperature rise. If they build the transformer with a
larger core that has a larger window opening, they can use larger wire
that drops less voltage and produces less heat at a given current,
raising the current rating. It turns out that the rated power (output
voltage times rated current) is roughly proportional to the weight of
the transformer.

You can get more current out of any transformer, if you accept the
lower voltage, and either do this briefly enough that the transformer
doesn't overheat, destroying the insulation, or are willing to put the
fire out.
 
On 29 Apr 2005 08:00:08 -0700, hartlyuk@yahoo.com (hartly) wrote:

Why does my 9v D.C. transformer only give a max. amps. of 0.22A?.
I've looked inside it and it doesnt seen to have any resistor in
series to restrict the current.
A 6v. battery charger can give out 4 amps.,which proves its nothing
to do with the voltage of input or output.
Thanks.

Its price or its weight are pretty good indicators of how many watts
it can deliver.

John
 
kell wrote:
I have a cordless floor sweeper that's missing the wall wart power
supply it came with. Inside the sweeper there is a 7.2 volt nicad pack
with just a power resistor to limit current to the battery when
charging (very primitive). The resistor is something like an ohm, and
the size is a watt or two. I don't know what voltage the wall wart
was, but it was probably unregulated. The sticker on the sweeper says
that when new, and charging the (empty) battery for the first time, to
let it charge for 16 hours.
When experimenting I hook the sweeper up to wall warts that have 15 or
20 volts open circuit things get hot -- the resistor in the sweeper,
and the wall wart.
I have a 7808 8-volt regulator I could wire up to a 12 or 15 volt wall
wart. My question, is 8 volts enough for charging a 7.2 volt nicad
pack? A really slow charge is good, this thing will sit in the hall of
the building where I live and probably just stay on the charger all the
time. If 8 volts is too low, I can get whatever voltage I need by
putting a diode or diodes in the ground terminal of the regulator to
bump the voltage up, or use a voltage divider like on a 317... I'm
really asking what fixed voltage would be best, with a 1-ohm resistor
in series, to leave a 7.2 volt nicad pack on all the time and not
overheat; it's okay if it takes a whole day to charge.
Use your 15V wall wart and put a light bulb in series. Size the light
bulb to get the current you want. I expect 100mA is good depending on
the size of the batteries. 12V automobile lamps are good for this.
Don't GUESS, don't use the rated current of the light bulb, you won't be
running it at rated voltage, MEASURE the current in your circuit.
mike

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<Randy2747@msn.com> wrote in message
news:1114788597.535746.173330@z14g2000cwz.googlegroups.com...
okay I have one more question about this led circuit. Can I use this
sound activated switch to hook up to my leds?

http://www.electronickits.com/kit/complete/elec/k126.htm
Yes. The relay output is rated for 12 VDC, 10 A, and the current drawn by
your LEDs is much less than that. Just wire the normally-open relay contacts
in series with the power supply that you use to run your LEDs; it works just
like a switch.

Jonathan
http://cq.cx/
 
<ty78weyu@yahoo.com> wrote in message
news:1114812669.609522.125760@o13g2000cwo.googlegroups.com...
I have electronics hobby and just start learning how oscillator works,
like the loop gain must be >= 1 and the phase between the input and
output should be 0 degree. The problem is, when I see the actual (even
simple) oscillation circuit, I am at a loss how it exactly works.

"http://img13.imagevenue.com/loc70/a2a_osc123.jpg"

For example, in Fig.1, what's the use of resistor R2, from oscillation
point of view (not from transistor DC basing point of view)? I figure
that, since C3 is basically to short AC (i.e. RF in this case), it
seems to be equivalent to place one end of the crystal directly to the
emitter, as shown in Fig.2. Unfortunately, it does not oscillate this
way. Likewise, since C3 is to short the AC, then it should also be OK
to remove R2 and C3 completely, and place the crystal in a way shown in
Fig.3. However, it does not oscillate, either.

(1) So what is the function of R2 here?
(2) If it is a must for the oscillator to work, how to choose it?
(3) How to choose the value of C3?

Your help is appreciated!
the circuit in fig1 works by feeding back the signal at the emiter to the
base through the emiter base capacitance, a tune circuit is formed between
the crystal wich will be slightly inductive and the emiter capacitance in
series with c3 wich should be slightly higher than the load capacitance
specified for the crystal, the resultant signal at the base gives positive
feedback.

if you short out the emiter to ground you remove the signal, also the dc
curent would no longer be controled exept by the hfe.

Colin =^.^=
 
aman wrote:
I was looking at some PC-Based Logic Analyzers. I came across this
one:

http://www.pctestinstruments.com/

Has anybody of you used it before. My requirement is fairly simple. I
have a 8-bit/16-Bit microcontroller at 12MHz. This one samples at 500
MHz. It has 34 channel. Seems a pretty good deal.

Any idea how good it works ?
What are you gonna use it for?
While it's easy to concoct a situation that requires 500MHz. sample
rate, as a practical matter, ain't nothin' much gonna happen on a 12MHz.
system that needs it.
If microcontroller means internal program store and memory, what exactly
are you gonna look at?

You're more likely to want very complex triggering arrangements to find
those subtle, intermittent bugs.
You're more likely to want a fast digital oscilloscope that can actually
see those ground bounces and glitches.

I used to design logic analyzers for a living. In the last 20 years,
the only time I've used one is to look at a RS232 signal when I couldn't
get my hands on a DSO.

If you'd said you were building Pentium processor boards or graphics
display subsystems or real-time FFT processors, my suggestion would have
been quite different.

Also, it's hard to imagine the hassle surrounding an instrument that
plugs into a general purpose PC until you actually try it. Unless you
permanentlly dedicate
the PC and the space it takes up, you'll often find that setting it all
up takes longer than to find the problem by inspection.

mike

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On Fri, 2005-04-29 at 14:59 -0700, aman wrote:
I was looking at some PC-Based Logic Analyzers. I came across this
one:

http://www.pctestinstruments.com/

Has anybody of you used it before. My requirement is fairly simple. I
have a 8-bit/16-Bit microcontroller at 12MHz. This one samples at 500
MHz. It has 34 channel. Seems a pretty good deal.
An article by Jack Ganssle is on embedded.com that compares a few
PC-based oscilloscopes/analysers. There's a second part to the article
that isn't out yet but that I'm waiting for.

Cheers.
 
neil wrote:

mnitin73@gmail.com> wrote in message
news:1114481060.688099.87690@l41g2000cwc.googlegroups.com...
Can I connect the port to the rx pin of the 9 pin serial port and make
any sense? Are the 1553B interface cards standard or does it have to be
custom made depending on the equipment
Excalibur make 1553 interface cards but they are about Ł1500 each.
The data (as I recall, no doubt someone will correct/clarify...)
is a serial differential transformer coupled stream, at 1Megbaud.
Each data word is 16 bits, with a couple of start and stop bits.
The words are grouped in messages, and sent under control of the bus
controller.
You may have a controller or remote terminal (slave device).
The voltage level ranges from 3 to 30 volts.
Probably difficult to get a com port to talk/listen.
hope this helps,
Neil
Neil C.? GE32?
 
Saran wrote:
Hello Everybody:

Good morning. I have a couple of basic MOSFET questions that I would
like to clarify. I have studied a lot of articles, asked a few people
about this doubt but could not find a satisfactory explanation. I am
hoping somebody will take the trouble of clarifying my doubt.

MOSFETs, as I understand, are voltage-controlled devices. Doesn't that
mean that they need a charge (voltage) applied to the gate to make the
device conducting? If this is the case, in some circuits I have found
in the process of learning analog VLSI, what is the meaning of a "gate
of a nmos/pmos transistor being driven by the output of a current
mirror?" I have read that the output current of the current mirror is
used to bias a next stage. What exactly does this mean?

Secondly, if the MOSFETs are voltage-controlled devices, why do we
most of the time deal with currents when talking about VLSI systems?

Hope, I am clear with my questions. Thanks in advance.

Regards,
Saran
If you force charge onto the gate of a MOSFET, you change the gate-source
voltage. MOSFETs are voltage-controlled devices because the drain current
depends on the gate-source voltage. It doesn't matter where the charge came
from.

The gate is basically a capacitor which obeys the rule Q=CV
 
spudnuty wrote:
I have a circuit that uses a light bulb as above and a LM317 voltage
regulator to limit charging current to 100ma and adjust the cutoff
voltage as required.
Richard
Tell us more.
How did you configure the light bulb as a current limit without
letting the voltage on the lm317 input collapse? I've never tried to
run a LM317 with in/out voltage less than the min spec.
What voltage do you use for cutoff and your rationale for choosing
that particular voltage?
Thanks, mike

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kell wrote:
It's a little unclear whether spudnuty used the 317 in the
constant-current config or in the usual voltage-regulating config and a
bulb in series with the output.
Mikes suggestion is the simplest but I want a setup where the charging
current drops off as the nicads voltage rises. That's why I was
thinking about using a comparatively low 8 or 9 volts, so that as the
batteries rising voltage approaches that, the voltage across the
current-limiting element (bulb or power resistor) will fall very low,
causing the charging current to diminish.
You can do that with lower differential voltage and lower voltage light
bulb.

Actually I know this is not
the best way to charge nicads; it is more how you would charge
lead-acid batteries. Maybe I should tape a thermistor to the battery
and use that to turn off the charging.
Temperature cutoff is the best way I know to cook the life out of nicads.

For my dustbuster, I use the supplied charger on a timer. Every night,
it charges enough to put back the average daily use plus some efficiency
factor. This works well even though my cells are old and have
significant self-discharge.
mike



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On 27 Apr 2005 12:53:19 -0700, "mjohnson" <crvmp3@hotmail.com> wrote:

Yup... that's the plan for phase 1. Plan for phase 2 (assuming phase 1
ever works) is to add another sensor to determine if there is actually
a car (or movement?) in the garage. If there is no car (or if there is
movement) then the remote won't activate.
---
For a solution for your phase 1 problem, click on this:

news:hc6871dqc4ir7ues829ntlac8h7p5cpepi@4ax.com

--
John Fields
Professional Circuit Designer
 
Saran wrote:
(snip)

MOSFETs, as I understand, are voltage-controlled devices. Doesn't that
mean that they need a charge (voltage) applied to the gate to make the
device conducting?
Charge and voltage are not the same thing. The gates of mos
transistors are nonlinear capacitors. To change the voltage of any
capacitors involves movement of charge, The rate of movement of
charge with respect to time is called current.

If this is the case, in some circuits I have found
in the process of learning analog VLSI, what is the meaning of a "gate
of a nmos/pmos transistor being driven by the output of a current
mirror?"
A current mirror provides a controlled current that passes charge into
or out of the gate. When the gate voltage reaches the saturation
voltage for the mirror, the current ceases, and the maximum or minimum
voltage that the mirror can drive into (called its compliance) has
been reached.

I have read that the output current of the current mirror is
used to bias a next stage. What exactly does this mean?
It means that the gate voltage is changed at a controlled rate by the
controlled current from a current mirror.

Secondly, if the MOSFETs are voltage-controlled devices, why do we most
of the time deal with currents when talking about VLSI systems?
(snip)

The speed of mos logic is limited by the current one stage can deliver
to change the gate voltage on the succeeding stage. Voltage may turn
the mos transistor on, but current determines how fast a mos
transistor can go from off to on or vice versa.

Have I misunderstood your questions?
 
"Jamie" <jamie_5_not_valid_after_5_Please@charter.net>
wrote in message news:G2Xce.7398$cZ6.3980@fe02.lga...
hartly wrote:

Why does my 9v D.C. transformer only give a max. amps. of 0.22A?.
I've looked inside it and it doesnt seen to have any resistor in
series to restrict the current.
A 6v. battery charger can give out 4 amps.,which proves its nothing
to do with the voltage of input or output.
Thanks.
size of the core, wire, etc...

the wire in the transformer is small, thus creating resistance of its own.
Yes.

Core saturation
No, core saturation does not act to limit the load current.
In fact, load current tends to reduce core saturation.

with inductance etc. etc. etc....
Yes.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com

Above views may belong only to me.
 
hartly wrote:

Why does my 9v D.C. transformer only give a max. amps. of 0.22A?.
I've looked inside it and it doesnt seen to have any resistor in
series to restrict the current.
A 6v. battery charger can give out 4 amps.,which proves its nothing
to do with the voltage of input or output.
Thanks.
size of the core, wire, etc...

the wire in the transformer is small, thus creating resistance of its
own.
Core saturation with inductance etc. etc. etc....
 
"js5895" <JoshTmp@nycap.rr.com> wrote in message
news:1114910484.108484.283140@z14g2000cwz.googlegroups.com...
Hi,
Hello.
Can I calculate the magnetic field around a wire in inches, I have
voltage and current.
Also, the magnetic field around a wire is cylindrical.
In air or vacuum, for an infinitely long wire, the strength
B of the magnetic field surrounding it at a radius r is:
B(r) = u0 * I / (2 Pi r)
where u0 is the permeability of free space (4e-7 * Pi),
B is in Tesla, r is in meters, and I is in Amperes. You
can do the conversions. The voltage is irrelevant.
The direction of the field is perpindicular to both the
wire and the normal from the given point to the wire.
This describes the static field for a DC current. For
AC currents and not r << wavelength, the above is
not the right equation, due to field propagation.

Thanks, for helping.
Welcome.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
 
Saran wrote:
So, let me restate what I understood. The current mirror's output
current is used to charge the gate of the transistor. So, depending on
the magnitude of the current, the gate voltage increaser faster or
slower (sorry about my layman language, I just want to be very very
clear).
That is the principle. For linear capacitors, the math is
I=C*(dv/dt), where I is amperes of current, C is capacitance in farads
and dv/dt is the the rate of voltage change in volts per second.
Unchanging gate voltage requires only leakage current to be
maintained, and that is a very small current for mos gates.

If I understood this correctly, then:
1) If a current mirror is charging the gate of a transistor, or in
other words, changing its Vgs, the maximum Vgs value is determined by
the saturation voltage for the mirror.
Yes.

2) We are controlling the rate at which gate is charged in a circuit.

Have I understood this correctly? Thanks in advance.
You have restated what I said, so if you are wrong, we are both wrong.

The logic speed of a mos system is based on both the current the
drains can deliver to the next stage gate, and the capacitance of that
gate.
 
"fpd" <foreignpigdogAThotmailDOTcom@nospam.com> wrote in message
news:42730cad_1@newsfeed.slurp.net...
neil wrote:
mnitin73@gmail.com> wrote in message
news:1114481060.688099.87690@l41g2000cwc.googlegroups.com...
Can I connect the port to the rx pin of the 9 pin serial port and make
any sense? Are the 1553B interface cards standard or does it have to be
custom made depending on the equipment
Excalibur make 1553 interface cards but they are about Ł1500 each.
The data (as I recall, no doubt someone will correct/clarify...)
is a serial differential transformer coupled stream, at 1Megbaud.
Each data word is 16 bits, with a couple of start and stop bits.
The words are grouped in messages, and sent under control of the bus
controller.
You may have a controller or remote terminal (slave device).
The voltage level ranges from 3 to 30 volts.
Probably difficult to get a com port to talk/listen.
hope this helps,
Neil

Neil C.? GE32?
I'm Neil from Chatham in Kent, England.
Work (currently) for BAE Systems.
Not sure what the GE32 is, so it's probably not me you're thinking of.
 
Saran wrote:
So, let me restate what I understood. The current mirror's output
current is used to charge the gate of the transistor. So, depending on
the magnitude of the current, the gate voltage increaser faster or
slower (sorry about my layman language, I just want to be very very
clear).

If I understood this correctly, then:
1) If a current mirror is charging the gate of a transistor, or in
other words, changing its Vgs, the maximum Vgs value is determined by
the saturation voltage for the mirror.
2) We are controlling the rate at which gate is charged in a
circuit.

Have I understood this correctly? Thanks in advance.
Yes. You are correct.
 

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