Chip with simple program for Toy

[John Bokma]

Bob Monsen wrote:

colin2000@talk21.com wrote:

find out the Vforward and Iforward for the LEDs you are going to use


(12 - x * Vf)
R = -------------
If

--
John


thanks.....one thing though
could you elaberate on Vforward and Iforward ((If Vf)?!


Put them all end-to-end (in series) and put a 1k resistor in series too.
That should light them up with 3.6mA. If you want them brighter, use a
lower value resistor, but too low will cause them to fail. The lowest
you probably want to go is 220 ohms.

Hence: look up Vf and If, and *calculate*

--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html

 

[Tom Biasi]

<upgrdman@mindspring.com> wrote in message
news:1116537008.278997.182100@g43g2000cwa.googlegroups.com...

Thanks yet again

A few more questions, hope you don't mind...

By 5% and 10% I assume you mean the carbon film percentage? What does
this mean, and how does it effect the results?

And what exactly is "drop" voltage? The amount used by the LED?

Thanks,
--Farrell F.

Hi,
"Drop" is a term meaning the voltage across a device caused by the current
going through it times the resistance.
If you have 1 volt across the resister in this series circuit the LED will
"see" 1 volt less than the supply, hence the term "drop".
The percentage is how close to the specified value the resister really is.
Ex.: A 100 ohm resister at 5% may really be 95 to 105 ohms. A 100 ohm
resister at 10% may be between 90 and 110 ohms.
(5% of 100 is 5 ohms 100 plus or minus 5 ohms.) Don't concern yourself with
this for this application.
Regards,
Tom

 

[Peter Bennett]

On 19 May 2005 14:10:08 -0700, upgrdman@mindspring.com wrote:

Thanks yet again

A few more questions, hope you don't mind...

By 5% and 10% I assume you mean the carbon film percentage? What does
this mean, and how does it effect the results?

The 5% or 10% is the tolerance on the resistance - a 5% resistor may
be up to 5% above or below its marked value (but I think most metal or
crabon film resistors will be within 1% of the marked value,
regardless of the indicated tolerance)

And what exactly is "drop" voltage? The amount used by the LED?

Yes.

Thanks,
--Farrell F.

--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[Ban]

u035m4i02@sneakemail.com wrote:

I should have said for the general case, like

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampvar6.html#c2

That is what I was referring to. Are you able to follow my explanation? I
tried to solve the different possibilities.
--
ciao Ban
Bordighera, Italy

 

[Tom Biasi]

<hartlyuk@yahoo.com> wrote in message
news:1116603805.581934.101220@o13g2000cwo.googlegroups.com...

I have an aero. search and rescue beacon transceiver from a Govt.
surplus store.
When I connect the homemade(bike spoke) ant. to the casing(at the same

time it is connected to the proper ant.input site) it changes to
receiving a different air to ground conversation,than if I don't
connect it to the casing.I presume it must be changing the frequency of

transmission received.
Why/how does this occur?.
Thanks.

Hi,
You may be experiencing detuning of the front end circuit due to antenna
mismatch.
The antenna will appear inductive or capacitive if the circuit is not loaded
properly. Use an antenna of the proper length for that unit.
I would expect a unit of that type to be tolerant of this however. Perhaps
you have some internal problems that can not be seen from where I sit.
Regards,
Tom

 

[Rich Grise]

On Fri, 20 May 2005 13:14:18 -0700, Andy Ladouceur wrote:

Hi everyone,

I took a basic electronics course a few years back, otherwise only know
what I've found online, and am having one hell of a time trying to work
out the right voltages / resistances for a project I'm working on.

I have 2.4V @ 24mA coming off an output from an IC, that I'd like to
use to power an LED. The LED needs 3.3V, 20mA.

Well, as you seem to have already found out, 2.4V isn't enough to power
a LED that needs 3.3V.

Exactly what IC is this? We need to know this before we go any further.

The rest of the post shows only that you're getting yourself all tied
in knots trying to do design principles that aren't even applicable
here.

I'm able to vary the 24mA output from the IC in 16 1.5mA steps. (From
1.5mA -> 24mA)

Please, what IC is this? It's now sounding like an analog IC. Please,
please, tell us the part number of the IC - you don't need anything
anywhere near what you're using here.

Can you make the output of your IC zero? Then that would be your "off,"
and then you just drive a transistor as a switch.

But first, we have to know exactly what this "2.4V @ 24mA...[variable]
from 1.5 mA -> 24 mA" is coming from.

Thanks,
Rich

I could be mistaken, but I'm quite sure what I'm looking for here is a
NPN transistor. I just don't know what values of.. -anything-, I'd
need. I had it calculated as:

NPN Transistor, Beta 50.

Power to collector: 12V, with a 420 Ohm resistor.
Power to base: 2.4V
Power to emitter: 0V (Ground), with an 80 Ohm resistor.

This would give me 3.25V at 20mA coming out of the collector. (Er..
right?) Which would work perfectly, if I didn't plan on varying the
current. But, I'm not sure how to calculate what the current going to
the LED would be when the base current is dropped to 1.5V. What I'd
ultimately like is to be able to vary the current and have 1.5mA
pretty much be 'off' for the LED, and 24mA at its brightest.

With my current setup, is this how it would work, or am I right in
assuming most of its current comes from the 12V collector?

Could someone please advise me of a setup that'd let me do the above,
if my current(sorry) one doesn't allow for it? Thanks in advance for
any help you might provide. I'd like to be able to understand the
workings of a transistor enough to be able to figure this out myself,
but it's been two hours, and it seems that's not happening, right now.


-Andy

 

[Rich Grise]

On Thu, 19 May 2005 15:05:17 -0700, Peter Bennett wrote:

On 19 May 2005 14:10:08 -0700, upgrdman@mindspring.com wrote:

Thanks yet again

A few more questions, hope you don't mind...

By 5% and 10% I assume you mean the carbon film percentage? What does
this mean, and how does it effect the results?

The 5% or 10% is the tolerance on the resistance - a 5% resistor may
be up to 5% above or below its marked value (but I think most metal or
crabon film resistors will be within 1% of the marked value,
regardless of the indicated tolerance)


And what exactly is "drop" voltage? The amount used by the LED?

Yes.

Well, the voltage across the LED is the voltage "dropped" by the LED;
the resistor will have its own voltage drop. It's not like volts get
dropped off a cliff, or down a well, or something - it's exactly
analogous to "pressure drop", which is used in pluming and HVAC
all of the time. It's the differernce in pressure between one end
of a component and the other. Volts don't get "used" other than
to provide the pressure that pushes the current through the circuit.

Think of a tall water pipe, with ports with pressure gauges arranged
along its height. At the top, there's zero water pressure, and at
the bottom, there's some amount of pressure just caused by the weight
of the column of water. This is how altimeters work - at altitude,
there's less air on top pressing down on itself. (and on you, and
your altimeter).

Now, if you have a 33ft. column of water, the water pressure at the
bottom will be 15 PSI. The water pressure half-way up the column (16 1/2
feet) will be half that, or 7.5 PSI. So there's a pressure drop of
7.5 PSI from the top of the column to the middle, and another 7.5 PSI
pressure drop from the middle to the bottom.

That's all voltage drop is - a difference in pressure, which is called
"electromotive force" - and as you remmeber from high school, force
is just a push - it's there whether the object is moving or not.

Hope This Helps!
Rich

 

[Michael A. Terrell]

John - KD5YI wrote:

I read that page with interest. I was a bit surprised to note that the
author, in the next-to-last paragraph, refers to Horowitz and Hill and that
their hybrid circuit doesn't work. Do you suppose Win is aware of that? I
don't know how to get in touch with him.

He is on news:sci.electronics.design and
news:alt.binaries.schematics.electronic quite often.

--
Former professional electron wrangler.

Michael A. Terrell
Central Florida

 

[Ban]

Andy Ladouceur wrote:

Hi Rich,

Sorry for not including that in my original post. The IC is a
MAX6956.

http://www.maxim-ic.com/quick_view2.cfm/qv_pk/3503/ln/

Direct link to PDF:
http://pdfserv.maxim-ic.com/en/ds/MAX6956.pdf

Andy, if you want to use the current adjustment possibility of your IC, it
would be better to supply +5V to it, then you can control the LEDs without
any external circuitry in 16 brightness steps. If the other ICs work on 3.3V
you can use a simple level translator to fit to the 5V logic, but they can
probably be interfaced just straight.
In the Datasheet is written that output level is 0.7*Vcc, which translates
to 4.7V to get 3.3V out. There are also tolerances of the LED forward
voltage, so 5V will give some margin. With 20 LEDs full on simultaneously
the dissipation in the IC will be almost 1W, so it will need a small
glued-on heatsink and/or a large plain copper area on the PCB to radiate
that heat. If only part of the LEDs are on or the current is set to lower
values, you do not need this.

--
ciao Ban
Bordighera, Italy

 

[Rich Grise]

On Sat, 21 May 2005 12:19:50 -0700, albertleng wrote:

Just a very basic question:

Can anyone share with me what's the benefit of using transistor as an
electrical relay compared to mechanical relay when the incoming signal
is from a computer?

Is it related to pricing, input voltage/current, power rating and etc?

No moving parts.

Cheers!
Rich

 

[Rich Grise]

On Sat, 21 May 2005 09:59:16 -0700, denidoank@gmail.com wrote:

i want to try isp in at89s8252 but i don't know how to make connection
between computer and device. I mean in avr system there is dongle to
make connection between computer and system. Is there any device to
make connection between computer and mcu?

Yes. It's called the "telephone".

Good Luck!
Rich

 

[Rich Grise]

On Sat, 21 May 2005 04:34:26 +0000, Ban wrote:

Andy Ladouceur wrote:
Hi Rich,

Sorry for not including that in my original post. The IC is a
MAX6956.

http://www.maxim-ic.com/quick_view2.cfm/qv_pk/3503/ln/

Direct link to PDF:
http://pdfserv.maxim-ic.com/en/ds/MAX6956.pdf


Andy, if you want to use the current adjustment possibility of your IC, it
would be better to supply +5V to it, then you can control the LEDs without
any external circuitry in 16 brightness steps. If the other ICs work on 3.3V
you can use a simple level translator to fit to the 5V logic, but they can
probably be interfaced just straight.
In the Datasheet is written that output level is 0.7*Vcc, which translates
to 4.7V to get 3.3V out. There are also tolerances of the LED forward
voltage, so 5V will give some margin. With 20 LEDs full on simultaneously
the dissipation in the IC will be almost 1W, so it will need a small
glued-on heatsink and/or a large plain copper area on the PCB to radiate
that heat. If only part of the LEDs are on or the current is set to lower
values, you do not need this.

Thanks, Ban and Chris, this is pretty much what I was going to say, but
you have done so much better than I could have.

Thanks!
Rich

 

[Harold Ryan]

It's none of the items mentioned. The signals from the PC may be on a
separate ground than the other electronic circuitry that is receiving the
signal. In this case, isolation is required by either by a mechanical relay
or an opto coupler. Just a transistor is not adequate.

Harold


"Chris" <cfoley1064@yahoo.com> wrote in message
news:1116715409.256914.69340@f14g2000cwb.googlegroups.com...

Rich Grise wrote:
On Sat, 21 May 2005 12:19:50 -0700, albertleng wrote:

Just a very basic question:

Can anyone share with me what's the benefit of using transistor as
an
electrical relay compared to mechanical relay when the incoming
signal
is from a computer?

Is it related to pricing, input voltage/current, power rating and
etc?


No moving parts.

Cheers!
Rich

Yup.

A basic question, indeed. Let's count the ways.

* Tranistors switch in microseconds, relays in milliseconds.

* Transistors switch directly on without contact bounce, relay
contacts smack and bounce together for a millisecond or so, causing
intermittent on/off for that time until the contacts settle in.

* Relays are mechanical devices with moving parts -- they will wear
out. A typical relay is rated for somewhere between 10 thousand and 10
million operations before failure. And relay contacts usually wear out
first, if they're swicthing at anywhere near rated load. Actually,
they're supposed to.

* Relay contact bounce can cause EMI which can cause problems.
Although you generally have to be more careful with transistors when
switching inductive loads, there isn't a spark being created with
switching like with relays. That EMI can cause the computer to spit
up.

* A typical high current transistor will usually cost much less than a
relay rated to switch equivalent current.

* A transistor will typically require much less power to operate than
a relay coil, especially if you use a darlington transistor or a
MOSFET.

I guess there are more reasons, but these are the ones that come to
mind first.

Good luck
Chris

 

[Ban]

Nikolas Britton wrote:

Hello all, I need some advice on the correct way to ground a standard
LM317 circuit. Right now I have the base of the transformer and the
negative side of the DC output connected to earth ground. To me it
seems weird having to earth ground the DC negative side, but if I
don't do that I'll get 12.5Vac / 20mA from the DC negative post to
earth ground. I thought the DC section was suppose to be isolated
because of the transformer?

Parts list (so far):
*The transformer is 60Vac with a center tap to give me 30Vac, pulled
from a dead Sony amp.
*KBPC10 Bridge rectifier, metal casing, I think it's rated at 10A /
50V.
*Sprague powerlytic 2600uF 150V filter cap.
*230 ohm 1/4 watt resistor.
*10K potentiometer.
*LM317.
*Big heat sink.

BTW, IIRC, when I first tried grounding it I connect the earth ground
to the heat sink that had the rectifier and the LM317, plus side of
the DC output, mounted on it but then I got 12.5Vac / 7A on the
negative side to earth ground....

Niko, with 30Vac you will get more than 42Vdc across the cap, which is above
the absolute max. rating(input/output differential) of the LM317. This means
when a short occurs on the output the regulator might blow. There is a
LM317HV version available, which is good for 60V.
Now to the connections:
1. connect the center tap to the neg. side of the cap and leave the - tab of
the rectifier open, so only 2 of the diodes are used. you do not need to
connect a earth. View the diagram with fixed font.

30V+30Vac
-. ,--------.
)|( | ____
)|( | | | +1.25...37V
120Vac )|(---. | +-+-------o------|317 |-------o--o
)|( | | A A | |____| |
)|( | '-+ | |+ | ___ |
-' '---)------(-+ === .----o----|___|-o
| A A /-\ | | |
| +-+-- | |+ .-. ---
| - open | === | |<-. ---
| | /-\ | | | |0.1u
'----------------o | '-' | |
| | | | |
'---o----o---o------o--o
10u/50V
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)
Be aware that your rectifier is not rated for 60Vac, better to take some
discrete diodes with 3A/200V(1N5402) rating.

--
ciao Ban
Bordighera, Italy

 

[Peter Bennett]

On 21 May 2005 23:50:07 -0700, "Nikolas Britton"
<nbritton@nbritton.org> wrote:

BTW, IIRC, when I first tried grounding it I connect the earth ground
to the heat sink that had the rectifier and the LM317, plus side of the
DC output, mounted on it but then I got 12.5Vac / 7A on the negative
side to earth ground....

Thanks.

If you look at the LM317 datasheet at
http://www.national.com/ds/LM/LM117.pdf, you will find that the
mounting tab is connected to the output terminal - you DO NOT want to
ground that tab, or the heat sink the part is mounted on (unless you
take measures to insulate the tab from the heatsink.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[tempus fugit]

What DC output are you trying to get from the regulator?

"Nikolas Britton" <nbritton@nbritton.org> wrote in message
news:1116744606.977085.289160@g14g2000cwa.googlegroups.com...

Hello all, I need some advice on the correct way to ground a standard
LM317 circuit. Right now I have the base of the transformer and the
negative side of the DC output connected to earth ground. To me it
seems weird having to earth ground the DC negative side, but if I don't
do that I'll get 12.5Vac / 20mA from the DC negative post to earth
ground. I thought the DC section was suppose to be isolated because of
the transformer?

Parts list (so far):
*The transformer is 60Vac with a center tap to give me 30Vac, pulled
from a dead Sony amp.
*KBPC10 Bridge rectifier, metal casing, I think it's rated at 10A /
50V.
*Sprague powerlytic 2600uF 150V filter cap.
*230 ohm 1/4 watt resistor.
*10K potentiometer.
*LM317.
*Big heat sink.

BTW, IIRC, when I first tried grounding it I connect the earth ground
to the heat sink that had the rectifier and the LM317, plus side of the
DC output, mounted on it but then I got 12.5Vac / 7A on the negative
side to earth ground....

Thanks.

 

[Bob Monsen]

eeh wrote:

Hi,

I have seen some RF transmitters (315MHz) PCB that there are some
inductor components which uses bare wires of 2.5 turns and 3mm
diameter. What are their functions? I have calculated the inductance to
be 13nH which is quite small.
Thanks!

If you build a tank out of that inductor, and a cap of 20pF, you have a
resonance frequency around 315MHz.

f = 1/(2.pi.sqrt(L*C))

---
Regards,
Bob Monsen

 

[Rheilly Phoull]

One day Yeongja_Choi@yahoo.com got dressed and committed to text

How Dare Could America Industrial Property Office Be In Conspiracy
With Jungang International Patent Office To Make An Extravagant
International Crime ?

Hmm, makes a change for an american to steal an idea from Asia for a change

After all the Asian industries have been copying western designs for years.

--
Regards ..... Rheilly Phoull

 

[Roger Johansson]

Chris wrote:

Jack// ani wrote:
Can I use 12V, 2amp centre tapped transformer in place of a

12*2=24Watt

24V/2amp.

24*2=48Watt

I
will not use the tapped wire, so I'll get 24volts across its two
ends.

You cannot use a 24 Watt transformer in place of a 48 Watt transformer,
unless you know it will not have to deliver more than 24 Watt.

Nope. The center tap provides a voltage tap halfway between the two
legs, like this (view in fixed font or Notepad):
` 12VCT Secondary Transformer
` o-----. ,--------o12VAC
` )|(
`Primary )|(
` ) ,--------o 6VAC
` 120VAC )|(
` )|(
` o-----' '--------o 0VAC
`
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Can't get there (24VAC) from here (12VAC). Whether you use the center
tap or not, you've only got 12VAC. Sorry.

Good luck
Chris

Many transformers are made in a way which makes it possible to split
the winding.
Look at the middle contact point in the secondary winding above, the
point marked 6VAC, look at the same point on the realworld transformer.
It is often a sling of copper wire coming out of the winding, and is
soldered to a post.

Desolder it from the post, cut the sling in the middle, connect the two
wires to two separate contacts.

Now you have separated the two 6 Volt windings. The wire is covered
with enamel, so they are isolated from each other even if it may not
look right. Test with a resistance meter that there is no contact
between the two.


--
Roger J.

 

[Roger Johansson]

kell wrote:

Radio Shack has (had?) a TO-220 mounting hardware with a mica
insulator, washers etc. I saw them as recently as last year. But
only a large, fully-stocked RS would have them.

The alternatives:

Isolate the heat sink from the chassis, and leave a warning label
inside saying the heat sing is not isolated from the circuit.

Isolate the chassis against outer touch, without reducing its heat
emissing properties too much. A thin layer of plastic, maybe?

Choose another regulator circuit, where the collector of the power
transistor remains at ground potential.

I have designed several such power supplies, The main part is a npn
power transistor sitting between ground (output ground) and the
negative end of the rectifier bridge, which is NOT grounded anywhere.

The positive end of the rectifier becomes the positive output from the
regulator.

Now we only need to control the base of the power transistor so the
voltage over the transistor is what needs to be subtracted to get the
output voltage we want between the ground (the collector of the
transistor) and the positive supply.

First I add a series resistor in the emitter of the power transistor,
and a smaller transistor which controls the current by pulling down the
base of the power transistor. This sets the maximum current this
circuit will allow through the power transistor.
The small transistor, e to minus, c to base of power transistor, b to
the top of the current sensing resistor, the emitter resistor of the
power transistor. 0.6V/1Amp=0.6Ohm, for example.

rectifier plus Output plus

-----------------------------------------------------------------

Output ground

-------------
|
|
|/
--------------------|
| |
\| |
|----------------|
| .-.
| | |
| | |
| '-'
---------------------------------|------------------|
rectifier minus

Then I need to compare the output voltage with the reference voltage.

The reference I use is often a zener, or led's used as zeners.

The output voltage is floating compared to the regulator circuit and
that poses a little problem.
The sensor or output voltage needs to float with the output, but send
control signals to the regulator part, which is completely under
ground, so to speak because it is below the zero voltage of the
output.

This is not so difficult, we can use current or even light to control
it.

The "sensor" must take a sample of the output voltage difference, and
use it to control a current in a wire to the regulator unit.

There are many fine engineers here who can help you with the rest if
you want to continue this lesson in power supply design.

Well, even if we complete this design and build it the chassis will be
grounded, and some care is needed for that chassis not to get in touch
with anything which it could short. We prefer boxes which are not
electrically active in any way.

For example a toaster where the chassis is grounded, and you poke some
bread crumbs out with metal wire. If you hold the other hand around the
toaster's chassis you can die if the chassis is grounded metal. If it
is isolated, and even better, made out of plastic, you have a better
chance of survival.

Sometimes we need direct contact between the metal chassis and the air
so much that we can sacrifice the rule that a chassis should be
isolated.

I built a computer monitor once, designed a 95 Volt 1A power supply as
described above, and bolted the output transistor directly to the sheet
metal outer chassis I made for the tube.
Its collector was the ground point in the regulator I built so there
was no need for isolated mounting, maximizing the possible power and
reducing the working temperature.



--
Roger J.