Chip with simple program for Toy

[Rich Grise]

On Monday 04 October 2004 07:54 pm, The little lost angel did deign to grace
us with the following:

On Mon, 04 Oct 2004 12:32:59 -0700, John Larkin
jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

Why don't governments tax the useless crap, and give people a break on
productive/educational stuff?

Maybe because those people selling useless crap like cigarettes make a
lot lot lot more money and therefore can afford to influence the
decisions? PpPpp

It's because there's so much more useless crap sold, and the ones

who buy it aren't in a position to bankroll lobbyists.

Cheers!
Rich

 

[CFoley1064]

Subject: Taking the 'hummm' out of a circuit
From: "anonymous" anonymous@catfarm.com
Date: 10/9/2004 5:09 PM Central Daylight Time
Message-id: <hkZ9d.212267$MQ5.19587@attbi_s52

Im working on a power supply for some musical bits. Unfortunately its
producing a great deal of "hum" in the circuit. Is there any way to filter
out residual AC in the DC line?

What I have:

+18V (wall wart) ---->3500uF
(electrolytic) ---->.22uF--->7809--->.1uF--->out
|
| |
ground
ground ground

the ground is common between the wall wart and all components. There are 4
7809s in parallel at the output.

You didn't mention if your wall wart is a switcher or a linear supply. If it's
relatively light, it's probably a switcher. Could be a lot of problems there.
But if it feels like it's got some iron in it, it's probably a linear. That
would narrow down your possibe problems.

Even if you're drawing an amp from the unregulated linear, you shouldn't have
more than 2V peak-to-peak of AC. If it's much greater than that, you've
probably got a bad 3500uF cap. Try replacing that first.

If you're still looking at problems with 2V p.p. of AC, your stuff may be very
sensitive to ripple in the power supply (the 7809s will probably cut that by at
least 60dB, which would give you less than 20mV ripple on the 9V output). If
you're measuring more than 20mV of ripple at the output of a 7809, you might
want to look at replacing them.

The noise on the power supply we've been talking about is 100Hz or 120Hz from
the wall wart. If it's higher frequency stuff, the output cap might be bad,
and you could try replacing that. If not, you might be getting crosstalk
between power supplies. A 7809 is very good at cutting down 100/120 Hz ripple,
but it's not very good at audio frequency. If that's an issue, you might want
to try putting an isolating resistor on each input to the 7809 like this (view
in fixed font or M$ Notepad):


Multiple 7809s ____
___ | |
o------o---o--.o-|___|-------o-----|7809||--o------o9V
| | | 10 ohm | |____| |
| | --- | ---
3500uF --- | --- | ---
--- | | === |
| | | GND |
o------o | === ____ ===
| | ___ GND | | GND
| o---o-|___|-------o-----|7809|----------o9V
=== | 10 ohm | |____| |
GND | --- | ---
: --- ---
: | |
| | |
: : === === ===
GND GND GND
:
:
____
| ___ | |
'---o-|___|-------o-----|7809|---o------o9V
10 ohm | |____| |
--- | ---
--- ---
| |
| | |
=== === ===
GND GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

These are the usual suspects.

Good luck
Chris

 

[Dbowey]

Steve (the OP) posted:

<< On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
<w9wi@invalid.invalid> wrote:

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

Sure, I'm aware of all this basic crap, but every reply I've had to
this question has thrown up a different answer and none of them make
much sense. Can't you guys come up with something in common that adds
up?
Thanks,

Two little problems here:
1. You should have posted more of the schematic so a conlusive evaluation
could be made.
2. You aren't bright enough to understand the offered insight.

Don

 

[Michael Black]

Steve Evans (smevans@jif-lemon.co.mars) writes:

On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
w9wi@invalid.invalid> wrote:

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

Sure, I'm aware of all this basic crap, but every reply I've had to
this question has thrown up a different answer and none of them make
much sense. Can't you guys come up with something in common that adds
up?
Thanks,

Steve

Then rephrase the question.

Are you wondering why the base of the transistor is at DC ground, or
wondering what the RF choke does to ensure that the base of the transistor is
not at ground for RF?

People have answered what they think you are asking, which isn't really clear.
Clarify that, and you might get an answer you want.

Michael VE2BVW

 

[tempus fugit]

Is this to power guitar effects by any chance? If your components are
working properly, you might want to look into the possibility of a ground
loop somewhere. For instance, if you have your amp plugged in one outlet,
and your effects power supply in another one on the other side of the room,
you're almost guaranteed to create a ground loop. Try plugging all the gear
your plugging into into one outlet or power bar and see if the hum goes
away. Alternately, try unplugging (from the wall) each piece of equipment
one at a time to see if the hum goes away. If it does, you've found the
problem.


"anonymous" <anonymous@catfarm.com> wrote in message
news:hkZ9d.212267$MQ5.19587@attbi_s52...

Im working on a power supply for some musical bits. Unfortunately its
producing a great deal of "hum" in the circuit. Is there any way to filter
out residual AC in the DC line?

What I have:

+18V (wall wart) ---->3500uF
(electrolytic) ---->.22uF--->7809--->.1uF--->out
|
| |
ground
ground ground

the ground is common between the wall wart and all components. There are 4
7809s in parallel at the output.

 

[Peter A Forbes]

On Sat, 09 Oct 2004 23:23:36 GMT, Rich Grise <null@example.net> wrote:

On Saturday 09 October 2004 10:07 am, R.Lewis did deign to grace us with the
following:
...
Best guess is that re-wiring the plug merely made good a poor contact.
Household mains powered equipment does not have a 'polarity'.

Oh, it most certainly does! It's just not "plus" and "minus," it's "hot"
and "neutral." Reversing them is a safety issue.

Good Luck!
Rich

Same in the UK but for slightly different reasons. Our Neutral connection is now
bolted up to the incoming mains Earth, so it doesn't float above earth as it
used to years ago.

The USA 120V is parts of a centre-tapped 240V if my memory serves me correctly?

Peter

--
Peter & Rita Forbes
Email Address:
diesel@easynet.co.uk
Web Pages for Engine Preservation:
http://www.oldengine.org/members/diesel

 

[me]

Maybe because those people selling useless crap like cigarettes make
a lot lot lot more money and therefore can afford to influence the
decisions? PpPpp

For all the time cigarettes were being sold the government was getting more
than half the profits. But they are are the ones saying that they are
looking out for us by suing the cash cow for money...


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! >100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

 

[CWatters]

"Rich Grise" <null@example.net> wrote in message
news:Yp_9d.2785$ua2.366@trnddc09...

On Saturday 09 October 2004 10:07 am, R.Lewis did deign to grace us with
the
following:
...
Best guess is that re-wiring the plug merely made good a poor contact.
Household mains powered equipment does not have a 'polarity'.

Oh, it most certainly does! It's just not "plus" and "minus," it's "hot"
and "neutral." Reversing them is a safety issue.

I think that might depend on which country you are in but I'm not 100% sure.

 

[Paul Burridge]

On Sat, 09 Oct 2004 22:40:20 GMT, Steve Evans
<smevans@jif-lemon.co.mars> wrote:

On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
w9wi@invalid.invalid> wrote:

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

Sure, I'm aware of all this basic crap, but every reply I've had to
this question has thrown up a different answer and none of them make
much sense. Can't you guys come up with something in common that adds
up?

Perhaps if you posted the proper schematic it would help. What's the
transistor in question?
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Roy Lewallen]

Steve Evans wrote:
. . .
I'm not so dumb as to
realise that a choke passes dc but not the high frequncy RF. I guess
it boils down to this: how did the designer arrive at the given value
of 0.4uH for this inductor? Why not 4nH? Why not 40uH? Why not 100mH??
What's the deal with this value and since I've only got a 1uH in my
junk box, will that be okay instead?

I'm guessing you're not experienced enough to realize that inductors are
far from ideal. Among their many imperfections, the most problematic in
an application like this is shunt capacitance. This resonates with the
inductance to create a parallel resonant circuit. At resonance, the
impedance is very high -- higher than that of the inductance alone. But
above the self-resonant frequency, the impedance drops, and at some
point becomes less, then very much less, than the impedance of just the
inductance. Well above self-resonance, all you see is the self
capacitance -- it looks like a capacitor, not an inductor.

That's why a designer doesn't just use 100 mH for everything. Just about
any 100 mH inductor looks like a capacitor at 145 MHz, with a very low
impedance, much lower than an inductor with smaller inductance value.(*)
The trick is to choose an inductor that's below or at its self resonant
point while still having enough impedance so it doesn't disturb the
circuit it's across. A good rule of thumb is an inductor whose reactance
is about 5 - 10 times the impedance it's across. (In your case, this
might not be easy to determine. You might be able to get it either from
knowing someting about the previous stage, or from the S parameter
specfication of the transistor.) 5 - 10 times is usually enough, and if
you try for too much, it won't work any better and you run the risk of
being above the self-resonant frequency. Of course, an individual
inductor can be measured to make sure its impedance is high enough at
the frequency of use, if you have the equipment to make the measurement.

So, now, is your 1 uH ok? It depends on its shunt C, which depends on
its construction. If you can't measure it, just try it. The worst that's
likely to happen is that it'll kill the signal (due to low impedance).
At only 2-1/2 times the value of the original, there's a good chance
it'll work ok. If it doesn't, and if your 1 uH inductor isn't potted,
you can get the value down to 0.4 uH by unwinding about 1/3 of the turns.

(*) Even this explanation is highly simplified. At higher frequencies,
the inductor will exhibit additional series and parallel resonances due
to various parasitic capacitances and inductances and transmission line
effects.

Roy Lewallen, W7EL

 

[Paul Burridge]

On Sun, 10 Oct 2004 11:57:58 GMT, Steve Evans
<smevans@jif-lemon.co.mars> wrote:

Thanks Michael,
I'm sorry I can't post the full diagram as my scanner's bust. I'll try
to clarify. I know why the base of the transistor (it's a 2n5771, in
answer to Paul's question) is at DC ground. I'm not so dumb as to
realise that a choke passes dc but not the high frequncy RF. I guess
it boils down to this: how did the designer arrive at the given value
of 0.4uH for this inductor? Why not 4nH? Why not 40uH? Why not 100mH??
What's the deal with this value and since I've only got a 1uH in my
junk box, will that be okay instead?

Well then it appears Reg's hunch was right. The transistor in question
has an input capacitance of just under 3pF, so the 0.4uH inductor
forms a parallel tuned circuit with it at 145Mhz. This prevents the
input signal from partially shunted to ground via the input
capacitance were the inductor not there. The idea is to allow as much
of the input signal as possible to develop across the BE diode to
maximise input impedance and gain. The input capacitance is in
parallel with this diode and bypasses RF signals around it - which you
*don't* want. Will a 1uH work instead? Do the maths and find out. But
if you don't know the inductor's Q that probably won't help much...
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[anonymous]

Looks like you are trying to get too much from the wall wart.
What is wall wart current rating at 18 volts and how much is
the load current?

-Bill

The wall wart puts out 400 mA. For my test I was drawing ~5mA.

 

[Jim Pennell]

On Sat, 09 Oct 2004 22:40:20 GMT, Steve Evans
<smevans@jif-lemon.co.mars> wrote:

On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
w9wi@invalid.invalid> wrote:

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

Sure, I'm aware of all this basic crap, but every reply I've had to
this question has thrown up a different answer and none of them make
much sense. Can't you guys come up with something in common that adds
up?

The problem is, there are quite a few possibilities for what the inductor
is doing. For example, it might be providing a DC path to ground for the
base of the transistor, or it might be supressing a parasitic oscillation in
the amplifier.

Or, for that matter, it might be part of an impedance transform from one
section to another. Without the complete schematic, there is no way to
tell.

Jim
N6BIU

 

[Steve Evans]

On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
<pb@notthisbit.osiris1.co.uk> wrote:

Well then it appears Reg's hunch was right. The transistor in question
has an input capacitance of just under 3pF, so the 0.4uH inductor
forms a parallel tuned circuit with it at 145Mhz. This prevents the
input signal from partially shunted to ground via the input
capacitance were the inductor not there. The idea is to allow as much
of the input signal as possible to develop across the BE diode to
maximise input impedance and gain. The input capacitance is in
parallel with this diode and bypasses RF signals around it - which you
*don't* want. Will a 1uH work instead? Do the maths and find out. But
if you don't know the inductor's Q that probably won't help much...

Sorry, but none of this makes sense to me. There's no diode involved
so I don't know where you get that from. And what's "input
capacitance" and "Q"?
Do try to speak in plain English!

Steve

--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

 

[Steve Evans]

On Sun, 10 Oct 2004 05:41:23 -0700, Roy Lewallen <w7el@eznec.com>
wrote:

[snip]

(*) Even this explanation is highly simplified. At higher frequencies,
the inductor will exhibit additional series and parallel resonances due
to various parasitic capacitances and inductances and transmission line
effects.

<groan>
Dur, nope. I still don't get it. Can anyone explain this stuff *in
simple terms*? I'm really struggling here with some of the
terminology. :-(
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

 

[Brian]

Something you might try:
Set your 555 timer up to put out pulses near a square wave. Have this
pulse trigger two one-shots (something like a 74LS123). Have one of the
one-shots trigger on the positive going pulse of the 555 timer and the other
one-shot trigger on the falling side of the 555 timer pulse. Have each one
of the one-shots output drive a reed relay.

Brian

"Jag Man" <Jag_Man653R-E-MOVE@hotmail.com> wrote in message
news:1ziad.27606$QJ3.19083@newssvr21.news.prodigy.com...

I want to test an automotive ECU to be sure it's generating the correct
pulses to the fuel injectors.

Let me describe the wanted test pulses by telling you how the they are
generated in the car. A "trigger board" in the distributor has 2 Hall
effect
transistors onboard,
at diametrically opposed positions. The rotor has a small magnet
embedded
and is whirling around right above the trigger board, so each Hall
effect
transistor
is fired once per revolution of the rotor. Three wires feed form the
trigger
board
to the ECU, A B, and C. As a result of the Hall effect transistors, the
resistance
between A and C momentarily changes from infinite to 0, followed shortly
thereafter (i.e., 1/2 a rotor revolution later) by B and C doing the
same.
Thus
the A, B, and C are seen by the ECU as dual normally open switches that
each
momentarily close once per rotor revolution.

So the question is, what is the simplest way to simulate this, perhaps
with
ICs?
I'm thinking of perhaps a pair of relays to represent the switches. A
pulse
stream would be generated by a 555 timer IC, somehow feeding every other
pulse to alternating relays. I have a feeling a flip-flop could be used
beneficially, but can't quite see how to do it. We don't have to worry
about pulse width, as the actual width of pulses emitted by the ECU
are independent of the closure times seen on A, B and C.


I am coming at this as an amateur, but I have done some circuits using
the 555 timer IC.

Any helpful suggestions would be appreciated.

TIA

Ed

 

[Robert Monsen]

Peter A Forbes wrote:

On Sat, 09 Oct 2004 23:23:36 GMT, Rich Grise <null@example.net> wrote:


On Saturday 09 October 2004 10:07 am, R.Lewis did deign to grace us with the
following:
...

Best guess is that re-wiring the plug merely made good a poor contact.
Household mains powered equipment does not have a 'polarity'.

Oh, it most certainly does! It's just not "plus" and "minus," it's "hot"
and "neutral." Reversing them is a safety issue.

Good Luck!
Rich


Same in the UK but for slightly different reasons. Our Neutral connection is now
bolted up to the incoming mains Earth, so it doesn't float above earth as it
used to years ago.

The USA 120V is parts of a centre-tapped 240V if my memory serves me correctly?

Its 110V and 220V.

http://www.school-for-champions.com/science/acwiring.htm

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Robert Monsen]

Steve Evans wrote:

On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
pb@notthisbit.osiris1.co.uk> wrote:


Well then it appears Reg's hunch was right. The transistor in question
has an input capacitance of just under 3pF, so the 0.4uH inductor
forms a parallel tuned circuit with it at 145Mhz. This prevents the
input signal from partially shunted to ground via the input
capacitance were the inductor not there. The idea is to allow as much
of the input signal as possible to develop across the BE diode to
maximise input impedance and gain. The input capacitance is in
parallel with this diode and bypasses RF signals around it - which you
*don't* want. Will a 1uH work instead? Do the maths and find out. But
if you don't know the inductor's Q that probably won't help much...


Sorry, but none of this makes sense to me. There's no diode involved
so I don't know where you get that from. And what's "input
capacitance" and "Q"?
Do try to speak in plain English!

Steve

The transistor acts like a tiny capacitor from base to emitter, in
addition to its other jobs. Thus, an inductor in series with the base
will act like a bandpass filter at

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

An inductor and capacitor with the same values in parallel will act like
a bandstop filter at that frequency.

Q is the "quality or merit factor" of the inductor, and is computed by
dividing the 'reactance' (or AC resistance) of the inductor by the DC
resistance *at the frequency in question*; it will generally be
different at different frequencies. Q is really the ratio of reactive
power in the inductance to the real power dissipated in the resistance,
and can be computed by using the formula:

Q = 2*PI*f*L/ri

where f is frequency, and ri is the resistance of the coil at f. You buy
inductors with a given Q range.

For an LC resonant circuit, the Q of the inductor will affect the 'Q' of
the resulting resonance, which simply means that with a bigger Q, the
passband or stopband will be wider. The Q of a resonant circuit is
defined as the resonant frequency divided by the width of the passband.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Mike]

In terms of TV/VCR remotes and the like, they're not constantly powered-on,
so there'd be no reply signal to a device you would make. I think the same
applies to cordless phones.

Mike

"jm" <john_20_28_2000@yahoo.com> wrote in message
news:c67e4bdd.0410101942.64820fb2@posting.google.com...

I am going to build a device like a remote control finder. I think I
could eventually figure out a circuit that makes a noise, but what
would I look at for the wireless part of it? Should I just take apart
my cordless phone and look!

Thanks, don't know what to look at.

 

[Jag Man]

Thanks, Brian!

Guess I'll have to do some reading up on the 74LS123
to see how to tell it what side to trigger on.

Ed

>