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Watson A.Name - 'Watt Sun
Guest

Wed Sep 29, 2004 4:25 pm   



In article <3F0651DC.5FA9A926_at_charter.net>, dressel1_at_charter.net
mentioned...
Quote:
I figured that was the reason. I did put a 10 ohm resistor in series with the
3.6v source and the LED, but it still appeared brighter (of course, that is based
on my eyes, which are far from perfect).

For the series of 5, I figure I can get by with a 430ohm resistor (maybe a 390ohm
if necessary for brightness, though that will bring the current to around
21-22mA). Does anyone have a quantitative feel for how the higher current will
affect LED life? Based on the "perfect" scenario, they should last around
50,000-100,000 hours (these are probably standard numbers). I won't need them on
full time, so if I could get a quarter of that, it would more than suffice. Heat
dissipation should not be a problem.

For the white LEDs, I use a 33 ohm resistor for 4.5V = three AA cells.
That gives about 30 mA, and with fresh batteries, it's probably closer
to 40 mA. The lifetime may be shorter by 50%, but I figure that it'll
take a lifetime to put that many hours on them, being they get used a
few hours a year. So it really doesn't matter. And the tradeoff of
increased light output is well worth it.

I've read that some flashlight makers run their LEDs at 2 or 3x the
max, or something like 50 to 90 mA.


Quote:
later,
Kevin

"Watson A.Name - 'Watt Sun'" wrote:

In article <3F06264D.4BB9FCE5_at_charter.net>, dressel1_at_charter.net
mentioned...

[snip]

If you put the LED across the 3.6V, you should *still* put a current
limiting resistor in series with it. If you don't, then the battery
is probably pushing much more than the 30 mA maximum thru the LED.
Watch out for the overly hot LED.

Thanks,
Kevin

[snip]



--
@@F_at_r_at_o_at_m@@O_at_r_at_a_at_n_at_g_at_e@@C_at_o_at_u_at_n_at_t_at_y@,@@C_at_a_at_l@,@@w_at_h_at_e_at_r_at_e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t_at_h_at_e@@a_at_f_at_f_at_l_at_u_at_e_at_n_at_t@@m_at_e_at_e_at_t@@t_at_h_at_e@@E_at_f_at_f_at_l_at_u_at_e_at_n_at_t@@

G.T.W.
Guest

Wed Sep 29, 2004 4:25 pm   



There are 2 types of output on TTL chips.

1. Totem pole , with an upper and lower transistor output, which
switch between a High and a Low. This transistors are a series
circuit, stacked one of top of the other, hence the term 'totem pole'.
These chips already have internal path for power supply and ground. If
2 totem poles are connected together, and the upper transistor of one
output turns on while the lower transistor output of the other turns
on, THEN the power supply is shorted out.

2. Open collector, the 'collector' side of the upper transistor has
been omitted so that the output has no internal path to +5 V. Open
collector outputs can be connected together since there is no danger
of power supply being shorted. The outputs are tied to a common
resistor that is itself connected to 5 V. The resistor supplies the
output with a path, and 'pulls up' the ouput to a High ( 1 ) level .

If you are reading a schematic that shows a resistor in between the
Output and Vcc, it is a good bet that chip is an open collector type.

CMOS chips can also be open collector, but since they can operate with
a wider range of voltages (generally 3-15 volts), it is not uncommon
to see their pull up resistors connected to 12 V.

On Wed, 25 Jun 2003 15:32:12 +0100, "Neil C" <neilc_at_despammed.com>
wrote:

Quote:
The HC series needs a pullp if driven by LS coz the LS output high voltage
is only 3.5 volts or so, and the HC likes at least 4.9v.
HCT132 won't need it coz the 'T' means TTL comptible levels.
If not mentioned, devices will be active pullup, not open collector (like
the 74xx05, 06, 07)
hth
Neil
"Anders" <anders_at_lonnstam.com> wrote in message
news:bdc2ti$rd1e7$1_at_ID-198818.news.dfncis.de...
If the load on the CMOS is low, it should not be neaded with pullup. In
the
case with TTL to CMOS there should always be a pullup, because the
switching
level on the CMOS is higer than TTL.
Anders

"Baphomet" <fandanospam_at_catskill.net> skrev i meddelandet
news:vfj1jsjo0hjkbc_at_corp.supernews.com...

"Colin Green" <es2136_at_my-deja.com> wrote in message
news:5fcd815c.0306250055.2c9a313c_at_posting.google.com...
Hi there,

I was told that most fo the TTL chips (if not all) need pull-up
resistors, whereas CMOS dont. That is not true, is it?!? Couldn't find
that info on the net.

No. It is not true. Most TTL chips don't require pull up resistors. They
have an active transistor to accomplish this.

What I know is if the Data Sheet doesnt mention anything about 'It is
an Open Collector chip...'. Then the chip doesnt need a pull-up
resistor. (eg. MM74HC132)

I don't know if you can necessarily assume that. Most CMOS do require a
pull
up resistor and in any event, it wouldn't hurt to add one.







G.T.W.
Guest

Wed Sep 29, 2004 4:25 pm   



Your're on the right track. You can make an oscillator with one IC, a
resistor and a capacitor. If you use a Schmitt trigger, you can get
a stable digital pulse train at approximately 9,000 Hz. and at least
2.4 volts output.

Use a 7414 Schmitt Trigger. It has 6 op amps on one IC. It will invert
the output, so once your oscillator is going, all you need to do is
send the output of that first op amp to the input of the next one on
the IC, and you are back in phase.

Add a 1 KHz resistor to the output of the first op amp. At the other
end of the resistor, tie a .068 MicroFarad ceramic capacitor on to it.
Connect the other side of the cap to ground.
Next, send a jumper wire from between the R / C, back to the input of
the first op amp. Then you add one more jumper directly from the ouput
(before the resistor), and connect it to an input of the second op
amp. Take your output from the second 'gate'.

At those R/C values, you would get a nice, stable digital pulse train
at approximately 10KHz.

You can calculate the frequency with the following formula, assuming
a 1KHz resistor.

F = 0 . 000679 divided by Capacitor (.068 microFarads ) = 9,985 Hz

This only works for a resistor value of 1KHz, but you can change the
capacitance for a lower value if you want. Using a capacitor value
of .67 microFarads will give an output close to 1KHz.

Anyway, the Schmitt trigger will shape up the charging / discharging
capacitor into a nice square wave, and it will keep oscillating until
you turn the power off.

I don't want to assume, but don't forget to 'power' the IC .
Pin 7 is ground. Pin 14 is Vcc (5 volts).

I am sorry I don't know how to draw a schematic in a newsreader. I
hope my explanation isn't too poor. good luck.

On Sat, 05 Jul 2003 05:51:10 GMT, "Patrick Leonard"
<transactoid_at_rogers.com> wrote:

Quote:
I'm trying to figure out if its possible to build a super simple signal
generator uisng only one transistor and no other ICs.

The idea being you charge up a capacitor through a resistor then when the
cap is at an adequate potential, its voltage will turn on a transistor. The
transistor being on simply completes a circuit allowing the cap to
discharge. Whence the cap has discharged, the transistor turns off, and the
cap charges again. Repeat......

This seems logical to me, yet I cannot design a circuit that works this
way....it always seems like I'm missing something. Am I just not being
"clever" enough, or is such a circuit indeed impossible?

Thanks
Patrick



Kevin Aylward
Guest

Wed Sep 29, 2004 4:25 pm   



Patrick Leonard wrote:
Quote:
I'm trying to figure out if its possible to build a super simple
signal generator uisng only one transistor and no other ICs.

The idea being you charge up a capacitor through a resistor then when
the cap is at an adequate potential, its voltage will turn on a
transistor. The transistor being on simply completes a circuit
allowing the cap to discharge. Whence the cap has discharged, the
transistor turns off, and the cap charges again. Repeat......

This seems logical to me, yet I cannot design a circuit that works
this way....it always seems like I'm missing something. Am I just not
being "clever" enough, or is such a circuit indeed impossible?


A relaxation oscillator of this sort is not possible with a normal
transistor in a normal circuit, to my knowledge. You can do this with a
unijunction one. Well, actually, you might get one to work in avalanche
mode. You drive the CE through a resister, with a cap across the CE, and
let it break down, which discharges the cap. I have never tried this, so
there may be some catch.

I would move on to solving some more usefull problems.

Kevin Aylward
salesEXTRACT_at_anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

Mark
Guest

Wed Sep 29, 2004 4:25 pm   



The 7414 has 6 inverting gates with schmitt trigger inputs they aren't
opamps. It is the schmitt input that allows them to be used as oscilators. a
high input will not trigger until about 2v, it will stay in this state until
the input goes down to about .5v .


"G.T.W." <rampart_at_easynews.com> wrote in message
news:3f0668d6.6426629_at_news.ev1.net...
Quote:
Your're on the right track. You can make an oscillator with one IC, a
resistor and a capacitor. If you use a Schmitt trigger, you can get
a stable digital pulse train at approximately 9,000 Hz. and at least
2.4 volts output.

Use a 7414 Schmitt Trigger. It has 6 op amps on one IC. It will invert
the output, so once your oscillator is going, all you need to do is
send the output of that first op amp to the input of the next one on
the IC, and you are back in phase.

Add a 1 KHz resistor to the output of the first op amp. At the other
end of the resistor, tie a .068 MicroFarad ceramic capacitor on to it.
Connect the other side of the cap to ground.
Next, send a jumper wire from between the R / C, back to the input of
the first op amp. Then you add one more jumper directly from the ouput
(before the resistor), and connect it to an input of the second op
amp. Take your output from the second 'gate'.

At those R/C values, you would get a nice, stable digital pulse train
at approximately 10KHz.

You can calculate the frequency with the following formula, assuming
a 1KHz resistor.

F = 0 . 000679 divided by Capacitor (.068 microFarads ) = 9,985 Hz

This only works for a resistor value of 1KHz, but you can change the
capacitance for a lower value if you want. Using a capacitor value
of .67 microFarads will give an output close to 1KHz.

Anyway, the Schmitt trigger will shape up the charging / discharging
capacitor into a nice square wave, and it will keep oscillating until
you turn the power off.

I don't want to assume, but don't forget to 'power' the IC .
Pin 7 is ground. Pin 14 is Vcc (5 volts).

I am sorry I don't know how to draw a schematic in a newsreader. I
hope my explanation isn't too poor. good luck.

On Sat, 05 Jul 2003 05:51:10 GMT, "Patrick Leonard"
transactoid_at_rogers.com> wrote:

I'm trying to figure out if its possible to build a super simple signal
generator uisng only one transistor and no other ICs.

The idea being you charge up a capacitor through a resistor then when the
cap is at an adequate potential, its voltage will turn on a transistor.
The
transistor being on simply completes a circuit allowing the cap to
discharge. Whence the cap has discharged, the transistor turns off, and
the
cap charges again. Repeat......

This seems logical to me, yet I cannot design a circuit that works this
way....it always seems like I'm missing something. Am I just not being
"clever" enough, or is such a circuit indeed impossible?

Thanks
Patrick




G.T.W.
Guest

Wed Sep 29, 2004 4:25 pm   



Did you add the capacitor after the bridge rectifier?

Otherwise, the 4 diodes provide the full wave positive alternations,
but are still largely 'ripple' voltage. The filtering provide by
capacitors smooths the ripple so that your average DC voltage is more
consistant.
Heavy loading will decrease the cap's filtering effectiveness (unless
you increase the value of the cap).

On Mon, 30 Jun 2003 07:55:58 -0700, Mike C. <noemail_at_nospam.com>
wrote:

Quote:


I added a bridge-rectifier circuit to my project's power input.
I needed this because it uses "wall-warts" for supply and sometimes
the polarity would be reversed. Adding the bridge solved that problem,
but created a new one.

The main circuit uses a National-Semi LM2674 switch to convert 9 to
35VDC input to 5VDC out. It works fine in full range (9-35V) without
the bridge.

After adding the bridge, it works fine from 9 to 14V. Above 14V, the
output of the LM2674 drops every few seconds for a few milliseconds.
This causes my embedded processor to reset. I'm positive that the
bridge is creating this problem because if I bypass it, everything
works as it should.

The bridge design I copied used just four 1n4004 diodes and is like
the one found at
http://www.mitedu.freeserve.co.uk/Circuits/Power/overvolt.htm

Any suggestions as to why it drops over 14VDC?

Mike.


G.T.W.
Guest

Wed Sep 29, 2004 4:25 pm   



Thank you for that correction. I didn't want to assume the author's
experience with schmitt triggers, so I tried to stay familiar to a
transistor referance.
When I was learning basics, transistors and op amps were taught
together. If anyone had told me I needed a Schmitt trigger, I would
not have had a clue. They're part of another class that would come a
semester later.

On Sat, 5 Jul 2003 17:08:13 +1000, "Mark" <higham_at_optushome.com.au>
wrote:

Quote:
The 7414 has 6 inverting gates with schmitt trigger inputs they aren't
opamps. It is the schmitt input that allows them to be used as oscilators. a
high input will not trigger until about 2v, it will stay in this state until
the input goes down to about .5v .


"G.T.W." <rampart_at_easynews.com> wrote in message
news:3f0668d6.6426629_at_news.ev1.net...
Your're on the right track. You can make an oscillator with one IC, a
resistor and a capacitor. If you use a Schmitt trigger, you can get
a stable digital pulse train at approximately 9,000 Hz. and at least
2.4 volts output.

Use a 7414 Schmitt Trigger. It has 6 op amps on one IC. It will invert
the output, so once your oscillator is going, all you need to do is
send the output of that first op amp to the input of the next one on
the IC, and you are back in phase.

Add a 1 KHz resistor to the output of the first op amp. At the other
end of the resistor, tie a .068 MicroFarad ceramic capacitor on to it.
Connect the other side of the cap to ground.
Next, send a jumper wire from between the R / C, back to the input of
the first op amp. Then you add one more jumper directly from the ouput
(before the resistor), and connect it to an input of the second op
amp. Take your output from the second 'gate'.

At those R/C values, you would get a nice, stable digital pulse train
at approximately 10KHz.

You can calculate the frequency with the following formula, assuming
a 1KHz resistor.

F = 0 . 000679 divided by Capacitor (.068 microFarads ) = 9,985 Hz

This only works for a resistor value of 1KHz, but you can change the
capacitance for a lower value if you want. Using a capacitor value
of .67 microFarads will give an output close to 1KHz.

Anyway, the Schmitt trigger will shape up the charging / discharging
capacitor into a nice square wave, and it will keep oscillating until
you turn the power off.

I don't want to assume, but don't forget to 'power' the IC .
Pin 7 is ground. Pin 14 is Vcc (5 volts).

I am sorry I don't know how to draw a schematic in a newsreader. I
hope my explanation isn't too poor. good luck.

On Sat, 05 Jul 2003 05:51:10 GMT, "Patrick Leonard"
transactoid_at_rogers.com> wrote:

I'm trying to figure out if its possible to build a super simple signal
generator uisng only one transistor and no other ICs.

The idea being you charge up a capacitor through a resistor then when the
cap is at an adequate potential, its voltage will turn on a transistor.
The
transistor being on simply completes a circuit allowing the cap to
discharge. Whence the cap has discharged, the transistor turns off, and
the
cap charges again. Repeat......

This seems logical to me, yet I cannot design a circuit that works this
way....it always seems like I'm missing something. Am I just not being
"clever" enough, or is such a circuit indeed impossible?

Thanks
Patrick






Kevin Dressel
Guest

Wed Sep 29, 2004 4:25 pm   



I suppose I can see the need with the flashlights, being that the alternative is more
LEDs (more costly upfront / bulkier).

I agree that based on usage, if I can get 25,000 hours of use out of them, it'll be
more than adequate. Very likely that by the time the LEDs burn out, some new and
improved light will be out anyway.

"Watson A.Name - 'Watt Sun'" wrote:

Quote:
In article <3F0651DC.5FA9A926_at_charter.net>, dressel1_at_charter.net
mentioned...
I figured that was the reason. I did put a 10 ohm resistor in series with the
3.6v source and the LED, but it still appeared brighter (of course, that is based
on my eyes, which are far from perfect).

For the series of 5, I figure I can get by with a 430ohm resistor (maybe a 390ohm
if necessary for brightness, though that will bring the current to around
21-22mA). Does anyone have a quantitative feel for how the higher current will
affect LED life? Based on the "perfect" scenario, they should last around
50,000-100,000 hours (these are probably standard numbers). I won't need them on
full time, so if I could get a quarter of that, it would more than suffice. Heat
dissipation should not be a problem.

For the white LEDs, I use a 33 ohm resistor for 4.5V = three AA cells.
That gives about 30 mA, and with fresh batteries, it's probably closer
to 40 mA. The lifetime may be shorter by 50%, but I figure that it'll
take a lifetime to put that many hours on them, being they get used a
few hours a year. So it really doesn't matter. And the tradeoff of
increased light output is well worth it.

I've read that some flashlight makers run their LEDs at 2 or 3x the
max, or something like 50 to 90 mA.

later,
Kevin

"Watson A.Name - 'Watt Sun'" wrote:

In article <3F06264D.4BB9FCE5_at_charter.net>, dressel1_at_charter.net
mentioned...

[snip]

If you put the LED across the 3.6V, you should *still* put a current
limiting resistor in series with it. If you don't, then the battery
is probably pushing much more than the 30 mA maximum thru the LED.
Watch out for the overly hot LED.

Thanks,
Kevin

[snip]


--
@@F_at_r_at_o_at_m@@O_at_r_at_a_at_n_at_g_at_e@@C_at_o_at_u_at_n_at_t_at_y@,@@C_at_a_at_l@,@@w_at_h_at_e_at_r_at_e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t_at_h_at_e@@a_at_f_at_f_at_l_at_u_at_e_at_n_at_t@@m_at_e_at_e_at_t@@t_at_h_at_e@@E_at_f_at_f_at_l_at_u_at_e_at_n_at_t@@


Watson A.Name - 'Watt Sun
Guest

Wed Sep 29, 2004 4:25 pm   



In article <3F06C992.734E02F1_at_charter.net>, dressel1_at_charter.net
mentioned...
Quote:
I suppose I can see the need with the flashlights, being that the alternative is more
LEDs (more costly upfront / bulkier).

I agree that based on usage, if I can get 25,000 hours of use out of them, it'll be
more than adequate. Very likely that by the time the LEDs burn out, some new and
improved light will be out anyway.

From my own personal experience, by the time the LEDs burn out, I will
have lost the light and had to replace it with another. Last week I
lost one of my regular mini Maglites, so I'm only out ten bucks. Glad
it wasn't the one with the LED conversion, 'cause that cost another
$30. :-/


Quote:
"Watson A.Name - 'Watt Sun'" wrote:

In article <3F0651DC.5FA9A926_at_charter.net>, dressel1_at_charter.net
mentioned...
I figured that was the reason. I did put a 10 ohm resistor in series with the
3.6v source and the LED, but it still appeared brighter (of course, that is based
on my eyes, which are far from perfect).

For the series of 5, I figure I can get by with a 430ohm resistor (maybe a 390ohm
if necessary for brightness, though that will bring the current to around
21-22mA). Does anyone have a quantitative feel for how the higher current will
affect LED life? Based on the "perfect" scenario, they should last around
50,000-100,000 hours (these are probably standard numbers). I won't need them on
full time, so if I could get a quarter of that, it would more than suffice. Heat
dissipation should not be a problem.

For the white LEDs, I use a 33 ohm resistor for 4.5V = three AA cells.
That gives about 30 mA, and with fresh batteries, it's probably closer
to 40 mA. The lifetime may be shorter by 50%, but I figure that it'll
take a lifetime to put that many hours on them, being they get used a
few hours a year. So it really doesn't matter. And the tradeoff of
increased light output is well worth it.

I've read that some flashlight makers run their LEDs at 2 or 3x the
max, or something like 50 to 90 mA.

later,
Kevin

"Watson A.Name - 'Watt Sun'" wrote:

In article <3F06264D.4BB9FCE5_at_charter.net>, dressel1_at_charter.net
mentioned...

[snip]

If you put the LED across the 3.6V, you should *still* put a current
limiting resistor in series with it. If you don't, then the battery
is probably pushing much more than the 30 mA maximum thru the LED.
Watch out for the overly hot LED.

Thanks,
Kevin

[snip]

--
@@F_at_r_at_o_at_m@@O_at_r_at_a_at_n_at_g_at_e@@C_at_o_at_u_at_n_at_t_at_y@,@@C_at_a_at_l@,@@w_at_h_at_e_at_r_at_e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t_at_h_at_e@@a_at_f_at_f_at_l_at_u_at_e_at_n_at_t@@m_at_e_at_e_at_t@@t_at_h_at_e@@E_at_f_at_f_at_l_at_u_at_e_at_n_at_t@@

John Jardine
Guest

Wed Sep 29, 2004 4:25 pm   



Patrick Leonard <transactoid_at_rogers.com> wrote in message
news:iHtNa.58918$a51.44869_at_news02.bloor.is.net.cable.rogers.com...
Quote:
I'm trying to figure out if its possible to build a super simple signal
generator uisng only one transistor and no other ICs.

The idea being you charge up a capacitor through a resistor then when the
cap is at an adequate potential, its voltage will turn on a transistor.
The
transistor being on simply completes a circuit allowing the cap to
discharge. Whence the cap has discharged, the transistor turns off, and
the
cap charges again. Repeat......

This seems logical to me, yet I cannot design a circuit that works this
way....it always seems like I'm missing something. Am I just not being
"clever" enough, or is such a circuit indeed impossible?

Thanks
Patrick

Nope, it won't work!.

A tranny doesn't have a true switching action, it's linear from full off to
full on.
It will always be happy to self stabilise at a balance point, where the
tranny is part 'on' and the cap' is still being slightly charged. Need at
least two transistors to do this job.
regards
john

John Fields
Guest

Wed Sep 29, 2004 4:25 pm   



On Sat, 05 Jul 2003 07:12:13 GMT, rampart_at_easynews.com (G.T.W.) wrote:

Quote:
Did you add the capacitor after the bridge rectifier?

Otherwise, the 4 diodes provide the full wave positive alternations,
but are still largely 'ripple' voltage. The filtering provide by
capacitors smooths the ripple so that your average DC voltage is more
consistant.
Heavy loading will decrease the cap's filtering effectiveness (unless
you increase the value of the cap).

---
Wrong.

You either went to the link and didn't know the bridge was shown
incorrectly connected or you didn't go to the link at all, in which case
you're still wrong.
--
John Fields
Professional circuit designer
http://austininstruments.com

Henry Kolesnik
Guest

Wed Sep 29, 2004 4:25 pm   



GE used to recommend NiCd be stored discharged and I've notice most new
stuff is shipped with the batteries requiring at least 24 hours, so I guess
discharged is best storge method. Cool and dry also.
hank wd5jfr
"a.n" <recycle_at_btinternet.com> wrote in message
news:bdqk5a$24f$1_at_titan.btinternet.com...
Quote:
Hello everyone,
Question about rechargable batteries:
What is the best way to store AA rechargable batteries. Is it to simply
keep them charged and then discharge them and recharge them after
about 6 month, or is it better to store them discharged and then charge
them
and discharge them again after 6 months. Would it make any difference
if the batteries are nicad, ni-mh, or ultra capacity ni-mh?



John Larkin
Guest

Wed Sep 29, 2004 4:25 pm   



On Sat, 05 Jul 2003 05:51:10 GMT, "Patrick Leonard"
<transactoid_at_rogers.com> wrote:

Quote:
I'm trying to figure out if its possible to build a super simple signal
generator uisng only one transistor and no other ICs.

The idea being you charge up a capacitor through a resistor then when the
cap is at an adequate potential, its voltage will turn on a transistor. The
transistor being on simply completes a circuit allowing the cap to
discharge. Whence the cap has discharged, the transistor turns off, and the
cap charges again. Repeat......

What actually happens in most transistor-r-c circuits is a stable
equilibrium without oscillation.

Quote:

This seems logical to me, yet I cannot design a circuit that works this
way....it always seems like I'm missing something. Am I just not being
"clever" enough, or is such a circuit indeed impossible?

Thanks
Patrick


If a transistor is operating normally, I'd say it's impossible to make
an oscillator with one transistor, one capacitor, and one resistor.
Special cases are...

1. Many transistors oscillate, typically at hundreds of MHz, if biased
right (or maybe wrong); but it's not your RC that sets the frequency.
You can make a nice RF oscillator from a 2N2219 and one resistor.

2. Avalanche mode: leave the base open and charge the collector
through the r-c, with a high-voltage (100-500v depending on the
device) supply; you may get a nice sawtooth at the collector.

One transistor and three each r,c makes a phase-shift oscillator.

John

John Popelish
Guest

Wed Sep 29, 2004 4:25 pm   



Patrick Leonard wrote:
Quote:

I'm trying to figure out if its possible to build a super simple signal
generator uisng only one transistor and no other ICs.

The idea being you charge up a capacitor through a resistor then when the
cap is at an adequate potential, its voltage will turn on a transistor. The
transistor being on simply completes a circuit allowing the cap to
discharge. Whence the cap has discharged, the transistor turns off, and the
cap charges again. Repeat......

This seems logical to me, yet I cannot design a circuit that works this
way....it always seems like I'm missing something. Am I just not being
"clever" enough, or is such a circuit indeed impossible?

Thanks
Patrick

I don't think that the oscillator you describe is possible. There are
several steps in the oscillation process that you are missing. What
causes the transistor to suddenly switch on when the cap reaches some
arbitrary voltage, but stay on while the voltage falls below that till
some other arbitrary voltage is reached?

The closes thing I can think of is a phase shift oscillator (which
produces a sine wave rather than a pulse or saw tooth waveform. And
it takes a few more resistors and capacitors, but can run on a single
transistor.
http://www.tpub.com/neets/book9/35e.htm

--
John Popelish

nick
Guest

Wed Sep 29, 2004 4:25 pm   



i141802596_at_yahoo.com (nick) wrote in message news:<4f981f90.0307041718.22a99a5a_at_posting.google.com>...
Quote:
1.if i want to make a simple remote controller and reveiver ,e.g. when
i pressing button A from the controller,then LED A will turn on in the
receiver side,i don't want the use IR TX &RX,so i want to use RF, but
i have no idea to make it,so anyone can give me a simple diagram,or
give me some web sites about the RF.

2.i have found some diagrams about them ,i think they are very
complex for me ,in the diagrams a component called inductor,what is
this? , How to use it?

3.if i want to make a simple bug,is it principle like the remote
contoller? i have found some web sites only teach how to make the
transmitter,they use radio to work as the receiver, that means anyone
turn their radio to the same band ,then they also can receive the
singles?

any one ? can help me?

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