Chip with simple program for Toy

[Scott Wiper]

"Fred Bloggs" <nospam@nospam.com> wrote in message news:406EC8D0.1000403@nospam.com...


Scott Wiper wrote:
I have made some changes on RLY DRVR A (PIN 2 U8) and RLY DRVR B (PIN 5 U8). You will notice the changes on sheet A. This inclucdes U4/U5 and U6/U5.
I have fixed the other problem with the 4071 by using D3 and Q3 on the LS273
http://www.travel-net.com/~swiper/7000a.gif

While in the LIGHT state, IF the LDR comparator indicates a
LIGHT-to-DARK transition, then:
1) enable the 120 second counter, no other signal changes made
2) wait for 120s timeout and RST 120 second counter on any DARK-to-LIGHT
transition of LDR
3) IF LDR remains DARK throughout 120 second interval then put circuit
in DARK state ELSE circuit remains in LIGHT state

All of the chnages as you descirbed have been made on sheet C. U19 on sheet C can be tied back to the MASTER RESET. Keeping both timer sections (A and B) in the light state or the dark state. In the light state the totaliser will be advanced on counter section B. The counter section A will be held in reset intil transition. This a achieved on U19 and U20 tying PIN 1 to PIN 13 so there is a 120 second hold until reset. I cheated a bit here by taking the 1HZ and using U17 as devide by two function.

You see it here.
http://www.Travel-net.com/~swiper/7000c.gif

The circuit in DARK state for less than 5 hours is:
1) 1 second pulses to totalizer remain enabled
2) RLY DRVR A is OFF
3) RLY DRVR B is ON
4) 5 HOUR timer is taken out of RST and allowed to count
5) Hold this state until the sooner of 5.1) LDR indicates DARK-to-LIGHT
transition after passing 120 second timeout test, or 5.2) 5 HOUR timer
indicates 5 hour mark has been reached.

The circuit is in DARK state > 5 hours, then:
1) 1 second pulses to totalizer disabled
2) RLY DRVR A is OFF
3) RLY DRVR B is OFF
4) circuit remains in this state until LDR DARK-to-LIGHT transition
passes 120 second timeout test.

I have done this by ORing the two outputs from U8 and inverting this signal to U12's rest input. So when in light state reset on counter section B is held low enabling the count to the totaliser. Even when RELAY B is in timeout. After the delay cycle the inverter will place a high signal on the reset stopping the count and disabling the totaliser, But counter section A continues to until DARK to LIGHT transition.

Here at http://www.trvael-net.com/~swiper/7000b.gif


Keep plugging at this.

Oh I am still plugging at it. I am begining to grasp what you are saying.
Can you look at this design.

 

[Fred Bloggs]

Scott Wiper wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:406EC8D0.1000403@nospam.com...


Scott Wiper wrote: I have made some changes on RLY DRVR A (PIN 2 U8)
and RLY DRVR B (PIN 5 U8). You will notice the changes on sheet A.
This inclucdes U4/U5 and U6/U5. I have fixed the other problem with
the 4071 by using D3 and Q3 on the LS273
http://www.travel-net.com/~swiper/7000a.gif

While in the LIGHT state, IF the LDR comparator indicates a
LIGHT-to-DARK transition, then: 1) enable the 120 second counter, no
other signal changes made 2) wait for 120s timeout and RST 120 second
counter on any DARK-to-LIGHT transition of LDR 3) IF LDR remains DARK
throughout 120 second interval then put circuit in DARK state ELSE
circuit remains in LIGHT state

All of the chnages as you descirbed have been made on sheet C. U19 on
sheet C can be tied back to the MASTER RESET. Keeping both timer
sections (A and B) in the light state or the dark state. In the light
state the totaliser will be advanced on counter section B. The
counter section A will be held in reset intil transition. This a
achieved on U19 and U20 tying PIN 1 to PIN 13 so there is a 120
second hold until reset. I cheated a bit here by taking the 1HZ and
using U17 as devide by two function.

You see it here. http://www.Travel-net.com/~swiper/7000c.gif

The circuit in DARK state for less than 5 hours is: 1) 1 second
pulses to totalizer remain enabled 2) RLY DRVR A is OFF 3) RLY DRVR B
is ON 4) 5 HOUR timer is taken out of RST and allowed to count 5)
Hold this state until the sooner of 5.1) LDR indicates DARK-to-LIGHT
transition after passing 120 second timeout test, or 5.2) 5 HOUR
timer indicates 5 hour mark has been reached.

The circuit is in DARK state > 5 hours, then: 1) 1 second pulses to
totalizer disabled 2) RLY DRVR A is OFF 3) RLY DRVR B is OFF 4)
circuit remains in this state until LDR DARK-to-LIGHT transition
passes 120 second timeout test.

I have done this by ORing the two outputs from U8 and inverting this
signal to U12's rest input. So when in light state reset on counter
section B is held low enabling the count to the totaliser. Even when
RELAY B is in timeout. After the delay cycle the inverter will place
a high signal on the reset stopping the count and disabling the
totaliser, But counter section A continues to until DARK to LIGHT
transition.

Here at http://www.trvael-net.com/~swiper/7000b.gif


Keep plugging at this.

Oh I am still plugging at it. I am begining to grasp what you are
saying. Can you look at this design.

I am still finding errors in the divide ratios and interconnections,
with some counters unconnected. Below I have drawn three essential
counter types that the controller requires, and this includes the good
one you have been using. Try to draw up a Sheet D that shows the overall
circuit operation controlling your counters using the "high level"
blocks given. I have provided an example below with a single 1Hz clock
source that is sufficient for your purposes- you do not require two 1Hz
sources- you are using many more counters than you need, and you do not
require a single CD4017. Your Sheet D should show the counter TYPE
blocks and how your external signals like the LDR comparator go into
controlling various RST lines.
Please view in a fixed-width font such as Courier.



BACKGROUND COUNTER TYPES






TYPE 1 This counter divides input frequency by N
and outputs a pulse of duration equal to
CK IN '0' time. This counter is used through-
out existing arrangement.
CD4024
-or-
CD4040
-or-
CD4020
+-----+ 1/2
| CTR | CD4082
| |---|__
| |---| \ G1
| N | | >----------+
CK | |---|__/ | 1/2
IN | |---| | CD4013

-+--o>CK | +-------+
| | | | S | CLKOUT

| | | +---|D Q|-+--->
| | | | | | |
| | | +--->CK FF | | ______
| | MR | | | _| | CLKOUT
| +-----+ 'O' | Q|----->
| | | R | |
| | +-------+ |
| | | |
+---------------------------+ |
| |
+---------------------+ |
| |
+------------------|-----+
| 1/4 |
| CD4071 |
+-----\ \ |
| >--------+
RST>-------------/__/





TYPE 2 This counter divides input frequency by N
and outputs a pulse of duration equal to
M CK IN periods.

CD4024
-or-
CD4040
-or-
CD4020
+-----+
| CTR |
| |---|__
| |---| \ G1
| N | | >-----------------------+
| |---|__/ |
CK | |---| |
IN | | |

--+-o>CK | CD4082 |
| | | |

| | |---|__ 1/4 |
| | |---| \ G2 CD4071 |
| | M | | >-----\ \ |
| | |---|__/ | >-------------|-----+
| | |---| +--/__/ | |
| | | | | |
| | | | +---------+ |
| | | | | CD4013 | |
| | | | +-----+ +-----+ |
| | | | | S | | S | |
| | | | +--|Q D|-+-|D Q|--|---> CLKOUT
| | MR | | | | | | | | |
| +-----+ | | | CK<-+->CK | |
| | | | |FF1 | | |FF2 _| | ______
| | | | | |'O'| Q|--|---> CLKOUT
| | | | | R | | R | |
| | | | +-----+ +-----+ |
| | | | | | |
+------------------|----|-----+ +-----+
| | |
+-------------|----|---------------+
| | 1/4 |
| | CD4071 |
| +----\ \ |
| | >------+
RST>------------------+---------/__/



TYPE 3 This counter times out N cycles,sets DONE='1'
and remains locked in that state until RST is
reapplied.

CD4024
-or-
CD4040
-or-
CD4020
+-----+ 1/2
| CTR | CD4082
| |---|__
| |---| \ G1
| N | | >----------+
CK | |---|__/ | 1/2
IN | |---| | CD4013

----o>CK | +-------+
| | | S |

| | +---|D Q|-+--DONE
| | | | | |
| | +--->CK FF | |
| MR | | | _| | ____
+-----+ 'O' | Q|-|--DONE
| | R | |
| +-------+ |
| | |
+-----------------+ | |
| | |
+-----------------|---------+
| 1/4 | |
| CD4071 | |
+-----\ \ | |
| >-------+ |
+-----/__/ |
| |
RST>----+---------------------+



SCHEMATIC SYMBOLS USED ON HIGH LEVEL DIAGRAM



+-----CTR_TYPE1-----+ +-----CTR_TYPE2-----+
--o>CKIN CKOUT|-- --o>CKIN CKOUT|--
| N=XXXX | | N=XXXX |
| _____| | M=ZZZZ |
--|RST CD40YY CKOUTo-- | _____|
| | --|RST CD40YY CKOUTo--
+-------------------+ | |
+-------------------+



+-----CTR_TYPE3-----+ :
--o>CKIN DONE|-- INDICATES CIRCUIT SCHEMATIC ABOVE WITH
| N=XXXX |
| | CD4082 G1 INPUTS CTR STATE BINARY XXXX
| ____|
--|RST CD40YY DONEo-- CD4082 G2 INPUTS CTR STATE BINARY ZZZZ
| |
+-------------------+ CTR IC PART NO. CD40YY






-- COMPUTATION OF CTR STATE N/M BINARY INPUTS --



CD4024, CD4020, CD4040 Q outputs are numbered 1 onwards.

* 0 1 (k-1)
Q1 represents 2 Q2 represents 2 Qk represents 2


Example: compute CD4082 inputs for N=3600

11
Method: 3600= 1 x 2048 (= 2 or Q12 )
10
+ 1 x 1024 (= 2 or Q11 )
9
+ 1 x 512 (= 2 or Q10 )
4
+ 1 x 16 (= 2 or Q5 )


Therefore wiring would be like so:


+----------+ 1/2
| | CD4082
| N= Q5 |---|__
| 3600 Q10|---| \
| | | >--
CK | Q11|---|__/ FREQ IN
IN | Q12|---| FREQ OUT= -------

----o>CK | 3600
| |

| |
| |
| |
| MR |
+----------+


* Some counter diagrams may show Q0, Q1, etc...use your
common sense.





Example 1HZ free-running timebase:


150HZ 1HZ
-> ->
+--------+ +-------------+ +-----CTR_TYPE2-----+
| | | Q14|----o>CKIN CKOUT|->1HZ50%
| OUT|--o>CK | | N=150 |
| | | | | M=75 |
| XYSTAL | | CD4020 | | _____| ______
|2.457600| | | | CD4040 CKOUTo->1HZ50%
| MHZ | |MR | |RST |
+--------+ +-------------+ +-------------------+
| |
| |
POWER_ON_RST--+------------------+

 

[Sal Brisindi]

Spencer,
Are you using a 9 volt transistor radio type battery, if so the reason why the
voltage drops below 3 volts is because the battery cannot supply the current
demand of the motor.

Sal
www.tuberadios.com

Spencer wrote:

Could anyone please help me with the following: I have built a toy car
powered by a 3V DC motor? The motor turns a small gearbox which is connected
to the wheels by means of a pulley drive.

I would like to use a LM317T voltage regulator to take the 9 volt supply
voltage down to 3V. Am I correct to say that I should first put the motor
under load and then adjust the resistor connected to the adjusting leg
accordingly to change the voltage to 3V?

I am using an ordinary 9V cell, but I am really having difficulty getting
the voltage higher than 1, 9 volts under load. (no matter what value
resistor I use)

Secondly, what is the purpose of adding capacitors between the + and -
rails?

Thank you - I appreciate your time.

Spencer

[Image]

 

[petrus bitbyter]

"Spencer" <spence@downtown.co.nz> schreef in bericht
news:tU7cc.5773$d%6.99948@news.xtra.co.nz...

Could anyone please help me with the following: I have built a toy car
powered by a 3V DC motor? The motor turns a small gearbox which is
connected
to the wheels by means of a pulley drive.



I would like to use a LM317T voltage regulator to take the 9 volt supply
voltage down to 3V. Am I correct to say that I should first put the motor
under load and then adjust the resistor connected to the adjusting leg
accordingly to change the voltage to 3V?



I am using an ordinary 9V cell, but I am really having difficulty getting
the voltage higher than 1, 9 volts under load. (no matter what value
resistor I use)



Secondly, what is the purpose of adding capacitors between the + and -
rails?



Thank you - I appreciate your time.



Spencer

Spencer,

Your circuit is faulty. Download the LM317 datasheet, print it and have a
closer look.

Don't know what type of 9V power supply you're using but it might not
provide enough current.
(It's not the cause of the current problem but may become the cause of the
next.)

The LM317 may become hot enough to require some cooling.

If your 9V power supply is a kind of a battery an LM317 is a terrible waste.
You'd look for a switching regulator.

petrus





---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.639 / Virus Database: 408 - Release Date: 22-3-2004

 

[Rich Grise]

Well, I dunno. What's 3 v * 153 ma?

"Spencer" <vetmeisie@webmail.co.za> wrote in message
news:9rr9c.2334$Tf3.38912@news.xtra.co.nz...

Petrus,

I calculated that I need a 20 Ohm resistor. What should I do if the 0.25
Watt resistor can't rid of the heat? That would probably be the case with
this resistor.

Regards,
Spencer


"petrus bitbyter" <p.kralt@reducespamforchello.nl> wrote in message
news:Vzn9c.40708$s9.24523@amsnews02.chello.com...

Consider the moter a part of the voltage divider and forget the lower
(10k)
resistor. You only have to adjust the upper resistor to a value that
makes
you measuring 3V at the motor. Guess it will be much lower then 10k.
Beware
of the power dissipation. A 0.25W resistor may be to small to get rid of
the
heat.

petrus

 

[Ian Bell]

pil wrote:
snip

Which books would you recommend for a third year EE student who is very
much into amplifiers and sound? I am looking for a book which actually
have circuits that I can construct.

Anything by Douglas Self. Find him here
http://www.dself.dsl.pipex.com/ampins/ampins.htm

Ian

 

[Ian Bell]

Neil Preston wrote:

The Class D amps are certainly making many inroads in many consumer
products, but how fast it replaces analog power amplifiers remains to
be seen. For stereos sold at Wal-Mart, the transition may already be
100 percent, but I doubt you could find a 'Class D' amp in high-end
audio stores, or if so, that it sounds as good as any analog amplifier
in the store. Analog amps in that area (admittedly not a large
consumer market) will surely be around for a long time, unless Class D
amps show substantially more improvement over current state of the
art.


Pro audio is wholeheartedly embracing the fully digital amplifier. Crown
Audio makes the CE4000 and other models that are fully digital, and Peavey
has been making digital amps since the '80s. I'm sure there are
others.....

Please define 'fully digital'

Ian

 

[John - G0WPA]

Good Question,

To answer your question as its worded, the 317 is short circuit protected,
which means it will current limit if the load goes short. This doesnt mean that
the device itself wont ever go short circuit. A 317 is as liable to go short as
any other semi-conductor device.

But as to why your LM317 went short in your circuit? Its a good question. Did
you check that the cap wasnt actually holding a reverse charge, or even a high
forward voltage! Maybe the LM317 was short before you used it! Using a series
resistor, maybe 10ohm, 5W would have been better, and allow the cap to charge
in a controlled fashion for you, rather than relying on the 317s internal
current limiting, which is a fail safe, not a useable feature as such. As for
the heatsink and thermal protection, if it blew instantly, theses werent
factors. .. John.

Drop QRM off my addy to e-mail me.

 

[CFoley1064]

Subject: Why do "short proof" votlage regulator chips short out?
From: larrymoencurly@my-deja.com (larrymoencurly)
Date: 4/12/2004 7:42 PM Central Standard Time
Message-id: <755e968a.0404121642.47fd4345@posting.google.com

I was using an LM317 to charge up a big capacitor. I connected it
after I turned on the power, and the LM317 blew out instantly. I
realize that it was stupid to turn on the power first, but why didn't
the LM317's built-in short-circuit and thermal protection work?

The LM317 was hooked up in linear mode, and I had bypass capacitors on
the input and output, just as National Semiconductor recommends (.1 uF
ceramic in parallel with 10 uF low-ESR Sanyo OS-Con, plus 2,000 uF
filter on the input side). The regulator was mounted on a fairly
large heatsink, roughly 3" x 3" x .75", and I think that I followed
proper layout recommendations.

I'm assuming here that you mean you connected the cap to the circuit after you
turned the power on.

Is it possible that you had a fairly long wire going from your power supply to
the cap? All wires have inductance. At the moment the cap is starting to make
contact with the power supply, you're forcing current through the inductor to
the cap. If there's any contact bounce, your inductor is going to start trying
to backfeed the LM317, and the voltage at the output pin (V1) will rapidly
exceed the voltage at the input. That might have been what killed the LM317.
Can't do that. You'll notice when you touch a cap with a battery wire, you
frequently get a bit of a spark. That's what's going on here.

Switched Cap Load
____ SW1
.-------. | | V1 ___ _/
o-----o~ +o----o----| 317|---o----o---UUU--o/ o----.
| | |____| | | |
AC | BR1 | | + | | | |
| | ### | .-. | |
| | --- | | | | | +
o-----o | | | | | --- ###
|~ -o----o | '-' --- ---
'-------' | | | | |
| o------' | |
| | | |
| .-. | |
| | | | |
| | | | |
| '-' | |
| | | |
'------o-----------o-----------------'


created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Get another LM317, and next time, connect the cap before turning on power. If
you can't do that, get a 1N5402, and connect the anode to the LM317 output, and
the cathode to GND. Get another 1N5402 and connect the anode to the input pin,
and the cathode to the output. This will protect the LM317 against reverse
polarity as well as Vout exceeding Vin. Not bulletproof, but the protection's
a bargain for the price.

317 Protection
.-------|<--------.
| ____ |
| | | |
o----o------| 317|-----o-------o
|____| | |
| .-. |
| | | |
| | | |
| '-' |
| | |
o--------' -
| ^
.-. |
| | |
| | |
'-' |
| |
o-------------o----------------o

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Good luck
Chris

 

[larrymoencurly]

johnandsimone@aol.comQRM (John - G0WPA) wrote in message news:<20040412214847.08083.00000308@mb-m11.aol.com>...

To answer your question as its worded, the 317 is short circuit
protected, which means it will current limit if the load goes
short.

I didn't word the title right. My LM317 went bad, and apparently it's
output transistor is now shorted because it now always puts out almost
the full input voltage, regardless of the voltage setting.

Did you check that the cap wasnt actually holding a reverse
charge, or even a high forward voltage!

I shorted both its leads against the aluminum case of the power
supply.

Maybe the LM317 was short before you used it! Using a series
resistor, maybe 10ohm, 5W would have been better, and allow
the cap to charge in a controlled fashion for you, rather
than relying on the 317s internal current limiting, which
is a fail safe, not a useable feature as such.

This LM317 had been working normally the day before. I'm sure I
ruined it with the capacitor.

 

[Temporary FL@L&ER]

On 13 Apr 2004 10:10:50 -0700, docsavage20@yahoo.com (Doc) wrote:

I'm about to try soldering some pieces onto a circuit board for a
condenser microphone modification. The metallic area on the back side
of the circuit board where the solder joints are supposed to be made
looks like a fairly small target and the tip on my 30 watt iron looks
a bit big, i.e. it's rounded off sort of like a very large turntable
stylus, not sharply pointy. Should it be sanded/filed down to a fine
point?

Thanks for all input.
I agree with Jonesy - if you need a smaller tip, improvise. I make my

own tips with 10AWG copper down to 16 ga- whatever fits the iron/gun I
am using at the time. Lasts just as long as the original at almost no
$$ spent.


Use the usual techniques to reply via email.

Molon Labe!

 

[John Miller]

Doc wrote:

I understand that you're supposed to coat the tip of the iron with solder
to facilitate heat transferance, however I'm finding that the solder is
simply balling up and won't flow over the tip of the iron. What could be
causing this?

Do you have a wet sponge for wiping the (hot) tip on before tinning?

--
John Miller

HOW YOU CAN TELL THAT IT'S GOING TO BE A ROTTEN DAY:
#32: You call your answering service and they've never heard of you.

 

[Phrankesh Roman]

On Wed, 14 Apr 2004 12:01:48 GMT, "Doc"
<docsavage20@_remove_this_to_reply_hotmail.com> wrote:

I understand that you're supposed to coat the tip of the iron with solder to
facilitate heat transferance, however I'm finding that the solder is simply
balling up and won't flow over the tip of the iron. What could be causing
this?

It may be time for a new tip. Another problem may be that the heating
element is shot and needs to be replaced.


Phrankesh Roman

 

[David Wood]

This may help prevent tip oxidation in the future: When finished using
an iron or solder station, clean the tip and leave a small ball of
solder on it while it cools. This leaves the tip surface in an anoxic
environment during cool down and during the next start up.

 

[Terry Pinnell]

"krem" <news@rkremser.endjunk.com> wrote:

Hello, I have a motor lifting an object over a set range. Upon reaching a
set height i would like it to reverse and then start to lower, and again
once reaching a bottom position change direction and begin to rase again. I
know i'm going to have to end up using limit switches and would prefer to
use two push button ones if possible. (temperary on) The motor which needs
to be controlled is going to be running on either 5-6 volt dc supply and
doesn't pull a huge current. (typically runs on 4 AA batteries and lasts a
long time) I would like to have a solution which doesn't include and
microcontrollers as i've yet to take that plunge. Thanks in advance for any
input or suggestions

This naturally doesn't meet your requirement for constant unattended
operation. But it may be of interest to anyone who ends up here via
keywords like 'motor' and 'reversal'. Uses a 2-pole changeover relay
plus two N/C microswitches.
http://www.terrypin.dial.pipex.com/Images/AutoMotorExtremes.gif

I used it on a couple of projects, such as an automatic curtain
operator, and a window-operator. The latter (using a surplus
windscreen-wiper motor) would let me open the window after switching
off the bedside light.

Used it after reading, with window initially closed against
moths/bugs; avoided having to emerge from 'heavy eyelid mode'. Or
again in morning in heavy rain/wind. Its inherent 'locking' into the
extreme positions was a nice feature, but really strong winds
occasionally got the better of my primitive mechanics. Gave me years
of useful performance.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[CFoley1064]

Subject: Choosing a general-purpose soldering gun
From: wylbur37nospam@yahoo.com (wylbur37)
Date: 4/22/2004 6:10 AM Central Standard Time
Message-id: <8028c236.0404220310.7545b461@posting.google.com

When I was a teenager, I purchased a Weller 8200 dual-heat (100/140
watt) soldering gun. It served me well over the years. Then about ten
years ago, I went through a period where I didn't need to do any
soldering and decided to give it away as part of "uncluttering". (I
later realized that was a mistake).

Now, I need to do some soldering again and need a soldering gun. I
could purchase a new Weller 8200 through mail-order, which will cost
over $40. (I could also get one used via eBay for as low as $8, but I
don't trust buying used stuff from strangers).

I could also purchase a new soldering gun at Radio Shack, either a
dual-heat (150/230 watt) model (Cat. No. 64-2187) for $29.99, or a
single-heat (100 watt) model (Cat. No. 64-2193) for $12.99.

If I recall correctly, back when I still had my Weller 8200, the first
heat setting (100 watts) was usually enough for most wiring jobs. Very
rarely did I have to click it back to the second position (140 watts)
except to solder some thicker metal items.

I'm currently leaning towards getting the Radio Shack 100 watt model,
not only to save money but because the heavier model (150/230 watt) is
probably too hot anyway and might burn out some components. Another
reason I'm leaning towards getting the Radio Shack one is that there
are numerous Radio Shack retail stores near where I live (in New York
City) and I could just go and buy one over-the-counter without the
hassle of waiting for it to be delivered. (As for the Weller 8200,
it's more expensive and I don't know which stores in New York City
would sell it anyway).


I was wondering what your thoughts were on ...

a. whether 100 watts is sufficient for most soldering jobs
(involving ordinary stuff like LEDs, resistors, etc.).

b. whether you've had any experience using the Radio Shack soldering
guns mentioned above and whether you think they're any good.

c. whether you know of any retail stores in the New York City area
that sell a Weller 8200 soldering gun for $30 or less.


Thanks for your help.

The right tool for the job. Use a soldering iron for PC boards and small work,
and a gun for the big jobs. If you need to do the Radio Shack thing, your best
bet for electronics might be the "Soldering Work Station with Dual-Powered
Iron", RS p/n 64-2184. It's $21.99, and has a built-in stand and sponge and a
dual temp setup (diode with switch) to give you 20 or 40 watts.

Having said that, the Weller is a good gun, and worth the price. Of course,
you'll have to replace the tips, but they're consumables, anyway. A reliable
beast. You can also get replacement parts from the manufacturer. A reputable
eBay seller wouldn't deliberately risk negative feedback over something that
small.

Neither of the two Radio Shack guns you're looking at are the best choice,
although they'll do in a pinch. The one I used a number of years back got
rather hot to hold after a bit of use.

If you're going to use it frequently, you need to pump a lot of heat for a long
period of time, or if the difference between adequate and good is important to
you, the Weller is worth it. If you want something cheap, go to

http://www.harborfreight.com/

They have two cheapies (150 and 185 watts) for 10 bucks each. No guarantees
there.

Good luck
Chris

 

[John Popelish]

John wrote:

I have a output that provides a voltage of 5v when high/0v when low.

This output can go repeatedly high & low. I want to smooth the output, so
when in goes low it cannot go high again for at least 5 mins.

The timer needs to start when the input goes low, and not start the timer
again if the input goes high/low during the "off" time.

If the input is high but the output is low (off time), I want the output to
go high again at the end of the time period.

I have tried to do this myself, I could not get a 555 timer to take the
input going low as a trigger as it held off the timer completely which the
input was high.

My second attempt used two AND gates and two capacitors and a transisitor,
which sort of works except that at the end of the time period if the input
is still high the output only goes up to 2.5v (see schmatic in abse)

Any idea how do do this easily?

[Image]

I think you need both a timing function and a gate function (to
disable any inputs to the timer when it is timing). Do you also want
the output to remain low after the time interval, if the input is low
at that point? Should the output go high immediately, after that if
the input goes high?

--
John Popelish

 

[Walter Harley]

"John" <John.Mason1212@hotmail.com> wrote in message
news:VqRhc.1214$Aa7.1019@pathologist.blueyonder.net...

I have a output that provides a voltage of 5v when high/0v when low.

This output can go repeatedly high & low. I want to smooth the output, so
when in goes low it cannot go high again for at least 5 mins.

The timer needs to start when the input goes low, and not start the timer
again if the input goes high/low during the "off" time.

If the input is high but the output is low (off time), I want the output
to
go high again at the end of the time period.

I think what you're looking for is a "non-retriggerable monostable
multivibrator" (Google for that). Use the input signal as the input to the
multivibrator (you might need to invert it first); then use the output of
the multivibrator to gate the input signal with an AND gate (again, you
might need to use an inverter).

In other words, the logic is that when the input signal goes low you trigger
a 300-second pulse that can't be retriggered until it's done; whenever that
pulse is high, you force the output to be low, and otherwise you let the
output be a copy of the input.

You should be able to do this with two chips: they make CMOS
non-retriggerable multivibrators, e.g. the 74HC221, and then you'd use a
quad NAND gate to get all the rest of your logic (tie the inputs of a NAND
together to get an inverter; use an inverter on the output of a NAND to get
an AND; and so forth). The only question is whether you can get a stable
timing interval as long as 300 seconds; if not, your other recourses are
going to be more complicated.

 

[Tim Wescott]

Walter Harley wrote:

"John" <John.Mason1212@hotmail.com> wrote in message
news:VqRhc.1214$Aa7.1019@pathologist.blueyonder.net...

I have a output that provides a voltage of 5v when high/0v when low.

This output can go repeatedly high & low. I want to smooth the output, so
when in goes low it cannot go high again for at least 5 mins.

The timer needs to start when the input goes low, and not start the timer
again if the input goes high/low during the "off" time.

If the input is high but the output is low (off time), I want the output

to

go high again at the end of the time period.


I think what you're looking for is a "non-retriggerable monostable
multivibrator" (Google for that). Use the input signal as the input to the
multivibrator (you might need to invert it first); then use the output of
the multivibrator to gate the input signal with an AND gate (again, you
might need to use an inverter).

In other words, the logic is that when the input signal goes low you trigger
a 300-second pulse that can't be retriggered until it's done; whenever that
pulse is high, you force the output to be low, and otherwise you let the
output be a copy of the input.

You should be able to do this with two chips: they make CMOS
non-retriggerable multivibrators, e.g. the 74HC221, and then you'd use a
quad NAND gate to get all the rest of your logic (tie the inputs of a NAND
together to get an inverter; use an inverter on the output of a NAND to get
an AND; and so forth). The only question is whether you can get a stable
timing interval as long as 300 seconds; if not, your other recourses are
going to be more complicated.

For a five minute time interval you really ought to be looking at
something with a counter. For something with a counter the smallest
thing by far that you'll find is a PIC or an AVR microcontroller.

If you just can't stand the thought of an 8-pin microcontroller then
consider a 555 for timing, a 4040, 4020, 4060 for a divider, a 74HC76
for tracking the state, and some glue logic. That's only four chips and
40 pins...

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[John]

I think you need both a timing function and a gate function (to disable
any inputs to the timer when it is timing).

Do you also want the output to remain low after the time interval, if the
input is low at that point? Should the output go high immediately, after

that if the input goes high?

Yes, youve got it. If input goes high during the time period then the output
should go high after the end of the time period. If input is low after the
time interval then output should also be low. After the time period if the
input goes high the output should immediatley go high.

The output should be the same as the input - with the exception that once
input goes low it remain low for 5 mins regardless of what the input does.
After the time period it follows the input again. i.e stays low untill input
goes high (then output goes high), or goes immedialy high is input is high.
When input falls low then time starts/cycle starts again.

The 5 mins isnt critical if it varys it doesnt matter as long as it is say
within 1min either way. (i.e 20% tollerance)