looking for a 555 timer circuit
Wayne
will have 3 120VAC outputs that it alternates between. This would be
used to drive 3 powerheads (water pumps) on a fish tank to create some
wave motion in the water. I would think that the time it alternates
could be varied based on a rheostat. Also, if possible, I want to have
a pushbutton that would stop all 3 pumps for 30 minutes or so for
feeding (push the button, power is killed for 30 minutes to the pumps,
then they startup again automatically). So, probably would be 2
circuits. Anyone have a design for something like this or can help?
TIA
Rich Webb
Although the guys for whom solder is a favorite programming language may
have different ideas, I'd do this with a small microcontroller. One A/D
port to read the rheostat, a digital port to sense the pushbutton, and
three digital outputs to control the power to the pumps.
Could probably be done with an 8-pin ATtiny15 (same size as the 555)
which has an internal oscillator, a couple of timers, and four A/D
channels plus the rheostat, pushbutton, relay drivers, and relays.
--
Rich Webb Norfolk, VA
Wayne
hand to build this, just not the knowledge of where to put the parts.
Rich Webb
>I'm looking for the solder version :^) I have most of the parts on
>hand to build this, just not the knowledge of where to put the parts.
Take a look at the "555 Designer" at:
http://www.schematica.com/555_Timer_design/Timer_Comparison.htm The
"Pro" version isn't very expensive and has some helpful "wizards" that
will at least give you some ideas, including I/O circuits, long interval
timers, variable timers, and so on.
Bob Monsen
555 -> 4017 -> 2N4401 -> relay coil. Set up the 4017 so the Q3 output
resets it. Also, use one of the timer-counter chips, like a 4541, to
inhibit the entire circuit using yet another relay for your 30 minute
pause...
The datasheets for these chips and transistors are freely available, and
usually have application schematics in them that are close to what you are
interested in. If you need more help, post again.
Note that you'll need a DC supply for the logic, which won't run on AC.
--
Regards,
Bob Monsen
"I cannot persuade myself that a beneficent and omnipotent God would
have designedly created parasitic wasps with the express intention of
their feeding within the living bodies of Caterpillars."
-- Charles Darwin
Wayne
http://www.uoguelph.ca/~antoon/gadgets/555/555.html. I was thinking of
using the '10 minute timer' to power a second circuit consisting of the
'555 timer tester'. Instead of powering the red LED in the timer
circuit, I would have that power the tester circuit. In the tester
circuit, I would substitute solid state relays for the LEDs. Is this
feasible?
Assuming so: The timer circuit isn't exactly what I want, but close.
Can someone show me what to change to make it 'default' to power on
(across the red LED), and how do I calculate what value of resistor to
use in place of the 500k resistor to get approximately 30 minutes of no
power across the red LED?
In the tester circuit, if I substitue a pot for R2, would this allow me
to adjust how long the relays would be powered? I would want equal
times (or as close as possible) of about 30 seconds on and 30 seconds
off. Ideally, I would want 3 outputs instead of 2, but this is a
start.
Hope this makes sense, and thanks for any help anyone can offer.
Deefoo
> could be varied based on a rheostat. Also, if possible, I want to have
> a pushbutton that would stop all 3 pumps for 30 minutes or so for
> feeding (push the button, power is killed for 30 minutes to the pumps,
Why? Fish can't eat when there are waves?
--DF
Wayne
saltwater environments need more time to catch food before it's
filtered out.
John Fields
---
I'll help you, but I don't understand the pump timing. In your
earlier post you said that you want the pumps to turn on
sequentially, which I took to mean one at a time, like this:
_____ _____ _
PUMP 1 __| |____________| |___________|
_____ _____
PUMP 2_________| |____________| |_______
_____ _____
PUMP 3_______________| |____________| |_
However, above, you said that you want the pumps to be on for half
the time and off for half the time, so you'd be talking about
something like this:
_____ _____ _____ ____
PUMP 1 __| |_____| |_____| |_____|
_____ _____ _____ _____
PUMP 2 |_____| |_____| |_____| |__
_ _____ _____ _____
PUMP 3 |_____| |_____| |_____| |____
where more than one pump would be on at a time, and the circuit
would be much different than for the first case.
--
John Fields
Professional Circuit Designer
Rich Webb
An alternate approach that might turn out to be easier if you persist in
your perverse dislike of microcontrollers (insert smilies here as
required to assist any humor-impaired readers) is to drop back to an
older method of implementing long period timers: motors, gears (or
belts), and a cam shaft. Think of a music box or washing machine timer.
There would still be a 555 but this time (no pun intended) as a PWM
generator that controls the speed of a small DC motor. The motor,
through suitable gearing, turns a cylinder through one complete rotation
every "wave cycle." High spots on the cylinder press against
microswitches that in turn actuate the relays that handle the high
current needed for the main water pumps. Fine tune the rotation period
with a standard 555 PWM circuit.
The half-hour feeder timer could be a similar gizmo. Turn it to the
starting position that shuts a microswitch controlling a double throw
relay. The NO position of that relay would run another motor (and 555
tuner if required) until it reaches the low spot which opens the
microswitch, opens the relay, and stops the motor. The NC position would
be in the power path for the water pumps. While it's open, the pumps are
off. When the half-hour relay drops out, those contacts shut again and
the pumps start up.
With the cam setup, it would also be pretty easy to cut more complex
wave profiles with, e.g., more dwell time for the pumps on either end of
the tank than the one in the middle, simulating (roughly) simple
harmonic motion that "sticks" more at the extremes and moves more
quickly through the center.
Wayne
No, you are correct. What I want is what you show in the first timing
sequence.
John Fields
---
OK, it's on alt.binaries.schematics.electronic under "Looking for a
555 timer circuit".
The outputs are positive true and should be able to drive a
solid-state relay each.
Wayne
Thank you. I'll build this over the weekend and let you know how it
works!
John Fields
---
You're welcome. :-)
Just as an aside, you can use a 4020 or a 4024 instead of the 4060
that's on the drawing, as well as the HCMOS (74HC40XX) versions
(preferred) if you happen to have them around.
If you get into trouble and you need to do some troubleshooting to
bring it to life, here's how it's supposed to work:
U1 is a 7555 (CMOS 555) wired as a free-running 50% duty cycle
astable multivibrator, and it supplies clocks to U2 and U4 all the
time.
Pressing S1 momentarily will RESET U4 and the RS latch, U3B-U3C,
forcing u2-1 low, which will keep the outputs of U2 low (and the
motor driving relays inactive) until U4 times out.
U4 times out after counting 64 low-going clock edges, and when it
does, U4-4 goes high, SETting the RS latch and forcing U2-1 high.
U3A is a 2-input NOR gate, and since U2-2 and U2-7 are being forced
low while U2-1 is low, U3-1 and U2-4 will be forced high until the
first high-going edge of the clock gets to U2-9 after the latch is
SET by U4 timing out.
When that clock pulse gets to U2 it shifts the high on U2-4 to U2-2,
enabling OUTPUT1 _and_, forcing U3-1 low. U3-1 going low will
cause U2-2 to go low and U2-7 to go high after the next high-ging
edge of the clock. Since U2-7 will now be high, U3-1 will stay low,
and the next clock will make U3-2 low, U3-7 low, and U3-10 high,
Now, since U3-3 and U3-2 are both low, U3-1 will be high and the
three-phase cycle will begin anew with the next clock:
_ _ _ _ _ _ _ _ _ _ _ _ _ _
CLK _| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_
_____________________________________________________
U2-1 __|
___ ___ ___ ___ ___
U2-2 ____| |_______| |_______| |_______| |_______|
___ ___ ___ ___
U2-7 ________| |_______| |_______| |_______| |_______
___ ___ ___ ___
U2-10____________| |_______| |_______| |_______| |___
Finally, since you said you wanted about a 30 second 'ON' time for
each motor when the thing is operating and about a 30 minute pause
when you hit the PAUSE switch, that means there will be 1800 seconds
in each pause period and 30 seconds in each active period.
If we make the period of the clock 30 seconds, then the total number
of clocks we'll have to count up to get the 30 minute pause will be
1800s
n = ------- = 60
30s
Which is pretty close to what U4 counts up to, 64, so if you use the
pot to adjust the motor ON times to 30 seconds, the pause time will
be:
t = 30s * 64 = 1920s = 32 minutes.
Not bad.
John Fields
---
Aarghhh!!!
I just saw how to get rid of the 7555.
Use S1 to RESET the 4060 and the rest of the circuit, and then when
S1 is released, use the 4060's front end for the oscillator it's
supposed to be. Then use its LSB to drive the '175 and use its Q9
to drive the latch. Or something like that. I'll work it out and
post the new schematic tomorrow.
Wayne
Is C1 100 F, or should it be 100 uF?
John Fields
>John,
>
>Is C1 100 F, or should it be 100 uF?
---
100湩, but hang in there before you build anything, I'll have a new
schematic for you today, sometime.
Wayne
Wayne
John Fields
>Got the update. RevB is much cleaner. Thank you for doing this.
---
You're welcome. :-)
In going over the circuit for the last time, (LOL) I found a few
errors:
1. U2-5 should be connected to U1-3, not U1-1
2. C1 should be an 0.18湩 polyester cap.
3. R1 should be a standard 2 megohm +/- 5% 1/4W carbon film
resistor.
Wayne
don't understand something: how does it get powered? I see where one
side of S1 (which I'm assuming is NO) is connected to VCC, but see no
others. Is S1 normally closed, or am I missing something obvious?
Thanks
John Fields
No, S1 _is_ normally open and, by convention, (since it's common
knowledge and available on the data sheets) we usually don't show
the power connections, or we tabulate them on the drawing. Guess I
should have made a table!
Anyway, here it is:
+5V 0V
REF TYPE Vcc GND
PIN PIN
-------+------+-----+-----
U1 7555 8 1
U2 HC175 14 7
U3 HC02 14 7
U4 HC4060 16 8
Also not shown are the 0.1µF bypass capacitors that it's considered
"good practice" to install between Vcc and ground on each chip.
Cheap ceramics are fine.
Wayne
outputs are high and stay high (>2V) until I press the reset button, at
which time they all go low. I haven't waited the 30+ minutes yet after
resetting to see what would happen. I've double-checked all of my
connections and made sure I have what is shown in revB plus the changes
you mentioned earlier. I'm using a 5V DV power supply.
Wayne
you have in the schematic: Q11 doesn't exist on the data sheet (I've
looked at 3), and Q6 is on pin 4, not 6. I'm trying to figure out how
the 4060 works, but haven't yet. What should Q11 actually be?
John Fields
---
Did you make the changes I outlined in my last post:
"In going over the circuit for the last time, (LOL) I found a few
errors:
1. U2-5 should be connected to U1-3, not U1-1
2. C1 should be an 0.18湩 polyester cap.
3. R1 should be a standard 2 megohm +/- 5% 1/4W carbon film
resistor."?
In either case, it should be working, but if you didn't make the
changes the timing will be off.
I don't understand whether you're saying that it doesn't work at
all, or whether you're saying that you don't know whether it's
working properly because you haven't waited the half-hour or so to
make sure whether it's working or not.
In any case, to start troubleshooting it, disconnect it from the DC
supply and disconnect all of the loads. Then set R2 to its halfway
point and reconnect the supply. You should see a 5VDC square wave
with a period of 1/2 s (2Hz) on U1-7.
If that works, then the oscillator and the first three stages of the
internal divider chain in the 4060 are working. If it isn't, then
there's a wiring error or a bad chip somewhere. Also, if you press
the PAUSE switch, all of U1's outputs should go low for as long as
the switch is pressed. After it's released, pin 7 should toggle at
2Hz.
Try it and post back with what you find and what kind of test
equipment you have available, and we can devise a troubleshooting
plan.
BTW, if U3 comes up with the outputs all high, (which is entirely
possible and doesn't indicate an error of any kind) that means that
all three pumps will run at the same time for 30 seconds when you
first power up, after which they'll drop into their normal
sequential, mutually-exclusive mode. If that's a problem we can use
the spare gate, a resistor, and a capacitor to build a
power-on-reset which will guarantee that the U3's outputs come up
all low whether you press the PAUSE switch or not. Would you like
to do that?
Wayne
multimeter. I don't know if you saw my other post below, but the 4060
pinout is different than the 4040, and does not have a Q11 output. Q6
is on pin 4, so I changed that. I used Q12 for now instead of Q11.
What I'm seeing now: Pin 9 on U3 (HC175) is cycling from 0V to 5V
every 8 seconds or so. Pin 2 on U3 (HC175) is cycling from 0V to 2.2V
at a large time interval (unsure how long, but greater than 30
seconds), maybe due to me using Q12 on U1 (4060). I have not observed
Pin 7 or 10 on U3 switching states. I'm double checking my wiring
again to make sure I haven't missed something.
Would me using Q12 on U1 cause any problems other than a longer on/off
time?
John Fields
>I'm looking at the data sheet for the 4060, and it's not matching what
>you have in the schematic: Q11 doesn't exist on the data sheet (I've
>looked at 3), and Q6 is on pin 4, not 6.
---
Yes, I explained that lack of naming consistency between
manufacturers in an earlier post. Regardless of the names used for
the outputs, the pin numbers I called out on the schematic (modified
by the post I made about the error I made) are correct.
Here's the essence of the post:
" In going over the circuit for the last time, (LOL) I found a few
errors:
1. U2-5 should be connected to U1-3, not U1-1
2. C1 should be an 0.18湩 polyester cap.
3. R1 should be a standard 2 megohm +/- 5% 1/4W carbon film
resistor."
---
>I'm trying to figure out how
>the 4060 works, but haven't yet.
---
It's a ripple counter with an on-board oscillator
_ _ _ _ _ _ _ _ _ _ _
CP _| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_
___ ___ ___ ___ ___ _
Q0 ___| |___| |___| |___| |___| |___|
_______ _______ _____
Q1 _______| |_______| |_______|
_______________
Q2________________| |_____________
_____________
Q3________________________________|
and so on...
---
>What should Q11 actually be?
---
Depends on whose data sheet you've got. Which one is it?
Wayne
John Fields
---
Since everything on the 4060 is toggling and you're getting clocks
into the 175, the 4060 and all of its circuitry is working.
U3-2 should be cycling between 0V and 5V, so that 2.2V range
indicates that there's probably a wiring error there. I suspect
you've wired U3-2 to U2-1 instead of to U2-2.
As for the timing, I'll wait until I hear from you after you've
read all my other [recent] posts as well as this one, (so we'll be
in sync) and then we'll straighten out the timing, OK? :-)
Wayne
chip is SCL4060BE with a picture of a cube with 'S' on each side of the
cube.
John Fields
---
OK, here's a link to the data sheet I used:
http://www.semiconductors.philips.com/acrobat_download/datasheets/74HC_HCT4060_CNV_2.pdf
and a link to the final schematic:
news://d7vst11cbgr3d26pa...@4ax.com
If everything works like it should, whenever you power up or hit the
PAUSE switch, U3Q1, Q2, and Q3 should go low and stay low until U1-3
goes low. That should be about 30 minutes with the pot set at
mid-range. During that time, U2-10 should be low, U2-4 should be
high, and the outputs of U10 should be toggling, with U1-7
exhibiting a period of 4 seconds (two seconds high and two seconds
low).
When U1-3 goes high, the reset will be taken off of U3, and 30
seconds later its 3 phase output cycle will begin with U3-2 going
high.
Wayne
to go get one). I still have not managed to find what I have wrong at
this point. Right now, U3Q1 seems to be staying high (2.2V) and I have
not observed U3Q2 or U3Q3 switching. U1-7 does exhibit a period of 4
seconds as you mentioned, and U1-6 does toggle.
John Fields
---
If U3-1 goes low when you hit the PAUSE switch, U3-2, 7, and 10
should all go low (close to 0V). Does that happen?
Wayne
I'm starting to think that U3 is bad, though. I see the clock on pin
9, toggling. Pin 16 (Vcc) has 5V. Pin 1 (clear) is high (5v). Pin 2
(Q1) is 2.2V. Pin 4 (D1) is 5V. Pin 5 (D2) is 2.2V. Pin 7 is 0V,
observed over 60 seconds, never changes. Pin 9 (clock) is toggling
0-5V every 15 seconds or so. Pins 10, 12 and 13 never change from 0V
observed over 60 seconds. I verified that Pin 8 (ground) has 0 ohms to
ground.
On U2, Pin 1 is 5V, Pin 2 is 2.2V, pin 3 is 0v.
If I'm understanding the logic correctly, every time the clock goes low
to high on U3 pin 9, the output should change: Q1 -> Q2 -> Q3.
Any thoughts?
Wayne
Shouldn't I have ~5V across the board? Could this be part of the
problem?
John Fields
---
Is U2 an HC02?
Wayne
John Fields
>MM74C02N is the part number on the chip.
---
OK.
If U3-1 is high, and U3-9 is toggling, then U3-2 should be either
low (close to 0V) or high (close to +5V), depending on where in the
three phase cycle U3 happens to be.
That 2.2V is the problem, and since it's on U2-2, U3-2 and U3-5,
that indicates that those three points are connected, like they
should be. In my experience, a voltage on a CMOS output about
halfway between Vcc and GND more often than not indicates a high
output shorted to a low output. If that's the case, then the
chip(s) involved in the short should feel warmer than normal, which
is anything other than cool to the touch.
What I'd do is to go over the wiring one more time, making sure I
didn't have an extra wire in there going somewhere it wasn't
supposed to, or a solder bridge creating a short, or maybe even a
burned or cut wire with the exposed conductor touching something it
wasn't supposed to.
If that doesn't help, then I'd disconnect the wire(s) going from
U3-2 to U2-2 and U3-5, connect U2-2 and U3-5 to ground, power up,
and hit the pause switch. U3-2 should wind up low, and if it comes
up high or at 2.2V then U3 is bad.
If it comes up low, then the trouble lies somewhere else and we'll
have to run some more tests to find out where it is.
Post back with what you find and we'll take it from there...
Wayne
didn't have the 74HC02 in stock, so I got the 74C02.
John Fields
>What's the difference between a 74C02 and a 74HC02? My local store
>didn't have the 74HC02 in stock, so I got the 74C02.
---
Here are links to the data sheets:
For the 74HC02,
http://www.fairchildsemi.com/ds/MM/MM74HC02.pdf
and for the 74C02,
http://www.fairchildsemi.com/ds/MM%2FMM74C02.pdf
Wayne
being patient and helping me with this.
Now I just have to get it all into a box and wire in some recepticles
to the relays.
Thanks again!
Wayne
They are working, but in testing I noticed that they bleed through
about 8VAC from the 120VAC when they are 'off'. Is this normal? Is
there something that I can do to stop that.
John Fields
>Now that I've got this working: The SSRs I'm using are Teledyne 641-1.
> They are working, but in testing I noticed that they bleed through
>about 8VAC from the 120VAC when they are 'off'. Is this normal?
---
I don't know if 8V is normal, but some voltgage there is inevitable
because of the relay's leakage current.
When you measure the output voltage of the relay, you'll have a
circuit that looks like this:
120V>-----+----E1
|
[RELAY]
|
+----E2
|
[METER]
|
120VAC>---+----0V
And simplifying it:
120V>-----+----E1
|
[R1]
|
+----E2
|
[R2]
|
120VAC>---+----0V
Assuming that the impedance of your voltmeter is 10 megohms, we can
solve for the OFF resistance of the relay like this:
R2 (E1-E2) 1E7R (120V - 8V)
R1 = ----------- = ----------------- = 1.4E8R = 140 megohms
E2 8V
Now, assuming that your pumps draw 100mA from the mains means that
they have an impedance of:
E 120V
R = --- = ------ = 1200 ohms.
I 0.1A
Then with the motor connected to the relay instead of the meter, we
have:
120V>-----+----E1
|
[140M] R1
|
+----E2
|
[1200R] R2
|
120VAC>---+----0V
Solving for the voltage across the motor yields:
E1 R2 120V * 1200R
E2 = --------- = ----------------- = 1.029E-3V ~ 1mV
R1 + R2 1.4E8R + 1200R
and the current through it will be:
E 0.001V
I = --- = -------- ~ 8.3E-7A = 0.83 microamp,
R 1200R
so not to worry!-)
>Is there something that I can do to stop that.
---
Get relays with lower leakage, but there's no problem with what you
have now.
According to their data sheet, the leakage at 100C is only 1mA, so
you should be fine with what you've got.
John Fields
>Great news! I replaced U2 and U3 and it's working now.
---
Excellent!
---
>Thank you for being patient and helping me with this.
---
You're welcome.
---
>Now I just have to get it all into a box and wire in some recepticles
>to the relays.
>
>Thanks again!
---
My pleasure; it's been fun! :-)