# Fan Timer Circuit...

Discussion in 'Electronic Basics' started by pipco, Feb 13, 2006.

1. ### pipcoGuest

I've been doing a lot of reading today on 555 timer circuits, and I'm
not sure if this type of circuit is the best for my application. I'm
hoping someone can steer me in the right direction. I originally was
going to try and use a PIC, but I don't really want to invest in a
programmer and although I know I could probably do it that way, I'm
afraid the learning-curve would be beyond what I could handle right
now.

I'm trying to use a timer to turn a small computer fan on for a short
time period and off for a longer time period, then repeat. I'd like it
to be adjustable if possible (both the on and off periods). It would be
something like: on for 1 minute, off for 5.

My goal is to power the whole thing with a 9-volt battery.

I'm not really an electronics guy, although I can build a circuit from
a diagram. I think part of the problem I'm running into is trying to
figure out the values for the resistors and caps.

If someone could either point me to a diagram on the web or can talk me
through some component selections, I would appreciate it.

This is the closest thing I could find:

http://www.zen22142.zen.co.uk/Circuits/Timing/asym.htm

I understand that the time values will be appoximate, I don't need a
ton of accuracy.

Also, if someone thinks this would be better handled by a PIC and wants
to write and program a chip for me, I would be interested!

pip

2. ### John FieldsGuest

Try this: (view in Courier)

+V---+---------+-----+-------+-------------+----+-------+
| | | |8 | | |8
[10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
| | | 2|_ Vcc |3 | | 2|_ Vcc |3
+-[0.1µF]-+-----|--O|T OUT|-[0.1µF]-+----|--O|T OUT|--+
| | | 6| | | | 6| | |
| O | +---|TH 7555| | +---|TH 7555| |
| O | | 7|_ _| | | 7|_ _| |
| | +--O|D R|O-+V | +--O|D R|O-|-+V
| | +| | GND | | +| | GND | |
| | [100µF] +---+---+ | [510µF]+---+---+ |
| | | |1 | | |1 |
GND>-+---------|-----+-------+-------------|----+-------+ [0.1µF]
| | |
+---------------------------|-------------------+
| |
| +--------------+
| |
+--A | +V +V
|NAND Y--+ | | |
+-|--B | | +-------+ O
| | | | |K | |
| +--A | | [DIODE] [COIL]- - - -|
| |NAND Y--+ | | | O-> |
+-|--B | | +-------+ |
| | | | | |
| +--A | | C |
| NAND Y--+---|--[1000]--B NPN [FAN]
+----B | | E |
| | | | |
| A--+ | GND GND
+----Y NAND |
B------+

4. ### pipcoGuest

John,

Thanks for posting that... unfortunately, there are parts of the
diagram that I don't really understand. That obviously has nothing to
do with the diagram, but of my abilities at reading ascii schematics
and my knowledge of electronics. I think this is way beyond my
abilities. I'm looking for something simple and idiot-proof... you
know, Rat Shack shopping list and all. I'm guessing that kind of
hand-holding is not really possible in this kind of environment.

Thanks though!

pipco

5. ### bwGuest

John's circuit will do exactly what you want.
It's just a dual 555 with modified output to control a relay/switch for your
fan.
Not sure why he used the buffer, a transistor can drive the relay.
It should be possible to get a single 555 to do the same thing.
timer.
There are all kinds, Cole-Parmer company might have what you are looking
for.
Also try your local electrical supply house.

6. ### Bob MonsenGuest

I can do this for you. Send email. rc<surname>@comcast.net

Replace <surname> with monsen.

--
Regards,
Bob Monsen

I feel that you are justified in looking into the future with true
assurance, because you have a mode of living in which we find the joy
of life and the joy of work harmoniously combined. Added to this is
the spirit of ambition which pervades your very being, and seems to
make the day's work like a happy child at play.
Albert Einstein (1879 - 1955), (referring to America)

7. ### pipcoGuest

Thanks. I never thought of looking for a kit timer... that might be the
best option.

8. ### John FieldsGuest

---
Actually, It's two 7555s, A dual 7555 (7556) would work just as
well, but it would pin out differently:

+V---+---------+-----+-------+-------------+----+-------+
| | | |14 | | |
[10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
| | | 6|_ Vcc |5 | | 8|_ Vcc |9
+-[0.1µF]-+-----|--O|T OUT|-[0.1µF]-+----|--O|T OUT|--+
| | | 2| 7556| | | 12| | |
| O | +---|TH U1A| | +---|TH U1B| |
| O | | 1|_ _|4 | | 13|_ _|10|
| | +--O|D R|O-+V | +--O|D R|O-|-+V
| | +| | GND | | +| | GND | |
| | [100µF] +---+---+ | [510µF]+---+---+ |
| | | U1A |7 | | U1B | |
GND>-+---------|-----+-------+-------------|----+-------+ [0.1µF]
| | |
+---------------------------|-------------------+
| |
| +--------------+
| |
+--A | +V +V
|NAND Y--+ | | |
+-|--B | | +-------+ O
| | | | |K | |
| +--A | | [DIODE] [COIL]- - - -|
| |NAND Y--+ | | | O-> |
+-|--B | | +-------+ |
| | | | | |
| +--A | | C |
| NAND Y--+---|--[1000]--B NPN [FAN]
+----B | | E |
| | | | |
| A--+ | GND GND
+----Y NAND |
B------+
---
A transistor _does_ drive the relay. It's identified as "NPN" on the
schematic, and could probably be something like a 2N4401.

The buffer (three NANDs in parallel are just to increase the drive
capability to the base of the transistor and, more importantly, to
use up the spares. Depending on the OP's supply voltage, (which
can be anything up to 18V for the 7555) a 4011 wouldn't be a bad
choice for a quad dual-input NAND.
---

9. ### pipcoGuest

Email sent! Thanks.

10. ### John FieldsGuest

---
Sure it is. That's what this newsgroup is for, and we do it all the
time.

BTW, forget the circuit I posted. The circuit you referenced should
work, and here's how to figure out the timing components you'll
need:

From the referenced URL:

"The charge time (output high) is calculated by:

T(on) = 0.7 Ra Ct

The discharge time (output low) is calculated by:

T(off) = 0.7 Rb Ct"

The common element is the capacitor, Ct, and we have to guess at
what the values of the components will be to start,(!) so if we
start with the OFF time and guess at 500k ohms for the pot (1 megohm
set at the half-way point of the pot so you can go more than or less
than 5 minutes) we can take:

T(off) = 0.7 Rb Ct

and rearrange it to solve for C:

T(off) 300s
Ct = -------- = ------------ = 857µF
0.7 Rb 0.7 * 500kR

Now, one of the problems associated with RC timers is that for a
long timing interval (large C and/or R) the leakage current of the
cap and the current needed to trigger the chip might be large enough
to keep the cap from charging to the voltage needed to trigger the
timer or might be great enough to take a much longer than expected
time for the cap to charge up to the trigger point.

The 7555 needs 500pA for the TRIGGER input and 500pA for the
THRESHOLD input, so since they're parallelled for this application,
that'll be a total of 1nA needed for the chip's bias, which is
insignificant for what we're trying to do.

We won't have any problem discharging the cap, but the leakage
current problem will come up when we're trying to charge it up to
the threshold voltage, which will be 2/3 Vcc.

Assigning a Vcc of 9V to the circuit yields:

Vcc
9V
|
[Rt]<--+
| | Vth|
+----+----|TH
| | 6V|
[Ct] [RL]
| |
GND GND

Where RL is what looks like the cap's internal parallel resistance,
which will allow charge (leakage current) to flow "around" the cap
instead of being accumulated by the cap.

6V is the timer's threshold trip point, so with a 9V supply and no
leakage current, the current flowing through R1 at the instant E2
got to 6V would be :

Vcc = Vth 3V
I = ----------- = ------- = 6µA
Rt 500kR

Such will still be the case if there _is_ current in RL, but it will
take longer to charge Ct because the charge which it would
accumulate without RL there will be diminished.

However, there will be a point where it will be impossible for Vth
to _ever_ get to 6V, and that will be when the value of RL is low
enough to keep Vth below 6V.

Looking at RT and RL as a voltage divider:

Vcc
|
[Rt]
|
+----Vth
|
[RL]
|
GND

The smallest value of RL which can be used to get a Vth of 6V can be
determined by:

Vth Rt 6V * 500kR
RL = ----------- = ------------ = 1E6R = 1 megohm
Vcc - Vth 9V - 6V

With Vth needing to get to 6V and RL equal to 1 megohm, that means
that when the cap starts charging up it'll start accumulating the
full 6µA of current because it'll look like a dead short. However,
as it accumulates charge, the voltage across it will rise and,
consequently, so will the voltage across RL. That means that as the
voltage across the parallel pair rises, RL will be diverting more
and more of the current away from Ct, as time goes by, with the
result that it'll take the cap forever to charge up to 6V!

One way around this problem is to find a low-leakage cap for Ct, but
for large capacitances that can get expensive.

Consider: Looking at the 857µF cap we got for your 5 minute OFF time
and trying to get the leakage down to even 10% of 6µA will leave you
looking for something like a cap with a leakage current of 0.6µA at
6V, if you're working with a 9V supply. Good luck...

The other way out is to put something reasonable in there for Ct and
then to use whatever you have to for Ra and Rb to get the ON and OFF
times to be what you want.

For your application, what I'd do would be to use a 1000µF +/- 10%
aluminum electrolytic for Ct and a couple of 1 meghohm pots set at
about the center of their travel for Ra and Rb, then fire the thing
up and see what happens.

If you wire the thing up right it'll work, but chances are the
timing won't be where you want it to be, so diddle with the pots
until you get what you want.

One caveat: You may want to put something like 1000 ohms in series
with Ra so that if you've got it cranked down to zero ohms
accidentally, when DISCHARGE (pin 7) goes low you won't be trying
to short the power supply to ground through the internal MOSFET
connected to pin 7.

11. ### John FieldsGuest

---
Error: with the cap looking like a dead short when it starts
charging, the initial current into the cap will be:

E 9V
I = --- = ------- = 18µA
R 500kR
---

12. ### pipcoGuest

John,

Thanks for the time and effort on this one. I think you've cleared up
most of my questions... I'll give it a shot and see what happens.

Thanks again!

13. ### Eric R SnowGuest

Greetings Pip,
I am a rank amateur when it comes to electronics. But I (or should I
say Velleman) may have an answer. They make a timing kit, P/N MK111,
or maybe PMK111, that has a relay and two adjustments. One is pulse,
the other pause. It runs off 12 volts. Your computer power supply can
power this. I have one of built one of these and it was easy. Follow
this link: http://www.gibsonteched.com/vmk111.html . The thing is
about 5 bucks. I have never bought anything from this outfit though, I
got my timer from a local store. Velleman kits are sold by many
companies.
Cheers,
Eric R Snow

14. ### pipcoGuest

Eric,

Thanks for the link. I've done a lot of searching for kits since bw's
post the other day, and I've seen this one and many other timing kits
based on the 555 chip. This is the one I've settled on:
http://www.canakit.net/Contents/Items/CK191.asp. The one you linked to
only has ON times up to 5 seconds (great for firing a camera). The
canakit one has both the ON and OFF times up to 30 minutes.

Of course, there is a possibility that I'll just build the circuit
myself based on my original post and the info from John... and there is
a possibility that I'll use a PIC based solution (with help, of
course).

Cheers!

pip

15. ### pipcoGuest

I've been trying to figure out some calculations to determine how long
this setup might last. According to zbattery.com, a 9V alkaline
averages 3.6 watt-hours. If I'm using a 12V .09Amp fan, but running it
at 9V, that comes out to .81watts. Does that mean that my fan set up
would last just over 4 hours running non-stop? Let's say that I run it
only 1/3 of the time, I'm looking at 12 hours of usage on one battery?
Or, are my calculations totally incorrect?

If I switched to a 5Vdc .15Amp fan running at 4.5V (3 AA batteries),
that would get me about 39 hours if I only ran it 1/3 of the time.

Someone tell me my calcs are off... otherwise, it doesn't make sense to
run a small fan off of a battery for any length of time!

16. ### ChrisGuest

Hi. Mr. Snow is a gifted amateur -- he's got a lot of common sense.

Your calculations are also good. But 3.6 watt-hours is optomistic, and
is probably based on a smaller load than 90mA. Your 9V battery
probably won't even last that long with a 90mA load.

If this is a PC, you have both 5V and 12V available on the power
connectors. If you can live with 5 seconds on, 25 seconds off, you can
use the Velleman MK111 kit (perfect choice for a newbie -- 1 soldering
iron difficulty level), the 12V supply, and drive a 12V muffin fan with
the relay output, and you'll be good to go.

Good luck
Chris

17. ### pipcoGuest

Thanks for the response. Actually this is fully battery powered. I'm
not using a computer or a computer power-supply. I'm not using a AC-DC
adapter either. Needs to be a self-contained battery powered unit. So,
this actually might not work anyway!

18. ### HanggliderGuest

Hey pipco, John Fields helped me out a week or two ago and is one of
the nicest, most helpful guy on this group.

John, thanks for taking the time to help us new guys out and for your
patience with our inexperience (or stupidity in my case!

19. ### Bob MonsenGuest

My emails to you have been returned as undeliverable, so I'll answer here.

If you use D cells, rather than a 9V battery, you can run your 12V
computer fan at a slower speed. I have a Cooler Master 12V 80mA brushless
fan that I've been playing with. I can get it to start at about 4V, and
run continuously down to about 2.3V. At 4.5V, the fan uses about 35mA. So,
if you just hooked up my fan to 3 D cells (MN1300 duracell cells, to be
exact, I'm looking at the datasheet) they would run for 380 hours.
However, as the voltage goes down, so does the current, so that gets
stretched out to possibly 450 hours. That is 19 days. It'll run down to
about 2.3V, which is what the graphs go down to.

Now, you could use my PIC circuit, to run it at 1/10 duty cycle (meaning 1
minute on, 10 off). Sadly, since the starting voltage is higher, the
circuit won't be able to start the fan once the batteries go below 4V,
which will happen fairly quickly. Darn. One solution would be an
additional battery (like your 9V) that got switched in just to start it,
then switched out. That would allow us to get out to about 180 days,
max... say 6 months.

You could also use a smaller fan. Your application requires only very mild
stirring of the air inside the box, so you could use a tiny heatsink fan,
for example. Or, you could forgo the fan entirely, and use a resistor to
heat up a bit of air at the bottom of the box, creating a convection
current. You won't need much energy to keep the air circulating inside. A
single D cell with a 220 ohm resistor across it will provide 10mW (down to
5mW at the end of its life) of heat. It'll last a year, won't heat the
inside too much, and will probably be enough to cause circulation to
prevent the humidity problem you've been seeing.

Also, you mentioned using a solar powered fan, since the top is clear
plastic. I've played with a little solar cell (4x4 in, 3.8V output) that I
can get to run the fan directly by shining one of those handheld 1,000,000
candlepower spots at. A bigger or more efficient solar panel would be even
better. Sadly, the light will heat up the interior of your box.

I'd try the resistor before anything else, since you can measure the
humidity at the top and bottom. If that doesn't work, try a smaller
resistor, maybe 100 ohms (22mW down to 12mW at end of life). If you want

--
Regards,
Bob Monsen

Nothing in the world makes people so afraid as the influence of
independent-minded people.
Albert Einstein (1879 - 1955), 1879-1955

20. ### Rich GriseGuest

It's not stupidity, unless it's born of stubbornness.

The only dumb question is the one you don't ask.

Cheers!
Rich