# 555 Monostable Timing Circuit Question

Discussion in 'The Projects Forum' started by jerseyguy1996, Feb 2, 2008.

1. ### jerseyguy1996 Thread Starter Active Member

Feb 2, 2008
206
9
Hello Everyone,

I am new to the forums and very much a novice when it comes to electronics. I am trying to build a simple circuit that will control a topoff pump for my fish tank. Currently I have the pump wired up to an electric relay which is attached to a float switch in the tank. The float switch is very sensitive to the water level which means the pump is going through very frequent start/stop cycles. The pump is typically on for about 10 seconds every 5 - 10 minutes. What I would like to do is have the float switch trigger a timing circuit that will keep the pump running for about 5 minutes at which point it will shut off and wait for the next trigger (i.e. the float valve lowering with the water level). I have been reviewing this page http://www.kpsec.freeuk.com/555timer.htm and it looks as though I need to build the circuit to the right of the heading "555/556 Monostable". What I am having trouble with is how to determine the capacitor and resistor values to achieve a time period of 5 minutes. Based on the formula T=1.1 X R1 X C1 I worked out that for a T = 300 Seconds I could use a 1000 uF Capacitor and a 270 Ohm resistor. Is this correct and is it really this easy?

2. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Well, it would be preferable to use a much lower value of capacitance, and a much larger resistor. Otherwise, it will draw quite a bit of current to charge up that capacitor.

But before you proceed with your timer idea - is there any way to decrease the sensitivity of your level sensor, or add hysteresis to it? Or, could you simply add another float switch to set your high and low levels?

What is the voltage and current requirements for the relay, and the pump itself?

3. ### jerseyguy1996 Thread Starter Active Member

Feb 2, 2008
206
9
The motor is 12V DC. Max current is 1.36 Amps. I'm not sure about the relay but I was thinking that I could do away with the relay and just use the output of the 555 chip and a transistor to run the pump.

I could do a low and high float switch but then I wouldn't have an excuse to tinker with a 555 timer.

So perhaps we could consider a 100 uF capacitor and a 2.2K resistor. That should give me about 4 minutes correct?

4. ### jerseyguy1996 Thread Starter Active Member

Feb 2, 2008
206
9
I don't understand

The formula states:
time period, T = 1.1 × R1 × C1

T = time period in seconds (s)
R1 = resistance in ohms ()
C1 = capacitance in farads (F)

C1 = 100 uF = .1 F
R1 = 2.2K ohms = 2200 ohms

T = 1.1 X 2200 X .1 = 242 seconds

What am I doing wrong here?

5. ### jerseyguy1996 Thread Starter Active Member

Feb 2, 2008
206
9

So it would be T = 1.1 X 2,200,000 X .0001 F

6. ### Audioguru New Member

Dec 20, 2007
9,411
896
The current in the 2.2M ohm resistor is very low, micro-amps. The leakage current of a 100uF electrolytic capacitor is high, micro-amps.

When there is too much leakage current then the timer will never time-out.

A CD4541 is a timer IC with a frequency divider that provides long time delays and uses a small low leakage capacitor.

7. ### jerseyguy1996 Thread Starter Active Member

Feb 2, 2008
206
9
Thank you audioguru for the suggestion. I am looking that over and it appears that it is much more suitable for the job (well I mean besides going the easy route and just putting a high and low level float switch). I am trying to figure out the formula for calculating the values in the RC network. It seems easy enough but so did the values on the 555 timer and obviously my amateurishness (is that even a word?) is poking out. The data sheet gives this information.

The RC oscillator, shown in Figure 2, oscillates with a
frequency determined by the RC network and is calculated
using:
f = 1/(2.3)(Rtc)(Ctc)
-----------------------------------
Where f is between 1kHz
and 100kHz
and RS>=10kOhm and roughly equal to 2RTC

My confusion is this. Does f represent frequency in Hz or frequency in kHz?

8. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
In this case, F would represent frequency in Hz - so as long as you are between 1kHz and 100kHz in calculated frequency, you should be OK. Best to aim for the middle though

I do like Audiogurus' suggestion about the alternate 4541 IC rather than a 555 timer. Be aware though, that EVERYTHING electromechanical WILL eventually fail. There is no "if", it's simply "when" that failure will occur.

I suggest that your "fail safe" should be a "NC" float relay posititioned in the tank just below it's maximum limits; and that it be in series with the pump's voltage supply.

If for whatever reason the timer circuit fails to function and leaves the pump running, you would not be happy to find a swamp in the room, and fish flopping around in the swamp.

Don't forget that CMOS circuits are sensitive to static/overvoltage, and that the coil of a relay can generate VERY high voltage spikes when it's current is cut off.

You should have a diode connected across the relay's coil in opposite polarity to the voltage supply (ie: cathode pointing towards the positive supply voltage; the cathode is the side with the band around it). A very small capacitor (around 500pF) across the diode will help a great deal to further reduce voltage "spikes" due to the time it takes for diodes to begin conducting.

9. ### jerseyguy1996 Thread Starter Active Member

Feb 2, 2008
206
9
Thanks Sgtwookie,

I do have a float valve mounted at the top of the tank that shuts off power to everything if the water level raises too high. A second safety feature is that the reservoir that the pump draws from is not large enough to overflow the tank (at least it shouldn't in most scenarios).

So just to make sure I have the formulas correct.

An RC network with a 220K resistor and a .001uF Capacitor should give a frequency of.
1/(2.3*.000000001*220000) = 1976Hz

With the CD4541 set for 16 stages that will give me a count of 2^16 or 65,536 so approx 65536/1976 seconds = 33 Seconds.

Do I have things right so far?