555 & 4060 repeatable timer help

AnalogKid

Joined Aug 1, 2013
10,986
There is another approach that can replace the 555 and its timing components with one diode. If you call a 40-second period 1 tick, then 4 hours equals 360 ticks and 6 hours equals 540 ticks. This means that 512 ticks, a nice round binary increment, equals a cycle time of 5 hours 41 minutes. If you adjust the frequency of the 4060 oscillator to 0.4 Hz (an oscillator period of 2.5 seconds), then a full cycle of Q13 equals the 5.7 hour period, Q14 going high signals the end of the delay and activates the output. Q5 has an 80 second period, 40 seconds off and 40 seconds on. Q14 ANDed with Q5 lets the valve stay on for 40 seconds, then resets the counter. This turns off the output and restarts the 5.7 hour delay period.

As in Wendy's circuit, R1 can be split between a fixed and variable resistor to tune the timing. Note that this circuit has a fixed relationship between the output on time and off time. For example, if you shorten the off time by 25% to 4 hours 16 seconds, the on time will decrease to 30 seconds.

Also, this schematic uses parts already in my design library. The caps do not need to be 50 V, the resistors do not need to be 1%, other diodes and transistors will work, etc.

ak
DogWaterTimer-1-c.gif
 

Attachments

Last edited:

Thread Starter

nottortoise

Joined Dec 14, 2015
17
Thanks for the help. Are there 2 transistors? or is that just how the TIP120 is drawn? How do you determine what gets connected before and after the diode to VCC? What is TAB?

Thanks for all of the help so far, you have all gone above and beyond
 

AnalogKid

Joined Aug 1, 2013
10,986
The TIP120 is a power darlington transistor in a TO-220 power device package. Everything in the symbol (signal transistor, power transistor, 2 base pulldown resistors) is inside the part, which has three pins like a normal transistor plus a metal tab to dissipate extra heat. If your water control device need an amp of current or more, a standard transistor will require more base current than a 4060 can supply. A darlington transistor has very high gain, an can easily sink 1 A of collector current with only 2 mA of base current.

As an alternative, you can replace R4 and Q1 with an n-channel power MOSFET.

ak
 

Wendy

Joined Mar 24, 2008
23,415
My final schematic...

4060 555 timer 2.png

With the circuit on post #21, use a non-polarized cap, since the chip will charge the chip both ways. It is the reason I have C1 and C2, that configuration will take polarized caps and make them non-polarized.

The advantage to this approach is both times are relatively independent to each other. The long duration is set by R2, the short by the 555.

You can also replace the relay with the solenoid. Keep the power diode next to the solenoid / relay.
 
Last edited:

crutschow

Joined Mar 14, 2008
34,282
Any event shorter than 0.0001 seconds will not show up. Run it and check. 0.1uF should be longer than that. However, in real life, your 4060 may not have enough current sinking capacity to pull such a big capacitor below Vcc/3 which is required to trigger the 555 timer. You will have to build it and check. Or, look at projects that other people have built to see how they did the capacitive link between the two devices.
The value of the capacitor does not affect the output current required from the 4060, it only affects the duration of the current.
The current is determined by the 10kΩ resistor.
 

GopherT

Joined Nov 23, 2012
8,009
The value of the capacitor does not affect the output current required from the 4060, it only affects the duration of the current.
The current is determined by the 10kΩ resistor.
When the output on the 4060 is in steady state (high), no current is flowing. Then the 4060 goes low (no resistor between cap and 4060 pin), the full charge of the cap drains into the 4060 pin. The only thing the resistor does is limit the additional current from flowing in to the 4060's pin after the cap is drained. The bigger the cap, the longer it takes to drain into the 4060's pin (resistance would depend only on the ON resistance of the MOS sinking the current, not on the 10k resistor - which is on the far side of the capacitor).
 

AnalogKid

Joined Aug 1, 2013
10,986
I don't think so. What you describe sounds more like a sawtooth oscillator than a square wave oscillator. The cap at pin 9 and the resistor at pin 10 both are connected to gate output stages that always are of opposite logic polarity. So the cap always charges and discharges through that resistor, with the currents in both directions coming from and going to a CMOS output. To the extent that the CMOS stages have symmetrical output impedances, and saturation voltages equidistant from the transition voltages, the charging and discharging currents are equal. That is why that resistor is the only one in the frequency equation, and it is equally weighted for both half cycles. The only problem with a large capacitor is leakage current skewing the time constant when the resistor also is very large.

Because of transient currents flowing in/out of the pin 11 input protection circuit, the frequency calculation is more accurate when the pin 11 resistor is 10x (or more) greater than the pin 10 resistor.

ak
 
Last edited:

AnalogKid

Joined Aug 1, 2013
10,986
In post #20, gopher made a comment about the charge/discharge current in the 4060 oscillator capacitor in both Wendy's and your schematics. Wally and I disagree with his analysis.

ak
 

AnalogKid

Joined Aug 1, 2013
10,986
Here is a tuned up version of the timer circuit. Since the timer period is almost 6 hours, this one has a heartbeat indicator that is on every few seconds. Also, this one shows the MOSFET output transistor option. I show a 2.2 uF unpolarized timing capacitor because I have a drawer full of them. Wendy noted that if you use an electrolytic cap, you should double the value and put two in series as in her drawings. This is the recommended practice, although I've never had a problem with just a single cap with a higher voltage rating (for example, a 50 V cap in a 12 V circuit).

ak
DogWaterTimer-2-c.gif
 

Attachments

crutschow

Joined Mar 14, 2008
34,282
When the output on the 4060 is in steady state (high), no current is flowing. Then the 4060 goes low (no resistor between cap and 4060 pin), the full charge of the cap drains into the 4060 pin. The only thing the resistor does is limit the additional current from flowing in to the 4060's pin after the cap is drained. The bigger the cap, the longer it takes to drain into the 4060's pin (resistance would depend only on the ON resistance of the MOS sinking the current, not on the 10k resistor - which is on the far side of the capacitor).
That is incorrect.
As AK noted, all the current through the capacitor is limited by the 10kΩ resistor.
The "full charge" of the capacitor must go through that resistor since it is in series with the capacitor. Where else could it go?
Perhaps you are confusing the fast change of voltage at one terminal of the capacitor with a fast transfer of charge. The charge transfer rate is always determined by the series resistance.
 

AnalogKid

Joined Aug 1, 2013
10,986
Perhaps you are confusing the fast change of voltage at one terminal of the capacitor with a fast transfer of charge. The charge transfer rate is always determined by the series resistance.
And that fast change of voltage happens at *both* terminals of the capacitor. Both ends are driven by CMOS output stages, similar to a BTL audio amp.

ak
 

Wendy

Joined Mar 24, 2008
23,415
Wendy,what software did you use to draw the schematics?
I use M/S Paint. I have also made a folder of templates to copy / paste from I call Paint CAD I have been polishing for 10 years. It is in the PD and can be downloaded from here...

Introduction and PaintCAD

Wendy's Index

Part of the reason I use it is M/S Paint is on every computer out there, no installation required.
 

Thread Starter

nottortoise

Joined Dec 14, 2015
17
All good AK as per post 23, I now see that those two wires go to the coil of my relay.

I'm sorry that my response has been so late. I've been a bit overwhelmed by the discussions.

I think I have my head around the schematics now, though nearly none of the theory. I will eventually try both circuits, I think I will start with AK's timer first. I still can't figure out what's happening with the wire between the resistors on pin 10 of the 4060 on Wendy's schematic.

Thanks to everybody with all of their help, you have all really helped me a lot. I will post some photos of the project once I have completed it.

Happy holidays,

Cam
 
Top