Astable 555 circuit with a period of up to 10mins

crutschow

Joined Mar 14, 2008
34,452
So to connect the 4060 output to the relay I need something like the one I have laid out below?
No.
There's not sufficient current going to the relay coil, which likely is in the neighborhood of 150Ω.

You put the relay is series with the transistor collector such as below:
The 2N2222 has a maximum current rating of 800mA, which likely will be sufficient for the relay coil current.
The value of R1 should give a base current of about 5-10% of the collector current to fully turn it on.

upload_2018-12-10_8-56-55.png
 

Thread Starter

JulesP

Joined Dec 7, 2018
383
No.
There's not sufficient current going to the relay coil, which likely is in the neighborhood of 150Ω.

You put the relay is series with the transistor collector such as below:
The 2N2222 has a maximum current rating of 800mA, which likely will be sufficient for the relay coil current.
The value of R1 should give a base current of about 5-10% of the collector current to fully turn it on.

View attachment 165504
Thanks for the info. To save power I’m looking for a latching relay instead and once I know the coil current requirement I can be more precise.
 

ScottWang

Joined Aug 23, 2012
7,400
No.
There's not sufficient current going to the relay coil, which likely is in the neighborhood of 150Ω.

You put the relay is series with the transistor collector such as below:
The 2N2222 has a maximum current rating of 800mA, which likely will be sufficient for the relay coil current.
The value of R1 should give a base current of about 5-10% of the collector current to fully turn it on.

View attachment 165504
The output current of CD4060 (page 2) maybe not enough to strobe the 2N2222 to reach up to 80mA(12V/150Ω) to drive the coil of relay when using gain=Ic/Ib=10 ?

Or make the 2N2222 doesn't works in the saturated area then it will provides enough current to drive the coil of relay or choose a relay with lower current.
 

danadak

Joined Mar 10, 2018
4,057
Note if you ever decide to use a UP you can parallel outputs
to generate enough drive current to make sure the external
transistor is saturated. The limitation for this is a spec that
UP datasheet has for total I/O current allowed.

Or of course use a MOSFET where gate drive is a lot easier
than Bipolar, for DC type on/off work. AC high speed switching
applications quite another matter however.

Regards, Dana.
 

Thread Starter

JulesP

Joined Dec 7, 2018
383
No.
There's not sufficient current going to the relay coil, which likely is in the neighborhood of 150Ω.

You put the relay is series with the transistor collector such as below:
The 2N2222 has a maximum current rating of 800mA, which likely will be sufficient for the relay coil current.
The value of R1 should give a base current of about 5-10% of the collector current to fully turn it on.

View attachment 165504
Hi Crutschow,

I have changed the relay type to a latching version operating at 12V, to save wasting power holding it open for 5mins at a time. The coil resistance for the one I have in mind is 51R so what value of R1 would work best? I estimate about 1k?

I note that the Zenner diode you suggest (IN4148) is rated at 75-100V. What's its function here?

This particular relay has a max switching voltage of 28V so I'm not sure if it will work at 38V but at least I'm getting the principles.
Timer to Relay Circuit.jpeg
 
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crutschow

Joined Mar 14, 2008
34,452
I have changed the relay type to a latching version operating at 12V, to save wasting power
Why are you worried about wasting power?
Is this battery powered?
I would not go with a latching relay unless the relay power is a critical concern.
The coil resistance for the one I have in mind is 51R so what value of R1 would work best? I estimate about 1k?
A 51Ω coil resistance would mean a coil current of about 235mA, required an R1 of no more than 850Ω.
That might be marginal for the CD4060 output.
For that I would go with a MOSFET driver.

But latching relays are more complex to drive.
The come in single-coil and dual-coil (set and reset coils) design.
The single-coil must be momentarily pulsed with an opposite polarity to set and reset the relay.
The dual-coil design must have a momentary pulse applied to the appropriate coil to set or reset the relay.
So you will need additional circuitry to apply the appropriate pulse and polarity to the coil(s).
I note that the Zenner diode you suggest (IN4148) is rated at 75-100V. What's its function here?
That's not a Zener, it's a regular junction diode.
It's purpose is to suppress the inductive spike from the coil so it doesn't zap the relay.
 

Thread Starter

JulesP

Joined Dec 7, 2018
383
The output from the 4060 is in fact now going to go towards switching a smaller relay first that will then apply a full 12V supply to the larger relay for the batteries. The smaller relay allows me to synchronise another aspect of the circuit. See attached. That being so then I will have a full 12V supply for the larger relay.

Yes power maybe an issue over long periods. The smaller one is indeed a dual coil type so I think I get the principles of toggling the latch so it's just a question of building the appropriate circuitry.
 

Attachments

recklessrog

Joined May 23, 2013
985
If you type "long duration timer using two 555's" into google, there are dozens of circuits there that will do what you ask.
Or you could use a 556 which is two 555's in one package.
 

eetech00

Joined Jun 8, 2013
3,951
The output from the 4060 is in fact now going to go towards switching a smaller relay first that will then apply a full 12V supply to the larger relay for the batteries. The smaller relay allows me to synchronise another aspect of the circuit. See attached. That being so then I will have a full 12V supply for the larger relay.

Yes power maybe an issue over long periods. The smaller one is indeed a dual coil type so I think I get the principles of toggling the latch so it's just a question of building the appropriate circuitry.
Hi

So the point of using the CD4060B was to get a long, accurate time delay. This chip has the internal circuitry to do that with an appropriate timing circuit. Using a 555 for the timing circuit circumvents the purpose of using the CD4060. Just use RC timing components with the CD4060. If you need more accuracy, then use crystal based timing components. You don’t need the 555.

eT
 

Thread Starter

JulesP

Joined Dec 7, 2018
383
Yes I now see that. As my 555 is already in place I may still use it. At least I now know how to setup either :)
 

briankj

Joined Sep 21, 2018
5
I think you could make it work with the SE555 using a modified # 3 circuit. Values shown are estimated and may need some tweeking. Timing cycle is appx. = 0.7CR.
SG
View attachment 165376
That circuit doesn't make sense to me. I get using pin 3 instead of the usual Vdd to feed the timing capacitor thus getting your 50% duty cycle but what is with the diodes and parallel resistor branches? Remove one branch and both diodes and you get exactly the same thing. Also, would the timing be affected by the output load on pin 3? (obviously there exist a load that would affect the timing)
 

MrChips

Joined Oct 2, 2009
30,810
Why do you think that a load on the output would affect the timing?

The output stage is buffered. As long as the load does not affect the power supply or throw excessive noise back into the timing circuit, the output timing should be unaffected by the load.
 

AnalogKid

Joined Aug 1, 2013
11,044
I get using pin 3 instead of the usual Vdd to feed the timing capacitor thus getting your 50% duty cycle
Maybe. The bipolar 555 has asymmetrical output voltage overhead values. That is, the low state output voltage is closer to GND than the high state output is to Vcc. Depending on how heavily the pin 3 output is loaded and the value of Vcc, this can introduce a (worst case) 5-10% error in output symmetry; the high state will be wider than the low state. The CMOS 555 is better, but still not perfect. For a perfect 50/50 output (within a few nanoseconds), have at least one divide-by-two stage between the clock source and the output.

ak
 
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Thread Starter

JulesP

Joined Dec 7, 2018
383
Hi all,

So here is where I am up to with this 'battery swapper'. Since I already have a circuit board with the 555 timer chip on, and now some 4060s, I'm using the 555 to feed the 4060 and taking the output, with a period of 10mins, from that to a small relay which I need to synchronise another part of my whole generator and battery charging system.

From that small relay I will have the main 12V circuit feed available to power the large relay that I need to switch between the two 36V battery sets. I have also now chosen a type of latching DPDT relay that can switch 30A at 36V and with its single coil working at 12V (coil resistance is 75 Ohms). The next and last hurdle is to trigger this single relay coil which requires a pulse (not sure how long) with one polarity to activate it and then a pulse of the reverse to reset it.

My first attempt, as per the attached image, is not at all ideal as the tranny will be permanently on for the 5min 'high' so wasting power and also the polarity will not flip when the next 'low' comes along.

If anyone has any ideas as to how to produce the pulse required to flip the relay at the start of the 5 min 'high' before doing the same again in reverse at the start of the 5 min 'low', it would be appreciated. I can imagine an RC circuit being involved somehow?

Thanks

Jules
Relay circuit.jpeg
 

crutschow

Joined Mar 14, 2008
34,452
I'm confused. :confused:
You talk about two relays but you show only one.

For the reverse polarity pulse you could use an H-Bridge plus some logic circuitry, plus a one-shot such as a 555.
It's a lot of added complexity just to save a little power.
What is powering the circuit that you are so concerned about that power?
 

AnalogKid

Joined Aug 1, 2013
11,044
The 4060 squarewave output goes to the 1.2K base resistor *and nowhere else*.
Break the red line from the transistor collector to the 4060 output.
Break the red line from the suppression diode cathode to the 4060 output.
Connect relay pin "A" and the diode cathode directly to the +12 V source.
Replace the 1N4148 diode with a 1N400x diode.

To make the discussion easier, add reference designators to each component.

ak
 
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