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snarf
Apprentece
Malta
26 Posts |
 Posted - Sep 05 2014 : 08:22:03 AM
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quote:
Originally posted by pebe
I am drawing up a suitable circuit. Can you confirm the 2mA current for the LEDs? - 2mA seems a bit low.
I am attaching the file I use in 'Paint' for making circuits. If you more details, just shout.
Download Attachment:  Z-library.GIF
19.28 KB
Yes the 3mm LED’s I have are low current 2mA but if that’s a problem 5mm LED’s (20mA) can be used instead, maybe also because 20mA LED’s are more common to find.
Specifications for 3mm LED – 1.8 – 2V, 2mA
Specifications for 5mm LED – 1.7 – 2.6V, 20mA
Many thanks for the component symbols.
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 06 2014 : 04:24:45 AM
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quote:
• DC supply voltage - what you consider is best for this circuit. DC power source coming from mains power supply and not from batteries.
I would use a 12V supply and relay - but that's just my personal choice.
Here is a suitable circuit that meets your criteria. The two 555s are both used in monostable mode to generate timed pulses. IC1 makes the timer-delay pulse of 5secs, and IC2 makes the pulse that turns on the relay for 5secs.
Both 555s operate in the same way; ie. when pin2 goes below 4V (1/3 of the supply rail), the output at pin3 goes high. When the voltage at pin6 goes above 8V (2/3 of supply voltage), the output toggles low again. That action also connects Pin7 to 0V.
1. Delay pulse
At startup, C2 is discharged so pin2, being low, triggers the output high, starting the delay period and lighting the green LED. At the same time, C2 also keeps the reset pin (4) of IC2 low, preventing it from triggering if there is a glitch from IC1 pin3 during startup. During the delay pulse, C4 cannot charge because IC2 output is low and D1 clamps C4 to 0V.
After a short period, C2 will charge via R2, making IC1 trigger pin2 inoperative and taking IC2 reset pin high. IC2 is then enabled but will remain inactive.
After 5.1secs, C1 will have charged via R1 to 8V and the threshold (pin6) will come into effect. It will cause the output to go low and pin7 will discharge C1 ready for next time. The green light will turn off. The 5sec delay has ended.
2. Timing Pulse
IC1 Pin3 going low now takes IC2 pin2 low via C3. That triggers IC2 and its output goes high. It also removes the clamp of D1 and allows C4 to charge. The red LED turns on and the relay is energised. C3 charges up via R4 to take the trigger pin and so disable it.
After 5.1secs, C4 will have charged via R5 to 8V and the threshold (pin6) will come into effect. It will cause the output to go low and pin7 will discharge C1 ready for next time. The red light will turn off. The 5sec timer has ended.
As the trigger pins (pin2) have both gone high the pulses cannot repeat, and the timer is effectively dead. But if the power is removed and reinstated, this 2cycle operation will be repeated.
Other:
The 3K9 (4K0 exactly) will give 2mA through the LEDs (OP only goes high to about 10V (supply voltage minus 2V). Use 390ohms for 20mA.
The chosen relay will depend on the required number of contacts and their type. For a single contact, something like this would be OK:
http://www.bitsbox.co.uk/index.php?main_page=product_info&cPath=253&products_id=1920
Download Attachment: 555 delayed start timer 2.GIF
6.73 KB
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 06 2014 : 9:23:03 PM
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Thank you so much for the step by step detailed explanation and circuit.
I will do my best to understand your explanations and the circuit and then I’ll post again.
Meanwhile please confirm the following;
What’s the voltage for C1 and C4, should they be 25V?
Are C2 and C3 ceramic disc capacitors?
The indicated OKO 12V 360mW relay should have a 400R coil resistance and should draw about 30mA. What’s the relay coil current limit for the circuit?
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 07 2014 : 02:18:43 AM
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Yes, 25V is OK for C1 and C4 (ideally bead tantalums).
Yes, ceramic discs are OK for C2 and C3.
The maximum current is limited by 555 to 200mA.
Good luck. |
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 08 2014 : 3:31:16 PM
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It took me some time to understand a little bit of how the circuit functions. All this thanks to your kind assistance and well detailed explanations (spoon feeding), thanks.
I rigged the circuit up on a breadboard and all went well. Both timing sequences perform great. For the trial I used a 12V SPDT relay but I will have to get a DPDT relay so that the second contact will keep the circuit power on during the Timing Pulse (2) as the remote automated switch powering the timer circuit will open when the Delay Pulse (1) is over.
I’m very grateful for your help, thank you.
We have already gone through this before but some points are not clear to me, could you please explain further on the below points.
• How does one calculate the value of the diode D2 across the relay?
• The purpose of diode D2 is to absorb any spikes the relay may produce when turning OFF, right?
• The 555 datasheets state that timing delays can be from micro seconds to hours. Well up to seconds all is OK, but what happens when one needs minutes or hours? You have explained that it’s better that the resistance R1 determining the charging period for capacitor C1 is not greater then 2Mohms, but for say;
300 seconds (5 minutes) C1 = 136µF,
299.2 seconds = 1.1 * 2M * 136µF
and for,
3600 seconds (1 hour) C1 = 1636µF,
3599.2 seconds = 1.1 * 2M * 1636µF
In my opinion 136µF Tantalum capacitors are hard to find especially if the capacitor voltage is high (say above 10V) and a bit pricey, imagine 1636µF. Goggling a bit revealed 150µF 16V Tantalum at €21.57, £17.27, $27.84.
What would be the solution to a similar problem?
What determines the voltage of timing capacitor C1?
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Edited by - snarf on Sep 08 2014 3:34:33 PM |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Sep 08 2014 : 7:17:06 PM
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| Long delay times are simple if you use a digital timer. It uses a reasonable value resistor and a low value capacitor to oscillate at a fairly high frequency, then a digital divider divides it down to a long accurate time delay. A digital timer is a CD4060, CD4536 and CD4541. |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 09 2014 : 04:53:47 AM
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D2 is there to suppress the negative peak voltage as the relay switches off. Just about any diode will suffice.
As Audioguru said, a binary counter is to be preferred for longer times. In fact, a CD4060 could be used in your application. Arrange an oscillator frequency such that, say, Q10 goes high after 5secs. After 10secs it will go low again and Q11 will go high. Connect a diode from Q11 to pin11 so as Q11 goes high it will stop the counter. Take your output from Q10. |
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 09 2014 : 1:32:13 PM
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OK and thanks to both Audioguru and Pebe, shows how little I know about circuits. I checked with the local shop and they stock the CD4060 so it should not be a problem to get it.
quote:
After 10secs it will go low again and Q11 will go high. Connect a diode from Q11 to pin11 so as Q11 goes high it will stop the counter. Take your output from Q10.
I found these CD4060 pin outs, which is Q11?
Download Attachment: CD4060-CMOS-pinout.jpg
27.8 KB
So if I understood correctly with the CD4060 one will achieve the same functions as the Delayed Action Timer with 2 LM555’s?
In the Delayed Action Timer using LM555’s circuit please explain what determines the voltage for capacitors C1 and C4. |
Edited by - snarf on Sep 09 2014 1:37:17 PM |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 09 2014 : 3:34:00 PM
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OOPS! I picked the 4060 because you can access all pins of the oscillator in order to stop it. I didn't realize it had no output for Q11. So use instead Q9 and Q10, or Q12 and Q13.
It will have the same action as the 2 x 555 setup.
Using a 12V supply, the voltage across C1 and C4 can never exceed 8V, so a working voltage of 10V or more would be OK.
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 09 2014 : 4:07:26 PM
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quote:
Originally posted by pebe
OOPS! I picked the 4060 because you can access all pins of the oscillator in order to stop it. I didn't realize it had no output for Q11. So use instead Q9 and Q10, or Q12 and Q13.
OK no problem.
quote:
Originally posted by pebe
Using a 12V supply, the voltage across C1 and C4 can never exceed 8V, so a working voltage of 10V or more would be OK.
I was thinking on those lines and at the same time wondered why I was saying that C1 and C4 should be 25V. Tantalums at 25V are much more expensive.
quote:
Originally posted by pebe
Arrange an oscillator frequency such that, say, Q10 goes high after 5secs. After 10secs it will go low again and Q11 will go high. Connect a diode from Q11 to pin11 so as Q11 goes high it will stop the counter. Take your output from Q10.
From your information I will try to draw a circuit with the CD4060 and post it for your comments and/or corrections but it may take me a while to conclude it. Meanwhile thank you. |
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JuliaAllan056
New Member
Madagascar
1 Posts |
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 20 2014 : 5:18:32 PM
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quote:
Originally posted by pebe
As Audioguru said, a binary counter is to be preferred for longer times. In fact, a CD4060 could be used in your application. Arrange an oscillator frequency such that, say, Q10 goes high after 5secs. After 10secs it will go low again and Q11 will go high. Connect a diode from Q11 to pin11 so as Q11 goes high it will stop the counter. Take your output from Q10.
First I apologise for my delayed response.
Available locally is the HCF4060 and not the CD4060. From my lack of experience and knowledge I would at first glance say that the only difference is the manufacturer, but after reading both data sheets I realised that it’s not as I first thought.
The CD4060 data sheet states that T = 2.2 RX CX and that RS is to be 2RX to 10RX.
Whilst the HCF4060 data sheet states that F ~ 1 / 2.2 RT CT (at VDD =10V) and that RS should be ~ 10RT.
This and other information I found on the internet got me confused at first. Finally I followed the HCF4060 data sheet to resolve the frequency calculation.
For the timing calculation I used F = 2n / T.
T is in seconds
F is in hertz
R is in ohms
C is in farads
n is the Q number (2^n meaning 2 for n number of times)
I am aware that you know all this but I wished to explain my conclusions as this is all new for me. Please correct me if I’m wrong.
This is the circuit I ended up with and I’m not so sure if it’s right and if that is what you explained to me.
Download Attachment: 4060.JPG
41.37 KB
I used Q10 for output as advised so the oscillator circuit = 3.342Hz so Q10 should go high after 306 seconds and the relay should become energised for 5 seconds.
I’m not sure about D1 which you told me to connect form Q9 to pin 11 to get the 5 seconds delay during which the relay is energised.
I’m also not sure if C1 – R1 which should reset the counter when power is turned ON are correct.
LED D3 indicates the timer is active, and LED D4 indicates when the relay is active.
As for the two 555’s circuit when S1 is closed a delay of ~ 300 seconds is required before the relay is turned ON for 5 seconds only, so the relay should turn OFF after the 5 seconds pulse. Further more S1 will open after the 300 seconds that is why one set of the relay contacts bridges S1 during the 5 seconds pulse, therefore keeping the circuit active during the 5 seconds pulse. The relay should turn OFF when the 5 second pulse has elapsed.
I have read that a capacitor from pin 16 (VDD) to 0V should be included to avoid any affect to the timing cycle due to minute voltage changes. The information was brief and I don’t know if it’s correct, what is your advice?
Please correct and amend the circuit as required. Your explanations and advice on the above please.
Questions regarding the two 555’s timer circuit.
quote:
During the delay pulse, C4 cannot charge because IC2 output is low and D1 clamps C4 to 0V.
I am not understanding how D1 is clamping C4 to 0V, could you explain further please.
quote:
The chosen relay will depend on the required number of contacts and their type. For a single contact, something like this would be OK:
The circuit is OK with a relay coil of say 12V 400R or higher resistance but fails with a relay coil of say 12V 275R or lower resistance. Kindly explain why a lower resistance relay does not energise.
Timing capacitors C1 and C4 difficulty.
When testing with electrolytic capacitors of different values the timing changed accordingly and the circuit performed OK. When testing with tantalum capacitors again using various different values the circuit failed the timing pulse but when using the same value for C1 and C4 and changing the value of R1 and R5 the circuit performed OK. Could you please explain why this happens.
Please remember that I’m new to electronics and trying to learn something so please explain in as much detail as possible, thank you.
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Edited by - snarf on Sep 20 2014 5:25:56 PM |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 21 2014 : 3:43:38 PM
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4060 circuit.
I think that for someone who says he knows very little about electronics you have sorted out a circuit very well.
Congratulations.!
quote:
The CD4060 data sheet states that T = 2.2 RX CX and that RS is to be 2RX to 10RX.
Whilst the HCF4060 data sheet states that F ~ 1 / 2.2 RT CT (at VDD =10V) and that RS should be ~ 10RT.
In order to maintain oscillations, Pin 9 output has to be connected (via C2) to pin 11 to give the circuit positive feedback. As it oscillates, the voltage on the connection of C2/R2 swings outside the supply rails. But the input pin11 has protection diodes built in to prevent it going outside the supply rails, so RS is there to pass the pulses from C2 but limit the voltage swing at pin 11. As long as the resistance of RS is much greater than RX, it will not affect the frequency. Let me know if you want a better explanation.
As frequency is the inverse of the period of one cycle (F = 1/T), then if you ignore the differences in RS, both expressions are the same. So the HCF4060 can replace the CD4060.
quote:
For the timing calculation I used F = 2n / T ......
..... I used Q10 for output as advised so the oscillator circuit = 3.342Hz so Q10 should go high after 306 seconds and the relay should become energised for 5 seconds.
To calculate the oscillator you need to solve for T, rather than F. T will be the time interval between successive clock pulses to the counter. In your drawing Q9 and Q10 should be exchanged because Q9 will be the active pulse (output)which will start after 5sec and end after another 5sec when Q10 goes high and D1 stops the oscillator from counting further. So you need to calculate so that Q9 goes high after 2^9 pulse; T needs to be 5sec/512 = 9.77ms.
From that, C x R = .00977/2.2 = 4.44ms. Suitable values would be C2 = 100nF and R2 = 39K +5.6K. R3 could be anything from 470K to 2M.
quote:
I’m not sure about D1 which you told me to connect form Q9 to pin 11 to get the 5 seconds delay during which the relay is energised.
Yes, you have the correct orientation for D1, but as I explained it should be connected to Q10.
quote:
I’m also not sure if C1 – R1 which should reset the counter when power is turned ON are correct.
Yes, that's correct. At switch-on, C1 is discharged so pin 12 will go high with pin 16 and cause a reset. After few millisecs C1 will have charged via R1, and pin 12 will go low allowing the count to start.
quote:
As for the two 555’s circuit when S1 is closed a delay of ~ 300 seconds is required before the relay is turned ON for 5 seconds only, so the relay should turn OFF after the 5 seconds pulse. Further more S1 will open after the 300 seconds that is why one set of the relay contacts bridges S1 during the 5 seconds pulse, therefore keeping the circuit active during the 5 seconds pulse. The relay should turn OFF when the 5 second pulse has elapsed.
Are you saying that a delay of 300sec is needed before the relay switches on for 5sec? I must have misread your posting of Sept 3rd because I had thought you wanted a delay of 5sec, and that is what the circuit of 2 x 555s would give. I thought you said originally that there would be no S1; the presence of the 12V supply should start the timed sequence of OFF/ ON/ OFF.
Can you clarify? If you want 300sec off, 5sec on, and then off the circuit will need revising.
quote:
I have read that a capacitor from pin 16 (VDD) to 0V should be included to avoid any affect to the timing cycle due to minute voltage changes. The information was brief and I don’t know if it’s correct, what is your advice?
It is always advisable to fit a capacitor across the supply terminals to limit interference spikes; a value of 100µF being typical. In this case switching on the relay may cause a slight drop in supply voltage, but it could only affect the reset pin and it would need to be a considerable voltage drop. But to be absolutely sure, fit the cap.
Questions regarding the two 555’s timer circuit.
quote:
I am not understanding how D1 is clamping C4 to 0V, could you explain further please.
When first switched on, all capacitors are discharged. As TR of IC1 is low and TH is also low, the 555 OP will be high (555 is switched on). In the instant it takes to switch on, OP may be still be low and would trigger IC2 on (TH of IC2 is low). To prevent that happening, R of IC2 is kept low until C2 has charged up. After that initial transient, OP of IC2 will be low. As C4 tries to charge via R5, D1 will conduct and hold TH down to 0.7V. TH needs to be kept low until the end of the delay the delay period.
quote:
The circuit is OK with a relay coil of say 12V 400R or higher resistance but fails with a relay coil of say 12V 275R or lower resistance. Kindly explain why a lower resistance relay does not energise.
I can see no reason why a 275ohm coil does not energise. It only takes 34mA, but a 555 can supply (source) 200mA so it can easily cope with that. However, the output current when the OP is high is sourced from two transistors in a Darlington Pair arrangement and the voltage drop across these is about 2V -2.5V. So there is only 9.5 - 10V available for the relay. It may be that that particular relay will not pull in on that voltage.
quote:
Timing capacitors C1 and C4 difficulty.
When testing with electrolytic capacitors of different values the timing changed accordingly and the circuit performed OK. When testing with tantalum capacitors again using various different values the circuit failed the timing pulse but when using the same value for C1 and C4 and changing the value of R1 and R5 the circuit performed OK. Could you please explain why this happens.
I can see no apparent reason for it. Can you tell me which values of C and R were used when the timing failed, and also the values when it worked OK?
I hope that answers your questions in enough detail. If not just shout.
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 24 2014 : 9:18:16 PM
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quote:
4060 circuit.
I think that for someone who says he knows very little about electronics you have sorted out a circuit very well.
Congratulations.!
Thank you for the compliment but I can’t take all the credit. First I took your advice and used MS Paint along with the component file you kindly provided. Secondly you are explaining every step and although I may not be absorbing everything it’s a big helping hand. Although I know nothing when compared to you when I set myself to do something I do my utmost to make it presentable even though it may take me a long time to achieve it. By all means all the credit is rightly yours and devoted to your kind assistance and detailed explanations.
I felt awkward asking for another circuit therefore although I was scared of the idea, I had to try and do the circuit by myself even if I make a fool of myself.
True I did make the circuit, and it’s my first circuit, but this does not mean I am understanding everything. I did a lot of research to try and understand more and yes I did use these in the circuit.
I’ll explain how I put the circuit together.
The Oscillator R2 C2 and R3 I comprehended for Data Sheet and now I think I should be able to calculate the required values.
As for C1 R1 counter reset I took the values and layout form another circuit I came across. I found other circuits which had different values for C1 R1 and also circuits with pin 12 directly connected to +12V but somehow 10nF and 1M seemed more reasonable. Therefore I didn’t calculate the values of C1 R1 and I don’t know how to do that. Maybe there is a formula to calculate the values. Your explanation may help me understand better.
The values of R5 and R6 was easy 12V-1.8V = 10.2V. I prefer to keep the LED voltage a bit lower, 1.8V instead of 2V. So 10.2V / 2mA = 5K1. This is one of those simple things I can figure out.
Now for T1 and R4, well Aaron Cake replied to my post on 30th August and said.
quote:
If isolation is not necessary, a NPN transistor (2N2222A, 2N3904, etc.) switching pin 2 of the 555 to ground will work. Put a 1K resistor (pulled that value out of the air, it's common) at the base of the transistor.
At the time I was not understanding what this meant so I searched for information and learned that an NPN transistor is required to switch a load ON when sufficient Base Current is present and a Current Limiting Resistor is required. I found formulas to calculate the value of R4 but it seems that this value is not so critical and that 1K is commonly used in most similar circuits. According to the 2N2222 Data Sheet the specifications exceed the load (relay) requirements therefore adequate for this application.
As for D2 and as you had explained before 1N4148 is adequate for a 44mA relay.
Finally I was left with D1. From your explanation I understood that the circuit will be feeding a signal to pin 11 to stop the counter, therefore I concluded that that’s the correct diode orientation. Since the current passing through the diode is relatively low (always if I understood the Data Sheet properly) any other low current diode could be used.
All that was left was to put all this information together and draw the circuit. Now I realise that one important point skipped my attention and as you clearly pointed out the ON pulse has to come into effect after the delay period, therefore the Qn must be higher then output Qn.
I must also admit that some 45 years ago I was trying to study electronics but had to give it up after a couple of months due to other courses and studies I was undertaking at the time. At that time (45 years ago) valves were more commonly used and semiconductors were starting to take over. But it’s true I consider myself to know nothing about electronics, I’m familiar with a few components and their common application (example – diode = rectification) but a diode has many other applications which I’m not familiar with (example – clamping). Above all I lack practical experience which provides a great deal of knowledge. Having said that I don’t consider myself ignorant but it does take me a long time to start to comprehend some things. Obviously lack of knowledge and experience are a prime factor.
quote:
As frequency is the inverse of the period of one cycle (F = 1/T), then if you ignore the differences in RS, both expressions are the same. So the HCF4060 can replace the CD4060.
Sorry I’m not sure if I follow you correctly. You mean both formulas can be used for either IC?
If we put the values of C2 and R2 in the two different formulas we get different results.
OK so that’s what confused me, ‘T’ in the different formulas have a different meaning, right?
I have renamed ‘T’ for clarity – Ti = Time Interval and Td = Required Time Delay.
I am using the values from the circuit in my pervious post which should give a time delay of 306 seconds.
CD4060 formula – Ti = 2.2 R2 C2 = 2.2 x 200K x 680nF = 0.2992 sec
So ‘Ti’ = time interval and not timer time delay, right?
Now using the formula – Ti = Td / Qn
evaluates for – Qn = Td / Ti = 306 / 0.2992 = 1022.7 (1024 – Q10)
HCF4060 formula – F = 1 / 2.2 R2 C2 = 1 / 2.2 x 220K x 680nF = 3.342 Hz
now using the formula – F = 2n / Td
evaluates for time – Td = 2n / F = 2Q10 / 3.342 = 1024 / 3.342 = 306 sec
and here ‘Td’ = required time delay, right?
That is why I could not make heads or tails, I confused ‘T’ and the resulting time delays did not match. Or am I still missing something?
quote:
quote:
________________________________________
For the timing calculation I used F = 2n / T ......
..... I used Q10 for output as advised so the oscillator circuit = 3.342Hz so Q10 should go high after 306 seconds and the relay should become energised for 5 seconds.
________________________________________
To calculate the oscillator you need to solve for T, rather than F.
Sorry I meant to evaluate F = 2n / T to T = 2n / F.
quote:
T will be the time interval between successive clock pulses to the counter. In your drawing Q9 and Q10 should be exchanged because Q9 will be the active pulse (output)which will start after 5sec and end after another 5sec when Q10 goes high and D1 stops the oscillator from counting further. So you need to calculate so that Q9 goes high after 2^9 pulse; T needs to be 5sec/512 = 9.77ms.
From that, C x R = .00977/2.2 = 4.44ms. Suitable values would be C2 = 100nF and R2 = 39K +5.6K. R3 could be anything from 470K to 2M.
OK maybe now I am understanding a bit better, Q9 gives the 5 seconds delay before the relay becomes active and Q10 keeps the output pulse active for another 5 seconds. When Q10 goes high the diode across Q10 (pin 15) and pin 11 will stop the counter and the relay will deactivate.
I have amended the circuit accordingly and awaits your corrections and/or comments. Regarding R3 I opted for 470K which is the closest to 39K + 5K6 = 44K6 x 10 = 446K.
For circuit clarification I kept the same component references but added R7 as reference for the 5K6 added resistor.
To achieve the C2 and R2 component values required I feel more confident using F = 1 / 2.2 R2 C2 and T = 2n / F, maybe because I prefer to consider T as being the required time delay and not the time interval between successive clock pulses.
Download Attachment: 4060 Revised small.JPG
38.43 KB
quote:
quote:
________________________________________
As for the two 555’s circuit when S1 is closed a delay of ~ 300 seconds is required before the relay is turned ON for 5 seconds only, so the relay should turn OFF after the 5 seconds pulse. Further more S1 will open after the 300 seconds that is why one set of the relay contacts bridges S1 during the 5 seconds pulse, therefore keeping the circuit active during the 5 seconds pulse. The relay should turn OFF when the 5 second pulse has elapsed.
________________________________________
Are you saying that a delay of 300sec is needed before the relay switches on for 5sec? I must have misread your posting of Sept 3rd because I had thought you wanted a delay of 5sec, and that is what the circuit of 2 x 555s would give. I thought you said originally that there would be no S1; the presence of the 12V supply should start the timed sequence of OFF/ ON/ OFF.
Can you clarify? If you want 300sec off, 5sec on, and then off the circuit will need revising.
You are right as on 3rd September I did not state the delay period required, sorry it was not intentional. I took it more or less for granted as initially on 2nd September I had indicated a required adjustable time delay period from seconds to minutes, again my mistake. I didn’t emphasize on the time delay because you had given me the formula to work out the time delay and this enabled me to adjust the time delay to 300 seconds.
Yes that is what I have in mind, a delay of 300 seconds before the relay switches on for 5 seconds.
True that is what the circuit of the 2 x 555s gives but I adjusted the values for other timings. This is explained later when replying to the issues related to this circuit so as to keep things related.
Once again you are right and I am not trying to confuse by any means. No physical manual switch will be present but it’s obvious that something must turn the DC power source ON. As I tried to outline on 8th September S1 is an automated remote switch in another electrical (mains) circuit which will turn ON the timer power supply thereby starting the timing sequence. The switch S1 in the circuit is more or less symbolic and to show that one set of relay contacts in the timer circuit is bridging the remote switch S1 (relay contacts RL1-1) to maintain the DC power supply ON until the whole timing cycle has ended, because S1 will open once the second set of relay contacts in the timer circuit has given power to the LOAD (relay contacts RL1-2). I didn’t include the DC power supply in the circuit because I did not think it was necessary (now included).
Yes 300 seconds OFF, 5 seconds On and then OFF. In simple words it should do the same timing cycles as the 2 x 555s circuit. The only difference is that instead of a 5sec delay and a 5sec pulse it should be 300sec delay and a 5sec pulse. What are the circuit changes required?
For clarity following are the timer sequences explained in 4 stages and hope I will not confuse.
The DC power supply (now included in the circuit) is activated by a remote switch, I will call it S1 for reference.
STAGE 1 :
Normal state of S1 open – OFF.
Normal state of DC power supply – OFF.
Normal state of timer circuit – OFF.
Normal state of timer relay (RL1) - OFF
Therefore normal state of timer relay contacts (RL1-1 and RL1-2) open – OFF.
Normal state of load – OFF.
STAGE 2 :
Remote switch S1 closes therefore S1 – ON.
DC power supply – ON.
Timer circuit – ON.
Timer relay (RL1) NOT energised during this cycle – OFF.
Therefore relay contacts (RL1-1 and RL1-2) remain open – OFF.
Time delay starts – 300 seconds.
STAGE 3 :
Time delay ends.
Timer relay (RL1) energised therefore relay contacts close – ON.
Timer relay 1st set of contacts (RL1-1) bridging S1 closed – ON.
Therefore keeping DC power supply active – ON.
Timer relay 2nd set of contacts (RL1-2) closed, load activated – ON.
Remote switch S1 opens – OFF.
Timing pulse starts – 5 seconds.
STAGE 4 :
Timing pulse ends.
Timer relay (RL1) de-energised therefore relay contacts open – OFF.
Timer relay 1st set of contacts (RL1-1) bridging S1 open – OFF.
Therefore DC power supply – OFF.
Timer circuit – OFF.
Timer relay 2nd set of contacts (RL1-2) open, load deactivated – OFF.
Condition returned to STAGE 1.
If it’s not clear please ask and sorry if I don’t always manage to explain properly what’s on my mind.
Not so important but thought I point this out. Currently the local electronics shop is closed for summer holidays and I still have to get the HCF4060, maybe next week. Therefore I can not test my own circuit yet.
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I have read that a capacitor from pin 16 (VDD) to 0V should be included to avoid any affect to the timing cycle due to minute voltage changes. The information was brief and I don’t know if it’s correct, what is your advice?
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It is always advisable to fit a capacitor across the supply terminals to limit interference spikes; a value of 100µF being typical. In this case switching on the relay may cause a slight drop in supply voltage, but it could only affect the reset pin and it would need to be a considerable voltage drop. But to be absolutely sure, fit the cap.
OK that’s not a problem, the capacitor will act as a filtering and smoothing capacitor whilst maintaining the DC voltage stable, right? But my question is why state pin 16 and not across the DC rails? It did not sound right to me. Below is a picture from the article and the exact words.
Download Attachment: Image 1.JPG
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“It is necessary to add a capacitor close to pin 16 of IC so that minute voltage changes will not affect the timing cycle.”
Questions regarding the two 555’s timer circuit.
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I am not understanding how D1 is clamping C4 to 0V, could you explain further please.
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When first switched on, all capacitors are discharged. As TR of IC1 is low and TH is also low, the 555 OP will be high (555 is switched on). In the instant it takes to switch on, OP may be still be low and would trigger IC2 on (TH of IC2 is low). To prevent that happening, R of IC2 is kept low until C2 has charged up. After that initial transient, OP of IC2 will be low. As C4 tries to charge via R5, D1 will conduct and hold TH down to 0.7V. TH needs to be kept low until the end of the delay the delay period.
Thank you for re-explaining it’s clearer in my mind now but I need to understand the IC circuitry first to be 100%. At least with your kind help I have a good basic idea.
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The circuit is OK with a relay coil of say 12V 400R or higher resistance but fails with a relay coil of say 12V 275R or lower resistance. Kindly explain why a lower resistance relay does not energise.
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I can see no reason why a 275ohm coil does not energise. It only takes 34mA, but a 555 can supply (source) 200mA so it can easily cope with that. However, the output current when the OP is high is sourced from two transistors in a Darlington Pair arrangement and the voltage drop across these is about 2V -2.5V. So there is only 9.5 - 10V available for the relay. It may be that that particular relay will not pull in on that voltage.
That’s exactly what I thought, if the IC can handle 200mA why not power a 34mA relay. I did try different types of relays all 12V, 185R and 275R, same problem, so I came to the same conclusion as you, that if the IC is dropping around 2V the relay might not operate (even though the relay did energise when I connected it directly to a DC source of 8.5V). Therefore I tried a 9V relay 226R same problem. When I used a 400R or a 890R relay the circuit operation went OK.
The delay cycle performs well with any type of relay, it’s only when it comes to the 5 second ON pulse that it fails. Mind you it’s not such a problem because I can use a relay that works with the circuit but I wished to know why this happens and maybe learn something. I discovered that although the relay was NOT being energised C4 was charging properly and the time period to reach the full charge was right (5 seconds) but when it discharged the charging cycle started over again and kept repeating on and on. Please also note that the timer was not resetting either. I might have missed to mention this before, sorry.
I reckoned that maybe C4 being 10µF could be the problem so I used 100µF for both C1 and C4 changed the values for R1 and R5 and the problem was solved. I still don’t know why, maybe the discharge from the 10µF capacitors was not sufficient?
Capacitors used are tantalum bead type 100µF 16V. Those are the highest value I managed to find locally.
I know this is of little importance but all the components I am using are new and not salvaged from scraped material.
Maybe you have a better answer for this strange difficulty.
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Timing capacitors C1 and C4 difficulty.
When testing with electrolytic capacitors of different values the timing changed accordingly and the circuit performed OK. When testing with tantalum capacitors again using various different values the circuit failed the timing pulse but when using the same value for C1 and C4 and changing the value of R1 and R5 the circuit performed OK. Could you please explain why this happens.
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I can see no apparent reason for it. Can you tell me which values of C and R were used when the timing failed, and also the values when it worked OK?
That’s what I thought, it should not happen. Following are the values requested.
When testing with electrolytic capacitors I used the following values.
For C1 - 10µF, 22µF, 47µF, 100µF, 220µF all 25V, for R1 I used many different values to get a time delay ranging between a few seconds to 1200 seconds.
For C4 and R5 I used 10µF 25V and 470K for 5 seconds and 47µF 25V and 180K+13K for 10 seconds.
Please note that the problem is not present with electrolytic capacitors therefore the circuit always worked.
When testing with tantalum bead capacitors I used the following values.
For C1 – 10µF 16V, R1 – 470K and C4 – 10µF 16V, R5 – 470K. Circuit failed.
For C1 – 100µF 16V, R1 – 470K, 360K, 43K+2K4 and C4 – 10µF 16V, R5 – 470K. Circuit failed.
For C1 – 100µF 16V, R1 – 2M, 1M, 470K, 360K, 43K+2K4 and C4 – 100µF 16V, R5 - 43K+2K4. Circuit OK
Therefore with C1 = 100µF and R1 = 2M I get 220 seconds,
and with C4 = 100µF and R5 = 45K4 (43K+2K4) I get 4.994 (5) seconds.
By circuit failed I mean – the time delay always went OK but the timing pulse ‘ON’ was mainly repeating and relay not energised.
Please note I also changed the 555 IC’s to eliminate the possibility that they may be faulty.
As I said for the relay difficulty the problem seems to resolve by increasing the capacitance value for C1 and C4 but if you have an explanation why this happens I would appreciate it, so as to learn something else.
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 26 2014 : 4:34:03 PM
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Thanks for your reply. There is a lot for me to go through with the 4060; I'll deal with it as soon as I can.
Meanwhile for the 555 circuit, the only thing I can think of for the failures when you use tantalums is that they have a high leakage current. That's not a feature of tants - quite the opposite, in fact. Could you have inadvertently wired them in with reversed polarity? That would certainly cause a high leakage current.
The polarity marks are quite small, so here is a link explaining them:
http://www.learningaboutelectronics.com/Articles/Tantalum-capacitor-polarity-markings
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Time Delay Relay Question
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