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snarf
Apprentece
Malta
26 Posts |
 Posted - Aug 28 2014 : 08:27:23 AM
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Hi all! My first post. I built this Time Delay Relay and works OK but since I’m not able to design or modify circuits I require some help to modify this circuit.
1. The timer must be activated when 12VDC is present and not with a switch. Will reset when 12VDC in not present.
2. The relay must be energised after the time has elapsed.
I have seen Time Delay Relay II but can not try it as can’t find 4011 CMOS NAND Gate IC at local shop.
Spoon feeding please and a diagram will help me, thank you.
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Aaron Cake
Administrator
Canada
6718 Posts |
Posted - Aug 30 2014 : 10:28:40 AM
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The 4011 is about the most common IC there is beside the 555 time.
If you need isolation, a small reed relay can be used in place of the switch.
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. |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Aug 30 2014 : 12:12:47 PM
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Use pin4 (Reset) of the 555. That's what it's for.
The following assumes the 12V activation signal is the same as the 12V supply.
Connect a 10K resistor between pin4 and 0V. That will reset the timer and pin3 will go low.
If you now apply +12V to pin4, the 555 will come out of its reset state and start timing, taking pin3 high for the timed period. |
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snarf
Apprentece
Malta
26 Posts |
Posted - Aug 30 2014 : 8:33:56 PM
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quote:
Originally posted by Aaron Cake
The 4011 is about the most common IC there is beside the 555 time.
If you need isolation, a small reed relay can be used in place of the switch.
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.
Thanks for reply. I don’t need a manual or other type of switch to start the time cycle since the 12VDC source will not be present continuously therefore the time cycle has to be activated when 12VDC source is present.
If I understand correctly the NPN transistor will eliminate the need of switch S1, so I would connect the collector to pin 2, the emitter to ground and the base to +12VDC through a 1K resistor. Is that right? Sorry but as I said I’m very green.
Any advice on point 2? The circuit as it is energises the relay when the time delay starts and off when the time delay has elapsed, therefore OFF DELAY type. I need the relay to be energised after the time delay has elapsed, ON DELAY type.
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snarf
Apprentece
Malta
26 Posts |
Posted - Aug 30 2014 : 9:21:40 PM
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quote:
Originally posted by pebe
Use pin4 (Reset) of the 555. That's what it's for.
The following assumes the 12V activation signal is the same as the 12V supply.
Connect a 10K resistor between pin4 and 0V. That will reset the timer and pin3 will go low.
If you now apply +12V to pin4, the 555 will come out of its reset state and start timing, taking pin3 high for the timed period.
Thanks for reply. Yes the circuit needs to be activated by the 12VDC supply and not by a switch, so the 12VDC supply will be the activation signal for the whole circuit.
If I understand properly with the 10K resistor between pin 4 and 0V you are assuming that the 12VDC source is always present and a switch will apply the same 12VDC to pin 4 to start the time cycle. In any case I will try it out and see the difference.
Please note that the 12VDC supply will not be present continuously therefore the time cycle needs to be activated when 12VDC source is present without the need of a manual switch. Time delay on power up.
In the Time Delay Relay Comments I read that switch S1 can be replaced by a capacitor for the circuit to start time delay on power up. Is this a solution?
Do you have any advice about point 2?
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Aug 31 2014 : 11:29:11 AM
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| If I read you rightly, you need a circuit that will turn on a relay several seconds after power is supplied. Is that right? |
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 02 2014 : 10:23:41 AM
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quote:
Originally posted by pebe
If I read you rightly, you need a circuit that will turn on a relay several seconds after power is supplied. Is that right?
Yes that’s right. Without the need of a switch to start the time delay. The time delay period needs to be adjustable and would be from seconds to minutes.
Either a modification to the Time Delay Relay circuit or a new Circuit. Available I have the 555 IC.
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 02 2014 : 11:06:56 AM
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quote:
Originally posted by snarf
quote:
Originally posted by pebe
If I read you rightly, you need a circuit that will turn on a relay several seconds after power is supplied. Is that right?
Yes that’s right. Without the need of a switch to start the time delay. The time delay period needs to be adjustable and would be from seconds to minutes.
Either a modification to the Time Delay Relay circuit or a new Circuit. Available I have the 555 IC.
OK. Here is a circuit that uses a 555.
At switch-on, C1 is discharged so pin2 is less than 4V and the output, pin 3, goes high. That keeps the relay switched off.
C1 starts to charge up via R1 so the voltage on pin 6 rises. When it gets to 8V, pin3 goes low and switches on the relay. It will stay that way until power is removed, when C1 will discharge through D1, ready for the next time.
The time delay is 1.1 x C1 (in microfarads) x R1 (in megohms). So 1µF and 1megohm will give about 1.1secs delay. It's best to use a bead tantalum for C1 because aluminium electrolytics have a variable leakage current that will affect the timing.
Download Attachment:  555 delayed start timer.GIF
4.31 KB |
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 02 2014 : 5:51:17 PM
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quote:
Originally posted by pebe
quote:
Originally posted by snarf
quote:
Originally posted by pebe
If I read you rightly, you need a circuit that will turn on a relay several seconds after power is supplied. Is that right?
Yes that’s right. Without the need of a switch to start the time delay. The time delay period needs to be adjustable and would be from seconds to minutes.
Either a modification to the Time Delay Relay circuit or a new Circuit. Available I have the 555 IC.
OK. Here is a circuit that uses a 555.
At switch-on, C1 is discharged so pin2 is less than 4V and the output, pin 3, goes high. That keeps the relay switched off.
C1 starts to charge up via R1 so the voltage on pin 6 rises. When it gets to 8V, pin3 goes low and switches on the relay. It will stay that way until power is removed, when C1 will discharge through D1, ready for the next time.
The time delay is 1.1 x C1 (in microfarads) x R1 (in megohms). So 1µF and 1megohm will give about 1.1secs delay. It's best to use a bead tantalum for C1 because aluminium electrolytics have a variable leakage current that will affect the timing.
Download Attachment: 555 delayed start timer.GIF
4.31 KB
Thank you for the detailed reply and circuit.
I am assuming C1 is 25V and the Relay is 9V, right?
I also assume that R1 can be a potentiometer with the wiper also connected to the +12V rail, right?
What is the rule of thumb for the time delay formula? Would it be best to change C and R proportionally?
I will do some testing and get back with the result, meanwhile could you please explain why IC pin 5 Control V and pin 7 Discharge are not utilised.
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 02 2014 : 11:38:31 PM
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quote:
Originally posted by snarf
Thank you for the detailed reply and circuit.
I am assuming C1 is 25V and the Relay is 9V, right?
I also assume that R1 can be a potentiometer with the wiper also connected to the +12V rail, right?
Yes to all your questions.
quote:
What is the rule of thumb for the time delay formula? Would it be best to change C and R proportionally?
I will do some testing and get back with the result, meanwhile could you please explain why IC pin 5 Control V and pin 7 Discharge are not utilised.
You can change either R and C, or both. But don't use higher than 2Mohms for R1.
Pin 5 is an input to change the frequency if the 555 is used as an oscillator. In this case it's not needed. Some circuits fit a capacitor from pin5 to ground, but I have never found it to be necessary.
Pin7 is the open circuit collector of an internal transistor that normally discharges C1 when pin 6 reaches is threshold voltage (2/3 of supply voltage) when the 555 is used as an oscillator. Again, it is not required in this circuit.
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 03 2014 : 06:33:11 AM
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Addendum.
I don't know the coil resistance of the relay you are going to use, so best make D3 a 1N4001 or similar. |
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 03 2014 : 10:56:04 PM
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Many thanks for your detailed explanations and assistance.
Sorry for not giving the specifications for the relay. The coil resistance is 226R so at 9V it will draw 40mA, the IC’s I have are NE555. According to the NE555 datasheet it should handle up to 200mA so I feel it’s at a safe level, correct me if I’m wrong.
I replaced D3 as suggested, I used the 1N4007 (higher voltage) as I don’t have the 1N4001, although it might not be necessary as the 1N4148 are 75V 300mA therefore to my reasoning they should be OK for this relay. Correct me if I’m getting it wrong.
Since the NE555 has 5V – 15V operating range I’m thinking of getting some 5V or 6V relays with a lower current drain which should also enable the lowering of the 12VDC supply to 6VDC without changes to the circuit. Is that right?
Also because the 9V relays are hard to find here.
The only difficulty is with R1. The NE555 datasheet at the foot of page 4, Note 3 (which refers to Threshold Current) states :
This will determine the max value of RA+RB, for 15V operation, the max total R=10Mohms, and for 5V operation, the max. total R=3.4Mohms.
Which to me means that R1’s value is more limited on a lower voltage but it might not effect timing since at 5V operation the resistance values required are lower. Right?
This brings me to ask why should R1 not be higher than 2Mohms? Does the 555 become unstable at higher values?
Tried to attach the NE555 datasheet but it’s in pdf and I didn’t manage to attach it.
I did a lot of testing with different values of R and C and all went perfectly. The time delay formula is a great assist. I had to use electrolytic capacitors for higher timings (minutes) as currently the highest tantalum I have is 6.8µF. Obviously the timing was a bit off with electrolytes but at this stage I was not after precise timings. By connecting a 0.01µF disc capacitor from pin 5 to ground timing (minutes) was better for electrolytes. Did not notice any changes with the low value tantalums I currently have, lower timings (seconds). I kept R1 at a max of 2Mohms as advised. Your comments will be appreciated.
The circuit performs as expected “Powered by Power-On, Function by On-Delay” and that’s what I asked for.
May I take it a bit further and ask if another function can be added to the same circuit with the same features of “Power-On, On-Delay”. As it is the relay is energised after the set time delay and remains energised until power +12V is switched OFF. Could the output to the relay be a pulse of say 5 seconds? That is after the set time delay the relay becomes energised for a period of say 5 seconds and after the 5 second pulse the relay is de-energised.
Also when power +12V is applied to the circuit it needs to hold the +12V ON for the circuit and switch the circuit OFF after the 5 second pulse has elapsed.
Is this possible or am I asking for too much?
Also could you please inform me what I can use to draw circuits, a simple to use and understand program, thanks.
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Edited by - snarf on Sep 03 2014 11:05:55 PM |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 04 2014 : 09:54:49 AM
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quote:
Sorry for not giving the specifications for the relay. The coil resistance is 226R so at 9V it will draw 40mA, the IC’s I have are NE555. According to the NE555 datasheet it should handle up to 200mA so I feel it’s at a safe level, correct me if I’m wrong.
I was not concerned with the 555 max output, I was thinking of the current through D3. The 1N4148 is a signal diode rated at only 150mA absolute maximum current, whereas the 1N4001 is rated at 1A. Pin3 is the output pin and pin2 is the trigger, and due to the capacitive coupling between them it is possible for the trigger circuit to be operated by negative spikes as the relay turns off if D2 is not fast enough. D3 is often fitted to prevent that. With hindsight, you can leave it out in this circuit.
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Since the NE555 has 5V – 15V operating range I’m thinking of getting some 5V or 6V relays with a lower current drain which should also enable the lowering of the 12VDC supply to 6VDC without changes to the circuit. Is that right?
The coils for different relays will require the same power (assuming the same type of relay), so 6V coils will require more current than 9V coils. With a 12V supply, the 555 will give about 10V (with D3 deleted). A readily available 12v relay will normally operate at that voltage.
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This will determine the max value of RA+RB, for 15V operation, the max total R=10Mohms, and for 5V operation, the max. total R=3.4Mohms.
Which to me means that R1’s value is more limited on a lower voltage but it might not effect timing since at 5V operation the resistance values required are lower. Right?
No. The resistance values will not be lower because timing is unaffected by the supply voltage.
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This brings me to ask why should R1 not be higher than 2Mohms? Does the 555 become unstable at higher values?
Pins 2 & 6 are bipolar inputs and will have leakage currents. The Texas datasheet shows the average leakage current of the threshold, pin 6, as 0.1µA and the worst-case as 0.25µA, which does not accord with the 10M#937; max resistance they recommend for Ra. With Ra = 10M#937;, then as pin6 voltage approached its threshold voltage, Ra would be trying to charge the cap with a 0.4µA current, while the 0.1µA to 0.25µA leakage current would be trying to discharge it, and the timing accuracy would suffer. For that reason, I always use a maximum of 2M#937; for Ra.
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……….. Obviously the timing was a bit off with electrolytes but at this stage I was not after precise timings. By connecting a 0.01µF disc capacitor from pin 5 to ground timing (minutes) was better for electrolytes……...
Pin 5 is a tap on a chain of 3 x 5K resistors and has an input impedance of 3.3K#937;. Connecting a capacitor there should make little difference to the timing.
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Could the output to the relay be a pulse of say 5 seconds? That is after the set time delay the relay becomes energised for a period of say 5 seconds and after the 5 second pulse the relay is de-energised.
Yes, but you would need two 555s – one to provide each delay. I can give you a circuit if you want.
quote:
Also, could you please inform me what I can use to draw circuits, a simple to use and understand program, thanks.
I use Microsoft’s ‘Paint’ that came with the XP system on my computer. I have a file of component parts that I can copy and paste on to a new drawing that I then file as a .GIF file. It also allows be to take in someone else’s drawings to amend. If you want more info on it, let me know.
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 04 2014 : 10:03:12 AM
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| Sorry about the text display. 10M#937 should be 10Mohms. This site does not recognize the Omega symbol for ohms and has printed #937 instead. |
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snarf
Apprentece
Malta
26 Posts |
Posted - Sep 04 2014 : 10:55:01 PM
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Once again thank you for your detailed explanations, it helps me understand better.
quote:
quote:
Sorry for not giving the specifications for the relay. The coil resistance is 226R so at 9V it will draw 40mA, the IC’s I have are NE555. According to the NE555 datasheet it should handle up to 200mA so I feel it’s at a safe level, correct me if I’m wrong.
quote:
I was not concerned with the 555 max output, I was thinking of the current through D3. The 1N4148 is a signal diode rated at only 150mA absolute maximum current, whereas the 1N4001 is rated at 1A. Pin3 is the output pin and pin2 is the trigger, and due to the capacitive coupling between them it is possible for the trigger circuit to be operated by negative spikes as the relay turns off if D2 is not fast enough. D3 is often fitted to prevent that. With hindsight, you can leave it out in this circuit.
My intention is not to remove D3 or other components from the circuit, I was trying to explain my conclusions. Being that the 1N4148 should be OK for a 40mA relay or NOT?
1N4148 datasheet states : Average forward current - 150mA and Forward continuous current - 300mA
So it’s the Average forward current the maximum diode current and not the Forward continuous current, therefore I got it wrong when I quoted that the 1N4148 is 300mA.
quote:
Could the output to the relay be a pulse of say 5 seconds? That is after the set time delay the relay becomes energised for a period of say 5 seconds and after the 5 second pulse the relay is de-energised.
quote:
Yes, but you would need two 555s – one to provide each delay. I can give you a circuit if you want.
Yes please, thanks, no problem with using two 555s. Below I’ll again outline the timer functions, etc.
• DC supply voltage - what you consider is best for this circuit. DC power source coming from mains power supply and not from batteries.
• Timer function – Power On activates timer, On Delay for time cycle and one 5 second output pulse after time delay (i.e. relay is on only during the 5 second pulse). Timer circuit must turn OFF after the 5 second pulse.
• The timer circuit must hold the DC power ON for the timer circuit until the whole cycle has been completed and turn OFF the DC power to the timer circuit when the 5 second pulse ends.
• Include all the components you think best to have a reliable circuit with all the necessary protection.
• In the circuit please include LED indication one for timer ON and one for relay ON (i.e. during the 5 second pulse). LED specifications 3mm, 1.8 - 2V, 2mA.
• Relay – what is best for this circuit.
quote:
Also, could you please inform me what I can use to draw circuits, a simple to use and understand program, thanks. quote:
I use Microsoft’s ‘Paint’ that came with the XP system on my computer. I have a file of component parts that I can copy and paste on to a new drawing that I then file as a .GIF file. It also allows be to take in someone else’s drawings to amend. If you want more info on it, let me know.
It’s that simple, I know how to use MS Paint but from where can I find the component parts? I tried to Google but had no success. Sure any information is always helpful, thanks.
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Edited by - snarf on Sep 04 2014 11:01:16 PM |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 05 2014 : 06:53:52 AM
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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 |
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Time Delay Relay Question
555 delayed start timer.GIF
Z-library.GIF