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foggyeyes
Apprentece
United Arab Emirates
18 Posts |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Nov 02 2010 : 9:39:09 PM
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The Popular Electronic website does not work. Please attach the relay's datasheet to your reply instead of posting the link that does not work.
The circuit probably will not work because a single 4011 gate is not powerful enough to drive your big-looking and old fashioned relay. Its typical output current is 15mA with a 9V supply and with a 6V output and its minimum output current is only 7.5mA.
The circuit should connect all 4 gates in parallel so its output current is 4 times higher. then it might be abble to drive a small new low-current relay. |
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foggyeyes
Apprentece
United Arab Emirates
18 Posts |
Posted - Nov 03 2010 : 07:31:05 AM
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The specs are LE-6H 2PDT 6VDC, 10 ohm Coil Resistance I don't have a datasheet.
also, reply about the programming. |
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foggyeyes
Apprentece
United Arab Emirates
18 Posts |
Posted - Nov 03 2010 : 07:32:59 AM
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The site should be working, now. Or it is, for me. |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Nov 03 2010 : 10:43:29 AM
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Your relay is much too big and old for this weak little circuit.
Ohm's Law calculates the current required in the relay's coil to be 6V/10 ohms= 600mA. But the weak little gate from the 4011 IC has an output current of only 7.5mA to 15mA which is about 60 times too low. Even if you use a transistor at the output of the gate then the relay still might not be driven with enough current because a transistor with an input of 10mA will reliably have an output of only 100mA.
The 600mA for the relay's coil is also much too high for a little 9V "transistor" battery.
The 4011 is simply 4 logic gates that do not require programming. When an input goes high then the output goes low. But the outputs have low maximum current because they are designed to drive logic inputs that do not require any current. |
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foggyeyes
Apprentece
United Arab Emirates
18 Posts |
Posted - Nov 03 2010 : 11:02:14 PM
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I see your point. Sorry for the ignorance. I'm still a noob.
Suggest a relay that cam work with this |
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foggyeyes
Apprentece
United Arab Emirates
18 Posts |
Posted - Nov 04 2010 : 1:29:20 PM
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Also, in the diagram, the 4th pin of the IC is not mentioned. Do i ground this as well ? |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Nov 04 2010 : 3:55:13 PM
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quote: Originally posted by foggyeyes Suggest a relay that can work with this.
Relay contacts have voltage and current ratings that you should know what is needed but I don't know. I don't know why you selected a huge old relay (for a 1930 car?) that needs a very high coil current to make it work.
quote: Also, in the diagram, the 4th pin of the IC is not mentioned. Do i ground this as well?
Absolutely not! Look at the datasheet of the CD4011. Pins 3, 4, 10 and 11 are outputs. If you short an output to ground then you might cause the IC to BLOW UP! All inputs must connect to the supply voltage or to ground (as shown on the schematic) and unused outputs must be floating. |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Nov 04 2010 : 6:43:45 PM
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You are flogging a dead horse with that circuit. Try a 555 timer. |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Nov 04 2010 : 7:36:00 PM
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The max output current from a 555 timer is only 200mA. But it can drive a power transistor that will drive the very old high current relay.
I wonder why nobody said what the relay is driving. A high current welder? |
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foggyeyes
Apprentece
United Arab Emirates
18 Posts |
Posted - Nov 05 2010 : 02:13:43 AM
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Well, this ain't for a car! Its a science project. The relay drives 4 motors in parallel. I need the 4 motors to activate after 2.5 minutes. - Thats all there is to it! I've got a different, MUCH smaller relay. Ill upload the datasheet as soon as I find one.
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foggyeyes
Apprentece
United Arab Emirates
18 Posts |
Posted - Nov 05 2010 : 05:00:22 AM
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Also, the power source i am using is a 9V Battery. How do i connect this to the IC. The diagram only indicates connection to one pin - 14. Assuming the positive terminal is connected to that ( or negative ), where does the other terminal connect to ? |
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foggyeyes
Apprentece
United Arab Emirates
18 Posts |
Posted - Nov 05 2010 : 05:39:54 AM
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By floating you meant unattached to anything, right ? |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Nov 05 2010 : 10:02:57 AM
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The datasheet for a CD4011 IC shows that pin 14 is for the positive of the battery and pin 7 is ground which is the negative of the battery.
But I would use a CD4069 IC instead which also has the same pin 14 and pin 7 for the battery but has 6 inverters that can be connected in parallel and then the total will have a max output current of 36mA to 72mA to drive a relay.
Yes, floating means unattached to anything. |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Nov 05 2010 : 2:31:37 PM
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quote: Originally posted by audioguru
But I would use a CD4069 IC instead which also has the same pin 14 and pin 7 for the battery but has 6 inverters that can be connected in parallel and then the total will have a max output current of 36mA to 72mA to drive a relay.
There is a potential problem using that method. The transition voltages for each gate will be marginaly different. So with the voltage on the gate rising very slowly (as it will for this long time period) the gates may change over at different times causing a conflict of output states.
A better method would be to use some hysteresis feedback, or use one gate as the timer and let it drive the other five in parallel. |
Edited by - pebe on Nov 05 2010 2:33:37 PM |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Nov 05 2010 : 10:54:26 PM
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quote: Originally posted by pebe
A better method would be to use some hysteresis feedback, or use one gate as the timer and let it drive the other five in parallel.
I agree. |
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