Self-cutting power strip
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Korben Dallas wrote:For those who want it anyway, I sell 2 :-)
Take the Lejuste
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- coucou789456
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Good evening
I made myself a system based on a relay supplied by the voltage across a resistor in series with the multiple socket.
across the resistor, I put a rectifier bridge and a capacitor, then relay it across the capacitor ... thanks to the hysteresis effect, when we cut the power supplies, the relay does not resume its rest position until after a certain time to define with the value of the capacitor.
the resistance value must be defined to obtain for example a voltage of 5 V across its terminals when consumers are connected to it, via the multiple socket. once straightened, it will give about 7.5 v
for 500W of power, 2.5 ohm which will give a voltage drop to take into account in the power supplies.
when there is almost no more current consumed, for example 4 standby devices at 10 w each, i.e. 40 w, at the terminals of the resistor, there will be more than 0.4 v at the terminals of the resistor and the relay will return to its rest position.
to initiate the release of the relay, a push button with a series capacitor connected to another rectifier bridge across the relay is required, to charge the capacitor and allow time to connect the devices.
another advantage, in the event of an unplanned power outage, the relay will return to idle and this will prevent the devices from involuntary restarting.
relay system cost, 2.5 ohm 10W resistor, 2 diode bridges, capacitor 47 to 100 micro farad 63 V, ignition capacitor 0.47 micro farad non-polarized, push button, (by Zener diode security of 12 V due to overvoltages ) 1 ten euros
if a 7.5v relay is difficult to find, change the resistance value to obtain 9v, you will need a 12V relay (automobile)
for 500W consumed, it will give 4.7 ohm 20 W (P = R * I²)
the values of the resistors and capacitor are standardized and therefore easy to obtain, no active components. no current flows when the relay is at rest
jeff
I will try to make you a little diagram, it will be clearer
I forgot a detail, the relay contact (s) cut the circuit upstream of the system
I made myself a system based on a relay supplied by the voltage across a resistor in series with the multiple socket.
across the resistor, I put a rectifier bridge and a capacitor, then relay it across the capacitor ... thanks to the hysteresis effect, when we cut the power supplies, the relay does not resume its rest position until after a certain time to define with the value of the capacitor.
the resistance value must be defined to obtain for example a voltage of 5 V across its terminals when consumers are connected to it, via the multiple socket. once straightened, it will give about 7.5 v
for 500W of power, 2.5 ohm which will give a voltage drop to take into account in the power supplies.
when there is almost no more current consumed, for example 4 standby devices at 10 w each, i.e. 40 w, at the terminals of the resistor, there will be more than 0.4 v at the terminals of the resistor and the relay will return to its rest position.
to initiate the release of the relay, a push button with a series capacitor connected to another rectifier bridge across the relay is required, to charge the capacitor and allow time to connect the devices.
another advantage, in the event of an unplanned power outage, the relay will return to idle and this will prevent the devices from involuntary restarting.
relay system cost, 2.5 ohm 10W resistor, 2 diode bridges, capacitor 47 to 100 micro farad 63 V, ignition capacitor 0.47 micro farad non-polarized, push button, (by Zener diode security of 12 V due to overvoltages ) 1 ten euros
if a 7.5v relay is difficult to find, change the resistance value to obtain 9v, you will need a 12V relay (automobile)
for 500W consumed, it will give 4.7 ohm 20 W (P = R * I²)
the values of the resistors and capacitor are standardized and therefore easy to obtain, no active components. no current flows when the relay is at rest
jeff
I will try to make you a little diagram, it will be clearer
I forgot a detail, the relay contact (s) cut the circuit upstream of the system
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- coucou789456
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hummmmm!
the relay is supplied only when the devices are in operation. moreover it is supplied by the voltage across the resistor ...
it must be 50 to 100 ma, no more. I did not measure.
relay at rest, no more current flows in the assembly.
therefore no consumption
jeff
small additional detail, by mounting a voltage doubler instead of the bridge and the capacitor, it is possible to halve the value of the power resistor, and therefore to divide by 2 the power of said resistor. because in normal operation, the resistance will heat up
provide a 10 nF / 400 V capacitor across the relay contacts to absorb the electric arcs that will occur when switching off and on.
the relay is supplied only when the devices are in operation. moreover it is supplied by the voltage across the resistor ...
it must be 50 to 100 ma, no more. I did not measure.
relay at rest, no more current flows in the assembly.
therefore no consumption
jeff
small additional detail, by mounting a voltage doubler instead of the bridge and the capacitor, it is possible to halve the value of the power resistor, and therefore to divide by 2 the power of said resistor. because in normal operation, the resistance will heat up
provide a 10 nF / 400 V capacitor across the relay contacts to absorb the electric arcs that will occur when switching off and on.
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- Econologue expert
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- coucou789456
- Grand Econologue
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- Registration: 22/08/08, 05:15
- Location: Narbonne
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I found this link which explains how to make a master-slave socket for PC yourself:
http://dreu.free.fr/index.html?/blog/stopveille.html?http%3A//www.google.fr/search%3Fq%3Dmultiprise%252Binterrupteur%26start%3D0%26ie%3Dutf-8%26oe%3Dutf-8%26client%3Dfirefox-a%26rls%3Dorg.mozilla%3Afr%3Aofficial
http://dreu.free.fr/index.html?/blog/stopveille.html?http%3A//www.google.fr/search%3Fq%3Dmultiprise%252Binterrupteur%26start%3D0%26ie%3Dutf-8%26oe%3Dutf-8%26client%3Dfirefox-a%26rls%3Dorg.mozilla%3Afr%3Aofficial
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Hello,
I just discovered this subject. I made myself a power strip of this kind.
The principle: for each outlet, a current measurement is made, if it is below a so-called standby threshold (adjustable threshold) for more than a certain time, the outlet cuts off. The order of the sockets is made by optotriacs (therefore little consumption).
Yes, but once the plug is cut, how do we turn it back on?
We press a small button which allows to re-energize the socket.
But in standby, you have to power the detection system ... I opted for battery power (9V rechargeable). So in standby, zero consumption in the sector. The battery life is around 25000h (almost 3 years).
In operation, remember to recharge the battery (which in most cases discharges very little) and power the optotriacs, all inclusive and in the worst case, it takes about 100mW
As you can see, no master / slave system.
I just discovered this subject. I made myself a power strip of this kind.
The principle: for each outlet, a current measurement is made, if it is below a so-called standby threshold (adjustable threshold) for more than a certain time, the outlet cuts off. The order of the sockets is made by optotriacs (therefore little consumption).
Yes, but once the plug is cut, how do we turn it back on?
We press a small button which allows to re-energize the socket.
But in standby, you have to power the detection system ... I opted for battery power (9V rechargeable). So in standby, zero consumption in the sector. The battery life is around 25000h (almost 3 years).
In operation, remember to recharge the battery (which in most cases discharges very little) and power the optotriacs, all inclusive and in the worst case, it takes about 100mW
As you can see, no master / slave system.
0 x
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