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Old 4th November 2009, 06:40 PM   #1
Default voltage surge protector

Hi all
This is my first post in this forum.
I am building a voltage surge protector and I have finished assembling it.
However I am in doubt how it works. Please help me if any one knows anything about or if my circuit needs some tweaks.
Thanks all.
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Old 4th November 2009, 10:31 PM   #2
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What's that little coil just above the battery? What's the battery voltage?
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Old 5th November 2009, 01:21 AM   #3
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My first thoughs are:

Hellfire! who designed that circuit!

Quote:
However I am in doubt how it works.
So am I, the circuit is complete nonsense.

Quote:
if my circuit needs some tweaks
A complete redesign starting with a statement of requrements would be a good start.

JimB
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Old 5th November 2009, 02:06 PM   #4
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Thanks for all your support.
Sorry i forgot to mention the two elements.
the battery is 9v rechargable and the coil is a bulb.
the full circuit is actually surge protector cum lighting system.
I have got some idea about how it will work. My doubt is how to know when the relay will disconnect the circuit. How can I calculate the parameters of the relay ?
The parameters of the relay known are
6v,100Ω.
Please help me with my new queries.
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Old 5th November 2009, 02:59 PM   #5
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It seems that the bulb would light when an over voltage occures. I don't see anything that disconnects the circuit. The disconnect will only happen when the power is turned off I believe.
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Old 5th November 2009, 03:22 PM   #6
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I am not at all sure this thing is going to work. Once you turn it on and push SW1, the relay will pull in and connect output to input. In theory, the darling pair will provide a bucking current to turn the SCR and relay off when a surge comes along. The first problem is that this is going to be so slow that whatever is connected to the output terminals will probably be fried before the relay can drop. The second problem is that I don’t really see the path for the shutoff current nor do I see an absolute reference for it to reference. That means it will only work if there is a quick change in voltage, but the nature of the components won’t allow it to respond in time to do any good.
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Old 5th November 2009, 03:38 PM   #7
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thanks. Now the question is what I am supposed to add to the circuit to introduce the reference and make the circuit more responsive to the changes
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Old 6th November 2009, 06:56 AM   #8
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I just took another look at the circuit and noted that all the transistor circuit does is light the lamp showing power is applied. What this thing is doing is giving a delay when power comes back on. When the power grid goes down and then comes back, most motors start and put a load on the grid. This thing will not restart until you push the button so whatever is connected to it avoid that trash on the line.

Commercial units start with a fuse or circuit breaker followed either by a gas tube or, in the better ones, an avalanche breakdown diode. The gas tube will short the line in microseconds; an avalanche breakdown diode will have a response time around one nanosecond. That is followed by secondary protection in the form of a metal oxide varisiter (MOV) to handle the small stuff.
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Old 6th November 2009, 09:01 AM   #9
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I am still stuck in the old position. What am I supposed to add to the circuit to make the response quick and what should be replaced.
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