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DELETED (Inactive)
27 Posts |
 Posted - Jun 04 2003 : 08:13:59 AM
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please can some one help me in finding a working cut phone line circuite. I need it urgently.
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Aaron Cake
Administrator
Canada
6718 Posts |
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DELETED (Inactive)
27 Posts |
Posted - Jun 04 2003 : 1:04:30 PM
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quote:
sir i have tried this circuit, it does not work. i have put 1k resistor in series with LED
The lED is always ON even if i disconnecte
the phone line. I realy need your help. I have sent you an email regarding this. I need to make this work as it is my futur. please help
me.
Um, http://www.aaroncake.net/circuits
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YS
Nobel Prize Winner
USA
1132 Posts |
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n/a
DELETED (Inactive)
27 Posts |
Posted - Jun 04 2003 : 5:16:54 PM
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Thank you very much for replaying .Let me just brief you, I am 35 years old married with 2 kids, I just got a job. They asked me to develop a circuit to show when the phone line gets cut. They have giving me 3 months period to do this.. If I managed to do this circuit I will get the job. Please try to help me. I bought all the component, I have connected the LED in series with LED instead of the load.
quote:
I did not notice any LED there...
Are you asking about http://www.aaroncake.net/circuits/phonecut.htm ?
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Aaron Cake
Administrator
Canada
6718 Posts |
Posted - Jun 05 2003 : 11:34:12 AM
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I'm not sure if building a circuit you find on the Internet would be the same as "developing" in your employer's eyes. Especially since they have given you 3 months to do so. What happens when they ask you to design the next circuit, and you can't find one on the web?
At any rate, designing a cut-phone line detector is very simple. All you need is a circuit to detect the missing 40V (when the phone line gets cut) that has a high enough impedance as to not hold the phone "off hook".
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DELETED (Inactive)
27 Posts |
Posted - Jun 05 2003 : 12:38:52 PM
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The job I applied for is computer hardware technician. I have 3 years experience in this field. They don’t mind if I get it from the Internet, as I will not design any circuit in the future. There is a requirement for such a circuit in the company where they have 3 lines where their customer access to the servers. And when these lines get cut they have no way to find till the customer calls. I have done some basic electronics in high school like putting circuits together but not designing. I came up with is idea as part of initiative so I can get the job. I have bought the components and built the circuit on a white project board. And I connected an LED with 1 k resistor where it say load. But the LED is always on even when I disconnect the phone line. I have read you articles where you have said you don’t support, but I really need a favor for you. Please help me.
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Aaron Cake
Administrator
Canada
6718 Posts |
Posted - Jun 06 2003 : 09:27:34 AM
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What is the idle voltage of your phone line? What about when a phone is "off hook"?
What happens at the collector of Q1 when you disconnect and connect the circuit to the phone line? Voltage?
Same question as abvoe, but this time check the gate of the MOSFET.
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DELETED (Inactive)
27 Posts |
Posted - Jun 06 2003 : 8:02:35 PM
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Thank you sir very much.
the idle voltage for the phone line is 48.8v
when the phone is off hook it drops to 7.43v
when i connect the circuit to the phone line the voltage at the collector of Q1 is 4.42v
when i disconnect, it is 4.65v
when i connect the circuit the voltage at the gate of MOSFET is 6.85v
when disconnected it is 7.34v
keeping in mined i have put LED in series with 1 K resistor insted of the load.
]
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YS
Nobel Prize Winner
USA
1132 Posts |
Posted - Jun 09 2003 : 12:59:52 AM
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Well, as Aaron did not test this circuit, let's analyze it a bit. First of all, to have reliable switching, the base of Q2 should be connected to emitter of Q2 or to +9V via big resistor. 2.2M will work.
Second, the input voltage must go through 3x22M resistors to reach Q1. 7V/66M = about 0.1uA. If Q1 has B=50 (B is small at small currents), its collector current would be 5uA, which goes to Q2/base and Q2 emitter current is 250uA. 0.25mA*2.2M = 0.5V - so it is not enough to open MOSFET reliably.
Personally I would reduce R1,R2,R3 to 1M. Say, when phone rings, there is 100V signal there, producing only 30uA of current.. pretty safe.
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DELETED (Inactive)
27 Posts |
Posted - Jun 09 2003 : 07:27:16 AM
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[Thanks for answering and following up with me.
I will do so later on today and let you know with the outcome.
One thing which I need you to clarify, why the LED which I put in series with 1 k resistor is always on with the current configuration of the circuit.
]
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YS
Nobel Prize Winner
USA
1132 Posts |
Posted - Jun 09 2003 : 09:46:07 AM
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The circuit containing very high impedances (like 22M resistors and MOSFETS) may be extremely sensitive to certain conditions. For example, if it has long wires and bad grounding - both conditions usually exist when you make it using prototype board - it may just pick up radio or TV signals from nearby station. Then this signal is rectified by one of transistors and your LED is on! Another example - some non-removed dirt, like fingerprints, have smaller impedance than resistors, and leak current is enough to trigger the circuit. I do not mention the absence of resistor between Q2(base) and Q2 (emitter), which would help to turn LED off, as I have said above. There may be other considerations, I've listed just most obvious.
Reducing impedances will help a lot.
And, maybe MOSFET is damaged? Try it without what is on the left side: connect its gate to the ground and then to +9V and see if LED goes off and on.
Good luck.
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YS
Nobel Prize Winner
USA
1132 Posts |
Posted - Jun 09 2003 : 09:57:28 AM
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Oh, by the way. Looked again at schematic - you know, LED should be ON when tel line is disconnected; and should be off, when it is connected. So, partially it works - congratulations. But, when you connect to phone line, just because of very high values of R1-R3, there is not enough current to shut LED down. Look how it works:
When line is cut, Q1 is closed, Q2 is closed, gate voltage is high, LED is on.
When you connect line, and in right polarity, Q1, Q2 open, gate voltage goes down, LED is off.
Oh! by the way: if line is connected + to ground, - to resistor, the circuit would not react at all. Diode will only protect the circuit. As phone line polarity is unpredictable - moreover, telephone company has legal rights to change it at any time, the circuit should have a bridge rectifier on the input. Take 400V diodes (current does not matter: cheapest is 400V 1A rectifier) and connect AC inputs to tel line, (-) to ground, (+) to R1.
Also keep in mind that ground of this circuit is connected to tel line and therefore must be isolated from real ground. So use either LED or relay as a load to keep this isolation.
Best regards.
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DELETED (Inactive)
27 Posts |
Posted - Jun 09 2003 : 10:10:24 AM
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you mentioned to connect the base of
Q2 to emitter of Q2 or 9V via 2.2M. I just want to confirm if you meant base of Q1 to emitter of Q2 or base of Q2 to emitter of Q2. I m not doubting your solution, I just want to make sure I do the write move before I blow up the circuit
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DELETED (Inactive)
27 Posts |
Posted - Jun 09 2003 : 10:19:27 AM
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Thanks YS
Now I am a bit in better shape. What I will do first , I will reduce the impedance by putting 1M resistor.
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YS
Nobel Prize Winner
USA
1132 Posts |
Posted - Jun 09 2003 : 10:32:56 AM
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Khalifa,
thanks for double-checking. I mean base Q2 to emitter Q2. That produces discharge path for input capacitance of Q2 and makes its closing easier. If you connect base Q1 to emitter Q2, (it would create negative feedback) the effect will be different - these transistors will form a linear amplifier, which you do not need in your case.
You may not need this resistor at all; still, it is a good practice to use it. It will cost couple of cents, but added predictability may save you from some trouble.
Good luck, YS.
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cut phone line detector
