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I have another new project, the project is simple one when a call lands to the mobile an ac machine is triggered and it turns off when the mobile ring stop, for that I have used this circuit

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The issue is that I have looped out the connection from vibrator of an old cell phone to this circuit , I need to turn this ac machine continually till the call rings , I have no other option other than the vibrator output , let me know how to modify the circuit thereby the circuit remains on till the call ends

here the vibrator outputs 0---2.1 ---0----2.1----0------2.1 volts and the interval is around 1sec; so I need to implement another modification that the optocoupler will remain in ON state for 1sec, then the circuit will probably remain on for a time of 180 sec from the start of the call.

> My requirement is when the 2.1v comes across the vibrator output the circuit must be turned on and it must persist in the On state for 180 or (180 x 2 )seconds after that turns off automatically

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  • \$\begingroup\$ I added an answer showing a simple solution to allow the load relay to stay closed all the while the call is ringing. This will only persist till after the ringing has stopped by a short amount. If you wish the appliance to stay on after the ringing stops then you will need a different solution that includes a latch type circuit. That could be a flip-flop, a cross coupled set of transistors or a second set of contacts on the relay arranged to force the relay coil to stay on even when the 2N2222 has gone off. With latching like this you may need to devise a separate turn off strategy. \$\endgroup\$ – Michael Karas Mar 24 '13 at 15:36
  • \$\begingroup\$ @MichaelKaras hmm :) likes the answer , can you just provide up a simplest circuit to attain this , i mean to turn on the ac appliance , when the call starts ringing (the vibrator turns ON ) and remains in the ON state for around 180seconds. \$\endgroup\$ – cc4re Mar 24 '13 at 16:05
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You may be able to get the circuit to stay on all the while the call is ringing by adding a capacitor as shown below. In this application the capacitor may have to be multiple uF in size. Select the capacitor to suit the cadence of the phone vibrator signal.

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If the capacitor gets too large then it may take more than one ring cycle before the capacitor charges sufficiently through the 470 ohm resistor to turn on the 2N2222.

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  • \$\begingroup\$ let me know an appx value of the capacitor ,, is that 6v 0.1uF enough \$\endgroup\$ – cc4re Mar 24 '13 at 15:58
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    \$\begingroup\$ You'll have a better chance when you move the cap to the left (@emitter) as in the original situation the cap will not charge above \$U_BE\$. That means it will quickly stop conducting. In turn it is important to check what current the optocoupler can handle. Maybe best option is to divide the 470\$\Omega\$ resistor in two of 220\$\Omega\$ and attach the capacitor there. \$\endgroup\$ – jippie Mar 24 '13 at 18:14
  • \$\begingroup\$ @jippie - I agree about splitting the 470 into two 220Ω resistors and connecting the capacitor between them. Much better solution than I showed originally. I'll edit the picture to show this difference. \$\endgroup\$ – Michael Karas Mar 25 '13 at 1:47
  • \$\begingroup\$ @jippie let me know an appx value of the capacitor \$\endgroup\$ – cc4re Mar 30 '13 at 10:02
  • \$\begingroup\$ Get couple capacitors and use trial and error. Start with \$1\mu\text{F}\$. \$\endgroup\$ – jippie Mar 30 '13 at 10:20

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