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Mechanical free energy devices => mechanic => Topic started by: jasonspinmaster on February 01, 2008, 08:53:35 PM
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i am building a pulse motor and i am using an optic sensor as the trigger. i am using a TIP 120 NPN Darlington transistor, from a 12 volt motor bike battery. i am getting a pulse but a small one. can any one help or does any one of any good but basic circuits. thanks all
jason compson
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@Jasonspinmaster,
Attached is a simple circuit that uses a Hall sensor. You can probably use a opto
coupler instead of the Hall device. It is also possible to put a diode bridge over L1 to
harvest back emf power.
Groundloop.
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i am building a pulse motor and i am using an optic sensor as the trigger. i am using a TIP 120 NPN Darlington transistor, from a 12 volt motor bike battery. i am getting a pulse but a small one. can any one help or does any one of any good but basic circuits. thanks all
jason compson
Need more information, whats the DC resistance of the coil?
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i am building a pulse motor and i am using an optic sensor as the trigger. i am using a TIP 120 NPN Darlington transistor, from a 12 volt motor bike battery. i am getting a pulse but a small one. can any one help or does any one of any good but basic circuits. thanks all
jason compson
@jasonspinmaster
Try this thread : http://www.overunity.com/index.php/topic,3318.0/topicseen.html
Cheers
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the DC resistance of my coil is 003 ohms, i do not know if this sounds right or not. my coil is shaped like an ice cream cone, with a other wire wrapped around the primary coil for the back EMF (not connected )
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sorry i made a mistake my coil resistance is 10 ohms. thanks jason
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sorry i made a mistake my coil resistance is 10 ohms. thanks jason
The maximum current the base of your TIP120 will see is 12/10000, 0.0012 amps, the TIP120 has a maximum current gain of 1000, that is the collector will load a maximum of 1000x0.0012 = 1.2 amps, with your coil only being 10 ohms its drawing the max, 12/10 = 1.2 amps.
You might want to increase the number of turns on the coil, say double, making it 20 ohms.
There are other ways, but I would need to know your exact configuration.
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i can not get any more wire, as i do not have the money at the moment.
you say I would need to know your exact configuration
what do you need to know and i will give you the information
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i can not get any more wire, as i do not have the money at the moment.
you say I would need to know your exact configuration
what do you need to know and i will give you the information
Heres a couple of things you can try, I assume the diode connected to the emitter goes to ground or 0v line, if this is the case then it only serves to reduce gain by dropping about 0.6 volts across it, try bypassing it, secondly reduce the 10k resistor connected to the base to about half the value, the two closest value resistors I know of are 4.7k and 5.1k, you could start with a 6.8k and reduce further if needed, I do not know what your optical sensor is so I recommend you do not drop the resistance below 4.7k, reducing resistance will increase base current and thus collector current.
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THis one should do but I'm looking for a way to bring the power back to the Power batters
to recharge
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i have placed a 4.7k resistor instead of the 10k, and it is spinning a little faster.
my optic sensor is a OPB876 (N package)
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i do not understand you when you said that i should try and bypass the diode from the emitter. how do i do this. if i just connect up i wire the transistor over heats, very fast
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i have been able to drop the resistance down to a 2.2k resistor not the 4.7k
the battery is at 10.25 volts
placing red pin of meter on base i am getting 0.99volts
from emitter 0.52
collector 12.25 volts
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i do not understand you when you said that i should try and bypass the diode from the emitter. how do i do this. if i just connect up i wire the transistor over heats, very fast
Do you have a full diagram of the circuit, I need to know where the diode from the emitter goes?
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the diode goes from emitter to negative in the battery