Transducer

Discussion in 'General Electronics Chat' started by Tuck3rz, Nov 9, 2009.

  1. Tuck3rz

    Thread Starter New Member

    Nov 4, 2009
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    [​IMG]

    This is a Voltage and current Transducer Circuit, which is to read the voltage and current from the generator and battery which is then transfered to a ADC of a microcontroller and displayed on hyperterminal.

    It makes sense by looking at the voltage transducer, but I'm all confused once i read the current transducer.
    I do not understand why there is a load input and why a MBRI045 schottky diode is being placed ..
    If the load input is removed, what do i connect to the OUT6 of the LA25-NP..
    Can anyone help me with this?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I would assume the following:

    1. The generator output supplies the circuit at GEN [1=+ve;2=-ve]
    2. The battery output supplies the circuit at BAT [1=+ve; 2=-ve]
    3. The load draws power from the circuit at LOAD [1=+ve;2=-ve]

    Suppose the generator is operating and the battery is floating (charging mode) in parallel with the supply to the load. The generator current & voltage are monitored by the left hand side transducers and the battery current & voltage by the right hand transducers. With the generator supplying the load the MBRI045 diode is forward biased and current flows through it to both the load and the battery on charge mode. Current through the right hand LA-25NP will be flowing opposite to that which would be the case if the battery were supplying the load, so the transducer will indicate the battery charging condition and actual charging current value.

    If the generator drops out or is turned off, the battery will take up the load current demand. The diode is now reversed biased and prevents any likely current flow back into the idle generator terminals from the battery. Battery current is now flowing in the normal direction for the condition where the battery is supplying the load - this will be sensed by the change in current direction and magnitude in the battery current LA25-NP transducer.

    I don't follow your question about removing the load.
     
  3. Tuck3rz

    Thread Starter New Member

    Nov 4, 2009
    17
    0
    Thanks alot... understand your explainations..
    [​IMG]
    This image is the transducer board, the left side is for the output to the ADC of a microcontroller. The right hand side is the input for the top Battery, middle Load, bottom Generator

    Supposely, I wanted to measure the current and voltage of the generator, i connect the output of the generator to the input of the Generator part and Hyperterminal shows the values. Sames goes for the battery.
    What I mean was if I were to measure them individually why is there a need for a load input?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Without a load the generator won't have anything to power - apart from charging the battery.

    Your load terminals aren't an input as such. It is via this connecting point that current is delivered to the load. So it's more an output, with the generator and battery connections being regarded more as inputs - even though the battery is essentially a load when being charged from the generator.

    Sure you can remove the load, but you'll only measure the no-load voltage(s). Very little current will be monitored by either current transducer once the battery reaches a fully charged state.

    Not sure if you've clearly understood the board function - it is interposed between the power sources (generator & battery) and the load drawing power , so as to enable monitoring of the prevailing voltage and current conditions for the overall system.
     
  5. Tuck3rz

    Thread Starter New Member

    Nov 4, 2009
    17
    0
    Thanks alot for your information...
    I'll try my best to understand...
     
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