Team:Grenoble-EMSE-LSU/Project/Instrumentation

From 2013.igem.org

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                                         <p id="legend"><strong><em>Electronic circuit that enables us to control light intensity</em></strong></br>
                                         <p id="legend"><strong><em>Electronic circuit that enables us to control light intensity</em></strong></br>
This circuit stabilizes the amperage of the LED lamp at 0.5A thanks to a bipolar transistor, three diodes and the R3 and R4 resistors.
This circuit stabilizes the amperage of the LED lamp at 0.5A thanks to a bipolar transistor, three diodes and the R3 and R4 resistors.
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</br>The MOS transistor is controlled by Arduino and is used like a switch. It allows us to control the average light intensity of the LED lamp.                                      
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</br>The MOS transistor is controlled by Arduino and is used like a switch. It allows us to control the average light intensity of the LED lamp.</br></br>                                     
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</p>
<p>The first part of this circuit – all components above the MOS transistor BS170 - stabilizes the amperage of the LED lamp and the second – composed by the MOS transistor and Arduino - allows us to control the average light intensity of the LED lamp.</br>
<p>The first part of this circuit – all components above the MOS transistor BS170 - stabilizes the amperage of the LED lamp and the second – composed by the MOS transistor and Arduino - allows us to control the average light intensity of the LED lamp.</br>
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Since we cannot be sure that our alimentation is completely stable, we need to stabilize it thank to a bipolar transistor, three diodes and two resistors.</br></br>  
Since we cannot be sure that our alimentation is completely stable, we need to stabilize it thank to a bipolar transistor, three diodes and two resistors.</br></br>  
We know that:</br></br>
We know that:</br></br>
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In order for the resistor not to burn, the power dissipated due to the Joule effect need to be under 0.25W. But it is not the case here!  P=UI=RI^2=1.4 x 0.5² = 0.35W > 0.25W</br>
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<p align="center"><img src="https://static.igem.org/mediawiki/2013/3/34/Eq_i_transistor.PNG" alt="law intensity transistor" width="350px" /></br></br></p>
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<p>Therefore</br></br></p>
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<p align="center"><img src="https://static.igem.org/mediawiki/2013/a/ab/Eq_V_circuit.PNG" alt="Voltage applied resistors" width="350px" /></br></br></p>
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<p>The value of the resistor is also:<img src="https://static.igem.org/mediawiki/2013/9/92/Eq_R_circuit.PNG" alt="Determination_of_R" width="350px" /></br></br>
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In order for the resistor not to burn, the power dissipated due to the Joule effect need to be under 0.25W. But it is not the case here!  P=UI=RI²=1.4 x 0.5² = 0.35W > 0.25W</br>
To solve this issue we use a little trick. We put two resistors in parallel that have 2xR=2.8 Ω</br>
To solve this issue we use a little trick. We put two resistors in parallel that have 2xR=2.8 Ω</br>
So that we divide the power dissipated due to the Joule effect by two but keep the same intensity Ie.</br>
So that we divide the power dissipated due to the Joule effect by two but keep the same intensity Ie.</br>

Revision as of 23:48, 1 October 2013

Grenoble-EMSE-LSU, iGEM


Grenoble-EMSE-LSU, iGEM

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