Team:Grenoble-EMSE-LSU/Project/Instrumentation

From 2013.igem.org

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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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<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 align="center"><img src="https://static.igem.org/mediawiki/2013/3/34/Eq_i_transistor.PNG" alt="law intensity transistor" /></br></br></p>
<p>Therefore</br></br></p>
<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 align="center"><img src="https://static.igem.org/mediawiki/2013/a/ab/Eq_V_circuit.PNG" alt="Voltage applied resistors"/></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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<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"/></br></br>
In order for the resistor not to burn, the power dissipated due to the Joule effect need to be under 0.25W.</br>  
In order for the resistor not to burn, the power dissipated due to the Joule effect need to be under 0.25W.</br>  
But it is not the case here!  P=UI=RI²=1.4 x 0.5² = <strong><I>0.35W > 0.25W</I></strong></br>
But it is not the case here!  P=UI=RI²=1.4 x 0.5² = <strong><I>0.35W > 0.25W</I></strong></br>

Revision as of 23:51, 1 October 2013

Grenoble-EMSE-LSU, iGEM


Grenoble-EMSE-LSU, iGEM

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