12:30 why in the bridge rectifier the voltage out is Vs-2*Von althrough the current will go through by the order in the diagram the current firstly will go through the diode1 then the Vout then the diode2 so the shortage happened in the diode one then the voltage out is Vout=Vs-Von1 then the drop in the second diode by this i mean that the Vout must be Vs-Von.
Assuming what you are saying is true, then the sum of potential drops is Vout + Von1 + Von2 = Vs - Vno1 + Von1 + Von2 = Vs + Von2, which is greater than Vs !!! The truth is, (Vs - Von1) is the remaining potential across the resistor AND the second diode (Vout + Von2). Hence, Vout = Vs - Von1 - Von2
12:30
why in the bridge rectifier
the voltage out is Vs-2*Von althrough the current will go through by the order in the diagram the current firstly will go through the diode1 then the Vout then the diode2 so the shortage happened in the diode one then the voltage out is Vout=Vs-Von1
then the drop in the second diode by this i mean that the Vout must be Vs-Von.
Assuming what you are saying is true, then the sum of potential drops is Vout + Von1 + Von2 = Vs - Vno1 + Von1 + Von2 = Vs + Von2, which is greater than Vs !!!
The truth is, (Vs - Von1) is the remaining potential across the resistor AND the second diode (Vout + Von2). Hence, Vout = Vs - Von1 - Von2
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