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REDUCE THE SIMSPEED TO 50us/s AFTER 5s.
An Active Rectifier or Synchronous Rectifier is a circuit that uses a MOSFET instead of (more often in combination with) a rectifier diode. Diodes such as Shottkey diodes or Ultrafast diodes have a rather high voltage drop at high currents, and hence dissipated a lot of power and provide a reduced output voltage. Sure, they're convenient to use and are everyone's first pick, but their power loss is significant and is the limiting factor of a high-current circuit's efficiency. To combat this problem, MOSFETs are placed in parallel with a diode (usually Shottkey) in order to significantly reduce losses. How? Simple. A MOSFET has a very low ON resistance, and hence drops a much lower voltage at high currents. Therefore, if a MOSFET is made to takeover the task of rectification from a diode, a lot of power is saved.
The circuit works as follows:
(Note: the positive power supply of the op-amp must be connected to the output that feeds the load. This is not needed here, but is necessary in case of, say, a Boost Converter)
The MOSFET used is an IRF4905 P-Channel MOSFET. It is used for synchronous rectification. It's gate is connected to an op-amp. The op-amp is used to sense the voltage across the MOSFET and turn it on accordingly.
Let us start with all the currents and voltages set to zero. The supply voltage is a 10V square wave. When the supply outputs 10V, initially, current flows through the MOSFETs body diode (or parallel diode). When that happens, there is a drop across of 0.5 to 1.5V across the diode (depending on the type of diode used), which is sensed by the op-amp. The load-side voltage (sensed by the + input) is lower than the supply voltage (sensed by the - input). Due to this, the op-amp's output goes to zero (when the voltage at the + terminal is lower than the voltage at the - terminal, the output drops to zero), which turns on the MOSFET (because it's Vgs is now -10V). The MOSFET takes over and the drop across it's DS path is low, around 400mV in this case (20A current), but the op-amp's output still stays at zero since the voltage difference is sufficiently high. Since the MOSFET is conducting current, the power loss is significantly reduce to 8W. A diode would waste around 20W (assuming the drop is 1V at 20A). Ofcourse, choosing a MOSFET with a much lower ON resistance (maybe around a few milliohms) will reduce the power loss even further. Moving on, when the input supply voltage drops to zero, there is a higher voltage at the output (since the capacitor charges up). Because of this, the voltage at the + terminal is higher than the voltage at the - terminal and the op-amp shuts the MOSFET off. The MOSFET is off and the diode doesn't conduct, so a backflow of current is prevented. This is how a Synchronous Rectifier works.
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