I calculate my voltages with my DC-link voltage and the duty cycles, but neglect a part of the voltage drop over my MOSFET.
I have only given my R_mos, but there is more voltage drop I think, is there?
The problem is that Rds(on) changes a lot with junction temperature. For example you may expect 70% of increase from the value in the datasheet at 25ºC to 100ºC. But the resistance to temperature graph is shown in all datasheets.
in power electronics applications any switching device is operating either in cut off (OFF state) region or saturation (ON state) region. so in saturation region even for small amount change in gate voltage will effect large variation in current and it causes the variation in Ron of MOSFET and corresponding voltage drop
The voltage drop across MOSFET, apart from being temp dependent (with positive temp coefficient of resistivity), as Francesc has pointed out, is also dependent on the gate drive voltage. If the drive voltage is below about 8V, the MOSFET can move into the linear region and have a higher on-state voltage drop than expected. Drive voltage is preferred to be 10 to 12 volts.
Go through following link....
http://electronicdesign.com/boards/calculate-dissipation-mosfets-high-power-supplies
Calculating average voltage drop across MOSFET is the good way of measuring voltage drop across circuits involving both AC and DC. Calculate the maximum voltage crop across MOSFET during cutt off state of it which is maximum applied voltages in case if there is only one MOSFET in the loop. Calculate minimum voltage drop across MOSFET ie., Ron, it depends upon resistances in series such as resistance of the load. Voltage drop across MOSFET during triode region of operation if connected load is short circuited will be maximum. On the contrary voltage drop will be minimum if load is opened. Take average of the voltage drop across MOSFET under all the conditions by giving relevant weight.
Ron in the state of triode region can be found from the data sheet of MOSFET you are using.
I want to know how I calculate it precisely with given duty cycles for each phase, but I am near a solution. It is not about the data, the data miss i.e. the resistance of the diode, can I say it is approximately R_ds,on?
The reverse body diode comes into the circuit only if there is a reverse current flow through the MOSFET. In that case, it is represented by a voltage drop plus a series resistance. Its characteristics will be given in the MOSFET data sheet, in values or in a graph. Typically, the voltage drop is about 0.7V @25deg C and the resistance is such that the total voltage drop is about 1.3V at rated current @25deg C. Drop is lower at higher temperature. The values can depend on type of MOSFET.
If the gate of MOSFET is on, and the current flow reverse, can we say the current flow through the diode or through R_ds,on?
Check the V-I waveforms of the MOSFET. Then use the data sheets of your particular MOSFET to conclude the voltage drop.
If the MOSFET is subjected to a reverse bias and the gate drive voltage is applied, then the MOSFET structure will be ON ! Now, the reverse current will be carried by the MOSFET structure rather than the parallel connected Diode structure as the MOSFET drop is expected to be less than the diode cut-in voltage. This is the principle of MOSFET being used as a Synchronous Rectifier.
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