Hi everybody,
I have a problem with analyzing my circuit. I hope you can help me
My circuit is that there is fast recovery diode (trr=30ns, di/dt=50A/us) connecting in parallel with MOSFET (IRFP4242). I wonder is that true that the parallel diode will disable the antiparallel diode of MOSFET? I am quite confused because the voltage drop of parallel diode is higher than that of antiparallel diode
Thank for your nice help!
I have a problem with analyzing my circuit. I hope you can help me
My circuit is that there is fast recovery diode (trr=30ns, di/dt=50A/us) connecting in parallel with MOSFET (IRFP4242). I wonder is that true that the parallel diode will disable the antiparallel diode of MOSFET? I am quite confused because the voltage drop of parallel diode is higher than that of antiparallel diode
Thank for your nice help!
Thank you so much for your help, Mr.Pafi
I have read the topic Disable MOSFET antibody diode. Some guys said that
if the voltage of parallel diode is higher than antibody diode( in my case), current also can flow in parallel diode because of inductance, is that right?
Do you know how can I simulate this phenomenon for sure?
Thanks
I have read the topic Disable MOSFET antibody diode. Some guys said that
if the voltage of parallel diode is higher than antibody diode( in my case), current also can flow in parallel diode because of inductance, is that right?
Do you know how can I simulate this phenomenon for sure?
Thanks
You're welcome!
Yes, that's right, but that current is only a part of the whole current, and decreases very fast! For example, if the loop inductance is 15 nH, and Vf difference is 0.45 V, then di/dt=0.45/15n=30 A/us, which means that if diode current starts at 10 A, then it disappears is 333 ns. The initial value depends on the layout (current paths, coupled inductances). If current loop of FRED is significantly smaller than the FET's, then it can work, but you must keep the FET's loop small also to keep turn off peak voltage small!
Yes, that's right, but that current is only a part of the whole current, and decreases very fast! For example, if the loop inductance is 15 nH, and Vf difference is 0.45 V, then di/dt=0.45/15n=30 A/us, which means that if diode current starts at 10 A, then it disappears is 333 ns. The initial value depends on the layout (current paths, coupled inductances). If current loop of FRED is significantly smaller than the FET's, then it can work, but you must keep the FET's loop small also to keep turn off peak voltage small!
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Thank you for your clear answer. My parallel diode is EA61FC4( they said it is ultra-fast recovery diode and I do not know is that different from FRED or not?). According to the datasheet, the peak voltage of EA61FC4 is 1.25V( 0.25V higher than antibody diode).So if I calculate as you did the time for disappear is 600ns. I wonder is that good for MOSFET turn on?
FRED: Fast Recovery Epitaxial Diode (include very fast and ultrafast).
In reality the forward voltage is not a single value, but a function of temperature and current. Take the values from the detailed characteristics (Fig. 8)! Your aim is higher current on the diode then on the FET, so you should look at for example 10 amperes, and a realistic 60 C, so Vf of body diode will be around 0.65 V. And on the ultrafast diode you need for example 30 A to be efficient, and with this high current the voltage is unspecified, but probably more then 1.25 V.
Actually the situation is more complicated, because the channel of the FET also conducts, increases the current on the FET. This doesn't store charge, but when you turn off the FET, this current immediately goes to the body diode, and starts to store charge. If dead time is extremely short, then the stored charge won't be too much, but this is not typical.
In reality the forward voltage is not a single value, but a function of temperature and current. Take the values from the detailed characteristics (Fig. 8)! Your aim is higher current on the diode then on the FET, so you should look at for example 10 amperes, and a realistic 60 C, so Vf of body diode will be around 0.65 V. And on the ultrafast diode you need for example 30 A to be efficient, and with this high current the voltage is unspecified, but probably more then 1.25 V.
Actually the situation is more complicated, because the channel of the FET also conducts, increases the current on the FET. This doesn't store charge, but when you turn off the FET, this current immediately goes to the body diode, and starts to store charge. If dead time is extremely short, then the stored charge won't be too much, but this is not typical.
Thank you so much for your nice answer Mr.Pafi
I am so sorry for my late reply. Following is my circuit:
First, Q2 is conducting and after that it is turned off. I really want to know that will the current follow in parallel diode or antibody diode after Q2 is turn-off? If the current is in parallel diode first, then when Q1 is on, we can reduce the reverse recovery effect.
I am so sorry for my late reply. Following is my circuit:
First, Q2 is conducting and after that it is turned off. I really want to know that will the current follow in parallel diode or antibody diode after Q2 is turn-off? If the current is in parallel diode first, then when Q1 is on, we can reduce the reverse recovery effect.
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