Bonjour Pierre
Thinking all MOSfets with the same number are the same is like thinking all women called Michelle are the same. Take a look at the datasheets of the IRF640 from several different manufacturers. The specs - particularly dynamic performance specs, capacitance and charge values - are not all the same. Many times the different makers use different measurement setups and parameters so that direct comparison of the specs is not possible
In a conventional switchmode PSU there's usually plenty of dead time and the fs is relatively low. In addition many of the topologies such as two transistor diagonal forward converter do not have commutation between two driven devices, like the half bridge and the full bridge do, so designers can usually use a broad variety of MOSfets.
But Class D amps require very fast switching and ultra low dead times, and frankly, you pick a MOSfet from a specific manufacturer as wisely as you can, and then stay with it (Another analogy with women springs to mind...) And snubbers have to be designed around specific MOSfets, because of the variation in device capacitances from one manufacturer to the next changes ringing frequency. Snubbers are designed around a specific ringing frequency.
Measure your ringing frequency with the new MOSfets and then with the old ones. If the new frequency is lower than the old, then the new FETs have more Cds capacitance (probably other C's as well) and will be slower switchers.
Regards
John
Thinking all MOSfets with the same number are the same is like thinking all women called Michelle are the same. Take a look at the datasheets of the IRF640 from several different manufacturers. The specs - particularly dynamic performance specs, capacitance and charge values - are not all the same. Many times the different makers use different measurement setups and parameters so that direct comparison of the specs is not possible
In a conventional switchmode PSU there's usually plenty of dead time and the fs is relatively low. In addition many of the topologies such as two transistor diagonal forward converter do not have commutation between two driven devices, like the half bridge and the full bridge do, so designers can usually use a broad variety of MOSfets.
But Class D amps require very fast switching and ultra low dead times, and frankly, you pick a MOSfet from a specific manufacturer as wisely as you can, and then stay with it (Another analogy with women springs to mind...) And snubbers have to be designed around specific MOSfets, because of the variation in device capacitances from one manufacturer to the next changes ringing frequency. Snubbers are designed around a specific ringing frequency.
Measure your ringing frequency with the new MOSfets and then with the old ones. If the new frequency is lower than the old, then the new FETs have more Cds capacitance (probably other C's as well) and will be slower switchers.
Regards
John
Error in my previous post
Whoops!
In my previous post on snubber design I said that when the addition of the external Cext produced half the original ringing frequency, C = Cext (approximately). This is wrong. (But strangely it seems to work ever so well in practice?)
From three different references I looked up, came two different answers. So we do it the hard way:
if
f = 1/(2*pi*sqrt(L*C)) ; the ringing circuit on its own
and
(f/2) = 1/(2*pi*sqrt(L*(C+Cext))); with the added Cext
then arrange the first eqn i.t.o. C and the 2nd eqn i.t.o C+Cext and divide the two:
C/(C+Cext) = 1/4.
and C = Cext/3
What's the bet when I try this correct formula in practice it won't work as well as the wrong one did!
Whoops!
In my previous post on snubber design I said that when the addition of the external Cext produced half the original ringing frequency, C = Cext (approximately). This is wrong. (But strangely it seems to work ever so well in practice?)
From three different references I looked up, came two different answers. So we do it the hard way:
if
f = 1/(2*pi*sqrt(L*C)) ; the ringing circuit on its own
and
(f/2) = 1/(2*pi*sqrt(L*(C+Cext))); with the added Cext
then arrange the first eqn i.t.o. C and the 2nd eqn i.t.o C+Cext and divide the two:
C/(C+Cext) = 1/4.
and C = Cext/3
What's the bet when I try this correct formula in practice it won't work as well as the wrong one did!
Hi John,
Nice contributions.
I'm sure you'll agree that while you must choose all devices wisely, when it comes to mosfets, it's really critical. If you happen to find several of them however which meet your criteria, possibly even with similar characteristics, you'd want to sample them all and find the best one for the job, there'll be obvious temperature variations, coolest one seems wisest to me, then you can tweak it for thd and sound.
While you're optimizing you're sticking with the same mosfet, and the hunt for better ones will pay off for awhile to come yet, for later designs.
No bet here.
Regards,
Chris
Nice contributions.
I'm sure you'll agree that while you must choose all devices wisely, when it comes to mosfets, it's really critical. If you happen to find several of them however which meet your criteria, possibly even with similar characteristics, you'd want to sample them all and find the best one for the job, there'll be obvious temperature variations, coolest one seems wisest to me, then you can tweak it for thd and sound.
While you're optimizing you're sticking with the same mosfet, and the hunt for better ones will pay off for awhile to come yet, for later designs.
No bet here.
Regards,
Chris
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