Gibbo wrote:
D from BC wrote:
On Mon, 12 Feb 2007 00:01:56 +0000, Gibbo <gibbo@smartgauge.co.uk
wrote:
D from BC wrote:
On Sun, 11 Feb 2007 23:32:38 -0000, "john jardine"
john@jjdesigns.fsnet.co.uk> wrote:
I've never hit a sweet spot. I filter on voltage it can cope with,
then
current, then price, then buy a couple.
10nF, 1nF who cares? those gate C's are really big. Bang 'em with
gate
driver chips. Only killer is operating frequency.
God know why but I spent 2 hours last month with paper and calculator
looking at switch losses on a design using a pair of 80A FETs.
Bought the
FETs and the calculated heatsink, built it, run fine. Within 5
minutes and
for other reasons I'd changed the design to a higher Frequency and a
different switching arrangement. Those 2 hours now lost for all
time. Lesson
learned was don't waste valuable time with the sums, just select
on price
and keep an eye on the spice.
Interesting... Are you saying that it's so complicated that hit and
miss
determination is the faster method. Just pick the essential mosfet
specs and then compare in spice..
Select on price...Eye on spice
Nice electronic poetry...
D from BC
If you look at the Rds losses and the gate capacitance losses you'll
realise you can calculate the whole lot dead easily. Then as you're
doing that you'll realise it depends not only on the frequency but
also upon the "on time" of the fet. So you can calculate the losses
only as long as you know the average on time. eg on time = 100% = no
gate capacitance losses and Rds is dominant. On time = 1% and Rds
becomes almost irrelevant and the switching losses dominate.
The Mot rule of Rds losses = to G cap losses works quite well when
the the average on time is 50%. But it's only a rule of thumb to
give you a starting point.
Yep...I got my eye on the duty...
For a given mosfet driver, it seems like this:
For low duty, then pick low gate charge to reduce Vds transient time.
For high duty, then pick low Rdson to make an efficient switch.
Correct.
I just want to clarify that Mot rule.. (Also..what is Mot? Motorola?)
Yes. Just an abbreviation I used which I think appeared earlier in the
thread.
Is it... Rds pwr = Drain/Source transient switching pwr (or on/off
state change
power)
Yes. But the losses increase depending how well the driver operates
obviously due to drain/gate capacitance. So both are related.
Or Rds pwr = Drain/Source transient pwr + gate drive power required
D from BC
No that is added in separately.
I've read some of your posts so I know you're not daft.
Imagine trying to switch on a mosfet with a 1M source impedance. The
falling drain voltage would (due to drain/gate cap) hold the gate low
for [comparatively] ages thus increasing the switching losses
(negative feedback). Effectivley the gate drive would be slow so the
drain voltage would fall slowly (thus increasing normal mosfet
losses). But you already know this.
I imagine gnome knows all the formulae. Try a new thread with smps in
the title :)
dont forget when comparing FET Rdson to pick some Tj other than 25C.
different FETs have different Rdson-vs-Tk curves. I always normalise to
125C and go from there.
