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Gate drive in datasheet

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Tim Williams

Jan 1, 1970
0
Whoa, never seen that before.
http://www.ti.com/lit/ds/symlink/csd18501q5a.pdf
Switching times measured at "RG = 0".

I'd really like to have a copy of their drive circuit. I could solve a
lot of problems with unlimited current drive and zero inductance, even if
it's only 10V.

Tim
 
Whoa, never seen that before.

That's because you are not familiar with the struggle to minimize ringing in circuits containing these devices. The RG refers to the value of dampingbootstrap resistance in the gate drive circuit. Since it's application specific they revised the original datasheet from RG 2R to RG 0R. The user will have to make his own determination of best RG.
http://www.ti.com/lit/an/slpa010/slpa010.pdf
 
M

miso

Jan 1, 1970
0
I could have cut them some slack if the idea was they used a low
impedance driver rather than a function generator with 50 ohms
impedance. That is, maybe they meant Rgen is low, as in Rgen!-=50. .
However RG in the datasheet is the intrinsic gate resistance, so it can
never be zero.
 
T

Tim Williams

Jan 1, 1970
0
miso said:
I could have cut them some slack if the idea was they used a low
impedance driver rather than a function generator with 50 ohms impedance.
That is, maybe they meant Rgen is low, as in Rgen!-=50. . However RG in
the datasheet is the intrinsic gate resistance, so it can never be zero.

No, the driver source impedance. Internal RG is specified (0.5 to 1.5
ohms or whatever). Presumably that's what they mean, but that still
requires zero external impedance.

How would you create a true zero ohm gate driver circuit?

Tim
 
T

Tim Williams

Jan 1, 1970
0
So if it's about ringing, why are they testing it at a clearly impractical
condition?

The appnote below only says: "durr, sometimes you need to slow it down
with RG, but then it runs hotter". None of their waveforms even solve the
problem set out in the title -- they all ring! I'm kind of embarassed by
that whole appnote, frankly -- a component manufacturer's engineers should
know the secret to quiet, efficient inverters by now. And it ain't
"minimize inductance".

At any rate, this doesn't answer my question: how do you implement a true
zero ohm gate driver? Are they just out and out lying? I wouldn't put
that past a manufacturer like International Rectifier, but TI?

For one example, I've never seen an IXYS datasheet showing RG < 0.2 ohms
or so, which happens to be around the output resistance of their driver
chips. Makes perfect sense, and it's practical.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://seventransistorlabs.com

Whoa, never seen that before.

That's because you are not familiar with the struggle to minimize ringing
in circuits containing these devices. The RG refers to the value of
damping bootstrap resistance in the gate drive circuit. Since it's
application specific they revised the original datasheet from RG 2R to RG
0R. The user will have to make his own determination of best RG.
http://www.ti.com/lit/an/slpa010/slpa010.pdf
 
M

miso

Jan 1, 1970
0
No, the driver source impedance. Internal RG is specified (0.5 to 1.5
ohms or whatever). Presumably that's what they mean, but that still
requires zero external impedance.

How would you create a true zero ohm gate driver circuit?

Tim

I know that they mean driver impedance, but they should call it
something else besides Rg, which is the intrinsic gate resistance. This
is just a bad practice. It looks like apps and design were not coordinated.

You can't make a true zero ohm driver because it is an analog world! ;-)
Even your power supply has a finite impedance.
 
T

Tim Williams

Jan 1, 1970
0
In case anyone was wondering, contacting TI was not illuminating. They
said it simply means no external resistor was used. They didn't happen to
mention what the driver impedance was. I take that to mean, if you see
this figure in a datasheet, ignore it completely.

Tim
 
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