>From: Jack Olson <[log in to unmask]>
>Subject: [DC] Trace Inductance
>Originally to: "[log in to unmask]" <[log in to unmask]>
>To: [log in to unmask]
>
>I'm curious about something I read in an article on trace inductance. We
>design boards that are 1.9GHz, and the engineers are so worried about
>inductance that every decoupling cap (other components too) has two vias per
>pad. It seems to me that its not the two vias that make any improvement, but
>the fact that there are now two traces instead of one connecting the pad to
>the plane.
>But these traces are SO SHORT... is the inductance of a 20 mil trace that
>travels a short distance (maybe 10 to 100mils) something to worry about? How
>do you know when a trace has enough inductance to be a problem? (is there a
>short answer? grin...)
I'll try ...
There is one way to think about this. It's not a hard
fast rule, Some people estimate trace inductance to
be approx 20nH/inch. I've met people who have actually
measured board inductances at 5nH/inch.
Ok, now find out what the edge rate is, psec?
Then, find out what the current is, milliamps?
Voltage generated by a change in current is
calculated as
V = L * di/dt
IOW, voltage (V) generated by inductance (L) is
produced by a change in the current (di) within
a specific change in time (dt).
Let's "normalize" everything just for argument.
That means, let's see what happens when we set
everything to a value of 1. You'll see why I'm
doing this further on down.
L (trace inductance) = 1nH per inch.
di (current from zero to operating) = 1 mA
dt (time span for change in current) = 1 psec.
V (volts per inch) = (10^-9)*(10^-3)/(10^-12)
= (10^-12)/(10^-12)
= 1
V (volts per inch) = 1 volt PER INCH.
Now, if the inductance is REALLY 20nH per inch,
then the voltage is 20 volts per inch.
If the time is REALLY sub-picosecond, say 0.1 psec,
then the voltage 10 volts per inch.
If the inductance is really 20 nH per inch AND
the time is really 0.1 psec, then the voltage
is 200 volts per inch. A 2 inch trace would
produce 400 volts. A half inch trace would
produce 100 volts.
(Is it clear how I did that?)
So, if you're working with semiconductor devices
that trigger on only 1/2 or 3/4 of a volt, you
got a big problem. That's why some of the really
good EEs go into panic about trace inductance.
Putting two inductances in parallel cuts the
inductance in half and therefore cuts the level
of the spike in half.
In other words, some people think in terms of volts
per inch when talking trace inductances. And, if
everything is added up, it can add up to some
really huge spikes.
Regards, Doug McKean
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