HCM / LCM signalling

  • Thread starter Geronimo Stempovski
  • Start date
G

Geronimo Stempovski

I just read an interesting paper about high-speed I/O's power dissipation.
Unfortunately there is an equation I don't quite understand. Maybe someone
is in the mood for discussing and explaining the correctness of the equation
to me.
The formula I am talking about is (1) in the paper [
http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf ]

For high-common mode signaling (which standard would that be, anyway? TTL?
CMOS? SSTL?) it is assumed
P = V*Vswing/Z0 = V*Vrx/Z0*H(f)

For low-common mode signaling (LVDS? CML? LVPECL?) it states
P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

What I don't understand is the factor 2 (2*Z0) in the calculation of the
low-common mode signaling. Furthermore I'm not sure if the H(f)^2 is
correct.

Any help is highly appreciated! Thanks a lot in advance!

Regards, Gero
 
J

Jon Slaughter

Geronimo Stempovski said:
I just read an interesting paper about high-speed I/O's power dissipation.
Unfortunately there is an equation I don't quite understand. Maybe someone
is in the mood for discussing and explaining the correctness of the
equation to me.
The formula I am talking about is (1) in the paper [
http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf ]

For high-common mode signaling (which standard would that be, anyway? TTL?
CMOS? SSTL?) it is assumed
P = V*Vswing/Z0 = V*Vrx/Z0*H(f)


If you look at fig. 3 they show cmos.
For low-common mode signaling (LVDS? CML? LVPECL?) it states
P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

What I don't understand is the factor 2 (2*Z0) in the calculation of the
low-common mode signaling. Furthermore I'm not sure if the H(f)^2 is
correct.

look at fig 3.

Notice that they say that for LCM the voltage is dependent on V_swing^2.
Then right above equations 1 they say that H(f) is the attenutation. So they
basicaly substitute the ration V_RX/H(f) = V_swing


Any help is highly appreciated! Thanks a lot in advance!

Hope that helps,
Jon
 
G

Geronimo Stempovski

Jon Slaughter said:
look at fig 3.

Notice that they say that for LCM the voltage is dependent on V_swing^2.
Then right above equations 1 they say that H(f) is the attenutation. So
they basicaly substitute the ration V_RX/H(f) = V_swing

Thank you, Jon! That sure helps me in understanding where the H(f)^2 comes
from. But what about the 2*Z0 ? Why is it 2*Z0 and not just Z0 ? Maybe I
just can't see the wood for the trees...

Regards, Gero
 
J

Jon Slaughter

Geronimo Stempovski said:
Thank you, Jon! That sure helps me in understanding where the H(f)^2 comes
from. But what about the 2*Z0 ? Why is it 2*Z0 and not just Z0 ? Maybe I
just can't see the wood for the trees...

Because there are twice as many transistors instead of one? To be honest I
really don't know ;/ He doesn't define what Z0 is so I'm only guessing. It
could also be due to the (V_swing/sqrt(2))^2 = V_swing^2/2 where it was some
rms like value. This seems to be the likely reason as he says the voltage
used is proportional to V_swing but its definately not clear IMO and not
written well also(IMO Ofcourse, it might be completely obvious to someone
else). He doesn't seem to define the terms well and seems to take it for
granted that you already know what he's talking about.

For example, he uses V_S in the fig but I never see him mention what V_S is.
I can only conclude that its V_swing.

I would expect that V_swing is peak to peak and so he uses the average in
the power equation but I could be entirely wrong here.

IMO it is a poorly written paper. It could just be me though.

Jon
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Members online

No members online now.

Forum statistics

Threads
473,769
Messages
2,569,580
Members
45,055
Latest member
SlimSparkKetoACVReview

Latest Threads

Top