### Network

G

#### Geronimo Stempovski

Jan 1, 1970
0
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

O

#### operator jay

Jan 1, 1970
0
Geronimo Stempovski said:
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

I don't know the topic or the equation in question, however,

P = Vpeak/sqrt(2) * Vpeak/sqrt(2) / R = Vpeak^2 / 2 * R
would give power for a simple AC voltage applied to an R. Here the
factor of 2 is 'changing' the voltages to their RMS values. Could it
be similar in the equation you are looking at?

G

#### Geronimo Stempovski

Jan 1, 1970
0
operator jay said:
factor of 2 is 'changing' the voltages to their RMS values. Could it be
similar in the equation you are looking at?

Thank you, Jay. Unforutnately not. Could I mail you the paper and you have a

N

#### Noway2

Jan 1, 1970
0
Geronimo said:
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

NOTE: Cross posting linker were removed and this response was posted
only to sci.electronics.design

I am not familiar with the terms high-common mode and low-common mode,
but the methods you refer to in parentheses seem to suggest single sided
transmission with a common ground return versus differential.

If this is the case, then the 2X factor could be due to that you have
differential signaling and hence two conductors with currents of equal
magnitude and opposite direction flowing in both conductors?

J

#### Jim Granville

Jan 1, 1970
0
Geronimo said:
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!

That does seem mangled.
When in doubt, check the dimensions of the answer ?

Normally where frequency is used in power calcs, it is of the form
of power dissipation capacitance : W = Fo * Cp * Vcc^2

-jg

T

#### Terry Given

Jan 1, 1970
0
Jim said:
Geronimo said:
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!

That does seem mangled.
When in doubt, check the dimensions of the answer ?

Normally where frequency is used in power calcs, it is of the form
of power dissipation capacitance : W = Fo * Cp * Vcc^2

-jg

it looks like its analagous to an RMS calculation, but made on an
unknown dataset of a particular (hardware) flavour. its a mere proof by
blatant assertion, so it'd take a fair bit of digging or maths to figure
out where it came from.

ya gotta love professional comics.

Cheers
Terry

F

#### Fred Bloggs

Jan 1, 1970
0
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.

The background presentation in the paper is somewhat abbreviated because
the authors are implicitly assuming the reader is participating in an
ongoing discussion of fundamental limitations of I/O throughput due to
channel, architectural, and power efficiency considerations. Both power
equations are of the form V*I where V is DC power supply for the
individual channel circuit driver and I is the current switched into the
line termination *at the destination* , making this is a calculation of
power supplied by the DC supply that is dissipated in the termination at
the destination. In the case of HCM the channel driver is powered by
Vdd, making V=Vdd, and the current switched from it into the line is
I=Vswing/Zo, where it is assumed the line is terminated in a matched
resistive impedance Zo, so that P=Vdd*Vswing/Zo. In the case of LCM, the
channel driver is powered by a DC supply of magnitude Vswing and the
driver is a series terminated circuit, by the MOSFET triode region
channel resistance, putting Vswing/2 on the Zo termination at the
destination. Therefore, for LCM, V=Vswing and I=(Vswing/2)/Zo. This may
seem inconsistent with the HCM equation but it is not when you
understand that the assumption in both cases is that the receiver
equalization makes I*Zo=Vrx/H(f). This confusion is just a scale factor
anyway and does not affect the calculation in section II for minimum
energy per bit versus data rate gamma*f over channel H(f).

A

#### [email protected]

Jan 1, 1970
0
This is a nice explanation but still a bit puzzling. Especially
confusing is the fact that
Vswing has a different definition in the cases of LCM and HCM. Also,
according to the paper, the signaling power is that dissipated by the
termination and the lossy transmission line. If Vswing stands for the
voltage measured at the output pin of the transmitter and the line is
lossless, wouldn't the power always be Vswing^2/Z0 ?

Also, in the case of HCM, Vdd is not really the voltage across the
channel' so the product V*I is not really the power dissipated by the
channel+termination. If I have two equal resistances in a series and
put V volts across both resistors and try to measure the power
dissipated by the bottom one, it will not be V*V(across R)/R=V^2/2R
but rather V^2/4R. Why is this not an accurate model of the HCM
transmitter? Or is the load resistor not considered part of the
driver? Or (more likely) they just use their expression because it has
the dimension of power and is related to the actual power dissipated
at the receiving end?

Finally, their analysis does not really care about the 2 in the
denominator (or any other constants for that matter) but does care
about the power of H(f) there. I can belive the fact that LCM is more
sensitive to the frequency response since the HCM transmitter is a
current source and therefore in V*I, the I part does not depend on
H(f) (as noone looses current along the way in a transmission line).
Is this the idea of their analysis (greatly simplified, of course)?

In any case, thanks for the explanation. I would appreciate any
clarification on the questions above.
destination. Therefore, for LCM, V=Vswing and I=(Vswing/2)/Zo. This may
seem inconsistent with the HCM equation but it is not when you
understand that the assumption in both cases is that the receiver
equalization makes I*Zo=Vrx/H(f).

This part is somewhat unclear to me. Regardless of what happens at the
receiver end, wouldn't it always be that
I*Z0=Vout and Vrx=H(f)*Vout so I*Z0=Vrx/H(f) (provided the load looks
resistive and that's essentially what equalization will achieve)?
Essentially, equalization will just make the reactive component of the
load invisible' to the reciever but this is already assumed in all
the models above, so how does this relate to the factor of 2 in
equation (2) in the paper?
This confusion is just a scale factor
anyway and does not affect the calculation in section II for minimum
energy per bit versus data rate gamma*f over channel H(f).

Agreed.

Thanks again

Alex

F

#### Fred Bloggs

Jan 1, 1970
0
This is a nice explanation but still a bit puzzling. Especially
confusing is the fact that
Vswing has a different definition in the cases of LCM and HCM.

The definition is the same and the difference between HCM and LCM is
that HCM powers the channel drivers from the logic level power supply,
Vdd, and LCM powers its drivers from a low voltage power supply on the
order of Vswing.
Also,
according to the paper, the signaling power is that dissipated by the
termination and the lossy transmission line. If Vswing stands for the
voltage measured at the output pin of the transmitter and the line is
lossless, wouldn't the power always be Vswing^2/Z0 ?

Lossless or not, the transmitter puts Vswing across the line at its
output, so the power into the line will be Vswing^2/Zo is correct.
Also, in the case of HCM, Vdd is not really the voltage across the
channel' so the product V*I is not really the power dissipated by the
channel+termination.

V*I is the generic calculation of the power *supplied* by the circuit DC
power source. If the line draws current Vswing/Zo from the driver, then
the DC supply, Vdd, must provide Vswing/Zo amperes to the driver, making
the power provided by the DC supply Vdd*Vswing/Zo. The paper is
interested in computing the power supply requirements as well as the
driver circuit internal dissipation requirements as a function of data rate.

If I have two equal resistances in a series and
put V volts across both resistors and try to measure the power
dissipated by the bottom one, it will not be V*V(across R)/R=V^2/2R
but rather V^2/4R. Why is this not an accurate model of the HCM
transmitter? Or is the load resistor not considered part of the
driver? Or (more likely) they just use their expression because it has
the dimension of power and is related to the actual power dissipated
at the receiving end?

They break the total power drawn from the DC power supply into two
components: the power dissipated by the line and its termination; and
the power dissipated by the driver electronics and whatever internal
source termination it uses. The receiver Vrx, the line loss H(f), and
the characteristic impedance Zo, are all that are necessary to compute
the power required by the line and its termination at a particular
Nyquist frequency f. The driver internal dissipation is not so
straightforward and must be computed on a case by case basis as a
function of CMOS process and circuit topology.
Finally, their analysis does not really care about the 2 in the
denominator (or any other constants for that matter) but does care
about the power of H(f) there. I can belive the fact that LCM is more
sensitive to the frequency response since the HCM transmitter is a
current source and therefore in V*I, the I part does not depend on
H(f) (as noone looses current along the way in a transmission line).
Is this the idea of their analysis (greatly simplified, of course)?

Yes, it is all very straightforward. Once again:

for the HCM case:

the receiver requires Vrx at its input,

this means the transmitter must place Vrx/H(f) at its side of the line,

Vrx/H(f) across the line input requires (Vrx/H(f))/Zo amperes,

the (Vrx/H(f))/Zo amperes supplied by the driver, into the line,
ultimately is supplied by the DC power supply Vdd to the driver,

so the DC power supply Vdd is providing Vdd*(Vrx/H(f))/Zo watts to make
just the line portion of the bit transmission happen.

The LCM case is similar.

In any case, thanks for the explanation. I would appreciate any
clarification on the questions above.

This part is somewhat unclear to me. Regardless of what happens at the
receiver end, wouldn't it always be that
I*Z0=Vout and Vrx=H(f)*Vout so I*Z0=Vrx/H(f) (provided the load looks
resistive and that's essentially what equalization will achieve)?
Essentially, equalization will just make the reactive component of the
load invisible' to the reciever but this is already assumed in all
the models above, so how does this relate to the factor of 2 in
equation (2) in the paper?

First of all, the I at the receiver end is not the same as the I at the
transmitter end. A transmission line does not obey the laws of lumped
element circuit analysis because it is entails energy transmission by
wave propagation. In both LCM and HCM, you have (Vrx/H(f))/Zo amperes
must be injected into the line at the transmitter side. One difference
is that for HCM, Vswing=Vrx/H(f), and for LCM, Vswing/2=Vrx/H(f). The
second difference is that for HCM, the DC voltage supply for the circuit
is Vdd, and for LCM, DC voltage supply for the circuit is Vswing. The
resulting power required of the DC supply for HCM and LCM are then
Vdd*Vrx/(H(f)*Zo), and Vswing*Vrx/(H(f)*Zo), respectively. But because
Vswing/2=Vrx/H(f) in the LCM case, Vswing=2*Vrx/H(f), making the power
in terms of Vrx, 2*(Vrx/H(f))^2/Zo watts, and not (Vrx/H(f))^2/(2*Zo) as
they state in the paper.

A

#### [email protected]

Jan 1, 1970
0
Fred,

V*I is the generic calculation of the power *supplied* by the circuit DC
power source. If the line draws current Vswing/Zo from the driver, then
the DC supply, Vdd, must provide Vswing/Zo amperes to the driver, making
the power provided by the DC supply Vdd*Vswing/Zo. The paper is
interested in computing the power supply requirements as well as the
driver circuit internal dissipation requirements as a function of data rate.

The only remark I have is that this is not the power that goes into
signaling: some of it is spent on driving circuitry in the driver, say
the load resistor in the HCM driver but I guess it is a power metric
as this is the minimal power the power supply must provide for driving
the line.
First of all, the I at the receiver end is not the same as the I at the
transmitter end. A transmission line does not obey the laws of lumped
element circuit analysis because it is entails energy transmission by
wave propagation.

This is certainly true, however, conservation of charge will guarantee
that the integral of I is the same at both ends. This, with
transmitter pre-emphasis (that will take care of phase disparity) is
enough to make the power computation valid. Now, that I have reread
it, what you said makes perfect sense, except that you used

In any case, thanks.

Alex

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