Re: question about power dissipation

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.

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?

Agreed.

Thanks again

Alex

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.

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.

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.

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.

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.

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.

Fred,

Thanks for your explanati> V*I is the generic calculation of the power *supplied* by the circuit DC

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.

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 equalization instead of pre-emphasis.

In any case, thanks.

Alex