[SI-LIST] Re: IO Power Dissipation in SoC

  • From: David Utah <telegrapher9@xxxxxxxxx>
  • To: ajay.dhingra@xxxxxxxxx
  • Date: Sat, 2 Aug 2014 09:09:20 -0600

1) When charging a capacitor from zero volts the energy stored in that
capacitor is W = 1/2(V^2)C
2) The charge stored in the capacitor is Q = CV = i x t

3) The energy input to charging the capacitor is Wcharging = V x i x t

Let's run the numbers using V = 3.3 volts and C = 10 pF.

The energy stored in the 10 pF is 1/2(3.3^2)(10pF) = *54 pJ*

The charge stored is (10pF)(3.3V) = 33 pC, and i x t = 33 pC

Now we run the equation on line (3) and Wcharging = (3.3V)(33pC) = *108 pJ*

We see that the energy into charging the 10 pF capacitance is 108 pJ while
the energy stored in the capacitance is 54 pJ. Where did the "missing" 54
pJ go? It is energy dissipated in the charging resistance, which in this
case is the driver transistor.

     Dave Cuthbert


On Sat, Aug 2, 2014 at 5:50 AM, Ajay Dhingra <ajay.dhingra@xxxxxxxxx> wrote:

> With this basis in my opinion we can say IO driver doesn't really
> dissipate significant and most of the energy is just transferred to load
> cap.
> In heat calculation IO driver consumption should be negligible as energy
> is transferred with in few nano seconds and thus not enough time for
> increasing the device temperature.
>
> Sent from my Huawei Mobile
>
> "Mahesh@Reliant" <mahesh@xxxxxxxxxxxxxxx> wrote:
>
> >
> >I would apply basics as follows
> >
> >Consider a straight line at 70 degrees as rise time
> >
> >By basics, the voltage drop between Vcc and O/p is dropped in the IO
> >buffer and will account for heat.  So for the rise time, the actual heat
> >disspated would be an integral function
> >where Tr is the rise time
> >
> >heat =  Integral 0 to Tr  ((Vdd-Vo)*Io dt
> >this multiplied by the toggles
> >
> >
> >This would be quite complex to calculate, so an approximation of half
> >the area of the triangle will do
> >
> >
> >On 02-08-2014 13:50, Ajay Dhingra wrote:
> >> My fundamental question is if the power dissipation in buffer driver of
> a transmitter is dependent on source impedance ( drive strength)or not.
> >>
> >> Apparently all the literature takes into account only load cap but no
> one is talking about AC Current sourced by transmitter irrespective of load
> cap.
> >>
> >> I really wonder the current which passes through IO buffer of a driver
> would really convert to any kind of heat. Apparently current would be
> sourced just for few nano seconds, basically for duration of rise/fall time.
> >>
> >> While calculating power dissipation in a device can we neglect IO
> Buffer power consumption which would be VccQ*IccQ. Where IccQ would vary as
> per different drive strength settings.
> >>
> >> Thanks
> >> Ajay
> >>
> >>
> >> Sent from my Huawei Mobile
> >>
> >> David Utah <telegrapher9@xxxxxxxxx> wrote:
> >>
> >>> When charging/discharging a capacitive load the energy dissipated is
> >>> voltage squared times the load capacitance (plus the driver C). Note
> that
> >>> energy is dissipated both on charging and discharging the capacitance
> and
> >>> that's why the 1/2 in the equation you cite, goes away.
> >>> W = (V^2)C
> >>>
> >>> For a resistively terminated load use (V^2)/Rload + (V^2)Cdriver
> >>>
> >>>      Dave Cuthbert, NARTE Certified EMC Engineer
> >>>      Consultant for analog, instrumentation, power conversion, EMC
> >>>
> >>>
> >>> On Fri, Aug 1, 2014 at 3:10 AM, Ajay Dhingra <Ajay.Dhingra@xxxxxxxxxxx
> >
> >>> wrote:
> >>>
> >>>> Hi All
> >>>> I was wondering if someone can put a thought on how to calculate power
> >>>> dissipation in the I/Os(Buffer Drivers) of an SoC. Will it be
> >>>>
> >>>> Buffer I/O Supply Voltage(VCCQ) * I/O Current
> >>>>
> >>>> Where I/O Current is as per the drive capability of driver(drive
> strength).
> >>>>
> >>>> OR it will be
> >>>>
> >>>> =1/2 *{CVF}* V
> >>>>
> >>>> Where C is the load cap
> >>>> V is the I/O voltage amplitude
> >>>> F is the toggle frequency of I/O.
> >>>>
> >>>>
> >>>> OR
> >>>>
> >>>> Both the proposals are wrong. Kindly educate me.
> >>>>
> >>>> Point to note: in first proposal current is independent of Load
> whereas in
> >>>> second current depends on Load Cap.
> >>>>
> >>>> Thanks
> >>>> Ajay Dhingra
> >>>>
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