[SI-LIST] Re: IO Power Dissipation in SoC
- From: Anto Davis <antokdavis@xxxxxxxxx>
- To: "si-list@xxxxxxxxxxxxx" <si-list@xxxxxxxxxxxxx>
- Date: Sun, 3 Aug 2014 11:23:20 +0530
Hi,
I would like to add, two articles to this,
1. On Energy Efficiency of Switched-Capacitor Converters
<http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=6218201&tag=1>
(efficiency
is actually Vaverage/Vmax - when starting from zero it is 50%)
2. See fig.6 of Capacitors can radiate
<http://www.electroyou.it/tucchoy/wiki/articolo1> (plot for fraction of
energy radiated, dissipated and oscillating)
thanks,
Anto
On Sat, Aug 2, 2014 at 8:39 PM, David Utah <telegrapher9@xxxxxxxxx> wrote:
> 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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