[SI-LIST] Re: IO Power Dissipation in SoC
- From: Ajay Dhingra <ajay.dhingra@xxxxxxxxx>
- To: Wolfgang.Maichen@xxxxxxxxxxxx
- Date: Mon, 04 Aug 2014 18:43:06 +0530
thanks for your feedback guys.
collectively it cleared the theory.
Regards
Ajay
Sent from my Huawei Mobile
Wolfgang.Maichen@xxxxxxxxxxxx wrote:
>How much power is dissipated in the driver stage depends a lot on what load is
>there.
>
>For purely capacitive load (high-impedance receiver, low-loss transmission
>line between driver and receiver) the (reactive!) load obviously cannot
>dissipate energy. All(!) the energy (RMS voltage * RMS current) will thus be
>dissipated in the driver. E.g. see Dave's calculation below - there half is
>dissipated during the charging cycle, the remaining half will be dissipated
>when the driver switches low (discharging cycle).
>
>For a resistive load (e.g. receiver terminated with 50 Ohms) there is current
>flow even during time when no switching occurs. Here part of the power will be
>dissipated in the driver and part in the load. For a 50 Ohm driver and a 50
>Ohm load it will be half each. For a 7 Ohm driver (PECL for example) and a 50
>Ohm load the load will dissipate most.
>
>If you want to get into even smaller details, of course there will be some
>(usually very small) portion of energy be radiated out into space, and another
>portion will get dissipated (converted into heat) in the transmission path
>(due to DC resistance, skin effect, dielectric losses).
>
>Regards,
>
>Wolfgang
>
>
>
>
>-----Original Message-----
>From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
>Behalf Of David Utah
>Sent: Saturday, August 02, 2014 5:09 PM
>To: ajay.dhingra@xxxxxxxxx
>Cc: Mahesh@Reliant; si-list@xxxxxxxxxxxxx; Rel Ragupathy
>Subject: [SI-LIST] Re: IO Power Dissipation in SoC
>
>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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