[SI-LIST] Re: IO Power Dissipation in SoC
- From: Ajay Dhingra <ajay.dhingra@xxxxxxxxx>
- To: "Mahesh@Reliant" <mahesh@xxxxxxxxxxxxxxx>, si-list@xxxxxxxxxxxxx
- Date: Sat, 02 Aug 2014 17:20:57 +0530
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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