Hi all,

Please clarify my doubts about power budget calculation of board.

(1) total power = No. of devices x [ ( supply current x supply voltage) + ( leakage current x supply voltage) ] am I right ??:?:

(2) Whether I have to consider input/output current (Iol..Ioh..)???, specially in case of transceivers, buffers.

Thanks in advance.

 

(1) total power = No. of devices x [ ( supply current x supply voltage) + ( leakage current x supply voltage) ] am I right ??:?:

That only works if all the devices are identical and are used in an identical manor. Generally you need to look at each component that connects to a power supply rail and determine the current.

(2) Whether I have to consider input/output current (Iol..Ioh..)???, specially in case of transceivers, buffers.

Trickier, as you seem to imply these signals are going "off board." These currents need to be considered only if they come from your boards supply. If you have say a part that sinks current from another device and then returns the current thru a common ground then no, those currents are not in your boards budget.

Thanks in advance.

You're welcome

 

Thanks ErnieM,

>If I consider Icc, Vcc and I leakage in case of TTL, thats enough right??

>For any IC, to source a current it should take it from Icc only right?? in this scenario if I consider both Icc and Ioh/Iol for calculaton, my calculation will be doubled right??

 

Nobody is there to answer !!!!

 

Thanks ErnieM,

>If I consider Icc, Vcc and I leakage in case of TTL, thats enough right??

>For any IC, to source a current it should take it from Icc only right?? in this scenario if I consider both Icc and Ioh/Iol for calculaton, my calculation will be doubled right??

Generally Icc is the quiescent current, meaning the current the chip itself needs, not any current running in or out of any pins.

If one chip drives another the Ioh comes from one chip thru it's VCC, and back to ground thru the Vee of the next chip.

One detail that may be confusing is Ioh is spec'd as negative, and Iol as positive. This is due to the convention of defining current flowing into the device as positive, and current flowing out as negative.

Same applies to Iih and Iil (input currents). The output currents are the max the chip can source or sink, while the input currents are what the device will consume. So when looking to get a total (and maximum theoretical) current add up the input currents together with the supply currents.

It's a completely theoretical exercise to get a max value, and it also leaves out switching losses. Example: CMOS power is nil for the input but significant for the switch. For something like a digital circuit just making the actual measurement on a breadboard is far better then calculation.

 

Good explanation ErnieM.

ok lets simplify it with block diagram,

I have attached a picture showing 3 IC's connected to a single Vcc.
All i/o currents are generalized.

Now total power consumed by this block should be

P = ( Vcc*Icc1 + Vcc*Icc2 +Vcc*Icc3 + Some leakage current ).

then why should I bother about i/o currents. I/o currents are finally derived from Icc only right???

 

Wow, you sure do ask a lot of questions.

I looked up some stuff from TI on how they spec things: Understanding and Interpreting Standard-Logic Data Sheets

ICC Supply Current
The current into the VCC supply terminal of an integrated circuit. This parameter is the current into the VCC supply terminal of an integrated circuit under static no-load conditions. ICC is tested by applying the specified VCC level and measuring the current into the device.
The outputs of the device are left open, while all inputs—control and data—are biased to either VCC or GND. For CMOS technologies, this is done to eliminate any current that may be caused by any input conditions or output loads.

They don't tell you what the inputs do they do disconnect the outputs. That means there is no output current component in the Icc measurement.

So while it is completely true Ioh comes from Icc the spec sheet measurement is done when Ioh = zero (disconnected).

When you connect the output you have to account for the extra current you're now using.

Back in the olden days we had there things called "data books;" big heavy blocks of something called paper, no back lighting but when illiminated with adequate room light they would display both text and pictures. They held lots of good data on every part in the series, and the first few chapters had great general info on the series and well worth a read once or twice.

Now a days books are gone and you need to hunt down a Pee Dee Eff of the part, but that doesn't have the nice chapter in the beginning so you never think to look for it.