PCA9557: about PCA9557 TJ

Hello Vera,

Unfortunately, we do not specify the max operating junction temperature.  Instead we specify the Max Operating free-air temperature which uses the JEDEC standard for the PCB.  Max power consumption depends on a lot of variables.  First, how many of the devices will be inputs and how many will be outputs?  The outputs can drive up to 24mA depending on Vol.  Generally speaking the IO don't drive large loads, so the power dissipation doesn't normally high.  But if driving higher loads then you need to figure out the max power dissipations (Pdis) because it can add a large delta for air to junction temperature.

Where did you get the 0.079mW?  That comes out to 21.9 uA.  You could have much higher loads than that depending on how the part is used.

If the power dissipation is low then the temperature differential between the ambient air (Ta) and the junction temperature (Tj) will be small and you can approximate the max junction temperature to be the max operating free air temperature.   We guarantee our specifications based on max free-air temperature based on JEDEC standard.  I wished we do it this way but it has been done this way for a long time on these types of parts.  My opinion is that we should spec and test our parts based on max junction temperature because the thermal resistance from Air to Junction varies to much, which the thermal resistance from junction to case does not, and you can measure the case temperature with a thermal couple which is very accurate. 

Let me know if this is clear and if there is anything else you need clarification on.

-Francis Houde

hi Francsis,

thanks for the reply!

yes, I got the 0.079W from the Iq, so for I load of 24mA, could I calculate the power consumption in this way: 0.079W+3.6*24mA? 

actually I don't have specified operating condition, this is request by customer's sourcing team, they just want to do investigation on the chip. tks!

Hello Vera,
It isn't quite that easy. It really depends on what you are driving.

If you are driving into an input that is high impedance than the power consumption is very low.

If you driving something like an LED, say for example it is an LED that has a Vf of 2.1V and can handle 24mA that the Power dissipated would be (Vcc-2.1)V*24mA , say we have Vcc = 3.3V than we would have (3.3-2.1)V*24mA = 28.8mW , now than you would have to add the Pq (quiescent power) + other output pin loading. Remember that the power dissipated IN the part is the current going through the device multiplied the voltage from Vcc to that on the pin of the output of the pin. Theses are linear devices so that simplifies some of the math.

Does this make sense? Let me know if you have more questions.
-Francis Houde