Then the lowest quiescent current is when all input are at VCC or GND, and there is no load. It is 80uA max (I expect around 10uA typical. It increases with temperature)
But if there is load connected to GND, and the output is HIGH, then ICC is increased by the load current (and may e so e negligibke additional current).
If the load is connected to VCC, then there is no significant increase in ICC...independent of load current.
Yes.that will be added to the ICC of the IC?
Iconsume = ICC + ILOAD
Iconsume = 80uA + 40mA
Iconsume = 40.08mA
I don´t know where your "8uA" come from. But maybe 8uA is correct.
Load: Tell us what your load is. Then we can discuss about.
you say: "so that means the (+) of load is connected to GND?" so your load has a "+" pin? then usually the "-" pin is connected to GND.
In the datasheet specifications usually "load" is a resistor. (you will find this information in the datasheet).
either:
output pin --> load --> GND
or:
output pin --> load --> VCC
a "load" is a (at least) two wire device. One wire is connected to the desired output of your logic device. The other is connected either to GND or VCC.
If the load is connected to GND, then the load current is added to the ICC_quiet.
--> take a sheet of paper and draw the current flow.
You say: "50mA is its absolute max ICC rating right?" --> How can I know this? You posted only a small part of the datasheet.
Yes.
But only with the given conditions: all inputs connected to GND or VCC, no change in level. = DC, = no frequency.
Hi Klaus,
Oh, ok, now i get it clearly (somehow). The load can be an LED then. The output of IC would either light up LED or activate transistor to light up LED. Though i have been thinking also, would that be different if the output of IC would be an input to another IC, like input to an enable pin of other ICs??
The output current of the first IC = the input current of the scond IC.this second image is now related on what pops out on my mind, how could i solve the first IC's current consumption (also other ICs) mathematically then??
Hi,
The output current of the first IC = the input current of the scond IC.
If it's another '138 then it's about nothing. +/- 0.1nA ... +/-100nA at 25°C. This is the benefit of CMOS devices.
(And that's one reason not to leave the inputs unconnected ... the logic state is unknown)
Sorry. Now I've recognized the link to the datasheet.
Klaus
That will be added to the ICC of the IC?
Iconsume = ICC + ILOAD
Iconsume = 80uA + 40mA
Iconsume = 40.08mA
Yes.
But only with the given conditions: all inputs connected to GND or VCC, no change in level. = DC, = no frequency.
You urgently need to read through logic families.Also, how can I determine if the device is CMOS in the datasheet? this datasheet doesn't actually specify if it uses, at first, I thought it was TTL.
Generally yes. Again with the restriction for "steady state" = non switching.then summation all of it to get the total power consumption?? is that right??
Why the 4mA?Lastly, I'm looking again at the datasheet at the Function Table pg.2 - So since there can only be 1 IOL (1 Low output pin) on every input combination, Iconsume while on operation is always 4.08mA?
Totally wrong.Total IOH = -4mA x 7
Total IOH = -28mA
Total Iconsume = 40.08mA - 28mA
Total Iconsume = 12mA (this'll be like no current was consume or power used?)
You urgently need to read through logic families.
https://www.ti.com/sitesearch/docs/universalsearch.tsp?searchTerm=hct family#linkId=13
https://www.ti.com/general/docs/lit...yMatch=hct family&tisearch=Search--Everything
https://www.ti.com/general/docs/lit...yMatch=hct family&tisearch=Search--Everything
**broken link removed**
.. There are a lot more documents.
Reading is essential. Boring, but unavoidable. Even we professionals need to read a lot of documentations/specifications - almost daily.
Generally. the "C" in "HCT" tells that is a CMOS device.
H = high speed
C = CMOS
T = TTL input levels
Generally yes. Again with the restriction for "steady state" = non switching.
Why the 4mA?
It can source 4mA, but the real current depends on load - as already explained. (if your car can drive with 200km/h it doesn´t mean it always goes that fast)
Totally wrong.
Especially I don´t know how you came to: "Total Iconsume = 40.08mA - 28mA"
I already agreed to your statement: "Iconsume = ICC + ILOAD"
but now you used a "minus"....and the 40.08mA already include the output current.
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