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# Solve Current Consumption of IC in Datasheet

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#### mheruian

##### Junior Member level 3
Hi, I'm new here and maybe this is the best section where I could need help I've been interested in electronics lately (not schooling anymore, more on self-study, but I'm not that smart, just a dedicated one and i can't understand everything right through). I don't have any equipment yet like osci, etc. Then I thought of doing mathematics on electronic components like integrated circuits (IC) since even though ideal, math in some percentage of little marginal error, result should reflect close to reality (simple math is what I rely on).

Here's where I am at, I'm trying to know the current consumption (minimum and max) of this IC (https://www.ti.com/lit/ds/symlink/sn74hct138.pdf). Upon reading the documentation, I am at first thought it was easy since 80uA was brought up in the description but skimming on the electrical characteristics (image link below) looks like this is the max current consumption when there is no output current (Io = 0) on its decoding output lines.

That means the IC is in idle state right consuming current just for powering it self up and not for output/decode operation? Then i found the continuous current output and continuous current through VCC or GND? I got confused but still proceeded.

Here's what I computed for getting current consumption for output operation, idle operation, min and max consumption:

Iconsume = Icc + IOL
Iconsume = 80uA + 4mA
Iconsume = 80uA + 4mA
Iconsume = 4.08mA

Imin = 1000nA or 1uA (this is idle state also?)

Imax = 50mA (Maybe an addition how did they get this?)

If I can determine these currents in the 3 states of the IC, I can now know the power consumption of the IC. I hope someone could verify this or correct me and show me how even on a small example.

#### KlausST

##### Super Moderator
Staff member
Hi,

ICC is the current at VCC node.

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.

Then there is I_in. The current at the input pin. It is low compared to ICC.

If the input voltage is not VCC or not GND, but something inbetween, then ICC increases, but not necessarily I_in. This is delta ICC.
ICC is increased caused by internal crowbar currents from VCC to GND through the input stage transistors.

Then there is dynamic current when there is some input signal (frequency). Not shown in this piece of datasheet.

Klaus

mheruian

### mheruian

Points: 2

#### mheruian

##### Junior Member level 3
Great Day Mr.Klaus! Just got home and studied some technical terms on electronics.

I'm really thankful you had answer because I really like electronics and been new to this stuff, no one teaches me or anyone i can talk to with. So to have this confusion reading datasheet to end at least, you've been a great help. If I were able to understand fully this datasheet, then i wouldn't be having trouble understanding other datasheet right? specially what I really want to know is power consumption of ICs for now only.

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)

Ok, so at least on ambient temperature of 25°C or less, quiescent current would be 8uA.

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.

This is where I'm getting confused at. Well, i know now what the term load is (circuit, appliance, etc). What do you mean load is connected to GND? Load is powered up by connecting it's (+) to VCC and (-) to GND right? so that means the (+) of load is connected to GND? Output is High = ICC increased by load current - the current being supplied on the load? let's say current's load is 40mA (50mA is its absolute max ICC rating right?), that will be added to the ICC of the IC?

Iconsume = 80uA + 40mA
Iconsume = 40.08mA

is that how its math works? Sorry if i still can't grasp it yet. I would like to understand it well in simple math since I want to compute the 2 currents the IC would consume; Iconsume while on idle state, Iconsume while on output/operation state. I hope i was able to convey my thoughts properly here. Thanks a lot Mr.Klaus

#### KlausST

##### Super Moderator
Staff member
Hi,

***
I don´t know where your "8uA" come from. But maybe 8uA is correct.

***
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.

--> 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.

*****
that will be added to the ICC of the IC?

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.

Klaus

V
Points: 2

#### mheruian

##### Junior Member level 3
Hi Klaus,

I don´t know where your "8uA" come from. But maybe 8uA is correct.

Well it can be found at the datasheet of the IC (at the section where you can see the ICC) I mention to calculate. Datasheet link was given on my first post.

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.

--> take a sheet of paper and draw the current flow.

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??

You say: "50mA is its absolute max ICC rating right?" --> How can I know this? You posted only a small part of the datasheet.

Sorry for that, the link can be found on my first post before the added image.

Yes.
But only with the given conditions: all inputs connected to GND or VCC, no change in level. = DC, = no frequency.

Oh ok, "all inputs" you mean the 6 inputs of the IC (IC were discussing is a decoder) with exception of Vcc pin and GND pin (these are input pins also right? but not considered to use the Input current parameter on the datasheet, it should be ICC?)

Best Regards,
Mheruian

#### mheruian

##### Junior Member level 3
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 image above a drawing of my understanding on what you had taught to me, so the red circle drawing would have it's load current be added to the current consumption of the IC right? the next one, since load's (+) is connected to VCC, it's current shouldn't be added to the current consumption of 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??

#### KlausST

##### Super Moderator
Staff member
Hi,

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??
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

mheruian

### mheruian

Points: 2

#### mheruian

##### Junior Member level 3
Hi Mr.Klaus!

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

Ok thanks a lot, that would mean, if all these (3 ICs in the image) ICs would be CMOS devices, i just have to compute each of their ICC and VCC to get their power then summation all of it to get the total power consumption?? is that right??

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.

That will be added to the ICC of the IC?
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.

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? but this consumption will be decrease since the 7 remaining output pin would be High - IOH, that'll be:

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?)

Do i get that right?

#### KlausST

##### Super Moderator
Staff member
Hi,

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.
You urgently need to read through logic families.
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
.. 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
****

then summation all of it to get the total power consumption?? is that right??
Generally yes. Again with the restriction for "steady state" = non switching.
****

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?
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)
****

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?)
Totally wrong.
Especially I don´t know how you came to: "Total Iconsume = 40.08mA - 28mA"

but now you used a "minus"....and the 40.08mA already include the output current.

Klaus

mheruian

V
Points: 2

### mheruian

Points: 2

#### mheruian

##### Junior Member level 3
You urgently need to read through logic families.
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
.. 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

Hi Mr.Klaus this will be a great help, I would definitely read them all!

Generally yes. Again with the restriction for "steady state" = non switching.

Ok, got it. If there will be output switching of HIGH or LOW on IC (source), make sure to add up that current and consider the input current capable of the another IC (receving)??

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)

Oh ok, this would mean that the IC (based on absolute max rating), it can output source current up to 25mA for the load, it'll be 4mA if the load requires 4mA input? Then since the IC can only have 50mA max from its VCC to GND, it can only accommodate two (2) 25mA output source current or one (1) 25mA, one (1) 10mA, two (2) 5mA output source current regardless if its high or low?

Totally wrong.
Especially I don´t know how you came to: "Total Iconsume = 40.08mA - 28mA"

but now you used a "minus"....and the 40.08mA already include the output current.

Oh, i forgot to tell details, sorry for that i came up on the 40.08mA by thinking that since based on absolute max rating, IC can only consume max of 50mA (VCC to GND), i considered that there will be instance that IC would output 40mA and use 80uA only for the IC to operate. And since this IC will only have one (1) low output, that'll be like:
Iconsume1 = Iout + Icc
Iconsume1 = 40mA + 80uA
Iconsume1 - 40.08mA

and since there are seven (7) outputs that are high, thinking these outputs should be added to, thus i saw -4mA on datasheet at VOH.
Iconsume2 = -4mA x 7
Iconsume2 = -28mA

then thinking it'll be like this:
Total Iconsume = Iconsume1 + Iconsume2
Total Iconsume = 40.08mA + (-28mA)
Total Iconsume = 12.08mA or 12mA

then i guess, what i think above was wrong.

Thanks for being supportive. I am learning. And sorry if it takes time for me to get it right.
Mheruian

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