basic question about reading datasheets current values

Jan 14, 2009 5 Replies

Hello,



I want to design a small circuit, and want to power it with voltage regulator which is connected to a 9V battery. I was looking at a voltage regulator that can output 100mA of current... so I thought I would look at all my individual ICs in my circuit and find out how much current they need and make sure that everything will be ok.... but I got confused looking at the datasheets.



So, I have an LM555 timer chip... I pull up it's data sheet:



Power Dissapation = 600mW (at this point, I know I will be using Vcc = 5V, so I'm thinking P/V = I = 120mA... so that 100mA regulator wont do it)



Then I read on Supply Current (Low Stable) @Vcc=5V, Icc=6mA (for high stable Icc=5mA)... Now this is really different than 120mA.... so I'm thinking.... maybe my regulator will work.... I read on



Isource = 100mA, Isink=5mA Hmmmmm.... if I can only send it 100mA tops.... am I too close to this source level?



any help is appreciated, much thanks!


Power dissipation is a total of what the chip consumes by itself running itself,and power dissipated because of current you draw out when high, or put in when the output is low.

The chip will source up to 100 mA max from the output when high, and sink only up to 5 mA when low.

ok, so am I right then to think that the 100mA voltage reg. will not be enough to even power this one 555 timer properly?

Does the sourcing and sinking not have anything to do with this? Do I only need to be looking for power dissipation in order to know what current needs to be made available... .and no need to look any further in the datasheet after that?

Thanks

That's the maxinum power the chip can safely disspiate, not hom much is normally does disspiate.

Yup, the chip itself only needs 5 or 6 mA.

The chip can drive a load of up to 100 mA. The power supply has to furnish the 6 mA chip current plus any load current. How much it actually has to supply is up to you, since you determine the load.

If you're dumping 10 mA into an LED, the supply has to furnish 16 mA max.

What IS the load that you'll be driving>

John

It's true that the dissipation is a measure of how much power the chip *can* dissipate safely without damage. Your circuit probably will not be anywhere near that.

But I suggest you just build your circuit and power it with batteries or a bench supply and simply measure it rather than spending time trying to come up with a theoretical power number. Remember the power used by the circuit can't be determined by summing up the current specs for the individual IC's. You may have quite a bit of power being dissipated in other components like resistors etc, and the power that flows in to an IC may be dependent on how you are loading the outputs too. Most likely you are going to be fine with 100mA unless you are driving a big LED or something.

-Kevin

No.

It depends. For a chip which isn't being used to drive an external load, then its power *consumption* will tell you how much current it's likely to draw.

The 555 can drive a significant load (by IC standards) from its output pin. The cited maximum power dissipation is essentially telling you how large a load you can connect to it, not how much it will draw by itself when driving little or no load.

A 500W PSU is capable of providing 500W; that doesn't mean that it will actually use that much just by being turned on. The situation with the

555's power rating is the same.

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