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New Member
Hello everybody !

I am a newcomer to the electronics field. Please spare me if I have some questions so simple, but actually I have difficulties to understand the terms and notions in the electronics. I got a trouble in the understanding "the noise". What is it and how does it exist ?

Suppose the output voltage from a pin on the chip is 5 V and the output current is quite small 3 mA, then we connect the pin to a load with low resistance 100 ohm. According to Ohm's law, the current passing through the resistor is

5 / 100 = 50 mA

But the output current of the chip is about 3 mA. Is it possible for the current turning from 3mA to 50mA. Besides, in the DC series circuit, the current is determined by the source voltage and the equivalent resistors, hence we cannot change or turn the current intensity by someway or at anytime other than changing the resistor. It makes me very confused at this point, so thanx in advance for any helps.



Active Member
50mA would be able to flow IF it isn't already limited before the resistor.
If the chip (what chip are we talking 'bout by the way?) limits the output current at 3mA then there can only flow 3mA even if the resistor allows more.

However, if 3mA is the maximum the chip can deliver and there isn't a current limiting inside the chip then the chip could die because you're trying to get more current out of it then it can deliver.


New Member
Thank Exo,

I got the point that if there is no current limitter inside the chip then the chip could die because of the overflow of current ... isn't it ?

Assuming that if there is a current limitter inside the chip and the current is always limited under 3mA even though the resistor allows more. I have a contradiction here, according to Ohm's law, then the voltage across the resistor should be

100*3mA = 0.3 V

but in fact, the voltage should be greater than 4.5 V (suppose that the output voltage is high) -> If it turns the voltage drop down 0.3 V -> The output will be low -> So basically does the chip drive the current or the resistor drive the current ?

n How about the "electrical noise" ? Is it somekind make the signal (suppose voltage) fluctuate in a specific range ?


Active Member
The chip drives the current.
100*3ma = 0.3V, So there will be a 0.3V drop over the resistor, meaning that it will pull-down the output voltage of the chip to 0.3V (making it low, indeed)

This is only the case when a current limiter is in the chip, if not, you may see some *magic smoke* :lol:

noise in the supply is indeed a fluctuation of the voltage. On a 'scope the voltage wouldn't be a nice straight line but filled with little spikes and pits.


New Member
Hei, amazing ... it happened as you said ... now I already got that point. Thank much.

But it arised another trouble -> That is the high level output current ...

From the datasheet, I checked that is maximum -0.4 mA ... but when I make the experiment, the high level output current I measured is 46.4 mA.

Based on the values of the components everything is ok if the high level output current is 46.4 ... far different from the datasheet ?

The resistor is 12.15 ohm and the voltage I got when connect the resistor to the pin is 0.55 V (suitable for the calculation 46.4mA*12.15 = 0.56 V)

Without the resistor, the voltage I measures is 4.23 V.

Is the datasheet wrong or did I measure wrong ?


New Member
I would not say that the chip drives the current, but that the chip regulates the current supplied. Say your pin max output is 3mA (sourcing) and 6mA(sinking). If you are to connect a motor that uses 100mA then the motor will try to pull 100mA no matter what!!!!

If the chip has "overcurrent" protection of some sort then it will act kind of like a breaker and probably it will stop trying to deliver any current. But if it doesn't have a protection (which is likely it won't) then you will let the smoke out. (by the way how do they manage to put that smoke inside the chip :lol: )

BY you putting a 100ohm resistance to ground you are pullind down the pin. But 100 ohms is a small resistance to pull down the voltage because the amount of current it pulls. Instead if you were to put something higher than 1K even a 10K pull down resistor then you would not have that "load" trying to pull so much from your pin.

This pull down resistors will ensure that if you connect another circuit to the PIN of your chip, that other circuit will see a "low" in the case your chip's pin is "floating". If there is any noise (like generated from radio noise, etc) the connected ciruit doesn't mistake the noise for a signal from your chip. When your chip pin goes high then the pull down resistor will take very little current and the conected circuit will read a "high" and will use take some current from your pin. The current to be used has to be known as to not overload the chips PIN.

What chip are you using so we can follow better what you are saying.

By the way when you are using such low resistances, the lower you go the closer you will get to short circuiting. As a reference 1000ft cable 18 AWG (size) has a resistance of 6.64 ohms :wink:



New Member
smoking chips

:D hi there ivancho,

i like the question you made "by the way how do they manage to put that smoke inside the chip?"
:lol: :lol:
my answer is: "all chips are smoke machines" :roll: :wink: :roll: :wink:
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