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Topic: internal resistance of atmega 328 (Read 1 time) previous topic - next topic

chetan0412

I am using voltage regulater IC 7805 to give supply to arduino but it gives higher current supply.I want to know the internal resistance of atmega 328 so that I use resistance in between the digital pin and 7805 to lessen the current

AWOL

The device will draw as much current as it needs.
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.

majenko

The current rating of any power supply, or voltage regulator, is the maximum current it can provide.  A device connected to that power supply will only take as much current from the supply as it requires.  If it tries to take more than the supply can provide you risk breaking the supply, or the voltage "drooping" to a level that causes the device to stop working properly.

A constant-voltage supply never forces current.

You can think of it as a well, with the device being a bucket.  You can't put a bucket down the well that's bigger than the well opening, and you will only draw as much water as will fit in your bucket.  The well doesn't squirt water out at you.

Grumpy_Mike

#3
Jun 05, 2013, 01:00 pm Last Edit: Jun 05, 2013, 01:15 pm by Grumpy_Mike Reason: 1
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I want to know the internal resistance of atmega 328

No you don't want to know this. It is not a constant value it varies during operation of the device.

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so that I use resistance in between the digital pin and 7805 to lessen the current

No all you will succeed in doing is making it not work at all.
You can reduce the current a small bit by reducing the voltage, but if you reduce it too much it will stop working and you can't feed in voltages higher than the supply voltage.

chetan0412

Thanks to all for fast reply......

chetan

fungus

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

pito

#6
Jun 05, 2013, 02:02 pm Last Edit: Jun 05, 2013, 02:06 pm by pito Reason: 1
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Ohms law really works!

Do you mean: 5V/0.0095A (16Mhz/5V, active mode supply current) = 526ohm :) :)

fungus


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Ohms law really works!

Do you mean: 5V/0.0095mA (16Mhz/5V, active mode supply current) = 526ohm :) :)


0.0095mA?
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

pito


fungus


Do you mean: 5V/0.0095A (16Mhz/5V, active mode supply current) = 526ohm smiley smiley


Yes!

Disclaimer: Amps may go down as well as up.
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

MarkT


I am using voltage regulater IC 7805 to give supply to arduino but it gives higher current supply.I want to know the internal resistance of atmega 328 so that I use resistance in between the digital pin and 7805 to lessen the current


Ohms law doesn't apply to active devices, in general.  It applies to conductors, semiconductors (if uniform, no PN junctions), electrolytes. 
Batteries are basically made of plates and electrolyte and the internal resistance there is largely due to the ohmic resistance
of the electrolyte I think (although this is by no means constant with time / temperature or age).
[ I won't respond to messages, use the forum please ]

fungus


Ohms law doesn't apply to active devices, in general. 


They don't obey the laws of physics?

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

pito

Ohm's law and Maxwell's eq etc are not fundamental laws of physics. They are only an approximation to make EE happier. Instead of solving Schroedinger's equation for metals/semiconductors we simply use Ohm's law.. :)

Grumpy_Mike

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Ohm's law and Maxwell's eq etc are not fundamental laws of physics.

I think you will find they are unless you are taking a radical view of the word fundamental if so then neither is Sshrodinger.
Ohms law embodies the definition of the units of volts amps and ohms.

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