[Solved] Using a battery eliminator with rating 500mA to power analog sensor

Other Hardware General Electronics
1

Hello.

I wish to use Battery eliminator to supply 5V to wheatstone quarter bridge for a strain gauge. The output of bridge (in mV) is to be amplified by an instrumentation amplifier to make it to 0-5V so that it can be fed into Analog inputs of Arduino Uno.

But the battery eliminator says the supply current rating as 500mA. I read on Arduino.cc that maximum current to each analog pin is 40mA. Will it be safe to connect this bridge output to Analog input via Instrumentation amplifier?
I am very new to Electronics, so I fear if this might damage the Uno board.

Thanks.

2

gkks:
I am very new to Electronics

OK, the very first thing you need to learn is a thing called "Ohm's law".

Ohm's law says that if you fix two of the values in a circuit then the other value is forced by the laws of nature.

The resistance of an Arduino analog input is about 100,000,000 Ohms.
The voltage being applied is 5V

Knowing these two values, Ohm's law tells you that only a very tiny current will pass, on the order of microamps.

Your Arduino is safe!

(If this wasn't true then your computer would explode - it's connected to a megawatt power station)

3

Thanks for your reply. By Ohm's law, if the current that will pass through an Analog pin is in Microamperes, why is it that an analog pin has current input limit of 40 milliAmperes (when actual value is in micro)?

4

All you need to know, in answer to your question, is that 500mA is what the supply is capable of delivering, not what the connected device demands

5

gkks:
Thanks for your reply. By Ohm's law, if the current that will pass through an Analog pin is in Microamperes, why is it that an analog pin has current input limit of 40 milliAmperes (when actual value is in micro)?

Analog pins can be used as digital pins, too. In digital mode the pin resistance is a lot lower and you have to add an external resistor if your power supply can deliver 40mA (or more).

6

Its actually more complex that ohm's law suggests, as semiconductor devices are
non-linear, but its always the starting point for calculations!

When used as an input a pin has effectively infinite DC resistance (think << 1nA
at room temperature - leakage currents increase dramatically at high temperatures).

Input pins also have capacitance though, especially the ADC inputs, so currents
do flow with AC signals.

However if you take the pin outside the 0V .. 5V supply range the protection diodes
will start to conduct and they only tolerate a few mA or so - exceeding this can
damage the diodes, or in extreme cases put the chip into latch-up state (very bad).

When a pin is set as an output the output driver transistors have a maximum rating
of 40mA - they can conduct more than this but will overheat and degrade or be
destroyed if so. They have very roughly 30 ohms resistance so if you short an
output pin it can cause about 150mA to flow, way above safe limits.

Anyway the main thing is that if all the voltages are 0V..5V inputs are perfectly
safe.... If your instrumentation amp uses a higher supply range than this
you need to add some protection components (a 10k resistor in series with the
analog input will handle a lot of cases simply).

7

Just because your supply can produce 500 mA doesn't mean that it has to always push that through a circuit. What's important, in this case, is that the voltage doesn't exceed 5 V. Current is what causes the damage when it is excessive, but current is proportional to voltage. So the current that flows is determined by the voltage applied.

Just because a circuit in your home may be connected to a 20 A circuit breaker doesn't mean that 20 A MUST flow. It just means that 20 A is the MAX that can flow.

8

My thanks to everyone who replied, for your time and effort. I will mark this topic solved.