How to change amperage of a 9V for a component?

Disclaimer: Still a super noob with electronics so any help you offer please explain it in great detail :slight_smile:

So if i have say a 9v battery and say a component whos documentation says "Current :100mA".

How would i modify the amperage of the 9v so its at 100ma for the component? Quick note i have no idea what the normally amperage of a 9v is or how to find it. I thought it might be something like I = V/R. My understanding of ohms law is still shaky so if im missing something really obvious please point it out.

For the same component btw it has the voltage listed as "High:2.3V≤Vin≤5V Low:-0.3V≤Vin≤1.5V ". Does this mean i would also have to lower the voltage of the 9v somehow along with changing the amperage?

My solution (No idea if its right, looking for confirmation)

I would take my 9v battery and do 9 v/90 omhs = 100ma (V/R= I).

Most devices work on a voltage so you don't limit the current but you limit the voltage to the requirement. Is the device you have is 9V 100mA you're fine. It will just draw it's 100mA. A voltage controller device (like chips and most sensors etc) will only draw the current it needs as long as you don't exceed the voltage. If it needs less voltage you have to limit that.

All appliances in your home do the same :wink: Think about is, the socket in your wall is like a mega powerful source ("battery") and you just plug in devices and they take what the need.

A common example of a current driven device is a LED. Here you have te regulate the current through it and the voltage across is is a result of that. That's why the voltage of the LED is not defined as working voltage but as forward voltage.

h4344:
For the same component btw it has the voltage listed as "High:2.3V≤Vin≤5V Low:-0.3V≤Vin≤1.5V ".

This voltage reading does not give you the working voltage of the device. The device is clearly a digital device with a logic in. This specifies when the chip sees a input as LOW (when its between -0,3V and 1,5V) or HIGH (between 2,3V and 5V). Because the max voltage for a high is 5V the device is probably a 5V device. But this is listed somewhere in the datasheet. So what is this mystery device?

PS, so your solution is nowhere right :wink: But I hope you know now what the solution is :wink:

The current draw depends on Ohm's Law, E =IR, voltage is equal to the current times resistance.

This can be turned around to I = E/R, current is equal to voltage divided by resistance.

The only way to change the current is to change the voltage or resistance.

The 100mA is the maximum current the battery is rated to give. If there is a light load there will be far less current than 100mA. It is saying that you should not draw more than 100mA from the battery. The current is load dependent.

The voltage rating of you component is that the voltage should be between the voltages listed to measure a logical level of HIGH or LOW. If you put a voltage of between 2.3 and 5v on the input, then it will be read as a logical HIGH or a "1".

Using a 9v battery is only any good for a low power d project and for a shortish time. They do not have a high capacity and will go flat quickly.

If you want to use it on an Arduino, you must put it into the Vin connector so it is regulated to the correct voltage. If you put it in the wrong place, you will damage your board.

You must also ensure that any peripheral does not put move that 5v onto any of the pins (input or output).

If in doubt give us a drawing of what you are wanting to do. We will advise if it needs to change.

Weedpharma

@septillion Its an infrared proximity sensor :slight_smile:

@weedpharma So the whole idea sounds like connect the battery to the component and make sure the battery output voltage is the same as the component input voltage. So what about when an LED has 100mA being supplied to it rather than the recommended 20mA? Wont it blow despite current being load dependent? Or does the current reflect the voltage so 100mA means the voltage is probably too high too?

Sorry if my questions are starting to sound strange cant think of how to word it correctly. Been deleting paragraphs for 20 mins XD

weedpharma:
The 100mA is the maximum current the battery is rated to give.

Uhm, no... The 100mA he talks about is from a device (which device that may be...), not the battery!

And weedpharma, where do you live? I rarely see anyone use E for voltage.... U (in Europe) and V (in the VS) are more common. And although ohm's law is good to know I think it's useless here.

h4344:
@septillion Its an infrared proximity sensor :slight_smile:

Any link to it? Datasheet, sellers page etc? You did get your info from somewhere, why not share it with us?

h4344:
So the whole idea sounds like connect the battery to the component and make sure the battery output voltage is the same as the component input voltage.

Yes, but you still did not give us the supply voltage of the device. Based on the high/low rating it's probabl 5V but you can't be sure without just looking up the supply voltage.

h4344:
So what about when an LED has 100mA being supplied to it rather than the recommended 20mA? Wont it blow despite current being load dependent?

Did you even read my part about the LED?

h4344:
Or does the current reflect the voltage so 100mA means the voltage is probably too high too?

It might, but it's current driven, not voltage. So the voltage reflects the current, not the other way around. A change is current might almost not change the voltage. For example, when you drive a LED with 20mA the forward voltage might me like 3,20V but when driven at 100mA is maybe just risen to 3,25V. So don't concern about the voltage of the LED, as long as the current is in spec it's okay.

So voltage driven devices you drive with a voltage and the devices takes as much current it need (you don't regulate the current).
Current driven devices you drive with a current and there appears some voltage across them (you don't regulate the voltage across it).

I rarely see anyone use E for voltage.... U (in Europe) and V (in the VS) are more common.

Which goes to prove that the UK should not be in EU, because it is what is used UNIVERSALLY in the UK.

Its an infrared proximity sensor

We need to know what infrared proximity sensor please provide a link to it and then we can tell you how to wire it up.

Grumpy_Mike:
Which goes to prove that the UK should not be in EU, because it is what is used UNIVERSALLY in the UK.

Possibly the weakest justification for a Brexit, yet :slight_smile:

E = IR is how I learnt electronics 45yrs ago in Western Australia.

Weedpharma

E = Electromotive Force which also equals Voltage.

Ohms Law is pretty straightforward

MattS-UK:
Possibly the weakest justification for a Brexit, yet :slight_smile:

Maybe, but it is symptomatic of a lot that is wrong with the EU and that the UK should not be a part of it. There is a lot of bureaucratic nonsense, like the fact that the EU's rivers policy has stopped rivers being dredged since 2010 and now we are getting all this flooding.

infrared proximity sensor

pwillard:
E = Electromotive Force which also equals Voltage.

Not quite, the Electromotive Force is measured in volts which is not quite the same thing.

This is the device http://www.amazon.com/Geeetech-Infrared-proximity-compatible-Arduino/dp/B00AMC1V2C/ref=sr_1_11?ie=UTF8&qid=1452154579&sr=8-11&keywords=arduino+infrared

It says:-

Voltage :DC 5V

So you can not run it off 9V. If 9V is all you have then you need a voltage regulator to step down the voltage to 5V.
However you should be able to connect it to the Arduino's 5V pin and power it that way. As that pin is not a processor output pin it does not have the 20 - 40mA current limit and 100mA will be fine to draw from that pin.

These days E is often electric field strength, not EMF, as in Maxwell's eqns and things like E = V/d,
which I think means V is much more commonly used for EMF (aka potential difference). That's
my hunch.

Do note that most alkaline 9V batteries are rated at
somewhere between 300ma-hr to 500ma-hr at a discharge
rate of 100ma.
That means your toy will only run for between 3 and 5 hours
per battery.
Dwight