9V battery question.

[BayBayMan]

I am using a 6V power source for a gas sensor I'm using. Then I'm using the on board 5V to power an LCD I am using. So, I am using two power sources.

My question is if I wanted to use a 9V battery, would that be able to power both components without needing to use a USB cable? I want to make my gas sensor more mobile and portable. I don't know much about power sources, and whether or not something is getting enough power.

Thanks!

[Robin2]

If you mean a PP3 style of 9v battery the answer is NO. they cannot provide enough current.

But a pack of 6 x AA cells should be sufficient.

HOWEVER, if the gas sensor needs precisely 6v, or a max of 6v you will not be able to power it from a 9v battery without something to regulate the voltage.

Best thing is to post a link to the datasheet for your gas sensor.

...R

[BayBayMan]

http://eph.ccs.miami.edu/precise/GasSensorSpecs/Ozone.pdf

I do have a PP3 9V battery, but am looking into those 6xAA cell holder. I kinda figured it was common sense that 9V wouldn't support 6V + 5V = 11V. I just thought there might be some kind of extra black magic in there that would allow 9V to power 11V worth of components.

What would you suggest I use for 11V+ worth of power?

[CrossRoads]

A boost regulator. Take a look at regulators at www.pololu.com

[Robin2]

I kinda figured it was common sense that 9V wouldn't support 6V + 5V = 11V. I just thought there might be some kind of extra black magic in there that would allow 9V to power 11V worth of components.

What would you suggest I use for 11V+ worth of power?

You have no need for 11v worth of power. Adding the voltages as you have done is all wrong. If you had a 2-seater car and a 4-seater car you would not have a 6-seater car.

You have an sensor that requires a precise 6v supply for its heater and if you have an Uno it requires a supply voltage between 7v and 12v. In the circuit diagram in the datasheet they use a 7806 voltage regulator to produce the 6v supply and you could also use one of them with your 9v battery.

However the datasheet says it requires a circuit voltage of 5v and I cannot figure how that is used or whether it could be provided by the Arduino's 5v pin. Is that sensor suitable for use with a microprocessor?

...R

[BayBayMan]

Ah, thank you for that analogy. I was pretty ignorant in how voltage works. It seems I have to look up some tutorials on how exactly that power issue could work out. I am looking into a boost regulator, as CrossRoads suggested.

I believe the gas sensor is suitable for a microprocessor. This is the sensor: https://www.amazon.com/Sensitivity-Detection-Testing-Sensor-Module/dp/B01FJC12XW/ref=sr_1_2?ie=UTF8&qid=1484084779&sr=8-2&keywords=mq131

[MarkT]

I am using a 6V power source for a gas sensor I'm using. Then I'm using the on board 5V to power an LCD I am using. So, I am using two power sources.

My question is if I wanted to use a 9V battery, would that be able to power both components without needing to use a USB cable? I want to make my gas sensor more mobile and portable. I don't know much about power sources, and whether or not something is getting enough power.

Thanks!

The gas sensor alone is rated at 900mW, ie about 150mA, which is beyond the ability of a small 9V battery to
power properly (if at all).

[BayBayMan]

The gas sensor alone is rated at 900mW, ie about 150mA, which is beyond the ability of a small 9V battery to
power properly (if at all).

That's interesting. Most tutorials show the MQ series gas sensors connected straight to the Arduino board, using the on board 5V. What would you suggest I use to power this sensor? I want to be able to make this sensor portable and mobile.

Thanks for the info!

[MarkT]

It contains a 1W heater, it won't run for long battery powered without a decent sized battery.  It will also have
a minimum-time-till-stable after the power is applied for the sensor to heat up too.

[BayBayMan]

Alrighty. It seems I need to research some options for some portable power, and how to connect it to the arduino board. Thanks, Mark!