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1  Using Arduino / General Electronics / Re: Voltage Dividers on: December 31, 2012, 12:55:48 pm
What makes the voltage fall depending on whether I'm using 270s or 10ks?
The lower the resistance, the higher the current.
The higher the current, the faster the battery drains.

I wasn't clear, sorry...

Unloaded battery voltage: 5.8V
Loaded battery voltage measured with 10k resistors: 5.6V
Loaded battery voltage measured with 270 resistors: 3.5V

10k current, measured: 0.28mA, and calculated: 0.28mA
270 current, measured: 6.48mA, and calculated: 6.48mA

I understand that a higher current will run the battery down faster, meaning that it will eventually deliver 0V at any load, and we'll get there sooner with Rt=540 than with Rt=20k. But what I meant by my question was, why does it deliver lower voltage with a higher load right now?

I feel like I'm missing some important terms to express myself, I hope you can decode what I'm trying to say.
2  Using Arduino / General Electronics / Re: Voltage Dividers on: December 31, 2012, 12:36:46 pm
After switching to 10k resistors, the loaded battery voltage stopped sinking, and held steady at 5.28V. The Vo voltage measured in at 2.63V, which I would call pretty darn close enough. Thanks again!

What makes the voltage fall depending on whether I'm using 270s or 10ks? I went and did some reading on batteries just now, and it seems like the voltage should be more or less the same (this is an alkaline) regardless of load until the battery dies, at which point it decreases quickly. Is it because this particular battery is already down its death-curve quite a ways? In other words, if I use a brand-new 9V, will I see approximately equal loaded voltages regardless of what resistors I'm using?

These replies are really helping me reach an understanding of what's going on, and providing great pointers for further research. Thanks, lefty.

Irregardless, this 9V is toast, and it belongs in the battery recycler.
3  Using Arduino / General Electronics / Re: Voltage Dividers on: December 31, 2012, 11:55:22 am
Yes, the 5.6V is measured with no load.

To measure the loaded battery, I leave the circuit built, but measure across the +/- rails of the breadboard? That measurement is about 3V... and sinking constantly. is that the issue??? I feel dumb. Plugging that back in...

Vo = (Vs*R2)/(R1+R2) = (3*270)/(270+270) = 810/540 = 1.5V.

Measured: 1.3V for Vo, then back to the battery for a re-measure of 2.6V.

Thank you, James! I'll go find a healthy 9V for the remainder of my experiments.   smiley-roll-sweat

EDIT: Thanks, retrolefty, I'll switch them to 10k. How do I determine the right resistors to use for a given power source?
4  Using Arduino / General Electronics / Voltage Dividers on: December 31, 2012, 11:28:29 am
I'm working through Boyson/Kybett's book Complete Electronics Self-Teaching Guide With Projects (, and I'm stuck on voltage dividers.  Here's the circuit:

I built this circuit on a breadboard, using R1=270, R2=270, and Vs=5.6V (approximately - it's a dying 9V battery), because those were the resistors I had at hand. I calculated that my expected Vo would be:

Vo = (Vs*R2)/(R1+R2) = (5.6*270)/(270+270) = 1512/540 = 2.8V.

I then measured the voltage at Vo by placing my red multimeter lead where "Vo" is on the diagram and my black lead at ground. I got 1.8V.

What am I doing wrong? Here's a photo of the breadboard, showing a red + where the multimeter red lead connects, and a black - where the black lead goes. It's a little blurry, sorry.

Did I build this circuit wrong? Did I measure the voltage incorrectly? I'm getting all the questions in the text right, so I think my calculations are OK, but I can't get my calculations to match with my measurements, which is very frustrating.
5  Using Arduino / General Electronics / Re: Noob question about current on: December 30, 2012, 10:47:48 pm
So... based on your responses, sounds like I was shorting the circuit, then. smiley-roll Thanks for the info, I will avoid doing this in the future, and I'll definitely check out those book suggestions.

6  Using Arduino / General Electronics / Noob question about current on: December 29, 2012, 12:38:21 pm
Hello all, this is my first post, although I've searched and lurked the forum for a while. My electronics experience is very rusty and rudimentary, so I'm going to ask some very dumb questions. Please feel free to mock.  smiley

I'm trying to understand current by experimenting with my Arduino Uno and a digital multimeter, and there are a couple of things I don't understand:

1) If I switch my meter to A/mA and connect it to the 5V or 3.3V power pins on the Uno, I get constantly fluctuating numbers - so fast that I couldn't even tell you where the decimal point is. Is this because the multimeter is varying the drawn current to remove its own internal resistance from the equation? If I include a resistor in series, it behaves more like I expect: a 270-ohm resistor causes my meter to show 18.5mA on the 5V pin and 12.2mA on the 3.3V pin. From a straight equation standpoint, I=V/R with no resistor in series would yield I=5/0=undefined. But that can't possibly be true in the real world - can it?

(edit: the spec sheet says the 3.3V pin carries a current of 50mA, which doesn't clear up my mysterious current readings at all)

2) The spec sheet on the Uno says that the digital output pins generate 5V at 40mA. But if I write a simple sketch that sets the pin (I'm using #7, but it shouldn't matter) HIGH, my multimeter reads 5V at like 87mA. What gives? Technically, I'm setting it high for 10 seconds, then setting it low for 10 seconds in a loop. The multimeter readings vary between 87mA and approx 0.15mA.

As you may have guessed, I'm still having trouble wrapping my head around current vs voltage, and how it all works together in a circuit.

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