# Voltage Dividers

I'm working through Boyson/Kybett's book Complete Electronics Self-Teaching Guide With Projects (Complete Electronics Self-Teaching Guide with Projects: Boysen, Earl, Kybett, Harry: 8601406006782: Books), 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 = (VsR2)/(R1+R2) = (5.6270)/(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.

Your battery is dieing. Is the 5.6V what you measured with no load?

What does the battery measure now, while the resistors are connected to it?

You resistor values are too low for use with a dying 9vdc battery. You should be using much higher resistance values, say two 10K ohm resistors. And when you take your voltage measurements read both the battery voltage while it’s connected to the resistors as well as the Vo voltage reading. You should find that vo = 1/2 the battery voltage, any two equal size resistors will divide the applied voltage in half.

Lefty

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 = (VsR2)/(R1+R2) = (3270)/(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. EDIT: Thanks, retrolefty, I'll switch them to 10k. How do I determine the right resistors to use for a given power source?

EDIT: Thanks, retrolefty, I'll switch them to 10k. How do I determine the right resistors to use for a given power source?

It depends on the current capabilities of the voltage source. The total current flow is simply I = E/R where R is the combined series resistance of the two resistors, so the higher their value the less current they draw from the voltage source. Again any two equal sized resistors will form a voltage divider that outputs 1/2 the voltage source, so just for learning purposes the higher the resistor values the better if it's just for learning purposes.

Lefty

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.

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.

Batteries have a finite current capacity. As you approach it, their voltage drops. 9V batteries already have a ridicously low capacity as it is, so a dead one is even worse.

9V alkaline batteries are designed for small loads, less than 30mA. AAs and larger do much better with larger loads.

AWOL:

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.

The battery has internal resistance. This value is relatively low when the battery is fully charged and rises as the charge drops. Your load resistance is in series with this internal resistance and hence the battery voltage at the output terminals is dependant upon , not only the state of charge but also the load applied. Since there is no current flowing with the load disconnected, no voltage is "lost" across this internal resistance and so the voltage measured is sonmewhat higher than that with a load.

And because in spite of what you read a 9vdc battery will not hold a steady output voltage under a heavy load even if brand new. As was explained about if having internal series resistance the voltage under load will be lower then when not under load. The higher the load current the more the battery voltage will sag even if brand new and fresh.

Lefty

I'm getting all the questions in the text right, so I think my calculations are OK

Just imagine a resistor inside the battery. As you are an expert in Ohm's law:

``````V+  --  Ri --+-- R1 --+-- R2 ----GND     0.28mA
5.7V   330   |  10k   |  10k
|        |
5.6V     2.8V

V+  --  Ri --+-- R1 --+-- R2 ----GND     6.48mA
5.7V   330   |  270   |  270
|        |
3.5V     1.75V
``````

However the Ri size is not a real constant, just an estimate got from your actual numbers.
( As is the assumed "no load voltage" of 5.7V )

It should be clear that a with an internal resistance of about the same magnitude as the load, your 5.x V battery is far away from an ideal voltage source 