I understand that if you have resistors in series, you can tap into a smaller voltage inbetween them. For example, if you have two resistors of the same value, the voltage inbetween them will be halved. Though I'm still unsure how that works practically in a circuit.

See the attached diagram:

Assuming the resistors are the same value, the voltage at point A will be 9v, and will be 4.5v at point B. The voltage returned to the power source (point E) should also be 9v.

What are the voltages at the remaining points C and D?

Circuit Diagram

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    \$\begingroup\$ Points C, D and E are the same point. \$\endgroup\$ Nov 2 '16 at 4:05
  • \$\begingroup\$ To say the voltage at a circuit node is any particular value, you need to first define a reference or ground node. Which one is the ground node in your circuit? \$\endgroup\$
    – The Photon
    Nov 2 '16 at 4:05
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    \$\begingroup\$ As @ThePhoton says, you don't measure voltage at a point, but rather across some component(s). The voltage across A and E is obviously 9V. And points C and D are the same as point E as there are no components between them. \$\endgroup\$ Nov 2 '16 at 4:07
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    \$\begingroup\$ It depends on the diode's forward voltage. Is it conducting or not? \$\endgroup\$ Nov 2 '16 at 4:10
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    \$\begingroup\$ @Mark, that would be a convenient place to define as ground, but you could define it to be at any node you like. \$\endgroup\$
    – The Photon
    Nov 2 '16 at 4:11

People's pedantic comments about voltage references and diode forward voltage are entertaining to read but fail to recognize the fundamental problem with your circuit: you're assuming that node "B" will be equal to 4.5V no matter what is connected to it — this assumption is wrong. Voltage dividers only provide a reference voltage that you can "look at" — the moment you start pulling current from a voltage divider, it will start to droop. To understand why, back up and remember that the voltage across a resistor is always proportional to the current going through it. So, if R1 and R2 are the same, and nothing else is attached to the circuit (i.e., the diode is gone):

  • the current flowing through R1 and R2 will be the same (since there's nowhere else for it to go),
  • thus the voltage across R1 and R2 will be the same (via Ohm's law),
  • thus the voltage at node B will be 4.5V — half of the 9V supply.

But! The moment you introduce the LED to the circuit, some of R1's current will flow through R2, while the rest of R1's current will flow through the diode. Thus:

  • the current flowing through R1 and R2 will not be the same
  • thus the voltage across R1 and R2 will not be the same
  • thus the voltage at node B will not be 4.5V.

For practical circuits, voltage dividers are usually built with 10k-500k ohm resistors (to keep power consumption minimal). LEDs pull orders of magnitude more power than these resistors supply at "normal" operating voltages (such as your 9V supply). Thus, your voltage divider isn't going to just "droop a bit" — it will droop so far that I bet you won't measure more than a millivolt or so across the diode.

Using the Reference Voltage in Practical Circuits

Speaking generally, and ignoring the LED for now (which wouldn't typically be powered from a reference voltage), if you want to actually use the reference voltage you generate with a divider, you will need to feed that into some sort of buffer. This is the classic example of a unity-gain op-amp buffer circuit. Feed the voltage divider's output into the positive pin of the op-amp, and feed the output of the op-amp back into the negative terminal to provide feedback. Now, the op-amp's output will track the voltage divider perfectly — and you'll actually be able to source some current from the circuit to do something useful!

Powering LEDs

If your goal is simply to power an LED from the 9V battery, you'd typically use a series resistor (in R1's place) in the 330-1000 ohm range, and you'd have no need for a resistor in R2's spot. If this doesn't make sense to you, you should read up on nodal analysis, Kirchhoff's current and voltage laws, and Ohm's law.

  • \$\begingroup\$ This is fantastic Jay, thank you for putting the effort into such a detailed response. This is great information. My goal here is to understand how voltage dividers work in general; I'm still pretty new to this, and a lot of the articles I've read so far didn't quite address these things. I'll need time to fully understand this, but wanted to thank you nonetheless. Thanks Jay! \$\endgroup\$ Nov 2 '16 at 5:37
  • \$\begingroup\$ Thanks! Glad I could help. Please mark my answer if this sufficiently answers your question. \$\endgroup\$ Nov 2 '16 at 6:09

I think, your question is a good occasion to show how we can handle voltage divider circuits involving non-linear elements. If there are only linear parts (resistors) in the circuit, we can use the commonly known rules (Ohm, Kirchoff) for calculating voltages and currents. However, when non-linear relationships are involved, a graphical solution is best - as demonstrated in the attached drawing.

Description of the method:

  • Starting with the diode characteristics, we draw the known exponential I-V curve in an ID-VB diagram (solid curve).

  • Because the current I2 through R2 is driven by the same voltage VB we can draw the ohmic line I2=VB/R2 in the same diagram (solid line).

  • Because we know that ID+I2 gives the current I1, we construct the sum of both I curves (vertical addition) I1=ID+I2=f(VB) (see the dashed curve).

  • We have another expression for I1 (Ohms law): I1=(9V-VB)/R1. This line (with a negative slope) can also be drawn in the same diagram as a function of VB (line in bold).

  • Because there is only one common point for both I1 curves (both functions must be fulfilled at the same time), we can find the actual value for VB (operational point).

  • Starting from this point, we can derive all other unknown currents.

enter image description here

  • \$\begingroup\$ +1 for the nice drawing (but not only for these ones). :-) \$\endgroup\$
    – Antonio
    Nov 2 '16 at 11:16

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