Voltage divider
An unloaded tap on a true series string samples a predictable fraction of the input voltage; the divider relation follows directly from the common current and Ohm's law.
- Identify the topological condition that makes the simple voltage-divider relation valid.
- Derive Vout=Vin·R2/(R1+R2) from the common series current and Ohm's law.
- Use divider ratio, limiting cases and power checks to validate a result.
- Choose resistor ratios for a target unloaded output and understand why absolute resistance scale still matters.
- Recognize immediately how an attached load changes the divider topology and output.
Why does a pair of series resistors produce a predictable fraction of an input voltage at their midpoint?
The voltage-divider relation is not an independent law. In the unloaded circuit, topology forces one current through R1 and R2. Ohm's law then converts that common current into two voltage drops. The output is simply the drop across the lower resistor. Deriving the relation once makes its assumptions visible—and makes it much easier to recognize when a load invalidates the shortcut.
03.1Start with topology: what does “unloaded” mean?
Figure 3.1 — Unloaded divider: one series string with an output node
R1 and R2 form one unbranched path from VIN to ground. VOUT is only an observation node here: no external load draws current from it.
Figure 3.1 contains an ideal DC voltage source, an upper resistor R1 and a lower resistor R2. The node between the resistors is named VOUT. Merely naming or observing a node does not create an electrical branch. In the ideal unloaded divider, no external element draws current from VOUT.
Therefore the VOUT node connects only R1 and R2. Kirchhoff's current law (KCL)—the standard abbreviation for Kirchhoff's Current Law—forces the current that leaves R1 to enter R2:
This is the topological fact behind the simple divider formula. If a third conducting branch is attached to VOUT, current can split and this one-current derivation must be revisited.
03.2Derive the common current before deriving VOUT
Figure 3.2 — Positive reference for the series current I
I is defined from VIN toward VOUT through R1. Because no load branch leaves VOUT, the same current then continues through R2 toward ground.
Let VIN denote the source voltage measured from ground to the input node. Let I be positive from VIN through R1 toward VOUT and then through R2 to ground. With ideal wires and linear resistors, Kirchhoff's voltage law (KVL)—the standard abbreviation for Kirchhoff's Voltage Law—gives
Because the same current flows through both resistors, Ohm's law gives
Substitution yields
The divider current is therefore controlled by the sum R1+R2. We have not used a divider formula yet; this is only the series result from chapter 01.
03.3VOUT is the voltage across the lower resistor
Figure 3.3 — Output-voltage reference VOUT
VOUT is the output-node voltage relative to ground. The voltage arrow runs from the negative reference GND toward the positive reference VOUT, following ADR-0018.
Define VOUT as the potential of the midpoint relative to ground. In Figure 3.3 the voltage arrow points from ground, the negative reference, toward VOUT, the positive reference. This follows AoN ADR-0018. Since R2 is connected from VOUT to ground,
Insert the current derived in the previous section:
Therefore the familiar relation is
The resistor in the numerator is not “always R2” by magic. It is the resistance across which the requested output voltage is defined. If the output were defined across R1 instead, the numerator would be R1.
03.4The divider ratio is dimensionless and bounded
Define the unloaded divider ratio α (Greek letter alpha) as
Because α is a ratio of voltages, it has no unit. For ordinary positive resistors R1>0 and R2>0,
Hence a passive two-resistor divider cannot produce an unloaded output magnitude larger than the applied input when both voltages use the same polarity reference.
Several limiting cases provide immediate error checks:
- R2 → 0: VOUT → 0 because the output node is pulled toward ground.
- R1 → 0: VOUT → VIN because the output node approaches the ideal source node.
- R2 ≫ R1: VOUT approaches VIN.
- R1 ≫ R2: VOUT approaches 0.
If a calculated result violates these limits, inspect the topology, output definition and algebra before trusting the number.
03.5Scaling both resistors preserves the ideal ratio—but not the circuit behaviour
Multiply both resistors by the same positive scale factor λ:
The factor λ cancels from the ideal voltage ratio, so α is unchanged. But the divider current becomes
and the total resistor power becomes
Larger resistor values therefore reduce standing current and dissipation. However, a high-resistance divider is more easily disturbed by finite load resistance, leakage currents and measurement inputs. The ratio alone is not enough to design a real divider.
03.6Design a divider from a target voltage
Suppose VIN is known and the desired unloaded output is VOUT. First define
From α=R2/(R1+R2), solve for the resistor ratio:
This determines a ratio, not unique resistor values. Infinitely many pairs have the same ideal α.
Example: 5 V to approximately 3.3 V. The target ratio is α=3.3/5=0.66, so R1/R2≈0.515. Choosing R2=10 kΩ gives an ideal R1≈5.15 kΩ. A nearby common value R1=5.1 kΩ produces
That numerical closeness does not yet guarantee a good real design. Resistor tolerance and the load connected to the 3.3 V node must still be checked.
03.7Complete worked example: 12 V, 2 kΩ and 4 kΩ
For the verified divider of Figures 3.1–3.3, VIN=12 V, R1=2 kΩ and R2=4 kΩ.
Step 1 — common current.
Step 2 — output voltage across R2.
Step 3 — voltage across R1 and KVL check.
Step 4 — independent divider-form check.
Step 5 — power balance.
The absorbed powers sum to 24 mW, exactly matching the power delivered by the ideal source.
03.8Generalize: a tap on a longer series string
The two-resistor divider is one case of a longer series string. Suppose N resistors R1,…,RN are connected in series from VIN to ground. The common current is
If a tap voltage is measured relative to ground across a lower subset of the string, its voltage equals the common current multiplied by the sum of the resistances below that tap. For a tap after resistor m,
Again the numerator is simply the resistance across which the requested tap voltage is defined. This form is useful for bias ladders, reference ladders and multi-level sensing—provided the taps are not significantly loaded.
03.9The first loading warning: equal resistors do not guarantee half the input
Figure 3.4 — A load changes the topology at VOUT
RL connects from VOUT to ground. Current arriving through R1 can now split between R2 and RL, so R1 and R2 are no longer a simple series pair.
An unloaded 10 kΩ / 10 kΩ divider across 10 V gives 5 V. Now connect a load RL=10 kΩ from VOUT to ground. The VOUT node acquires a third branch, so R1 and R2 no longer carry the same current. The simple two-resistor derivation is no longer the correct topology.
R2 and RL are parallel, so their equivalent is
The actual output is therefore
The load did not “slightly perturb a formula”; it changed the electrical network. Chapter 05 will treat loading and equivalent resistance systematically.
03.10Practice for mastery
- An 18 V source feeds R1=3 kΩ and R2=6 kΩ. Find the unloaded divider current and VOUT across R2.
- Design an unloaded 10 V → 2.5 V divider if R2 is fixed at 10 kΩ. Find R1.
- Compare 1 kΩ/1 kΩ and 100 kΩ/100 kΩ dividers across 10 V. Find VOUT, source current and total resistor power for each.
- A 10 V divider has R1=R2=10 kΩ. Attach RL=10 kΩ from VOUT to ground. Find the loaded output.
- A 12 V source drives a 1 kΩ, 2 kΩ, 3 kΩ series string. Find the tap voltage at the node between the 2 kΩ and 3 kΩ resistors, measured relative to ground below the 3 kΩ resistor.
- A student writes VOUT=VIN·R2/(R1+R2) for a circuit in which an unknown sensor is connected from VOUT to ground. What must be checked before accepting that equation?
Solutions
1. I=18 V/9 kΩ=2 mA; VOUT=(2 mA)(6 kΩ)=12 V.
2. α=2.5/10=0.25. R1/R2=(1−0.25)/0.25=3, so R1=30 kΩ.
3. Both ideal outputs are 5 V. For 1 kΩ/1 kΩ: I=5 mA and Ptotal=50 mW. For 100 kΩ/100 kΩ: I=50 µA and Ptotal=0.5 mW.
4. R2∥RL=5 kΩ, so VOUT=10·5/(10+5)=3.33 V.
5. Total resistance is 6 kΩ; the tap is across the lower 3 kΩ, so Vtap=12·3/6=6 V.
6. Check the sensor's input/load resistance and the resulting topology. If it draws non-negligible current, R2 must be combined with that load rather than using the unloaded formula unchanged.
The next chapter turns the dual pattern around: in a parallel network the voltage is common, and branch current becomes the quantity that divides.