Two resistors in series across a supply give you any smaller voltage you want — that is the voltage divider, the simplest circuit in electronics. It sets reference voltages, scales sensor outputs, and biases transistor bases. But it has a trap: connect a load to the output and the voltage falls. This post gives the formula, works a complete example, shows exactly how much a load pulls the output down, and designs a divider for a real 5 V load.

Figure 1: The divider (a), and how a load pulls the output down (b). With no load the output is 8 V; a 2 kΩ load drags it to 6 V — a 25% error. (Schematic, not to scale.)

The formula

Two resistors R₁ (top) and R₂ (bottom) in series across a supply V_s. The same current flows through both, so the voltage splits in proportion to the resistances:

V_out = V_s × R₂ / (R₁ + R₂)

The formula only holds with nothing connected to the output. The moment a load draws current, R₂ is in parallel with the load, its effective value falls, and so does V_out.

Worked example — unloaded divider

A 12 V supply, R₁ = 1 kΩ, R₂ = 2 kΩ, no load.

Step 1 — total current: I = V_s / (R₁ + R₂) = 12 / 3000 = 4 mA.

Step 2 — output voltage: V_out = 12 × 2000 / (1000 + 2000) = 12 × 2/3 = 8 V. Check: the drop across R₁ is 4 mA × 1000 Ω = 4 V, and 4 V + 8 V = 12 V ✓.

The loading effect — the trap

Now connect a load resistor R_L across the output. R_L sits in parallel with R₂, so use R₂ ∥ R_L in the formula.

With R_L = 2 kΩ: R₂ ∥ R_L = 2000 × 2000 / 4000 = 1 kΩ. V_out = 12 × 1000 / (1000 + 1000) = 6 V — down from 8 V, a 25% error. The total current rises to 12 / 2000 = 6 mA, split 3 mA through R₂ and 3 mA through the load.

With R_L = 500 Ω: R₂ ∥ R_L = 2000 × 500 / 2500 = 400 Ω. V_out = 12 × 400 / 1400 = 3.43 V — the divider has nearly collapsed. The rule is simple: a divider only holds its voltage when the load resistance is much larger than R₂ — at least 10 times larger for the error to stay small.

Design example — 5 V for a 10 mA load

Design a divider from 12 V to power a load that needs 5 V at 10 mA (so R_L = 5 / 0.010 = 500 Ω).

Step 1 — choose the bleeder current: the current through R₂ (the "bleeder") should be about the same as the load current or more, so the load does not dominate. Take I_bleed = 10 mA, equal to the load current.

Step 2 — R₂: R₂ = V_out / I_bleed = 5 / 0.010 = 500 Ω.

Step 3 — R₁: R₁ carries bleeder + load current = 20 mA, and drops 12 − 5 = 7 V: R₁ = 7 / 0.020 = 350 Ω → nearest standard value 360 Ω.

Step 4 — check the real output: R₂ ∥ R_L = 500 ∥ 500 = 250 Ω. V_out = 12 × 250 / (360 + 250) = 4.92 V — only 1.6% below the 5 V target ✓. The load current is 4.92 / 500 = 9.84 mA, close to the 10 mA wanted.

Step 5 — power ratings: the current through R₁ is (12 − 4.92) / 360 = 19.7 mA, so P_R1 = 7.08 × 0.0197 = 0.139 W → use a ¼ W resistor. P_R2 = 4.92² / 500 = 0.048 W. The supply delivers 12 × 0.0197 = 0.236 W while the load only uses 0.048 W — about 20% efficient. A divider is a fine reference, but a wasteful power supply: for real power, use a regulator instead.

Key points to remember

• V_out = V_s × R₂ / (R₁ + R₂) — but only with no load. Worked example: 12 V, 1 kΩ and 2 kΩ → 8 V at 4 mA.
• A load sits in parallel with R₂ and pulls the output down: a 2 kΩ load on this divider drops it to 6 V (−25%); a 500 Ω load drops it to 3.43 V.
• Design rule: make the bleeder current at least as large as the load current, and make the load at least 10× R₂ if you want the voltage to stay put.
• Design example: 12 V to 5 V at 10 mA → R₁ = 360 Ω, R₂ = 500 Ω, actual output 4.92 V, R₁ needs a ¼ W rating (0.139 W).
• Dividers waste power (here ~20% efficient) — use them for signals and references, not for powering loads.