7 kΩ and 5 kΩ: the exact one
- Input
- 12V 7k 5k
- Expected output
- 5.00V
Exact 1.4 ratio, but 7 kΩ is not E24.
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12v to 5v voltage divider
A 12 V signal has to reach 5 V without spending an integrated circuit on it. A two-resistor divider does the job, as long as whatever reads the output draws little current. The tool opens in "Find the resistors" mode with 12 V in, 5 V out and the E24 series, and lists the best pairs grouped by the total of the resistors.
Exact 1.4 ratio, but 7 kΩ is not E24.
+0.65% error, 3.87 mA.
−0.69% error, 1.38 mA.
Same ratio, 0.39 mA.
The load pulls the output down by 7%.
Pair chosen with the load already counted.
Vout = Vin × R2 ÷ (R1 + R2), with R1 between the input and the output and R2 between the output and ground. For 5 V with R1 = 1 kΩ and R2 = 2 kΩ, Vout = 5 × 2000 ÷ 3000 = 3.33 V.
The R1 ÷ R2 ratio should be 1.4. In E24, 1.8 kΩ and 1.3 kΩ (5.03 V). In E96, 5.11 kΩ and 3.65 kΩ give 5.000 V. With a load connected, redo the math including it.
No. As soon as the circuit draws current the output falls, and the voltage difference turns to heat in the resistors. To power it, use a 5 V regulator.
They would give exactly 5 V, but 7 kΩ does not exist in the E24 series (it goes from 6.8 kΩ to 7.5 kΩ). You would have to combine two resistors, such as 6.8 kΩ and 200 Ω in series to make 7 kΩ.
Tolerance adds to the series error. Two 5% resistors can shift the ratio by up to about 10% in the worst case, and the output can end up far from the calculated 5.03 V.
R1 + R2 between 100 Ω and 1 kΩ
| R1 | R2 | Vout | Error | Current | Use |
|---|---|---|---|---|---|
| 180 Ω | 130 Ω | 5.032 V | +0.65% | 38.7 mA | |
| 510 Ω | 360 Ω | 4.966 V | −0.69% | 13.8 mA | |
| 470 Ω | 330 Ω | 4.95 V | −1.00% | 15 mA |
R1 + R2 between 1 kΩ and 10 kΩ
| R1 | R2 | Vout | Error | Current | Use |
|---|---|---|---|---|---|
| 1.8 kΩ | 1.3 kΩ | 5.032 V | +0.65% | 3.87 mA | |
| 5.1 kΩ | 3.6 kΩ | 4.966 V | −0.69% | 1.38 mA | |
| 4.7 kΩ | 3.3 kΩ | 4.95 V | −1.00% | 1.5 mA |
R1 + R2 between 10 kΩ and 100 kΩ
| R1 | R2 | Vout | Error | Current | Use |
|---|---|---|---|---|---|
| 18 kΩ | 13 kΩ | 5.032 V | +0.65% | 387 µA | |
| 51 kΩ | 36 kΩ | 4.966 V | −0.69% | 138 µA | |
| 47 kΩ | 33 kΩ | 4.95 V | −1.00% | 150 µA |
R1 + R2 between 100 kΩ and 1 MΩ
| R1 | R2 | Vout | Error | Current | Use |
|---|---|---|---|---|---|
| 180 kΩ | 130 kΩ | 5.032 V | +0.65% | 38.7 µA | |
| 510 kΩ | 360 kΩ | 4.966 V | −0.69% | 13.8 µA | |
| 470 kΩ | 330 kΩ | 4.95 V | −1.00% | 15 µA |
R1 + R2 between 1 MΩ and 10 MΩ
| R1 | R2 | Vout | Error | Current | Use |
|---|---|---|---|---|---|
| 620 kΩ | 430 kΩ | 4.914 V | −1.71% | 11.4 µA | |
| 680 kΩ | 470 kΩ | 4.904 V | −1.91% | 10.4 µA | |
| 750 kΩ | 560 kΩ | 5.13 V | +2.60% | 9.16 µA |
One group per range of total resistance. A low sum wastes current, a high sum lets the load pull the output down. The error is only series rounding; add the resistor tolerance on top.
The math runs in your browser; no value is sent to a server. A resistive divider is for signals and references, not for powering circuits that draw current.