A voltage divider is a passive circuit producing an output voltage that’s a fraction of the input voltage. In guitar pedals they’re often used to adjust signal levels or bias active components like transistors and op-amps.

Voltages can be divided via resistors, capacitors, inductors or a combination of those. This post is about resistive dividers formed by series resistors. We’ll keep those other types of voltage dividers for later.

Resistive dividers

A divider network can be formed by connecting two resistors in series. The input voltage Vin is applied across both and the output voltage Vout emerges from the connection between them.

[Schematic of a voltage divider]

[Schematic of a voltage divider]

The input voltage distributes proportionally over resistors connected in series. That means if R1 has twice the resistance of R2 (R1 = 2 × R2) then R1 will also have twice the voltage across it, V1 = 2 × V2. The voltage V2 across R2 is equal to the output voltage, Vout = V2. Both partial voltages add up to the input voltage, Vin = V1 + V2.

[Voltages across a voltage divider]

[Voltages across a voltage divider]

From that you can derive that the ratio of input to output voltage is equal to the ratio of the sum of both resistance values to the resistance of the second resistor. Vin / Vout = (R1 + R2) / R2.

Example 1: R1 = R2

In a voltage divider where both resistors have the same value the input voltage distributes equally. Thus, Vout = 1/2 × Vin = 4.5 V. Vin / Vout = 9 / 4.5 = 2 and (R1 + R2) / 2 = 2. That’s a common configuration for biasing op-amps.

[Example 1: a voltage divider with R1 = R2]

[Example 1: a voltage divider with R1 = R2]

Example 2: Vout = 3V

If the ratio of Vin to Vout is 3 then so is the ratio of (R1 + R2) to R2. If you want to get 3 V from a 9 V source start by picking e.g. R2 = 10 kΩ and then (after rearranging the formula a bit) compute R1 = R2 × (Vin / Vout - 1) = 20. And so R1 = 20 kΩ.

[Example 2: a voltage divider with Vout = 3]

[Example 2: a voltage divider with Vout = 3]

Example 3: Linear Power Booster

R1 = 430 kΩ and R2 = 43 kΩ are common values for biasing the transistor of the LPB1 — the Linear Power Booster. A ratio of ~10 is commonly used, which feeds 0.82 V to the base and keeps the transistor (e.g. a 2N5088) in active mode for the entire range of the input signal.

[Example 3: a voltage divider inside the LPB1]

[Example 3: a voltage divider inside the LPB1]

Picking resistor values

Theoretically, any two resistance values R1 and R2 can be used — as long as they establish the desired ratio of input to output voltage. In reality however there are a few things to consider.

Choosing resistor values too low will cause the divider network to pass a relatively high current. For R1 = R2 = 1 kΩ and Vin = 9 V for example, that would mean 4.5 mA (ignoring loading effects). That’s quite a lot if you’re planning to run off a battery.

Choosing resistor values too high produces more self-noise, which is relevant especially when designing high-gain circuits. That noise can be reduced by adding a capacitor as a power supply filter.

Loading effects

Another issue with higher value resistors is that they lead to a high source resistance. If presented with a low enough load resistance, Vout might drop lower than acceptable. That’s usually not a problem with setting bias voltage for op-amps but it can absolutely be if you insist on picking 10 MΩ resistors and no longer supply the required bias current.

“The trick is to size the resistors so that the amount of current you pull out or push in will not move the reference voltage around more than some allowable amount." (R.G. Keen — “Calculating Bias Divider Networks”).