**"Preferred Headphone Resistor Network" Circuit Analysis**

*or How the Resistor Network Works*

Using my HE-500 headphones we'll analyze the "Preferred Resistor Network" using Ohm's Law. Inputting the HE-500's specs of 38 ohms impedance and 89 dB/mW sensitivity into the Headphone Power Calculator spreadsheet shows:

So with my HE-500s connected directly to an amplifier providing 0.389 volts RMS you get 10.24 milliamps (0.01024 amps) of current flowing through the headphones and 95 dBSPL Vrms of loudness. That loudness level would be too loud for most.

Using our Resistor Network Calculator on this webpage we input Headphone Impedance = 38 ohms, and the standard "Preferred Network" resistor values of R2 = 6 ohms, R3 = 2 ohms:

The calculated "Effective Speaker Load" is the total Headphones + "Preferred Network" circuit impedance/resistance of 7.9 ohms and Attenuation is shown as -12.37 dB.

Now lets plug the 0.389 volts from the amplifier and the 7.9 ohms of total circuit resistance into our "preferred network" and analyze it using Ohm's Law.

First we calculate the total circuit current using Ohm's formula: Current = Volts / Resistance = 0.389v / 7.9ohms = 0.0492 amps of current (or 49.2 milliamps). We can also calculate the total circuit Power (watts) using: Watts = Volts^2 / Ohms = 0.389v^2 / 7.9ohms = 0.0192 watts (19.2 milliwatts). **So the amplifier is putting out 0.389 volts, 0.0492 amps and 0.0192 watts.**

Now we can calculate the voltage drop caused by resistor R2's 6 ohms. All 0.0492 amps of current run through R2 so we can use the formula Volts = Current * Resistance = 0.0492amps * 6ohms = 0.2954 volts (295.4 millivolts).

We can also calculate the watts dissipated by R2 using the formula Watts = Amps^2 * Ohms = 0.0492amps^2 * 6ohms = 0.0145watts.

Since our Headphones are in parallel with resistor R3 they will both get 0.2954 volts but the 0.0492 amps will be divided between them.

To determine how the current is split between R3 and the Headphones we must first calculate the resistance of R3 + Headphones in parallel. The formula is 1 / (1 / R3 + 1 / Headphones) = 1 / (1 / 2ohms + 1 /38 ohms) = 1.9 ohms.

Now we can calculate the voltage drop across R3 and the Headphones. The formula is: Volts = Current * Resistance = 0.0492 * 1.9ohms = 0.0936 volts (93.6 millivolts).

Knowing the voltage drop across R3 and the Headphones we can calculate the current split through them. Lets start with R3. The formula is Current = Volts / Resistance = 0.0936v / 2ohms = 0.0468 amps flowing through R3. For the headphones it is 0.0936v / 38ohms = **0.0025 amps (2.5 milliamps) flowing through the headphones at 0.0936 volts (93.6 millivolts)**.

Now we can also calculate the watts dissipated by R3 using the formula Watts = Amps^2 * Ohms = 0.0468amps^2 * 2ohms = .00438 watts.**The watts dissipated by the headphones is: 0.00246amps^2 * 38ohms = 0.00023 watts (0.23 milliwatts)**.

We can use Kirchhoff's Voltage Law to double-check our voltage calculations. The law states that the sum of all voltages around a closed circuit is equal to zero. Sure enough we add the output voltages of R2 + (R3 & Headphones) = 0.2954v + 0.0936v and subtract that from the amplifier output of 0.389 volts and you get 0.

**So our amplifier puts out 0.389 volts, 0.0492 amps and 0.0192 watts**

**And the resistor network drops that to 0.0936 volts (93.6 millivolts), 0.00246 amps (2.46 milliamps) and 0.00023 watts (0.23 milliwatts) through the headphones.**

We can plug our calculated **0.0936 volts at the headphones** into the Headphone Power Calculator spreadsheet:

With 0.0936 volts RMS getting through the resistor network to the headphones you get a nicely matching 2.46 milliamps and 0.23 milliwatts and a loudness of 82.64 dB. If we compare this loudness to the 95.0 dB of the headphones connected directly to the amplifier we get an attenuation of: 82.64dB - 95.0dB = -12.36 dB which matches the attenuation calculated by the Headphone Network Calculator on this webpage.

Edited by robrob - 12/19/13 at 11:49am