Speaker Impedance Calculator (Series and Parallel)
Speaker Impedance Calculator (Series and Parallel)
Total impedance of up to six drivers in series, in parallel or in series-parallel, the power each one takes from your amplifier and the total the amplifier is being asked for — with a warning when the load you have built goes below the minimum your amplifier is rated for.
Speaker network impedance
two 8 Ω drivers in parallel on an amplifier rated 100 W into 8 Ω, minimum load 4 Ω
Impedance, and what the amplifier then has to do
- Z
- the load the amplifier sees, in ohms
- V
- the amplifier’s output voltage at its rated power into its rated load — the voltage is what it tries to hold
- Pₖ
- in series, I²Zₖ (same current); in parallel, V²/Zₖ (same voltage)
- dB
- 10·log₁₀ of the power ratio; +3 dB is twice the power, which is only about a 3 dB rise in level
Worked example
two 8 Ω drivers in parallel on an amplifier rated 100 W into 8 Ω, minimum load 4 Ω
1/Z = 1/8 + 1/8 = 0.25, so Z = 4 Ω
The amplifier's output voltage at rated power is √(P × Z) = √(100 × 8) = 28.284 V
Into 4 Ω that voltage means 28.284² ÷ 4 = 200 W total and 7.071 A — twice the current of the rated load
Each driver sees the full voltage, so each takes 100 W; the level rises by 10·log₁₀(200/100) = 3.01 dB if the amplifier can actually deliver it
4 Ω is exactly the amplifier's stated minimum: legal, with no margin, and a real driver dips below its nominal impedance at some frequencies
What each wiring does to the load
| Drivers | Series | Parallel | Series-parallel |
|---|---|---|---|
| Two 8 Ω | 16 Ω | 4 Ω | — |
| Three 8 Ω | 24 Ω | 2.67 Ω | — |
| Four 8 Ω | 32 Ω | 2 Ω | 8 Ω |
| Four 4 Ω | 16 Ω | 1 Ω | 4 Ω |
| Six 8 Ω | 48 Ω | 1.33 Ω | 5.33 Ω |
Wiring drivers to an amplifier
Impedances in series add; in parallel their reciprocals add. Two 8 Ω drivers in series are 16 Ω, in parallel 4 Ω, and four of them wired as two series pairs in parallel come back to 8 Ω. That last arrangement is why series-parallel exists: it lets you hang four or six drivers on one amplifier without the load collapsing.
Why the impedance matters more than it looks. An amplifier is a voltage source: below clipping it holds its output voltage whatever the load, so halving the impedance doubles the current it has to supply and doubles the power it has to deliver. That is the whole danger. Two 8 Ω cabinets in parallel on an amplifier rated 100 W into 8 Ω ask it for 200 W and twice the output-stage current, and the output devices dissipate correspondingly more. A well-built amplifier will current-limit or go into thermal protection; a cheap one will fail. This page warns you when the load you have built goes below the minimum you enter, and that minimum should come from the amplifier’s specification, not from optimism.
The level you gain is small. Doubling the power is 10·log₁₀(2) = 3.01 dB, which is a just-noticeable change in loudness — the decibel calculator converts any power ratio the same way. Adding a second driver usually gains more than that, because you have added radiating area as well as drawing more power, but the arithmetic on this page only covers the electrical half.
Nominal is not actual. An “8 Ω” driver is not 8 Ω. IEC 60268-5 defines rated impedance as a resistance you substitute for the loudspeaker when stating the source power, and requires only that the lowest modulus of the real impedance across the rated frequency range is not less than 80% of it — so an 8 Ω driver may legally dip to 6.4 Ω, and many dip further. The real curve rises to a tall peak at the driver’s free-air resonance, where the moving mass and the suspension compliance form a resonant circuit (the LC resonant frequency calculator is the same mathematics), falls to a minimum above it, and then climbs again as the voice coil’s inductance takes over. A crossover network moves all of it again. So treat the number this page gives as the nominal load, and leave margin: a nominal 4 Ω load on an amplifier whose minimum is 4 Ω has none.
Practical points. Wire drivers in phase — one reversed driver in a cabinet cancels much of the bass. Series wiring shares the current so the drivers with the higher impedance take the most power, while parallel wiring shares the voltage so the LOWEST impedance takes the most; this page shows each driver’s share so mismatched drivers do not surprise you. Keep speaker cable short and thick, because its resistance is in series with the load and matters far more at 4 Ω than at 16 Ω — the voltage drop calculator will size it, and the RMS and peak converter relates the voltages here to the peaks the amplifier has to swing.
Frequently asked questions
What happens if I wire two 8 ohm speakers in parallel?
You get a 4 Ω load. The amplifier holds its voltage, so it has to supply twice the current and deliver twice the power — 200 W from an amplifier rated 100 W into 8 Ω. Only do it if the amplifier is rated for 4 Ω.
Series or parallel for speakers?
Series raises the impedance and is safe for the amplifier but delivers less power; parallel lowers it and asks the amplifier for more. Series-parallel — pairs in series, pairs paralleled — keeps four drivers at the same impedance as one.
What is my amplifier’s minimum impedance?
It is on the specification sheet, commonly 4 Ω for a domestic amplifier and 2 Ω for a good professional one. Below it the output devices run out of current capability and the amplifier protects itself, distorts, or fails.
Is a 8 ohm speaker really 8 ohms?
No. That is the nominal or rated impedance. IEC 60268-5 only requires the lowest modulus in the rated frequency range to be at least 80% of it, and the real curve peaks at resonance and rises again at high frequency with the voice coil’s inductance.
How much louder is doubling the power?
About 3 dB electrically — 10·log₁₀(2) = 3.01 dB — which is a small but audible step. Adding a second driver usually gains more than that because it adds radiating area as well.
Related calculators
References
- IEC 60268-5:2003. Sound system equipment — Part 5: Loudspeakers, clause 16.1: rated impedance is the pure resistance substituted for the loudspeaker when defining the available electric power of the source, and “the lowest value of the modulus of the impedance in the rated frequency range shall be not less than 80% of the rated impedance”.
- Self D. Audio Power Amplifier Design, 6th ed. Focal Press, 2013. Output-stage current and dissipation against load impedance, and why an amplifier’s rated minimum load is a thermal and safe-operating-area limit rather than a suggestion.
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Chapter 1: series and parallel combinations, and power in a resistive network.