I'd like to build my own Amplifier, and because I know that the current creates the magnetic field, I don't understand, how amplifying the voltage makes it louder. I mean with high voltage doesn't automatically come high current. For example if an Amp can only deliver 100V at 1A (I don't know whether such an amp exists but ok) (the reason for that is the Load Resistance RL, right?) , then the Output Power is 100W.

Would that make the same sound on the same speaker in same Conditions, like if I put 100A with 1 V through it?


Ohh thank you guys for that many answers in that short time :_D

How could a class D amplifier amplify with ~90% Efficeny, if 4,5W is the Maximum amount, you can get out of a 12V PSU? (4,5W/0,9 = 5W -> 5W / 12V = 0,417 A) That would mean, the PSU has 0,417A of output current, and so, because the output current can be calculated, the current doesn't play a role... How could that be? Or is that the reason, why you wrote "Rough Calculations"?

In case, higher Impedance means the Voltage of the Power can be higher: If i want to build a Subwoofer with Amplifier out of a Car-subwoofer with 4Ω , is it a must, to have relatively low Voltage audio signal and relatively High current? Then, depending on Phil G's Explanation, i have to build an Amplifier with high Current Output? How to calculate the Current and Voltage I need to achieve the wanted Power of 200W through that 4Ω Subwoofer?

My plan is, to RC Low-pass filter the Pure Audio Signal, amplify it and put it out on my Subwoofer, any Ideas? Ty for Answers in advance!

  • \$\begingroup\$ Generally yes P=V*I for fixed R=V/I \$\endgroup\$ – Sunnyskyguy EE75 Feb 19 at 19:44
  • \$\begingroup\$ From the equations that Transistor provided, you need about 7Vrms for 200W into a 4 ohm load - and about 20V peak to peak. That's assuming that the impedance of the speaker is 4 ohms, it varies with frequency, especially around the resonance where subs often run. education.lenardaudio.com/en/05_speakers_3.html You can do it with a bridged amplifier pair, where you drive both ends of the coil with antiphase signals, that gets you effectively double the voltage, but for high output powers you need really low impedance speakers, a boosted supply voltage, or matching transformers. \$\endgroup\$ – Phil G Feb 19 at 23:09
  • \$\begingroup\$ You can't put 100A at 1V and 1A at 100V through the same speaker. You can put the first through a 0.01 ohm speaker and you can put the second through a 100 ohm speaker. \$\endgroup\$ – immibis Feb 19 at 23:12

For a given voltage, the impedance of the speaker determines the current.

For a given current, the impedance of the speaker determines the voltage.

In order to get a given current with a given voltage, you need just the right impedance. So you can't drive 100A at 1V into an arbitrary speaker -- you would have to find a \$\frac{1}{100}\$-ohm speaker (and some really thick speaker cable).

So for a single 8-ohm speaker, the power you can drive into it depends on the voltage -- if you want more power, you need more voltage. This is why car speakers are often 4-ohms -- it's an easy way to get twice the power that you could with 8-ohm speakers, from the 12V that's available in the car.

  • \$\begingroup\$ Many car stereos use IC's with bridge type outputs to get something like 15 watts per channel into 4 ohms. THD is a bit high due to cheap topology and a low bias current that is not adjustable. \$\endgroup\$ – Sparky256 Feb 19 at 23:02
  • \$\begingroup\$ @Sparky256 true, but once you've gone to bridge outputs, a fixed supply voltage and a fixed speaker impedance is still going to give you a fixed power capability. If you want more you either need 2-ohm speakers (which I don't think is done, but my idea of fine car audio is to get a less restrictive muffler) or you to generate more than 12V somewhere in your amp so you have more overhead. \$\endgroup\$ – TimWescott Feb 19 at 23:15
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    \$\begingroup\$ Note that the impedance of the speaker is very much not a constant impedance. The nominal value should not be assumed to be the same at all driving frequencies. \$\endgroup\$ – Hearth Feb 20 at 15:28

Would that make the same sound on the same speaker in same conditions, like if I put 100 A with 1 V through it?

In general, yes.

Loudspeaker design is a balance between several factors which include mechanical strength, wire size, reasonable voltage (insulation) and reasonable current. The result is that speakers are readily available in 4 Ω, 8 Ω and 16 Ω versions.

The system designer now has to decide whether the amplifier will be a higher voltage, lower current or lower voltage, higher current design. This may be determined by the available power supply - 12 V auto systems being a good example.

I don't understand, how amplifying the voltage makes it louder.

As you suspect, you need to increase the power. This may be achieved by increasing the voltage, the current or, more usually, both.

I mean with high voltage doesn't automatically come high current.

Correct, though it does if the speaker resistance remains the same.

Rough calculations:

The max power output of the amplifier into a speaker of known resistance, R, can be calculated from the available DC voltage in the amplifier.

$$ P_{max} = \frac {V_{RMS}^2}{R} = ( \frac {V_{DC}}{2\sqrt 2} ) ^2 \frac {1}{R} $$

So for a 12 V supply and a 4 Ω speaker \$ P_{max} = \frac {V_{RMS}}{R} = ( \frac {12}{2\sqrt 2} ) ^2 \frac {1}{4} = 4.5 W\$ is the maximum available unless some other trickery is used.


The power delivered will be the same regardless of the voltage, but what is important is the load that it will be driving. Most loudspeakers (if that's what you are intending to drive) have an impedance of between 2 and 16 ohms, but there are PA amplifiers that do as you say and produce a limited current but much higher voltage output - which allows thinner wiring to be used over long distances, that are intended to drive high impedance speakers, which either have voice coils with many more turns of finer wire, or use step-down transformers at each speaker.


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