I want to create a simple sinusoidal oscillator (Wien-bridge with soft limiting), connected to a high gain amplifier.

I don't have much experience with audio, but I'm guessing a speaker needs a positive and negative voltage to function.

If I wanted to power this circuit with only a +5V rail, what's the easiest way to get the -5V rail (if needed)?


I don't have much experience with audio, but I'm guessing a speaker needs a positive and negative voltage to function.

A loudspeaker is (typically) a two-terminal device so it simply requires a voltage across the terminals.

What you want to be sure to do is to eliminate any constant voltage across; you want AC voltage only across the speaker terminals.

Thus, for example, you can have a single supply amplifier coupled to the loudspeaker with an appropriate capacitor. For example:

enter image description here

In summary, you do not need bipolar power supplies but you must make sure to remove the DC component from the output via, e.g., a coupling capacitor.

how is a decoupling capacitor able to produce a voltage lower than 0V

In the quiescent (no-signal) state, there is a voltage across the coupling capacitor. Assume for concreteness, that you're using a single +5V power supply and that, when there is no signal, the voltage at the output of the amplifier is +2.5V.

The coupling capacitor charges to this voltage when the amplifier is powered up so that the voltage across the speaker is 0V.

Now, if we assume that the capacitance is large enough such that the sinusoidal signals of interest do not significantly charge or discharge the capacitor, the voltage across the coupling capacitor is effectively constant.

Thus, if the output of the amplifier 'swings' down to, say, +1V, the voltage across the speaker is:

$$v_{sp} = 1V - 2.5V = -1.5V$$

In other words, the quiescent voltage across the coupling capacitor is subtracted from the amplifier output voltage to find the voltage across the speaker.

Under the assumptions above, the capacitor is acting like a 2.5V battery and this is the source of the negative voltage.

  • \$\begingroup\$ Forgive my lack of understanding of basic circuit theory, but how is a decoupling capacitor able to produce a voltage lower than 0V? \$\endgroup\$ – tgun926 Jun 30 '14 at 12:23
  • \$\begingroup\$ @tgun926, see update. \$\endgroup\$ – Alfred Centauri Jun 30 '14 at 12:35
  • \$\begingroup\$ @tgun926 It just is! Think about the moment in time when the capacitor is, say, charged to +5V, and the speaker is at near 0V. Then suppose the amplifier-side voltage quickly drops to 0V. The capacitor does not discharge instantly. For an instant, it continues maintains a 5V potential. This is now subtracted from the amplifier's 0V to create a negative voltage on the opposite terminal of the capacitor: the one connected to the speaker. \$\endgroup\$ – Kaz Jun 30 '14 at 19:11
  • \$\begingroup\$ @tgun926 With capacitors you can do funky things, like build voltage doublers and triplers. Or charge pumps that provide a positive and negative voltage, given a single voltage supply. It's all based on the lagging behavior: build a charge in the capacitor and then flip the voltage, like an electronic sleight-of-hand, or juggle. \$\endgroup\$ – Kaz Jun 30 '14 at 19:13

Yes, speakers work best when no DC component is included in the signal they are being driven with. There are several ways to drive a speaker from a single supply:

  1. For small currents, just do it anyway. You only have a 5 V supply, so putting the speaker between a transistor and the supply will work OK for a simple test or homebrew project. For a typical 8 Ω speaker, you'll have a few 100 mA of quiescient current. That's not great, but for a quick one-off experiment will work.

  2. Put a capacitor in series with the speaker. You will need a low impedance output that can drive both ways. It should sit at mid supply and be able to swing most of the way to the supply and ground rails. The speaker is connected between this point and ground with a capacitor in series.

    The big issue here is that the cap needs to be fairly large. To get 8 Ω at 20 Hz requires 1 mF. That's doable. For something like this you probably don't care about HiFi, so you can relax the low frequency requirement and get away with a smaller capacitor. Note that the cap can be electrolytic since the voltage will always have only one polarity.

  3. Use a transformer. There are audio transformers designed for this purpose. They have enough magnetic headroom in the core to tolerate some bias current and still be able to provide enough swing in both directions. The output of the transformer is inheretly AC, which is then used to drive the speaker.

    A side advantage of this scheme is that you get a impedance conversion for free. You don't have to use a 1:1 transformer, so the impedance the amplifier has to drive (into the transformer primary) doesn't have to be the impedance of the speaker.

  4. Use a bridge output. This is two power amps with the signal inverted between the two. One drives one side of the speaker high while the other drives the other side of the speaker low the same amount at the same time. This method has the additional advantage of having 4x the power capability into the same speaker. This is one reason this method is common in low voltage systems.

    With a single-ended output and 5 V supply, the speaker can at most be driven ±2.5 V, which is 1.8 V RMS. That into 8 Ω is only 390 mW. With a bridge output, the speaker can be driven to ±5 V, which is 3.5 V RMS, which can deliver 1.6 W into the same 8 Ω speaker.

  • \$\begingroup\$ I have created a new question regarding a bridged output - would appreciate your insight. \$\endgroup\$ – tgun926 Jul 1 '14 at 8:17

You are correct that it is usually not a good idea to apply DC to a speaker. However, you can avoid the need for a negative supply by using a transformer to couple the output of your amplifier to the speaker. The transformer will also allow maximum power transfer by maatching the speaker impedance to the amplifier.

  • \$\begingroup\$ Is using a transformer a common solution in real high powered amplifiers? \$\endgroup\$ – tgun926 Jun 30 '14 at 12:24
  • 1
    \$\begingroup\$ @tgun: Generally no, at least not today. High power amplifiers are designed with the right power rails in the first place. Back in the pleistocene before the dawn of transistors, transformer-coupled outputs were common, whether high power or not. However, this was mostly due to tubes being higher voltage and higher impedance devices, and thereby poorly matched to 8 Ohm loads. \$\endgroup\$ – Olin Lathrop Jun 30 '14 at 12:34

In general, the various answers here are quite correct. The zero-voltage issue with a speaker is mechanical - the cone is built to move both forward and back within the magnet, and applying only positive voltage will halve the available motion and push the rest into a high-mechanical-resistance band as the cone approaches either it's forward stretch limit or hits the magnet behind, depending on which way you hook it up.

However, you are clearly not building a hi-fi home theatre nor preparing for the next Spinal Tap concert - 5 volt amps don't go to 11. So I would simply toss in a convenient (and optional) capacitor and overspec the speaker. At 5v almost any speaker in your nearest catalog will either be way over what you need or cheap enough that you won't care.


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