0
\$\begingroup\$

The last stage of a power amplifier to an 8 ohm speaker is usually a common collector. Now many textbooks mention that it is used as a buffer as it has a low output impedance compared to the speaker (Zout << Z speaker). To me this is what people call voltage impedance matching as the full voltage of the amplifier is applied to the speaker. But I always thought that what is of interest is the maximum power transfer. So what matters is power impedance matching which implies that Zout =Z speaker. So which one is true : voltage or power impedance matching in the last stage of a power AMPLIFIER?

\$\endgroup\$
  • \$\begingroup\$ I think you are confusing impedance bridging with impedance matching. \$\endgroup\$ – Bart Jan 24 at 9:58
0
\$\begingroup\$

Think about it this way.

If you had a 12 volt light bulb that has a resistance of 12 ohms and you applied it to a 12 volt power supply, it would produce a certain amount of brightness based on the power consumed.

Then, you insert a 12 ohms resistor in series with the bulb and looked at the bulb’s brightness.

In which scenario is the bulb the brightest?

In your question you are putting the cart before the horse and assuming maximum power is obtained when the source impedance matches the load impedance. In fact it’s when the load impedance matches the source impedance that you have the maximum power scenario.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

In a power amplifier the designer is typically trying to present a perfect, undistorted voltage signal to the speaker. The output voltage should simply be the input voltage multiplied by the gain of the amplifier. You get this in part by having your source impedance be much lower than the load impedance.

Without this buffer there would be some voltage dropped across the power transistor and some across the speaker. The amount of drop across the transistor would be dependent on the instantaneous current through the transistor. Since this current varies from one instant to another, the voltage across the transistor would also vary from moment to moment, causing distortion in the output voltage.

As a side note: Zout << Zload results in high efficiency as opposed to maximum power transfer. As Zout drops, more power is dissipated in Zload, less in Zout.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ OK. So if I understand correctly, in the last stage of a power amplifier we have impedance bridging( so as to have the maximum voltage applied to the speaker) not impedance matching ( so as to have a maximum transfer of power from the source to the speaker). \$\endgroup\$ – Joel Jan 24 at 15:55
  • \$\begingroup\$ Yes, that's a fair way to say it. \$\endgroup\$ – 65Roadster Jan 26 at 1:35
0
\$\begingroup\$

You're confusing two slightly different things.

When you're using an amplifier with a certain output impedance then you can extract the maximum power from it when the load impedance has the same value.

Example: a signal generator (like this one, they're used in electronics labs), these have "50 ohm" outputs and only when you load them with 50 ohms will you get the maximum power. A much higher or lower load will also just work but the power will always be less.

Audio amplifiers generally have a very low output impedance, like in the order of less than 1 ohm. If you impedance matched that with a very low impedance speaker you might think that you would get maximum power. Theoretically, yes you would BUT I have yet to see an amplifier that behaves that way (theoretically ideal I mean).

Real world amplifiers cannot deliver more than a certain amount of current and a certain amount of voltage. Since power = voltage x current, for maximum power you want voltage and current to be as large as possible.

That's where the recommended impedance (4 ohms, 8 ohms) comes in, that is the load impedance that will make the amplifier operate at it's maximum voltage and current that it can handle. Use a lower impedance and the current will become too large. Use a higher impedance and the current will be lower than the amplifier can deliver (you'd be wasting that current capability).

I'm deliberately not writing: "Use a higher impedance and the voltage will become too high" as that doesn't happen. Audio amplifiers are practically all voltage output amplifiers which means that they amplify the input signal (also a voltage) to an output voltage. The current will depend on the load. Without a load, you still get the voltage. With a load that is too low, too much current will flow and the amplifier will clip (distort) and/or switch off to protect itself and/or can be damaged.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.