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I am about to use this TPA0211 chip to amplify my audio signal. Supply voltage will 5V and load resistor is 4Ω. According to figure 6 page 6, Output power will be around 2W. However, I need 1.5W as output power.

I see two solutions:

  1. Low supply voltage thanks to diode
  2. Increase speaker's internal resistor

Here is my questions: How is it possible to increase Rl (still in figure 6 page 6)? Will a simple serial resistor work (Radd in the next figure)?

enter image description here

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You need to understand that the power that the amp outputs does not depend on the power rating of the amp (I know, this seems a paradox). Just like a resistor rated for 10W will not always dissipate 10W, the amp power rating is just the maximum you can achieve with it.

The amount of power the amp actually outputs depends on three things:

  • The voltage level of the input (this is a parameter only you can know).
  • The gain of the amplifier (in your specific case, it depends on RI and the SE/BTL configuration - see datasheet).
  • The impedance of the speaker.

Let's do a bit of math: P = U*I = U²/RL (RL being the speaker impedance). If we say U = G*VIN (G being the gain) and G = 125kΩ/RI (assuming BTL mode - see datasheet), we have:

\$P = \dfrac{(V_{IN}\cdot\frac{125\text{k}\Omega}{R_I})^2}{R_L}\$

Therefore, in order to not exceed the given Pmax power, you must choose:

\$R_I > \dfrac{V_{IN}}{\sqrt{P_{max}\cdot{R_L}}}125\text{k}\Omega\$

So, if you know what maximum voltage you have at the input, you know what is the RI you need.

For example, if you have a line input level of 1V, you need to choose RI = 51kΩ to get 1.5W power output into 4Ω (these are all instantaneous max values).

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  • \$\begingroup\$ Thanks for this explanation, by the way, what is the difference between SE and BTL mode, I asked wikipedia, but it's still not very clear in my mind. In the datasheet, the SE/BTL mode just change gain calculation... \$\endgroup\$ – M.Ferru Jul 28 '16 at 10:05
  • \$\begingroup\$ The datasheet explains it in detail starting from page 13. Basically, BTL (Bridge-Tied Load) is appropriate for a speaker, because it allows more output power by doubling the output voltage, driving both poles at 180° phase shift. But this means output is not referenced to ground. SE (Single-Ended) is more appropriate for a jack output, where you need a reference to ground (but the power is less). Actually, I should have assumed using BTL in my example, it would have been more appropriate for your case. I'm editing my answer. \$\endgroup\$ – dim Jul 28 '16 at 10:21
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Output power will be around 2W. However, I need 1.5W as output power.

I presume from your question that you have a 1.5 W loudspeaker and you are worried that a 2 W amplifier may damage it.

You don't need to do anything except turn down the volume on the audio input. You will hear distortion (a harsh sound) from your loudspeaker before it gets damaged. When you hear distortion you know that you need to turn the signal volume down.

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  • \$\begingroup\$ Thanks for your fast answer. Then I will keep my current setting. Just by curiosity, will adding a resistor going to work? \$\endgroup\$ – M.Ferru Jul 28 '16 at 8:45
  • \$\begingroup\$ Yes, it would work. On a hi-fi system we try to drive the loudspeaker directly with very little resistance including cables so that we overcome the inductance of the speaker. I guess this is a much simpler, low-quality audio so that would not be a concern. \$\endgroup\$ – Transistor Jul 28 '16 at 8:48
  • \$\begingroup\$ You can add a resistor on the input, so that max volume is not actually max. \$\endgroup\$ – Vladimir Cravero Jul 28 '16 at 9:04
  • \$\begingroup\$ Yes, I'll put a resistor on the input to adapt the gain. The overall volume level will be monitored by the embedded Linux. Thanks again for your fast accurate answer :) \$\endgroup\$ – M.Ferru Jul 28 '16 at 9:16

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