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I have a TPA6111 stereo audio power amplifier part (http://www.ti.com/lit/ds/symlink/tpa6111a2.pdf) that has this typical application (from the datasheet): enter image description here

My application requires a mono signal amplification and I'd like to know if there's a way to sum the audio input&output signals. Alternatively I could use just one channel I think.

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  • \$\begingroup\$ What load are you driving? i.e., Speaker, headphones, further amplifier, lights? \$\endgroup\$ – Transistor Jan 16 '16 at 10:31
  • \$\begingroup\$ It's headphones I'm driving. \$\endgroup\$ – florind Jan 16 '16 at 19:58
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You can feed both channels from the same mono signal to have both channels amplify the signal and drive the left and right channels of the output speakers or headphones. Alternately, you could just use one channel to have one channel produce output.

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  • \$\begingroup\$ Thanks for the answer @Tahmid. When feeding both channels from the same signal, what would be the values of Ri, Rf and C(c)? \$\endgroup\$ – florind Jan 16 '16 at 18:40
  • \$\begingroup\$ Rf and Ri are used to set the gain. The capacitors are part of the input and output high pass filters used to block DC. Pages 15 and 16 in the datasheet give more details. You'd set these based on the gain you're looking for. Setting the capacitors is easier since the cut-off isn't super strict - you just need to be careful to pick large enough capacitances that the signals aren't attenuated. \$\endgroup\$ – Tahmid Jan 16 '16 at 19:29
  • \$\begingroup\$ Makes sense. So say for a gain of -2 (?) I'll need a single pair of Ri=Rf=20k plus one Ci<=1uF with both IN1- and IN2- tied together, right? \$\endgroup\$ – florind Jan 16 '16 at 19:58
  • \$\begingroup\$ Given that the 2 amps have their non-inverting inputs shorted, you could probably use a single Ri and a single input cap. You should still use 2 Rf's to individually set the gains. However, the best would be to have 2 of each as shown as the schematic in the original post and tie together the two audio inputs. \$\endgroup\$ – Tahmid Jan 16 '16 at 20:06

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