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I'm using this module based on the popular bluetooth module CSR8635 https://www.tinyosshop.com/index.php?route=product/product&product_id=875

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I was wondering why, in the right-hand side of the circuit, there are two capacitors before the speaker.

In building this circuit, I have found that leaving out these capacitors result in no sound whatsoever. Tinier capacitors give less sound while higher valued capacitors give more sound. In fact, it seems the audio quality also rises with the size of the capacitor? Why is this?

Why does the audio signal become more "intelligible" after the capacitor? How does the capacitor value affect the frequencies that resonate inside the speaker? Shouldn't the capacitor act as a LPF and filter out any high frequencies?

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The two large capacitors are acting as AC coupling capacitors. A capacitor in series with a signal acts as a high pass filter (if you're curious about that, look into the impedance model of capacitors).

Basically, the smaller the value of the capacitor, the higher the cutoff frequency and so more of the low frequency signal (in this case, the audio bass) is going to be filtered out.

The reason why the capacitors are so large is so that the cutoff frequency is lower, filtering out the DC bias while also keeping more of the low frequency signal. The DC bias would just drop across the speakers and dissipate extra power.

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  • \$\begingroup\$ ah, good catch. yes, you are correct \$\endgroup\$
    – lemonlime
    Commented Nov 12, 2017 at 20:06
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The capacitors are there to block the DC voltage from the Bluetooth module. The BT module is unipolar, that is it does not have positive and negative supplies. If you allow this DC voltage to reach your headphones, it will pull them to one side and is likely to cause damage.

The capacitor forms a high pass filter, not a low pass filter. The headphones act as a resistor to ground, and so high frequencies pass through the capacitor but low frequencies are blocked. The sound changes because of the capacitor's affect on the filter. Low values of C will move the frequency roll off to higher frequencies and so you'll lose the bass tones.

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  • \$\begingroup\$ The DC bias (and low f signals) are blocked. What's to change? \$\endgroup\$
    – awjlogan
    Commented Nov 12, 2017 at 19:32
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    \$\begingroup\$ You are als stating that the headphones provide a path to ground, which they don't. \$\endgroup\$
    – user207421
    Commented Nov 12, 2017 at 20:10
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    \$\begingroup\$ @EJP - actually they probably do provide a path to ground, only it's missing from the diagram. Or else they've been miswired - the two outputs shown connected aren't meant to work against one another. \$\endgroup\$ Commented Nov 12, 2017 at 20:39
  • \$\begingroup\$ @ChrisStratton Slightly awkward wording, changed. \$\endgroup\$
    – awjlogan
    Commented Nov 12, 2017 at 21:07
  • \$\begingroup\$ @EJP Yes, the ground path is missing from the diagram. \$\endgroup\$
    – awjlogan
    Commented Nov 12, 2017 at 21:07
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Just to pass (filter out) only the high frequency components i.e., AC signals and to block the DC component. For a given frequency of the incoming AC, capacitive reactance decreases with increase in capacitance ie., increase in size. Hence you obtain better sound output.

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The lower the capacity of the capacitor, the higher the frequencies allowed to pass while blocking lower frequencies. A large value capacitor would pass most audible frequencies, so you would indeed hear 'better sound quality'. The capacitor in line would be to block the DC bias. If there is enough DC bias on there, it would lock the headphone driver in place and prevent much sound from being produced.

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Do the math. 10uF into a 32 ohm earphone cuts frequencies below 500Hz. 100uF cuts frequencies below 50hz. 220uF cuts frequencies below 23Hz.

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