I am using a PAM8610 break out board for a single channel application (mono). Therefore only the RINN is being connected to the audio source and also only the ROUT is connected to one 4 Ohm speaker. Looking at the data sheet of the PAM8610, page 15, chapter "differential Input" I am confused on how I must connect the left and right channel. The part which confuses me says: "ac-ground the INP input through a capacitor equal in value to the input capacitor on INN and apply the audio source to the INN input". Especially the second part of the statement is confusing.
Could maybe somebody provide a simple connection diagram showing what this instruction means. Thank you very much.
Could maybe somebody provide a simple connection diagram showing what this instruction means. Thank you very much.
Attachments
If you are using a mono source, the capacitor connectiing the INP to ground must equal the feed capacitor of the source.
1uF is shown for reference as an ideal coupling capacitor.
Connect your source to INN with INP grounded through the capacitor.
1uF is shown for reference as an ideal coupling capacitor.
Connect your source to INN with INP grounded through the capacitor.
Thank you for the quick reply. I still do not understand. The first part is clear but the second part of the instruction makes no sense to me. If I connect my source to INN with INP grounded by the capacitor, I understand my signal will be grounded? I have attached a simple connection diagram on how I understand this should be done.
What is this "audio source"?
If everything is inside the box and permanently connected, it may work out that all the DC levels agree. But in general, to be sure, you must put a capacitor on every input pin, either to a source or to ground.
