No. You’re over thinking it.
Why not do a few practical tests of current usage with your DMM?
Maybe you spelled it out and I missed it, but why this preoccupation with using the channel outputs instead of the simple pair of speaker connections? Output is perfectly loud enough to fill a room. Also try two identical speakers used in series.
And on that note, how do I know how much current the amplifier itself will draw from the Arduino? All I know is the max voltage it draws from my Arduino, 5 or so volts max, though I can give it anything I want. I know nothing of its resistance, and no schematic seems to tell me how much it has. How do I know it won't draw so much current so as to fry my arduino?
How do I know it won't draw so much current so as to fry my arduino?
That will actually vary, depending on the setting of the volume control but we can figure out the worst case.
Pot resistance = 10K
PAM8403 input resistance =18K
Input resistor on PCB = 10K
The PCB resistor is in series with the input resistance so we have 28K
28K is in parallel with the 10K pot at maximum setting, so the total resistance is 7.37K ohms
Current = 5V/7.37K = 0.678mA
where did you find this number for the input-resistance?
Edit. I found something in the datasheet
which talks about 18 kOhm as input-resistance. OK.
The datasheet says
Now that is unprecise.
What is a "not too large" input-signal in numbers?
The maximum gain is 24 dB. What does this mean for the maximum input-voltage?
I expect that a voltage of 0.7V will be below "a not too large input signal"
What I do not know is:
can this input voltage be calculated with the gain 24 dB and the supply-voltage?
I'm using a datasheet downloaded from diodes inc.
In the application information there is a section about maximum gain and they refer to the diagram on page two. I'm assuming the R=18K is the input resistor shown in that diagram.
Ups are answers and edit overlapped about the 18 k
The datasheets are very poor.
The PAM products were acquired by Diodes inc when they bought Power Analog Micro and as usual the data sheets suffered a lot of cuts/pastes/deletions when Diodes made them into their own image.
With a resistance that freakishly high, how does anything come out of the amp? Shouldn't such a high resistance severely limit the current from both the MCU and the 5V external power supply?
7.37K is far from being high
Shouldn't such a high resistance severely limit the current from both the MCU and the 5V external power supply?
From the MCU, yes but I would hardly say it's severe, from the power supply, no.
dear @pwmpin,
you seem to think very quick without analysing the
situation.
You will accelerate your learning if you start to
in a certain manner:
If a question arises inside your head. Start thinking about a possible answer yourself.
As you have the question:
start thinking over it all.
"there are so many amplifiers working very good all around the world"
where could they get the power from?
And maybe after some time the word "power-supply" comes to your mind.
Then draw a hand-drawn rough picture of the components and how these components are wired together.
Then ask yourself from where could a high current for the high power come from?
As a last celebrating of beeing a lazy thinker / Laziness of thought
here is a picture that explains it.
I recommend from now on you should start thinking as described above or at least search for explanatory videos and websites
Hey really sorry! I seemed to have missed these replies, just saw them now. What do you mean by channel outputs exactly? I didn't get you. The amp only has outputs for each channel right? Do you mean that I should connect both speakers instead of just one to the amp?
Hey sorry, missed this message too! Just noticed this one now too
The AUDIO-inputs named Left Channel Input / Right Channel Input have a
HIGH input-RESISTANCE. The current that can flow in there is limited down to microamperes throughg the big internal input-resistance.
Right, so I decided to read up on input resistance. Had a bit of trouble understanding it at first (I had trouble when I briefly read up on these several month back in school) but I think I got it now after some more reading. Would it be right to say that the input resistance being high causes a low current draw, thus preventing any potential current overdraw when I connect the MCU to the amp?
Am I understanding this concept right?
Yes you understand it right
Now all you have to worry about is damaging the PAM8403 and your speakers
I decided to do some extra reading on pots before jumping into this. Just acquired a few potentiometers from a friend, though they're the linear kind
From what I have read and understood, a potentiometer allows me to manipulate the resistance at that point in the circuit based on how much I have the potentiometer dial turned. But we are trying to use it as a volume regulator here right?
From what I can think of, to regulate volume, you 'scale down' the wave, or I reduce the voltage supplied from the MCU. But given this device only lets me increase the resistance from 0 to whatever the rated resistance is, won't I just lower the current reaching the amplifier and not the voltage?
Everywhere I have read states that these are used as volume controls, but I can't seem to make sense of it. We control volume by regulating current instead of voltage? How does that work? Or have I gotten something here wrong? Nothing I have read seems to answer that particular query of mine
I have read about these things called potential dividers or voltage dividers. If I replace one of the resistor components of such a divider with my potentiometer, could I then get a setup that allows me to regulate voltage instead of current?
The pot acts as a voltage divider.
For example, refering to the diagram, if the input voltage is 5V and the wiper (sliding contact) is right in the middle then the output voltage will be 1/2 of the input voltage.
Make sense?
This is a good explanation
Xou seem to be interested in understanding it all
So I recommend that you work through this video-channel
