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it's the first time I encounter this in a schematic, and I'm wondering at how to reproduce this in Eagle.

PAM8900 Example usage

So first number 1. As per description, the volume pin is "Internal Gain Setting Input Connect to VDD which Set Max. Gain = +20dB ", this is the first part I didn't get along with that schematic symbol. Do I have to connect a button there? If so, how to connect a pair of buttons to increase/decrease volume? My educated guess is that it means it's a potentiometer? (Maybe I could use a digital pot here then if that is really the case)

Now, number 2 seems to me even more strange, there isn't a node connecting the lines, and again that arrow. Is there a button here again to switch modes from BTL to SE? (If that is even possible to be switched like that...) and as the arrow goes through, is it still a pot? What is that? Also it's description is "Output Mode Control Input High for SE Mode and Low for BTL Mode ", so it pretty much is a pot right?

Thanks everyone! Tried to be as clear as possible here, let me know if I forgot something.

PAM8900's datasheet here:

https://www.diodes.com/assets/Datasheets/PAM8009.pdf

Edit:

Looks like BTL are speakers and SE are headphones, so how exactly is that part of the schematic going to be hooked up to choose between the two of them? At first I thought that simply by hooking up the headphones the IC would switch to SE mode...

Edit 2:

I think I may be overthinking this, would a simple jack like this and the PAM 8407 I was using do the trick of disconecting audio from speakers?

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    \$\begingroup\$ The SE/~BTL pin should be connected directly to either Vcc or Ground - don't know why that resistor would be shown there. #1 - that should be a 50K pot to conrol the volume. \$\endgroup\$ – Peter Bennett Aug 15 '19 at 19:30
  • \$\begingroup\$ Probably there is a connection between inserting the headphone and connecting then it to either VCC or gnd? So the volume I will really go with the digital pot option, I was using a PAM 8407 which already has volume functionality with switches, but being able to use a headphone is equally as important to me. ( I had already sent my pcb to production when I remembered I forgot to add a headphone jack =( ) \$\endgroup\$ – Luis Aug 15 '19 at 19:53
  • \$\begingroup\$ If you go with a digital pot be very careful selecting one. Most of them only handle current in the nA and uA range. Just make sure you select a pot that can also handle the current. \$\endgroup\$ – Ron Beyer Aug 15 '19 at 20:06
  • \$\begingroup\$ Thanks Ron, will definitely check that out if I got through with this chip. \$\endgroup\$ – Luis Aug 16 '19 at 0:32
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The pot between Vdd and GND to pin 7 is to provide a DC voltage to control the volume, with the gain responding in 64 steps, as detailed in the table in the datasheet. I don't see an input resistance or bias current spec in the datasheet, but the use of a 50K pot implies that it's pretty high impedance, so a 10K to 50K (linear) digital pot could be used. You might also be able to use PWM from a microcontroller provided you filter it to low enough ripple and maybe buffer it. Obviously, a voltage-output DAC would work too.

The 100K is a pullup for the input. The series resistor is probably for isolation to control noise or transients. It's shown in the datasheet for the PAM8019:

enter image description here

Again, it's not clear how much leakage can be present, nor did I see it explicitly stated that there is no pullup on the chip, but probably 10K - 50K is a safe range of values. You can always measure a sample to confirm no pullup.

I'm not convinced this datasheet (or at least the info on it) originally was written in the English language, so there may be other things missing or confusing.

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  • \$\begingroup\$ Yeah, on this datasheet for the 8019, you can see the dot, on the 8009 one there is none, at least for me it could cause a bit of confusion wether they forgot it, meant another thing... I can now see more clearly this schematic thanks to all the responses and overall I have learnt some good new things today, but as I stated in an edit, wouldn't it be much easier instead of reengineering (Wow hard word to write) what I have done, to simply use a headphone jack to cut/let pass audio to the speakers? \$\endgroup\$ – Luis Aug 15 '19 at 21:38
  • \$\begingroup\$ The speakers use a bridged amplifier configuration (4 wires) to get that power from relatively low supply voltage. The headphone amplifier, and, more critically, the headphone itself, uses 3 wires including the ground that appears to be missing from the datasheet schematic so it's not trivial. That's why they did it in the chip. If you have a socket with a switch contact it may be possible to control the SE/nBTL line with that. \$\endgroup\$ – Spehro Pefhany Aug 15 '19 at 21:56
  • \$\begingroup\$ Oh I see... thought it would be easier and simpler like: Amp chip > hp Jack > amplifier. If the hp is in the Jack, it would close the way to the amps, if it's not, signal would flow through it to the amps. In my mind the hp would use the same wires as the amps, since both amps share a "ground/negative" line (Here it's the kind of place where I cannot tell if the negative and ground lines could be the same thing, I know they come from a different pin on the chip) and the remaining 2 lines would be the same to be used on the hp. Hp = headphone for short. \$\endgroup\$ – Luis Aug 16 '19 at 0:31

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