Help decyphering a spec sheet.

[Buk]

I'm trying to add an off switch to my amp. The board has a jumper described as "Power switch":

And another described as "Mute".

But the datasheet for the PAM 8610 chip has 3 pins that could conceivably be used to implement either of these two functions:

8     FADE    Input for controlling volume ramp rate when cycling SD or during power-up.
                      A logic low on this pin places the amplifier in fade mode. A logic high on
                      this pin allows a quick transition to the desired volume setting.
           
25     MUTE   A logic high on this pin disables the outputs and a logic low enables the outputs

29    SD        Shutdown signal for IC (low= shutdown, high =operational).
                      TTL logic levels with compliance to VCC.

Pin 25 is almost certainly used to provide the "Mute switch" jumper on the board, but which of the other two is used for the power switch is less clear?

The datasheet also carries this information:

High level input voltage: SD... 2.0V to Vdd. MUTE, FADE ... 2.0V to 5V.
Low-level input voltage: SD ... 0 to 0.3V. MUTE, FADE ... 0 to 0.3V

Apart from the fact that the jumpers are only single position, which I think means they are left floating rather pulled down when left open; there is the difference in the max input voltage.

I have a DPDT push switch I intend to use (one pole of) as the on off switch and was considering tying the pin to ground when off, but the chip has several "grounds"; and there is the problem of working out which of the two jumper pins is connected to the SD or FADE pin on the chip. The 6x6mm chip is completely obscured by a 10x10mm heat sink; and the tracks and vias on the underside don't match up with the physical pin layout shown in the datasheet:

Perhaps I should just use a single throw and forget grounding?

Thoughts?

Cheers, Buk.

 

[Pommie]

My guess would be that the power switch connector goes to pin 29. The other pin will be connected to GND. Are you able to check continuity between the pins and Ground?

Mike.

 

[Visitor]

Looks pretty clear to me. There's a jumper installed in the power switch connector. Presumably, that jumper makes the board work if no external power switch is used. If you want to add a remote power switch, remove the jumper and connect the power switch across those two pins.

It might be connected to a logic pin on the chip. More likely, it's simply an interruption in the power supply line to the chip. At any rate, a switch connecting those two pins will turn the amp on. Opening the switch will turn it off.

 

[KeepItSimpleStupid]

If you want to add a remote power switch, remove the jumper and connect the power switch across those two pins.

let's make it a bit clearer. The jumper is what's known as a shunt. When you remove it, you have a connector. Add the mating connector, attach it to a remote power switch -- instant power switch.

 

[Buk]

More likely, it's simply an interruption in the power supply line to the chip.

I considered that; but neither pin appears to be directly connected to the positive terminal of the DC input plug.

(If I move to a high enough setting, both pins of the jumper appear to have some connection to both pins of the DC input plug, in the order of 100s of kOhms, but that presumably is just leakage.

At any rate, a switch connecting those two pins will turn the amp on. Opening the switch will turn it off.

I get that; but isn't it generally considered bad form to leave an IC gate floating?

As I had a suitable double throw switch to hand I thought it made sense to ground pin when in the off position. maybe I'm just being over caucious.

 

[Visitor]

> Connect power. See if the amp works.

> Remove the jumper. Verify the amp doesn't work.

> Replace jumper. Verify amp works.

> Disconnect power. Substitute switch for jumper.

>> Switch on, amp on.

>> Switxh off, amp off.

Despite anything you read here (and the xenophobic biases shown by a few members), the modules coming out of China are generally solid designs. If a module has a 2 pin connector labeled "power switch", directly connecting a switch across it will probably be the right answer. That's far more likely than needing to solder wires to get a ground connection, etc.

 

[Buk]

My guess would be that the power switch connector goes to pin 29. The other pin will be connected to GND. Are you able to check continuity between the pins and Ground?

Neither jumper pin appears to be connected directly to any of the obvious ground points on the top or bottom of the board, or to the negative pin of the DC input plug. (Move to a high enough resistance setting and everything appears to be connected to everything at some level.)

The first thing is that the pics above were taken from the sellers web pages on Amazon, and appear to be from a different revision of the board to the one I have. Hence I've added a pic of my board here:

I've boxed the "power switch" jumper pins in Blue front & back.

The right-hand pin (from above) I've traced in yellow (on the back) connects through a via to the R15 boxed yellow. I can't make sense of where the other end of R15 connects.

The left-hand pin (from above) I've traced in lurid green to a via that comes through the legend "C32" (Boxed in green). That trace disappears under the edge of the heat sink (which is glued to the chip.)

Looking at the chip pinout, if the chip is oriented with pin 1 adjacent to the "C1C16" legend (which seems logical) then the "C32" via is roughly adjacent to the SD shutdown pin 29. If that is the case, then as pin 29 has to be taken high to enable the chip, the other jumper pin would need to connect to Vdd, which tallies with my not being able to trace it to ground.

If my logic makes sense to you, would you bother (risk?) connecting pin 29 to ground when in the off position; or just leave well alone and leave it open?

Thanks.

 

[Visitor]

If the switch input is connected to a logic pin, it will have a pull-up or pull-down resistor. If the jumper isn't in place, the resistor pulls the pin to one state. When the jumper (or switch) is in place, a connection is made to pull the port pin in the other direction.

 

[Nigel Goodwin]

Be nice if the sellers included a schematic in their advert - but it's fairly rare on these cheap Chinese modules.

Mind you, a quick google finds one who does:

PAM8610 Amplifier Module

Mini-kits stock the popular PAM8610 class D 2x 10 watt Audio amplifier modules that are ideal for battery operated projects.

www.minikits.com.au

 

[Buk]

Mind you, a quick google finds one who does:

Strong, The (google) fu is, in this one. I did look, but obviously not in the right place

Many thanks for this.

 

[Buk]

If the switch input is connected to a logic pin, it will have a pull-up or pull-down resistor. If the jumper isn't in place, the resistor pulls the pin to one state. When the jumper (or switch) is in place, a connection is made to pull the port pin in the other direction

Ooo arrr! That sounds like electrickery to me 'andsome 'ead. I'll 'avta be puttin' on me thinkin' 'ead.

 

[Buk]

If the switch input is connected to a logic pin, it will have a pull-up or pull-down resistor. If the jumper isn't in place, the resistor pulls the pin to one state. When the jumper (or switch) is in place, a connection is made to pull the port pin in the other direction.

Okay. See if I've got this right.

According to this diagram (thanks Nigel Goodwin)
https://www.minikits.com.au/image/cache/catalog/modules/pam8610-aud-01c-640x640.jpg

The 100k R15 and the unnumbered 47k (pull down?) resistor form a voltage divider between ground and the Vdd (12v in my case) so the when the jumper is closed, 47k/147k * 12V = 3.85V is presented to pin 29; and when open 0V?

 

[audioguru]

The PAM8610 datasheet is labelled "Not Recommended for New Design, Use PAM8006A" but Digikey still has some in stock. Maybe the PAM8610 has a problem.
PAM8006A is available since 8 years ago and PAM8620 is nearly the same but is not available.

 

[Visitor]

Beat me to it.

 

[Visitor]

The PAM8610 datasheet is labelled "Not Recommended for New Design, Use PAM8006A" but Digikey still has some in stock. Maybe the PAM8610 has a problem.
PAM8006A is available since 8 years ago and PAM8620 is nearly the same but is not available.

12AX7s aren't available either. Perhaps they had some long-secret fault too?

Sorry you see crooks and thieves everywhere you look.

There are roughly 9,000 PAM8610 chips available from reputable suppliers TODAY. Perhaps it's been upgraded to a newer version (as chip manufacturers do), but the ONE in hand should be sufficient.

Perhaps you have some documentation showing it's inferior? Please provide it if so.

 

[Buk]

Beat me to it.

Wow. I got it right.

Now I've learnt electronicals; I think my next project will be a long range 6G transmitter; see if I can't beat the Chinese to it

 

[Nigel Goodwin]

Perhaps you have some documentation showing it's inferior? Please provide it if so.

No documentation, BUT a common reason for replacing a chip with a newer version (other than a bug fix) is a cheaper manufacturing process, meaning a lower price or higher profit. Hence you discontinue the old one, and only the old stock is out there - unless of course you sub-contracted the original manufacture to China, where they pirate the design and continue selling it