PCB reverse engineering: TK2050 Amplifier

[hitech95]

Hi, it's been a while that I have this idea in my head.
I wanted to ask some advice: figure out a circuit board.
I haven't physically the board (but I may have it in the future), for now I only have a few pictures. In addition I couldn't remove the components.
The card in question is an amplifier with TK2050 chipset. I noticed that there are severe changes in implementation than the Datasheet, I would understand why.

Do you have any idea on how to make it easier to map the traces?
I wanted to ask why choices were made by "design".

Thanks, hitech95.
Gallery: http://imgur.com/a/Oodxl
EDIT:
11/25/15: Update thread title, added gallery.

[dmills]

With switching power amplifiers like that the board layout is the other half of the schematic, so reversing the schematic is only half the battle.

I generally find that there are usually only a few bits of such things are in any way interesting, and they are generally the bits where the interesting tradeoffs are, so for a switching amplifier, that typically means the output filter and (in some designs) the feedback network.
Understanding why the designer made a particular design call is often problematic (How do you know that inductor was used because the one the designer really wanted went on long lead time after the gerbers were sent out and the replacement was the first thing a search on Mouser found that would fit? It happens more often then you would think).

Component choice in this sort of thing is often driven as much by BOM cost, house standard parts, fashions in marketing, availability, and simple designer preference as it is by actual engineering.

73, Dan.

[Shock]

https://www.eevblog.com/forum/eda/pcb-reverse-engineering/

There is another post somewhere with photos showing how to take front and back photos of the PCB flip one and change them to transparent and create a layout, trivial if you have ever used a graphics editor before. From there you could import into layout software and add the components and nets.

I believe there is also new software that will also create a schematic from a photo but yeah good luck with that.

[hitech95]

@dmills
I understand, what made me think it is how they did to drive a 12V relay with 24V, it does not seem that there are resistor dividers. The single regulator is for 5V.
What I do not understand is the output stage, is full of optocouplers.
@Shock
Oh, I had not found it with the search. 

[EEVblog]

[hitech95]

Oh wow. Thank you.
How I missed this video? (I'll check the subscription on Youtube ...)

[hitech95]

This is the final filter, that's what I'm trying to understand.
Damned blue solder mask

[hitech95]

Sorry for the double post.

Question how is generated the 13V rail? (I have double checked with the multimeter and is 13V  ) R21 (100Ohm) and R23 (82Ohm) are really big resistor (1W or 2W)

[dmills]

If the 13V rail is just used to pull in the relays then a simple dropper resistor is all that is really required.

If you are dropping from 24V to 13V, you are dropping 11V across 100 ohms, so the current on the 13V rail is a  tadge over 100mA, power dissipated in the resistor is 1.2W, sounds broadly reasonable for a relay supply (Relays are not fussy).

The resistor in series with the LM317 is to move some of the power dissipation away from the linear regulator

The opto couplers will be for DC fault detection on the speaker outputs, the speaker outputs are bridged so the easiest way to do this is with a lowpass filter feeding an AC input opto, if the output of the lowpass exceeds a volt or so the optos led will be biased on and I assume that opens the relays somehow (Q2 and frends?). 

Note that the lowpass could well be single ended and just floating on the other side of the speaker output.

Not the most efficient design in history, but I see no superficial reason to think it would not work, and the parts are cheap.

Regards, Dan.

[hitech95]

If the 13V rail is just used to pull in the relays then a simple dropper resistor is all that is really required.

If you are dropping from 24V to 13V, you are dropping 11V across 100 ohms, so the current on the 13V rail is a  tadge over 100mA, power dissipated in the resistor is 1.2W, sounds broadly reasonable for a relay supply (Relays are not fussy).

The resistor in series with the LM317 is to move some of the power dissipation away from the linear regulator

The opto couplers will be for DC fault detection on the speaker outputs, the speaker outputs are bridged so the easiest way to do this is with a lowpass filter feeding an AC input opto, if the output of the lowpass exceeds a volt or so the optos led will be biased on and I assume that opens the relays somehow (Q2 and frends?). 

Note that the lowpass could well be single ended and just floating on the other side of the speaker output.

Not the most efficient design in history, but I see no superficial reason to think it would not work, and the parts are cheap.

Regards, Dan.

Yes, the rail is used to drive the relay.
And thanks for the explanation.

Optocouplers trigger the relay, I have not finished the wiring diagram, I miss this part.

WORK done:
http://imgur.com/a/Oodxl

[hitech95]

Double post. Forgive me. 

I finished drawing the pattern. I hope it is correct; I checked several times.
Now I do not understand the operation. Specifically, the part of the optocouplers. From what I understand they serve to trigger the relay if the output voltage is too high. I could be wrong.

The NPN transistors are in Darlington configuration. (I think.)
I am looking in the books the configuration of the capacitors, but nothing comes out. 

EDIT: missing image:

[timb]

@Shock
Oh, I had not found it with the search. 

Check out the first three pictures in this thread I posted a few months back:
https://www.eevblog.com/forum/index.php?topic=52614

Basically, I took clear pictures of the front and back of the board. Then I converted the picture of the back to grayscale, upped the contrast and used an intensify filter to make the traces well defined. After that it was a simple matter of using the magic wand tool to select all the traces, fill the selection in a solid color on a new layer and flip it horizontally.

Now that you've got a clear set of traces, simply paste it onto the photo of the board front and adjust the opacity. Easy peasy.


Sent from my Smartphone

[hitech95]

@Shock
Oh, I had not found it with the search. 

Check out the first three pictures in this thread I posted a few months back:
https://www.eevblog.com/forum/index.php?topic=52614

Basically, I took clear pictures of the front and back of the board. Then I converted the picture of the back to grayscale, upped the contrast and used an intensify filter to make the traces well defined. After that it was a simple matter of using the magic wand tool to select all the traces, fill the selection in a solid color on a new layer and flip it horizontally.

Now that you've got a clear set of traces, simply paste it onto the photo of the board front and adjust the opacity. Easy peasy.


Sent from my Smartphone

Thanks, I wrote the circuit. Now I do not understand how it works.

[dmills]

I am fairly sure you have the relay contacts the wrong way around.... (It should be wired normally open).

Each pair of speaker leads has a lowpass filter R43/C50/C51 wired across it, (C50 + C51 are back to back because this was apparently cheaper then a non polar cap), so the audio will be substantially rejected by this filter.

Following the filter we have a pair of opto couplers with the leds wired back to back (presumably cheaper then an AC input coupler), such that if the output of the lowpass filter exceeds 1.2V or so in either direction the leds will start to conduct and turn on the phototransistor.

Turning on the phototransistor in one of the couplers pulls the left hand end of R41 down, which (if it persists for long enough) will pull the base of the darlington below ~1.4V which will turn the relay off disconnecting the speakers.

On switch on there is a short delay before the relay pulls in (R39 charging C37), which helps to avoid startup thumps.

Regards, Dan.

[hitech95]

I am fairly sure you have the relay contacts the wrong way around.... (It should be wired normally open).

Ops... you are right.

Each pair of speaker leads has a lowpass filter R43/C50/C51 wired across it, (C50 + C51 are back to back because this was apparently cheaper then a non polar cap), so the audio will be substantially rejected by this filter.

How do I calculate the filter frequency: how do I find the equivalent value of C?

Following the filter we have a pair of opto couplers with the leds wired back to back (presumably cheaper then an AC input coupler), such that if the output of the lowpass filter exceeds 1.2V or so in either direction the leds will start to conduct and turn on the phototransistor.

Turning on the phototransistor in one of the couplers pulls the left hand end of R41 down, which (if it persists for long enough) will pull the base of the darlington below ~1.4V which will turn the relay off disconnecting the speakers.

On switch on there is a short delay before the relay pulls in (R39 charging C37), which helps to avoid startup thumps.

Regards, Dan.

I understand the use of the delay but not that optocoupler logic. 
So, as I understand it is used to disconnect the speakers if there is a voltage higher than a X value. (R42 & R43 are 3K, caps are 220uF)
Now I would understand if the voltage of the cutoff voltage depends on the supply voltage or if there is a safety threshold to be respected in any case.


All this work is because the sound chip can run at higher voltages (up to 30V) and I would like to understand why this board is limited to 20V.

Regards, Nicolò.

[dmills]

The amplifier channels are full bridge topology, and this thing is used to disconnect the speakers in the event of a failed mosfet in the power chip forcing one side of the speaker to either rail.

Look at OK1 and OK2, you can see that the leds are wired anti parallel but the transistors are wired in parallel, now consider what happens if a fault takes OUT_LP_3.6A to the supply rail and holds it there...

OUT_LP_3.6B will still be at about mid rail, so there will be 10V or so of DC at the input to the lowpass filter, in fairly short order OK1 will start to conduct and will trip the speaker DC fault protect relay. The same thing happens if the short is to ground, via OK2, if you work thru the combinations you will find that any condition putting significant DC on the speaker terminals will trip the relay. 

As to why 20V? Well, heat in the regulators and their dropper resistors is one possible reason, but actually a 30V amp running at 20V is almost always a better design then a 30V rated part running balls out.
Heat in the amplifier chip is another possibility as is core saturation in the inductors, or even over voltages caused by the ringing on the switching nodes (That whole layout dependence again).

Regards, Dan.

[hitech95]

The same thing happens if the short is to ground, via OK2, if you work thru the combinations you will find that any condition putting significant DC on the speaker terminals will trip the relay. 

I did not understand this part.

As to why 20V? Well, heat in the regulators and their dropper resistors is one possible reason, but actually a 30V amp running at 20V is almost always a better design then a 30V rated part running balls out.
Heat in the amplifier chip is another possibility as is core saturation in the inductors, or even over voltages caused by the ringing on the switching nodes (That whole layout dependence again).

Regards, Dan.

The chip has the highest maximum 40V (STA508), I believe that the real limitation is due to the 5V regulator and the "divider" for 12V.

Could you explain me what you mean by:

... as is core saturation in the inductors, or even over voltages caused by the ringing on the switching nodes (That whole layout dependence again).

Thanks, Nicolò


EDIT:
Quick note: How read the resistances R24, R25, R27, R26? They seem to start with Black.  And the "gray" seems more silver than gray. So 0.10Ohm?

[hitech95]

I take this opportunity to ask you another question: can you explain what is the purpose of Q1 in the following scheme?
Schematic

Thanks, Nicolo.

[dmills]

That is a P channel mosfet, which switches on when the gate goes negative with respect to the source.

That circuit fragment can be split into three parts:

• U2 and friends which form the battery charger for the lithium pack.
• U4 and friends which are the voltage regulator.
• Q1,R7,D3 which form the switch that disconnects the battery from the load when external 5V is supplied, the arrangement acts kind of like a diode but with much lower losses.

The cases to examine are USB Power & Battery power.
When the USB power is present, the main circuit is powered via D3, and the mosfet gate is positive by one diode drop with respect to the drain, so the P channel mosfet is pinched off.
The diode drop does not matter in this case as a few hundred mV drop from a 5V rail when the regulator output is ~3.6V is not too significant.
The battery is also being charged by U2 if needed.

When there is no external power, the gate is pulled down via R7, biasing the mosfet on and connecting the battery to the load with very low drop (In this case voltage drop does matter).

It is actually a fairly standard idiom for this sort of thing.

Regards, Dan.

[hitech95]

That is a P channel mosfet, which switches on when the gate goes negative with respect to the source.

That circuit fragment can be split into three parts:

• U2 and friends which form the battery charger for the lithium pack.
• U4 and friends which are the voltage regulator.
• Q1,R7,D3 which form the switch that disconnects the battery from the load when external 5V is supplied, the arrangement acts kind of like a diode but with much lower losses.

The cases to examine are USB Power & Battery power.
When the USB power is present, the main circuit is powered via D3, and the mosfet gate is positive by one diode drop with respect to the drain, so the P channel mosfet is pinched off.
The diode drop does not matter in this case as a few hundred mV drop from a 5V rail when the regulator output is ~3.6V is not too significant.
The battery is also being charged by U2 if needed.

When there is no external power, the gate is pulled down via R7, biasing the mosfet on and connecting the battery to the load with very low drop (In this case voltage drop does matter).

It is actually a fairly standard idiom for this sort of thing.

Regards, Dan.

Oh, OK thanks.

And for the other question?

The same thing happens if the short is to ground, via OK2, if you work thru the combinations you will find that any condition putting significant DC on the speaker terminals will trip the relay. 

I did not understand this part.

As to why 20V? Well, heat in the regulators and their dropper resistors is one possible reason, but actually a 30V amp running at 20V is almost always a better design then a 30V rated part running balls out.
Heat in the amplifier chip is another possibility as is core saturation in the inductors, or even over voltages caused by the ringing on the switching nodes (That whole layout dependence again).

Regards, Dan.

The chip has the highest maximum 40V (STA508), I believe that the real limitation is due to the 5V regulator and the "divider" for 12V.

Could you explain me what you mean by:

... as is core saturation in the inductors, or even over voltages caused by the ringing on the switching nodes (That whole layout dependence again).

Thanks, Nicolò


EDIT:
Quick note: How read the resistances R24, R25, R27, R26? They seem to start with Black.  And the "gray" seems more silver than gray. So 0.10Ohm?

[hitech95]

The same thing happens if the short is to ground, via OK2, if you work thru the combinations you will find that any condition putting significant DC on the speaker terminals will trip the relay. 

I did not understand this part.

As to why 20V? Well, heat in the regulators and their dropper resistors is one possible reason, but actually a 30V amp running at 20V is almost always a better design then a 30V rated part running balls out.
Heat in the amplifier chip is another possibility as is core saturation in the inductors, or even over voltages caused by the ringing on the switching nodes (That whole layout dependence again).

Regards, Dan.

The chip has the highest maximum 40V (STA508), I believe that the real limitation is due to the 5V regulator and the "divider" for 12V.

Could you explain me what you mean by:

... as is core saturation in the inductors, or even over voltages caused by the ringing on the switching nodes (That whole layout dependence again).

Thanks, Nicolò


EDIT:
Quick note: How read the resistances R24, R25, R27, R26? They seem to start with Black.  And the "gray" seems more silver than gray. So 0.10Ohm?

up? And sorry for all these questions. 

[hitech95]

Could you explain me what you mean by:

... as is core saturation in the inductors, or even over voltages caused by the ringing on the switching nodes (That whole layout dependence again).

I hope that someone can explain this to me, I did not understand.   

Bye, hitech95.