Boom box modification

[Califauna]

Hi,

I have a W-King D8 Bluetooth boom box which I'd like to modify with some external speaker output jacks and some buttons to switch the internal speakers on and off.

From factory the specs of the boom box are:

-Power Input: DC 5V
-Speaker Output Power: 2 x 25W ( confirmed by Youtube channel Hartdegen review as 12.5W RMS per channel)
-Drivers:
-2 x Woofer, 70mm, 10Ω, 70mm,15W
-2 X Tweeter, 8Ω, 30mm,10W x 2
-Frequency Response: 60Hz - 20kHz
-Battery Type: Li-ion ( 2 x 4000mAh)
-Battery Voltage: 7.4V ( I measured 8v)
-True Wireless Stereo (TWS): Support
-Technology: BT4.2, A2DP, AVRCP, HSP, HFP.

-The measured resistance at each amplifier output (amp side, speakers disconnected) is 1.2 million ohms.

The tweeters have a resistor in parallel creating a passive high-pass filter.

Here are the circuit diagrams from factory and for the modification I'd like to make.

Factory:
Diagram 1.

And here with my modification:
Diagram 2.

As you can see, I have added:

1. Jacks to connect to external speakers.
2. A higher level high pass filter function on each side (HPF #3 and #4), within which you can see that the factory high-pass HPF #1 and #2 are also nested.
3. Switches 1 and 4, which (I hope) allow the built in speakers to be switched into the passive high pass filter configuration which I have added , using capacitors 2 and 3. This could be handy when connecting to external speakers which can better handle the bass part of the audio. For calculating the value of capacitors 2 and 3 in these added HPFs, I have considered the added external speakers (connected through the jacks) to be the 'resistor', and that this external speaker nominally has an impedance of 4 ohms. Regarding this I have some questions below.
4. Switches 3 and 4, which allow the built-in speakers to be turned off entirely (eg. when I want only external speakers to be used).

Regarding the HP filter, typical diagrams and explanations are that a resistor is ALWAYS necessary in parallel with the load (in this case a speaker through which passes filter audio), sometimes with a connection to ground. Eg:

Diagram 3.

Firstly, I do not understand why a resistor (external speakers) is always necessary for a filter like this to work and wonder whether it could work without them. Indeed, according to this book, the simplest form of a high pass filter has no resistor:

Diagram 4.

Among the explanations of why a resistor is included in such filter circuits, I have read that 'it is to avoid 'thump' when connecting a load, and to allow a steady state at all times' (+ capacitor capacitor always able to discharge) . However, I am not concerned about an initial 'thump' when connecting a speaker. Also I don't see why this 'steady state', or continually being able to discharge/charge is necessary or advantageous. Actually I see that as a disadvantage potentially, as the amplifier/battery power is being left to drain away through such a resistor completing the circuit (?), even when the external speaker is not connected.
FFI also see explanations indicating that a resistor helps prevent current leaks to ground, as in the picture above with ground connection. However, as it is a portable speaker in this case, there is no ground connection, so why would a resistor be necessary there?

Another explanation I've read is that "if you want to predict the filter's performance you need to know how much resistance is in series with the signal. A theoretical source with zero impedance will not be affected by the capacitor. If the signal is coming from something which has significant output impedance, then an additional resistor may not be needed."

Regarding that explanation, I DO know the resistance which is in series with the signal, (4ohms - not a theoretical 0 ohms), so why the need for an additional resistor in parallel with it?

I also have the following questions:

I figure that the addition of capacitors C3 and C4 will continuously affect the cutoff frequency of HPF#1 and #2 (factory installed HPFs), which filter the signal to speakers 1 and 4, because the resistance of the parallel 'resistor' each sees will vary. I figure this because, taking HPF#3 for example, the 'resistor' it sees when speaker 5 is connected and switch 1 is open will no longer be just speaker 2, but the combined resistance of speaker 2, capacitor 3, and speaker 5 all in parallel. Thus the 'resistor' of HPF#1 will vary continuously as the frequency of the signal into capacitor 3 changes (the changing audio signal). Perhaps it will also be affected slightly by varying amounts of back EMF produced by speakers 2 and 5? How much is all this likely to affect the filtered output of speaker 1 in this scenario?

Will typical 3.5 mm audio jacks be able to handle the current flowing through speakers 5 and 6? I also have phono plugs, but there isn't much space to work with inside the enclosure part where the electronics are, so I'd like to go with smallest components if possible. I'm also not sure what rating the switches would have to meet. When S1 and S2 are closed and SP5 is connected, I have calculated that theoretically the peak current through switch 1, switch 2, and jack 1 (when all the signal is above the cut-off frequencies), would be I=V/R = 7.4v / (1/((1/8Ω)+(1/10Ω)+(1/4Ω)) = 3.515a. This is ignoring any back EMF induced in the speakers (if indeed this is a real issue).

I notice the rating for switches is often higher when used in 110 Volts circuits than when the same switch is used in a 240v circuit. Is this something to do with the rating being partly determined by the power (P=IV), not just the current flow through the component?
Am I right in thinking capacitor 3 will do nothing (remain infinite resistance) when switch 1 is closed?

Thanks for your explanations, corrections, suggestions, and possible additional ideas/functionalities that might be useful and could be added while the box is open.

 

[unclejed613]

Firstly, I do not understand why a resistor (external speakers) is always necessary for a filter like this to work and wonder whether it could work without them. Indeed, according to this book, the simplest form of a high pass filter has no resistor:

the driver is an inductor as well as a resistance, and this is why there's no resistor on the driver side of the capacitor.

 

[audioguru]

Why do you think the amplifier can survive a load of 4 ohms? Maybe the DC voltage boosting circuit also might not survive.
The woofer and tweeter are never in parallel, the crossover capacitor blocks the tweeter at low frequencies and a series inductor or the woofer's inductance blocks it at high frequencies.

A series LC is a dead short at its resonant frequency when the speaker is disconnected from it. It will destroy the amplifier if that frequency occurs..

 

[Califauna]

Why do you think the amplifier can survive a load of 4 ohms? Maybe the DC voltage boosting circuit also might not survive.

It's designed to handle a load of 4ohms. This is the combined resistance of the drivers it comes installed with. Is there possibly something else I should be considering?

The woofer and tweeter are never in parallel, the crossover capacitor blocks the tweeter at low frequencies and a series inductor or the woofer's inductance blocks it at high frequencies.

Which woofer, tweeter and crossover capacitor are you referring to? Something in the factory diagram, my modification diagram, or both? If it's one of those two, could you refer me to the component labels (eg. C1, SP1, etc.)?

A series LC is a dead short at its resonant frequency when the speaker is disconnected from it. It will destroy the amplifier if that frequency occurs..

Which series LC are you referring to? Are you saying that in my modification (diagram 2) , when S1 is open and S2 is closed and the external speaker is not connected, a series LC circuit is created by SP2 and C3 which therefore will short at a certain frequency? I don't understand how that could be the case, because the driver itself always has its own 10 ohms resistance (resistance of it's coils).

 

[Califauna]

the driver is an inductor as well as a resistance, and this is why there's no resistor on the driver side of the capacitor.

So, in other words, the diagram 4 circuit will work as a high pass filter and does not require an additional resistor because the driver provides resistance and the Vout in that circuit is therefore not a theoretical load of 0 ohms. Is that right?

 

[audioguru]

It's designed to handle a load of 4ohms. This is the combined resistance of the drivers it comes installed with. Is there possibly something else I should be considering?


Which woofer, tweeter and crossover capacitor are you referring to? Something in the factory diagram, my modification diagram, or both? If it's one of those two, could you refer me to the component labels (eg. C1, SP1, etc.)?



Which series LC are you referring to? Are you saying that in my modification (diagram 2) , when S1 is open and S2 is closed and the external speaker is not connected, a series LC circuit is created by SP2 and C3 which therefore will short at a certain frequency? I don't understand how that could be the case, because the driver itself always has its own 10 ohms resistance (resistance of it's coils).

The woofer and tweeter are never in parallel. The woofer's series inductor or its internal inductance prevents it from being connected at high frequencies and the fragile tweeter's series capacitor prevents it from working and being destroyed by low frequencies.

See the word "crossover"? The power amplifier is "crossed over" to the woofer or the tweeter depending on the frequency.

A simple crossover is a series capacitor on the tweeter. A more complex crossover adds inductors.
Your circuit with a series capacitor and an inductor to ground is a dead short to the amplifier at its resonant frequency if the tweeter is removed. It is a highpass filter when the tweeter's resistance is parallel with the inductor.
Your simple capacitor in series with the tweeter blocks low frequencies but passes high frequencies to the tweeter that has a load of 8 ohms in your figure #5.
Here is a graph of the impedance of a woofer caused by its internal inductance. It resonates at 30Hz, it is 7 ohms at 150Hz, it is 10 ohms at 400hz and it is higher impedance at higher frequencies where it does not load the amplifier.

 

[Califauna]

I don't know which woofer and tweeter you refer to above. Also, do you mean the 'ground' shown in diagram 2? Isn't my circuit (diagram 2) a V+ and V-, or a '+' and '-'? Ground, I thought, is differnet to +/- when it comes to circuits (?).

Do you mind explaining again using (or adding into the explanation above) the labels I have put on the compents and circuits in diagram 1 and diagram 2 (so I know exactly which circuits/components you are referring to)?

 

[audioguru]

Sorry, when I said "ground" I should have said "common" or the - amplifier output since the amplifiers are probably bridged.

Your boom box has a tweeter and a woofer for each stereo channel. The tweeter has a series coupling capacitor as a simple highpass filter as the crossover network as shown in your diagram #4 and to prevent low frequencies from damaging it. If you add an inductor parallel to the tweeter as in your diagrams #2 and #3 then the low frequencies are cut more.

 

[Califauna]

If you add an inductor parallel to the tweeter as in your diagrams #2 and #3 then the low frequencies are cut more.

I am trying to achieve precisely this (cut more low frequencies out to SP2 and SP3). For this reason I have tried to create a HPF on each side by adding (on the left stereo side for example) C3 and S1 and a switch (S1) to enable it. I have selected 132uF as the C1 capacitance as I understand this creates a cut-off of 300Hz, using the formula fc= 1/(2πRC), so that when an external speaker is connected at jack 1 and S1 is open and S2 is closed, SP2 receives signal but with frequencies below 300Hz attenuated (highpass filter).

Are you saying the circuit I have there in diagram 2 achieves all this correctly?

I haven't tried to add any inductors (unless you are referring to the external speaker?). Do you mean capacitor? I did add that into the internal circuit diagram.

 

[audioguru]

A single series capacitor highpass filter cuts low frequencies gradually at 6dB per octave. Adding an inductor cuts low frequencies much more at 12dB per octave. Adding a second capacitor to the capacitor and inductor cuts low frequencies at 18dB per octave. The inductance of a speaker is too small to be in an audio filter.

The amplifiers in the boom box probably do not pass the low frequencies to your additional speakers.

I have a cheap clock radio that had a 3" speaker that sounded awful. I replaced the little speaker with a very good sounding hifi speaker I made and the radio still sounded squeaky with no bass. I increased the values of three coupling capacitors in the radio and now it sounds hifi.

 

[Califauna]

I don't know how to claculate the values of components in the 18db filter you describe, or where exactly in the circuit diagram they would go.

I haven't actually done any modifications yet. Just made diagrams. I want to know if it will work first.

The amp passes a lot of bass. Signal as low as 50Hz. Signal below that greatly attenuated (

).
This boom box has serious bass (for it's size). The drivers are small but good with it.

 

[audioguru]

The highpass filter is part of a passive crossover for a woofer. There are online articles with the calculations.
It is connected from the amplifier output to the speaker that you do not want to play low frequencies.

6dB per octave is a very gradual slope:
300Hz= -3dB which is barely noticeable drop in level.
150Hz= -6dB to -9dB which is a small drop in level.
75Hz = -12dB which is a little more drop in level.
37.5Hz = -18dB a little more drop in level but still very obvious.

Then you will have this speaker playing at the same time as the added woofer that might overload the amplifier unless you add a complementary matching lowpass filter to the added woofer. Usually the highpass and lowpass are at the same -3dB frequency so that they add perfectly to no gain and no loss at the crossover frequency.

Just now I played a 50Hz sinewave on my small computer speakers. It had a lot of 100Hz and 150Hz harmonic distortion.
50Hz is not a low frequency since we can hear down to 20Hz. 25Hz is one octave below 50Hz.

 

[Califauna]

50Hz is not a low frequency since we can hear down to 20Hz.

True. I will hook it up to a decent speaker to test frequencies output by the amp below 50hz.

The highpass filter is part of a passive crossover for a woofer. There are online articles with the calculations.
It is connected from the amplifier output to the speaker that you do not want to play low frequencies.

Yes I knew this, if you mean the general concept of what a passive crossover for a woofer is. I added (I hope correctly) HPF#3 and HPF#4 when I came up with diagram 2.

I will look at third order HPF filters and propose a circuit diagram for it.

Then you will have this speaker playing at the same time as the added woofer that might overload the amplifier unless you add a complementary matching lowpass filter to the added woofer.

I like this idea. I was already thinking of adding any external filters which might be needed by the external speakers to the external speakers themselves. Seems logical.

Assuming a low pass filter for the external speaker is added at the external speaker, can you confirm if my modification will work, based on the diagram (diagram 2) with the components and wiring as they are indicated, with the first order HPFs I've added there? Or help out with the other questions in the original post?

 

[audioguru]

Do you have subwoofers that can play lower than 50Hz?
Will the amplifiers in the boom box survive the doubled current when the subwoofers are 4 ohms?

 

[Califauna]

Do you have subwoofers that can play lower than 50Hz?
Will the amplifiers in the boom box survive the doubled current when the subwoofers are 4 ohms?

I made the circuit design with regular external hifi speakers in mind. Typical frequency response of 40Hz-15000Hz, something like that.

Why will the current be double? Could you show me how you came to this calculation and what configuration you mean?

If you mean the configuration where only the external speakers are connected, I figure that when external 4ohm speakers are connected and the internal speakers disabled the max current will be 7.4/4 = 1.85 amps on each side. The amp is designed to handle 1.665 amps, given that 7.4 / (1/((1/8)+(1/10))) = 1.665a. How long would it be able to handle the 0.185a difference assuming I had the thing at max volume? I don't know. Will the battery will reach it's power supply limit before the current is enough to fry the amp anyway? I don't know.

In a scenario with all three speakers connected on each side, including external SP5 and SP6, and no LPF on the external speakers, I figure the max current could potentially be 7.4 / (1/((1/8)+(1/10)+(1/4))) = 3.515a. So, more than double. Could it handle it? I don't know. I figure it would be a pretty strange signal to reach that current, as it would all need to be above the cut-off frequencies of the filters on each side, eg. HPF#1 and my HPF#3 idea.

In a scenario with all three speakers connected on each side, including external SP5 and SP6, and an 6db (for the moment) LPF on the external speakers with a cut-off of 300Hz, I'm not sure how to calculate that accurately, as the resistance would be changing all the time as the signal changes, and whatever (frequency) the signal is, it will meet resistance in one or other of the filters.

As I say, to be able to have a go at answering these questions I need to know the answer to the questions I posted in the thread started, eg.

I figure that the addition of capacitors C3 and C4 will continuously affect the cutoff frequency of HPF#1 and #2 (factory installed HPFs), which filter the signal to speakers 1 and 4, because the resistance of the parallel 'resistor' each sees will vary. I figure this because, taking HPF#3 for example, the 'resistor' it sees when speaker 5 is connected and switch 1 is open will no longer be just speaker 2, but the combined resistance of speaker 2, capacitor 3, and speaker 5 all in parallel. Thus the 'resistor' of HPF#1 will vary continuously as the frequency of the signal into capacitor 3 changes (the changing audio signal). Perhaps it will also be affected slightly by varying amounts of back EMF produced by speakers 2 and 5? How much is all this likely to affect the filtered output of speaker 1 in this scenario?

 

[audioguru]

You said that the original speakers are 8 ohms and 10 ohms and have a simple single 3.3uF capacitor for each tweeter as the crossover. Then a 4 ohm external speaker on each channel will overload the amplifiers. A 4 ohm speaker draws double the current of an 8 ohm speaker. The datasheet of the amplifier IC, its power supply voltage, its amount of heatsinking the speaker impedance determine the maximum allowed current.

Two Li-Ion cells produce 8.4V when fully charged, not 7.4V and the voltage is stepped up so that the output is 12.5W (according to the video) for each channel which is 25W peak for each channel. 25W peak into an 8 ohm speaker is a peak current of 1.77A.

Your 132uF capacitor per channel has an impedance of 4 ohms at 300Hz which is much lower than the 6.06kHz at -3dB highpass of the 3.3uF capacitors with the tweeters.
The SP2 or SP3 woofer's impedance of maybe 50 ohms or more at 6kHz due to its inductance will barely cause it to load the amplifier so the amplifier's load at high frequencies is only the SP1 or SP4 tweeter.

The amplifier will be overloaded at all frequencies if it has an external woofer and tweeter with crossover circuit connected at the same time the original speakers are connected.

 

[Califauna]

Two Li-Ion cells produce 8.4V when fully charged, not 7.4V and the voltage is stepped up so that the output is 12.5W (according to the video) for each channel which is 25W peak for each channel. 25W peak into an 8 ohm speaker is a peak current of 1.77A.

According to the specs in the video and the manual, the output voltage from the battery is 7.4v, not 8.4. I'm not sure where you got 8.4v from. Please see 4:13 in the video. Could you link me to the source/time where it says the battery output is 8.4v?

You said that the original speakers are 8 ohms and 10 ohms and have a simple single 3.3uF capacitor for each tweeter as the crossover. Then a 4 ohm external speaker on each channel will overload the amplifiers. A 4 ohm speaker draws double the current of an 8 ohm speaker. The datasheet of the amplifier IC, its power supply voltage, its amount of heatsinking the speaker impedance determine the maximum allowed current.

Two Li-Ion cells produce 8.4V when fully charged, not 7.4V and the voltage is stepped up so that the output is 12.5W (according to the video) for each channel which is 25W peak for each channel. 25W peak into an 8 ohm speaker is a peak current of 1.77A.

Why are you using a value of 8 ohms for that calculation? 8 Ohms is the value of SP1 and SP4 but the amplifier doesn't see a resistance of 8 ohms on each side of any of the circuits I have uploaded.

From factory (diagram 1), I calculate that the resistance of SP1 and SP2, which are in parallel in the speaker, is 1/((1/8)+(1/10)) = 4.4 ohms. Therefore, using I=V/R there is a max current of 7.4/4.4 = 1.681. This ignores and resistance potentially offered by C1.

P=I*V does not seem a very useful formula for me for comparisons with my proposal (diagram 2) because I can only calculate the theoretical max
current for my diagram 2 using the values of the components I use there, and NOT the power output (since it is unknown to me), and from that calculation the power output, but for what its worth, using P=I*V, the current for factory design (diagram 1) using P=I*V, therefore I=P/V, = 25w/7.4v = 3.378 amps.

In my diagram (2), If speaker 5 is connected and S1 is open AND S2 closed, the amplifier outputs will see SP1, SP2, and SP5 in parallel. Thus the minimum resistance it can see will be 1/((1/8+(1/10)+(1/4)) which equals 2.10 ohms. Maximum current (assuming no filter is placed on SP5) will therefore be, according to my calculations, 7.4/2.10 = 3.515 a. This also ignores potential resistance from C1 and C3.

I do realize that with S5 in parralel, there will be a lower resistance seen by the amp than if there were just two drivers. The questions regarding the above calculations, and the issues you raise in your last post, are:

1. The resistance will not be constant. C1 and C3 will affect this, as well as ( I presume) back EMF/inductance from the drivers themselves. What formula can I use to factor this in to the above calculations to calculate a typical current draw at a given moment, and calculate the maximum current draw possible.
2. When I have and know the above, will the amp likely be able to handle the additional draw?

 

[Califauna]

According to the code on the circuit board, the amplifier is this:

https://www.ti.com/product/TPA3118D2

 

[Nigel Goodwin]

According to the specs in the video and the manual, the output voltage from the battery is 7.4v, not 8.4. I'm not sure where you got 8.4v from. Please see 4:13 in the video. Could you link me to the source/time where it says the battery output is 8.4v?

Li-Ion batteries are 4.2V each when fully charged, so two in series is 8.4V when fully charged - as simple as that. The 7.4V is 3.7V per cell, which is fairly flat.

 

[audioguru]

A lithium battery's life is shortened if it is always at a full charge of 4.2V per cell therefore they are sold, stored and labelled at a half charge of 3.7V per cell (7.4V for your battery). The manufacturer usually says, "charge before use". A Lithium battery cell is almost completely dead at 3.2V then 3.7V is a half charge.

Speakers are used with AC, not with DC. The impedance is measured with AC and resistance is measured with DC. Speakers have inductance that increases their impedance at high audio frequencies as I showed in post #6. Your tweeters are never parallel with the woofers to avoid destruction of the tweeters. They have a 3.3uF capacitor in series to block low frequencies. A capacitor has a high impedance at low frequencies. At high frequencies the capacitors are like a dead short then the tweeters and woofers are in parallel but the impedance curve of a woofer shows a high impedance at high frequencies then the tweeters get all of the power.

The speaker system has a DC voltage booster so that the amplifier IC produces up to 12.5W of real power per channel which is 25 Whats of advertised power per channel.