how to isloate audio ground from the amplifier

[Adhith]

I just used a multi meter ohm function and found that the dc jack ground and the left channel ground is having a resistance of 350 ohm.

The left channel has no ground. Nor does the right channel.  No speaker is connected to ground.  Both terminals of the speaker are driven by push-pull outputs. In this way the reason they label them "+" and "-" is for phasing.  If you connect speakers backwards you will still hear music it will just be out of phase with the other channel, it has nothing to do with ground reference.

sorry sir my mistake. i meant the wire other than the +Ve output of the amplifier. SO what should I call that terminal. the lower side terminal will do??

[Adhith]

Ill connect  the ground to the vu meter let you guys know about it.

Yes, that will work.

As this is a class D amplifier, you might find the existing filter is inadequate at removing all the high frequency content, so the VU meter might indicate there's a signal, even though there isn't one. Another RC low pass filter might be required, before the VU meter.

thank you again. could you suggest me a RC low pass filter for it, the values of R, C??. If i'm using it then there is no need of an additional DC blocking capacitor right??
I have attached the schematics of my VU meter circuit and the circuit do have a 2.2mf electrolytic capacitor at the input and I was under the assumption that its a filter for the input. Is this the case or I'm wrong??

[Adhith]

I just fed the ground near the DC jack and the +ve audio output of the left channel to the VU meter input (without any RC filter), but the led was not even lighting up. There is no problem with the VU meter circuit since I have tested it with the previously mentioned LA4440 amplifier board and it works fine. So what would be the reason that the board is not working with my class D amp board?? While measuring the voltage between this +ve left channel output and the ground near the DC jack it is found that it is around a constant of 6.4V. its not changing a little according to sound or increasing or decreasing with the volume or even when there is no audio playing. The voltage appears at these terminals as soon as the amp is powered, afterwards its moreover a constant. So why is it so??. does this shows the ground of the amp is some other terminal??

[Zero999]

I just fed the ground near the DC jack and the +ve audio output of the left channel to the VU meter input (without any RC filter), but the led was not even lighting up. There is no problem with the VU meter circuit since I have tested it with the previously mentioned LA4440 amplifier board and it works fine. So what would be the reason that the board is not working with my class D amp board?? While measuring the voltage between this +ve left channel output and the ground near the DC jack it is found that it is around a constant of 6.4V. its not changing a little according to sound or increasing or decreasing with the volume or even when there is no audio playing. The voltage appears at these terminals as soon as the amp is powered, afterwards its moreover a constant. So why is it so??. does this shows the ground of the amp is some other terminal??

That's to be expected. The output of the amplifier will be biased at half the supply voltage, hence the 6.4V. I don't know why the VU meter isn't responding at all. It could be the RF parts of the signal.

The VU meter needs to be AC coupled to the amplifier. Try adding this circuit between the amplifier and VU meter. It both blocks DC and greatly attenuates the ultrasonic class D switching frequency. If it doesn't work, then you might need to build another op-amp based amplifier, connected to the input of the class D amplifier and drive the VU meter from that.

[Audioguru]

Your VU meter circuit is completely wrong:
1) The LEDs are backwards with the anodes at ground and the cathodes driven to +12V.
2) The input diode passes DC to the input of the LM3915 and ruins the accuracy of the lower LED turn on voltages. The diode and the 2.2uF capacitor it charges form a very poor peak detector that shorts the amplifier output badly each time the input goes 0.7V or more positive that could blow up the amplifier and/or the diode.
The datasheet shows a simple opamp peak detector circuit that I have used and it works perfectly. It is accurate and has an input capacitor to block the 6.4VDC from the amplifier output. The opamp needs a dual polarity supply or an opamp that has inputs that work all the way down to ground like one of the two opamps in an LM358.
You cannot add a series capacitor to your horrible diode circuit.

Since the amplifier output is +6.4V then it must have a +12.8V supply. Then if the horrible unregulated pot in your circuit is set so that the 10th LED lights when the input peak is 10V, the series diode's voltage drop prevents the 1st and 2nd LEDs from lighting when they should. A peak detector with an opamp will allow all LEDs to light when they should.
In your circuit, the +6.4VDC will cause the 8th LED to light in the DOT mode or cause LEDs 1 to 8 to light in the BAR mode with no signal.
The horrible pot in your circuit is its unregulated reference voltage but the LM3915 alredy has a regulated reference voltage at pin 7 that is set with two resistors on pin8.
You show 2N3906 transistors with a maximum allowed current of only 200mA driving LED strips that might draw 500mA each?

You have each output of the LM3915 driving 1k resistors in series with the base of the transistors. Then when all LM3915 output are driving these resistor the LM3915 will be too hot and will fail. The datasheet recommends reducing the output currents so the heating is not too high. How much current does each LED strip need?

Here is my peak detector circuit:

[Adhith]

I just fed the ground near the DC jack and the +ve audio output of the left channel to the VU meter input (without any RC filter), but the led was not even lighting up. There is no problem with the VU meter circuit since I have tested it with the previously mentioned LA4440 amplifier board and it works fine. So what would be the reason that the board is not working with my class D amp board?? While measuring the voltage between this +ve left channel output and the ground near the DC jack it is found that it is around a constant of 6.4V. its not changing a little according to sound or increasing or decreasing with the volume or even when there is no audio playing. The voltage appears at these terminals as soon as the amp is powered, afterwards its moreover a constant. So why is it so??. does this shows the ground of the amp is some other terminal??

That's to be expected. The output of the amplifier will be biased at half the supply voltage, hence the 6.4V. I don't know why the VU meter isn't responding at all. It could be the RF parts of the signal.

The VU meter needs to be AC coupled to the amplifier. Try adding this circuit between the amplifier and VU meter. It both blocks DC and greatly attenuates the ultrasonic class D switching frequency. If it doesn't work, then you might need to build another op-amp based amplifier, connected to the input of the class D amplifier and drive the VU meter from that.

Thank you very much again. Ok let me add the circuit before the amp and check if its working or not. I let you know as soon as I have done it.

[Adhith]

Your VU meter circuit is completely wrong:
1) The LEDs are backwards with the anodes at ground and the cathodes driven to +12V.
2) The input diode passes DC to the input of the LM3915 and ruins the accuracy of the lower LED turn on voltages. The diode and the 2.2uF capacitor it charges form a very poor peak detector that shorts the amplifier output badly each time the input goes 0.7V or more positive that could blow up the amplifier and/or the diode.
The datasheet shows a simple opamp peak detector circuit that I have used and it works perfectly. It is accurate and has an input capacitor to block the 6.4VDC from the amplifier output. The opamp needs a dual polarity supply or an opamp that has inputs that work all the way down to ground like one of the two opamps in an LM358.
You cannot add a series capacitor to your horrible diode circuit.

Since the amplifier output is +6.4V then it must have a +12.8V supply. Then if the horrible unregulated pot in your circuit is set so that the 10th LED lights when the input peak is 10V, the series diode's voltage drop prevents the 1st and 2nd LEDs from lighting when they should. A peak detector with an opamp will allow all LEDs to light when they should.
In your circuit, the +6.4VDC will cause the 8th LED to light in the DOT mode or cause LEDs 1 to 8 to light in the BAR mode with no signal.
The horrible pot in your circuit is its unregulated reference voltage but the LM3915 alredy has a regulated reference voltage at pin 7 that is set with two resistors on pin8.
You show 2N3906 transistors with a maximum allowed current of only 200mA driving LED strips that might draw 500mA each?

You have each output of the LM3915 driving 1k resistors in series with the base of the transistors. Then when all LM3915 output are driving these resistor the LM3915 will be too hot and will fail. The datasheet recommends reducing the output currents so the heating is not too high. How much current does each LED strip need?

Here is my peak detector circuit:

Thank you very much Sir for your detailed suggestions. I now understand that I should have taken much more seriously in understanding the faults in the circuit. the circuit was found from the internet and I had build another project based on it and it works, so i didn't go in depth in understanding the concept. I now totally know that I shouldn't have gone blind over the circuit.

1) yes the leds are connected backwards but I have connected connected correctly in my project, sorry about that mistake
2) I'll remove the diode and capacitor from the circuit
3) In my setup i"m using two voltage input for the VU meter board. the voltage fed to the LEDs during the transistor switching is given by the +ve output of a WiFi led controller. The voltages to the pin 3 and and the input to the 47K potentiometer is fed by a 12V buck/boost converter.
I'm already using these boards in the same project for other functions,so just used these to use on VU meter board also for a steady supply.
4)I'm using 12 V RGB led strips for the 10 outputs of the VU meter . the strips are have three individual leds per section. I'm using a continuous 4 sections for each of the outputs of the VU meter board, so that makes 4x3=12 individual leds in each row of the 10 output sections of the VU meter output line.
5) So is it over 200mA current limit?? but I wont be using it in too brightness.
6) The LM358 is a single supply op amp right?? is it capable of working in the output ranges of the of the amplifier?? in the datasheet its written that it could withstand upto 32V
7) Yes i remember some heating issues with the IC in my previous circuit based on the same IC and also after few months 3 strips were not working, but back then I thought it was normal for these ICs to heats a little on continuous use and was under the assumption that it might be the loose wiring to the leds that makes it not to light up. Now I totally understand the case. So what should be done to reduce the heating??

[Adhith]

Are the grounds the same in the circuit??

[Adhith]

just powered a similar RGB led strips and used multi meter to find the current draw. For the white light at maximum brightness it shows a total draw of 860 mA for the strip that contain 38 sections( each containing 3 leds). So 860/38 give around 23mA. I'm using a continuous 4 section of these strips in each output lines of the VU meter. So 23x4=92mA will be the total current draw through each transistors in the VU meter right??. So there is no problem of overloading the transistors right??

[Audioguru]

You are powering the LEDs from +12V so the base current of the transistors is (12V - 0.7V)/1k= 11.3mA. If all 10 transistors are turned on then the total output current from the LM3915 is 113mA and since it is powered from +12V then its heating is 113mA x 12V= 1.36W plus it operating power= more than its absolute maximum allowed heating. It is designed to drive one ordinary 3V LED from each output. Then its LED supply voltage can be 5V then if the LEDs use 10mA each the total heating is (5V - 2V) x 10mA x 10= 0.3W which makes it just a little warm. You can double the LED current to 20mA each then it gets only a little warmer but far from its maximum.

Of course in the peak detector circuit I showed all grounds are connected together and to the ground of the amplifier power supply. The opamp in my peak detector cancels the forward voltage drop of the rectifier transistor so that the lowest LEDs on the LM3915 light when they should light. The transistor allows the peak holding capacitor to charge very quickly without loading down the opamp.

[Adhith]

You are powering the LEDs from +12V so the base current of the transistors is (12V - 0.7V)/1k= 11.3mA. If all 10 transistors are turned on then the total output current from the LM3915 is 113mA and since it is powered from +12V then its heating is 113mA x 12V= 1.36W plus it operating power= more than its absolute maximum allowed heating. It is designed to drive one ordinary 3V LED from each output. Then its LED supply voltage can be 5V then if the LEDs use 10mA each the total heating is (5V - 2V) x 10mA x 10= 0.3W which makes it just a little warm. You can double the LED current to 20mA each then it gets only a little warmer but far from its maximum.

Of course in the peak detector circuit I showed all grounds are connected together and to the ground of the amplifier power supply. The opamp in my peak detector cancels the forward voltage drop of the rectifier transistor so that the lowest LEDs on the LM3915 light when they should light. The transistor allows the peak holding capacitor to charge very quickly without loading down the opamp.

So inorder to double the LED current to 20mA each i should chage the base resistors right?? from the calculations (12V-0.7V)/20mA = 565 ohms. so I should change from the 1K resistor to 565 ohm resistor right??
OK now things are clear with the peak detector circuit.  but just one thing, It is fed from the amplifier output right?? not input right??

[Audioguru]

So inorder to double the LED current to 20mA each i should chage the base resistors right?? from the calculations (12V-0.7V)/20mA = 565 ohms. so I should change from the 1K resistor to 565 ohm resistor right??

No. The LED strips have resistors inside that set the current to 23mA each but you have four sets of LEDs then their current is 92mA. The transistor base current must be 92mA/10= 9.2mA. But then the LM3915 must produce total output current of 92mA to drive the transistor bases and it overheats. In my VU meter I use a huge resistor in series with the LEDs (no transistors) to share the heat with the LM3915. you can use darlington transistors instead of transistors since their base current can be 25 times less.

OK now things are clear with the peak detector circuit.  but just one thing, It is fed from the amplifier output right?? not input right??

Feed its input from one speaker wire and ground.

[Adhith]

No. The LED strips have resistors inside that set the current to 23mA each but you have four sets of LEDs then their current is 92mA. The transistor base current must be 92mA/10= 9.2mA. But then the LM3915 must produce total output current of 92mA to drive the transistor bases and it overheats. In my VU meter I use a huge resistor in series with the LEDs (no transistors) to share the heat with the LM3915. you can use darlington transistors instead of transistors since their base current can be 25 times less.

So considering the heating issues as a major drawback I'll go with changing the transistors. One of my major concern on moving to darlingtion is the size, like I said before the vu meter was the final thing in my project thus I cant think of having larger area for the PCB. So could you suggest some commonly used darlington transistors with the same dimensions as that of the 2n3906 transistor ??

[Audioguru]

So could you suggest some commonly used darlington transistors with the same dimensions as that of the 2n3906 transistor ??

I looked in Google for "PNP darlington in TO-92 case" and found the MPSA64. Look at its datasheet to see how much base current for a reasonable saturation voltage loss.

[Adhith]

I looked in Google for "PNP darlington in TO-92 case" and found the MPSA64. Look at its datasheet to see how much base current for a reasonable saturation voltage loss.

Thank you very much sir for your sincere help. Could you please help me with these last things
1) I looked over the datasheet and got a little confused with the beta value. in the graphs two beta values are shown 100 and 1000. we usually take the lowest value right?? to avoid complications, i'm not sure of it.

2)so for considering the calculation for base resistor i just need some help too. The 10 led outputs of the LM3915N is having a 5v output right??( in the datasheet under output voltage its written 5v)
3)while calculating the voltage drop we usually take drop across the resistor and a 0.7v at the pn junction right?? since this is a darlington I should take its double right?? coz of two junctions.

below is my small calculation for the base resistor, it may be a total blunder sometimes but it would be a great help if you could read it and point out the mistakes

So assuming the beta value as 100, rounding off the maximum current for each strip to be around 150mA and base voltage as 5v and taking two junctions drop as 1.4V(0.7x2)

So for an output current of 150 mA, base current = 150/beta value= 150/100 = 1.5mA

base current = (5v-0.7v-0.7v)/Rb = 1.5mA   

So Rb = (5v-0.7v-0.7v)/1.5mA = 2400 ohms(2.4k)

so base resistor = 2.4k ohms

[Audioguru]

Beta is current gain and is used when a transistor or darlington is an amplifier with plenty of collector to emitter voltage, not when it is a saturated switch. The datasheet says that its maximum saturation voltage loss is 1.5V when its collector current is 100mA and its base current is 0.1mA but the saturation voltage loss is less when the base current is higher, use 1mA. If the saturation voltage loss is 1.5V then the LED string gets only 10.5V when the supply is 12V and it might not light. The graphs on a datasheet are for a "typical" device that you cannot buy, use minimum or maximum printed spec's.

The outputs of the LM3915 are collectors of transistors that go near 0V when activated. They have no voltage, your darlingtons provide the voltage. Their output current is adjustable but the calculations for the output current affects the reference voltage. Your supply is 12V and the darlington base-emitter voltage is about 0.8V and the output of the LM3915 is about 0.1V so for 1mA darlington base current the resistance should be (12V - 0.8V - 0.1V)/1mA= 10.2k, use 10k. The darlingtons will also need a resistor from the base to emitter to turn them off, use 10k.

The first page of the datasheet shows the reference pin7 set to about 10V with a 1.28k resistor and a 8.06k resistor. Use 1k and 6.2k instead to make a regulated reference of about 9.5V.

[Adhith]

The outputs of the LM3915 are collectors of transistors that go near 0V when activated. They have no voltage, your darlingtons provide the voltage. Their output current is adjustable but the calculations for the output current affects the reference voltage. Your supply is 12V and the darlington base-emitter voltage is about 0.8V and the output of the LM3915 is about 0.1V so for 1mA darlington base current the resistance should be (12V - 0.8V - 0.1V)/1mA= 10.2k, use 10k. The darlingtons will also need a resistor from the base to emitter to turn them off, use 10k.

The first page of the datasheet shows the reference pin7 set to about 10V with a 1.28k resistor and a 8.06k resistor. Use 1k and 6.2k instead to make a regulated reference of about 9.5V.

Thank you again sir for your kind help. So things are clear with the base resistor.
1)In the circuit which I'm using the pin 7 is only grounded through a 1K resistor. it doesent have a second resistor and its connection to the pin8 instead, pin 8 is directly grounded. Seems like it doesn't create any voltage reference i guess ?? how bad is this to the performance of the VU meter??
2) you said about that the darlington also need a resistor from base to emitter to turn them off  right?? in the circuit which I have attached there is already two resistors from the transistor. one from the base and another from the emitter and its other ends are connected to the VU meter output pins. So the resistor from base to emitter for turn off is included in this right?? So replacing the two 1k with 10K will do right??
3)I have pointed out the corrections in the circuit which is attached below. could you please have a look and see its correct? the IN4148 diode and the 2.2mf capacitor at the audio input will be removed in actual circuit

[Audioguru]

You have the regulated reference voltage not used then the input voltage that turns on each LED depends on the regulation of the power supply feeding the pot.
The current of each output can be as high as 10 x (1.25V/1k)= 12.5mA.
The transistor base current is 11.2mA and the darlington base current is 1.07mA.

[Adhith]

You have the regulated reference voltage not used then the input voltage that turns on each LED depends on the regulation of the power supply feeding the pot.
The current of each output can be as high as 10 x (1.25V/1k)= 12.5mA.
The transistor base current is 11.2mA and the darlington base current is 1.07mA.

1)I'm using a 12v regulated supply from the buck/boost converted to feed the pots, anyhow I'll use the reference voltage of 9.5v with the resistors like you have suggested to be on the safe side.
2)also the 100K and 220 ohm resistors connected to the pin 5 of the LM3915N is not needed right?? since we are feeding the pin 5 from the peak detector
I'll make the circuit today itself and let you know about the results. Thank you once again

[Adhith]

unfortunately the MPSA64 transistor was not locally available. So I had given an order for it and it would take around 5 days to reach me. So I let you know after i get it and finish buildinding my circuit.

[Audioguru]

I live in North America so I buy Canadian and American things here, not in India. Can't you look for Indian things and buy them there?

[Adhith]

I live in North America so I buy Canadian and American things here, not in India. Can't you look for Indian things and buy them there?

yes sir, I normally do buy things that are commonly avaliable here. The place where i'm coming from have very limited options in electronics sir. when I went to look for darlington today I understood that the shopkeepers haven't seen those type of transistors, even the buck boost modules was available only since last year. Things are tough here and we dont even have hobby shops, radioshack and things like that. just a few small electronic shops thats all .So the case is that if i look for an alternate component also it wont be available here sometimes. I have a good contact with a shop, where they take orders from shenzhen, china so they are usually helping me out to find the unavailable components with very cheap price due to my interest in electronics. I dont mind struggling, i just somehow need to learn electronics ,thats why

sir, what about the 100k and 200k resistors at the pin 5 of the VU meter that I mentioned in my previous post?? i dont need those right?? just feed pin 5 dirlectly from the peak detector output right??

[Audioguru]

Disconnect everything from the pin 5 input of the LM3915 and connect it to the output of my peak detector circuit.

Many cheap electronic parts from China are fakes or manufacturer's rejects. There are many videos on You Tube showing garbage put inside cheap fake lithium batteries from China.
I buy genuine North American or European parts that are inexpensive and in stock from a reputable local parts supplier.

[Adhith]

Disconnect everything from the pin 5 input of the LM3915 and connect it to the output of my peak detector circuit.

Many cheap electronic parts from China are fakes or manufacturer's rejects. There are many videos on You Tube showing garbage put inside cheap fake lithium batteries from China.
I buy genuine North American or European parts that are inexpensive and in stock from a reputable local parts supplier.

Ok sir I'll disconnect everything from pin 5 and connect to the output of the peak detector.
yes I admit that there are lots of fakes here and is a serious problem. Most of the time the shops sells two variants of IC , a cheap variant and other costs a couple of bucks extra. I always buy the good one but I know that extra money is not a surety for the original product but it seems to have a good casing when compared with the lower cost variant and thus move on with it. But the batteries sold here especially 18650s have lot of fakes with different labelled capacities than the company actually produces.

 In my previous post I mentioned that I got a 6.3V between the left channel positive output and the ground near the DC jack just after switching on the amp, what is this voltage actually??. I know that we are taping from a bridged amplifier so half of supply voltage (12v/2 = 6v) is seen,but it could be seen even without a music playing so makes me little confusing that I'm i using the wrong terminals or not. Also this peak detector could block this signal right??

[Audioguru]

All audio amplifiers that use a single positive power supply voltage bias the input so that the output is at half the supply voltage so that it can swing equally up to near the supply voltage and down to near 0V.
The peak detector has an input coupling capacitor that passes the audio but blocks the DC.

Audio amplifiers that use a positive and negative supply bias the input so that the output is at 0V and can swing equally up to near the positive supply voltage and down to near the negative supply voltage.

An ordinary amplifier with a single positive supply has DC on its output that causes trouble if it feeds a speaker so it has a coupling capacitor to feed audio to the speaker but block the DC.
A bridged amplifier has two amplifiers, each one feeds one speaker wire. Since both amplifiers in your bridged amplifier have a single positive supply then both have their outputs at half the supply voltage so a coupling capacitor is not needed to feed the speaker.