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how to isloate audio ground from the amplifier
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Adhith:

--- Quote from: Audioguru on March 26, 2018, 09:48:13 pm ---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.

--- End quote ---
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:

--- Quote from: Adhith on March 27, 2018, 05:07:52 am ---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??
--- End quote ---
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.


--- Quote ---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??
--- End quote ---
Feed its input from one speaker wire and ground.
Adhith:

--- Quote from: Audioguru on March 27, 2018, 03:55:24 pm ---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.

--- End quote ---
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:

--- Quote from: Adhith on March 27, 2018, 04:50:37 pm ---So could you suggest some commonly used darlington transistors with the same dimensions as that of the 2n3906 transistor ??
--- End quote ---
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:

--- Quote from: Audioguru on March 27, 2018, 05:40:48 pm ---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.

--- End quote ---
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


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