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LED VU meter - two questions

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drabina

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I am going to build a VU meter based on the following schematic by audioguru:

vu meter schematic

I took it from this post.

My knowledge about electronics is very limited but enough to be able to actually convert this schematic to working project.

I have two questions though:

1. Is there a way to add switch to pin 9 to enable dot mode on the LM3915 or schematic would have to be redesigned?

2. If I want to connect two LEDs to each channel (in series) so my VU meter is brighter, would that work without making any changes?

Thanks.
 
I am going to build a VU meter based on the following schematic by audioguru:
1. Is there a way to add switch to pin 9 to enable dot mode on the LM3915 or schematic would have to be redesigned?
Look on the datasheet for the LM3915. It shows the DOT mode occurs when pin 9 is disconnected from the positive supply.

2. If I want to connect two LEDs to each channel (in series) so my VU meter is brighter, would that work without making any changes?
The datasheet shows (the output dropout voltage) that outputs of the LM3915 need a supply voltage that is at least 1.5V more than the LED voltage. So if you use a 9V regulated supply (not a 9V battery that drops to only 6V over its life) then two 3.7V blue LEDs can be in series at each output.
 
Thanks. I was a little confused when I read in another post that in dot mode the chip will run hot. I guess I can add heatsink to it if required.

I guess I forgot about one question. If the leads to the LEDs will be about 2.5 feet long (longest) and 10" (shortest), would I need to modify the value of the 100uF capacitor?
 
Thanks. I was a little confused when I read in another post that in dot mode the chip will run hot. I guess I can add heatsink to it if required.
In the bar mode the chip might overheat when all 10 LEDs turn on with 18mA each (total current is 180mA) and with 8.2V across the LED drivers (10V supply - 1.8V red LEDs) resulting in heating of 180mA x 8.2V= 1.5W. The max allowed power is 1.37W when the chip is 150 degrees C inside. It cannot easily be heatsinked. Instead, simply reduce the supply voltage to the LEDs or add a resistor in series with all the LEDs to share the heat.

I guess I forgot about one question. If the leads to the LEDs will be about 2.5 feet long (longest) and 10" (shortest), would I need to modify the value of the 100uF capacitor?
The datasheet discusses a few ways to prevent oscillation of the outputs if the LEDs are away from the IC in the Application Hints section.
 
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I read thru the datasheet but since most of it is way beyond my knowledge, I still have some questions. Given that I will use 10 LEDs (not 20) with 3.5V forward voltage and 20mA current:

1. What resistor value should I use? Looks like this combination will result in 1.64W of heat generated by the controller. Is there a formula that I could use to calculate the resistance needed?

2. The decoupling capacitor as stated in the datasheet should be between "0.05 µF to 2.2 µF". Given that I will stick with 1 LED per pin and the longest wire from Vled to anode will be 3 feet (down to 10" for the closest one), how do I calculate the actual value of the capacitor?

Sorry if those questions are very basic but like I stated, my electronics knowledge is very limited.

Thanks.
 
I will use 10 LEDs (not 20) with 3.5V forward voltage and 20mA current:

1. What resistor value should I use?
The 560 ohm resistors set the LED currents to be about 18mA each. If you have a 9V regulated supply then the current in the LEDs and output transistors is 180mA when 10 LEDs are turned on and the voltage across the output trabsistors is 9V - 3.5V= 5.5V. A resistor with 2.5V across it and 180mA in it is 2.5/0.18= 13.9 ohms. Use 15 ohms/1W to feed the voltage to the LEDs. The resistor will heat with 0.18A squared x 15 ohms= 0.49W and a 1W resistor will get pretty warm. The IC outputs will have 9V - 3.5V - 2.7V= 2.8V across them so thier max heating is 0.18A x 2.8V= 0.5W which is fine.

2. The decoupling capacitor as stated in the datasheet should be between "0.05 µF to 2.2 µF". Given that I will stick with 1 LED per pin and the longest wire from Vled to anode will be 3 feet (down to 10" for the closest one), how do I calculate the actual value of the capacitor?
You might need only one 2.2uF capacitor at the 15 ohm resistor or you might need ten 2.2uF capacitors with one at each output pin of the IC. Try it to see.
 
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Thanks a lot. I think I am ready now to build the circuit. I will report back when I have it up and running.
 
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