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Help with LM 3915 VU Meter

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haxor12

New Member
I'm trying to make a VU meter using a lm3915 dot/bar driver using this circuit:

8472-14.gif


A few questions:

* This schematic is recommended by NI, but it appears some people modify it:
* This meter looked like it works well: YouTube - Indikátor vybuzení s LM3915 but the schematic is quite different... Google Translate If I followed that schematic, do you think everything would still work?

Thank you for your help!

Joe Green
 

audioguru

Well-Known Member
Most Helpful Member
You should use a peak detector circuit at the input so that momentary peak voltages light the LEDs long enough to appear bright, instead of the dim blur or not seen.
These circuits have peak detector circuits added. The peak detector circuits are shown in the datasheet of the LM3915.
 

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haxor12

New Member
Thanks, but why? The video shows that one working just fine so I guess I don't see the need for more parts? Or is this circuit significantly better?
 

audioguru

Well-Known Member
Most Helpful Member
The LEDs for the max signal level are frequently very dim or are not lit.
I use a peak detector circuit in all my VU meter circuits and all the LEDs are the same brightness including ones that are pulsed for only a moment.
 

audioguru

Well-Known Member
Most Helpful Member
The one with the opamp has gain if you need it. The MC34071 or MC33171 opamp must be used.
 

BigE4u

New Member
I made one myself not too long ago using the LM3915, not my schematic, but i use it and not pictured, i added a 68Ω 1/2w resistor between the anode end of the LEDs and the positive power source. I also substituted the shown 2.7k resistor for a 1.2k resistor. For demo purposes due to the size limitation of the PCB, i used 3mm LEDs, hooked up for single channel, runs off a +9v battery and i incorporated an off/on switch.

EDIT: Video: YouTube - vu meter for pc project #3
 

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audioguru

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Most Helpful Member
I made one myself not too long ago using the LM3915
The passive peak detector using a diode ruins the low signal levels that have a voltage less than 0.6V peak so the circuit does not show the low signal levels.

The datasheet shows a peak detector that uses a PNP transistor to cancel the voltage loss of the diode.

I have made peak detectors with a rail-to-rail opamp that also cancels the voltage loss of the diode.
 

BigE4u

New Member
The passive peak detector using a diode ruins the low signal levels that have a voltage less than 0.6V peak so the circuit does not show the low signal levels

Unfortunately true... but i wanted to use as little parts/space as possible due to the size of the PCB i used and it works well for me. For the moment, my circuit is hooked up to the "lineout" of the sound card and to get more than enough signal strength, i max the sound in windows. With the 10k pot, i can always adjust the sensitivity to the IC so the LEDs are not peaking half the time causing a quicker battery drain. I use amplified stereo speakers that has volume control so it doesnt matter and i prefer using the preamped output because the vu meter still displays no matter what volume level on the speakers. I may be wrong but im sure this circuit was more designed towards hooking up directly to speakers wheres theres more peak power to overcome the 0.6v drop of the diodes.

Ps- I'll be looking forward trying the gain 10 peak detector circuit you posted next.
 

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audioguru

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The peak detector with an opamp can have any amount of gain from less than 1 to 300 if the opamp is a wider bandwidth MC34071.
 

BigE4u

New Member
Sounds good... i think i'll give the transistorized peak detector a try 1st since i have those parts available over the op amp version.

Ps- I currently have the 471 op amp... can i still use the other parts listed along with the substituted op amp?
 

audioguru

Well-Known Member
Most Helpful Member
There is no 471 opamp. But the lousy old 741 opamp is horrible for audio and will not work in that circuit.
 

haxor12

New Member
Yeah...the peak detector circuit doesn't work. The one with the transistor, not the other one. Just causes the chip to get super hot and light up most of the LEDs. I could've set it up wrong, but I triple checked...
 

BigE4u

New Member
The IC is dissapating too much power when everything is going full tilt, you need a 68Ω 1/2w resistor between the anode ends of the LEDs and the positive power source.

Ps- Adjust R2(10K) for less sesitivity so you dont have full display all the time.
 

audioguru

Well-Known Member
Most Helpful Member
Yeah...the peak detector circuit doesn't work. The one with the transistor, not the other one. Just causes the chip to get super hot and light up most of the LEDs. I could've set it up wrong, but I triple checked...
The transistor peak detector circuit works perfectly if the transistor and diode are not connected backwards. If the wrong diode is used (a 1N400x rectifier) then a few LEDs might light without an input signal. A 1N4148 diode is the same as a 1N914 diode.

The chip does not get too hot if its supply voltage is 9V and less and the resistor from pin 7 to ground is 1k ohms.

EDIT: Here is the schematic:
 

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audioguru

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The IC is dissapating too much power when everything is going full tilt, you need a 68Ω 1/2w resistor between the anode ends of the LEDs and the positive power source.
No.
68 ohms will have 8.16V across it when all LEDs are lighted. Then the resistor dissipates 0.98W and it will smoke.

Ps- Adjust R2(10K) for less sesitivity so you dont have full display all the time.
R2 has nothing to do with sensitivity. The pot reduces sensitivity.
 

BigE4u

New Member
I didnt have a 68Ω 1w resistor handy but i did have a 1/2w one handy, so i used it and it worked.

adjust for less sensitivity and pot reduces sensitivity seems like they do the same thing to me.
 

audioguru

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Most Helpful Member
I didnt have a 68Ω 1w resistor handy but i did have a 1/2w one handy, so i used it and it worked.
then maybe your supply was lower than 12V.

adjust for less sensitivity and pot reduces sensitivity seems like they do the same thing to me.
No.
R2 in the peak detector ciruit is the emitter resistor of the transistor which is an emitter-follower with a gain of 1. R2 does not affect sensitivity.
 

haxor12

New Member
I used a 1N4148 and i double checked it was placed in the proper direction...
Supply was 13v
I'll try building the circuit again...
 
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