Frankenstein audio booster circuit

[captainate]

Hey all,

I'm trying to adapt a circuit for a basic ~27 dB boost intended for use in a guitar pedal. I would like to build it, with these modifications: on/off LED (easy), and 5 LED VU meter, which reads the OUTPUT (post-boost effect) at a reasonable level (highest input volume from unamplified guitar signal reaching red LED at max boost). The VU meter would be a subjective measure (strictly serving the purpose of quick-glance notification of boost level) and will probably require an internal trim potentiometer for fine tuning.

The only LED VU schematics I could find were for 10 LED's, and I assumed (perhaps wrongfully) that simply leaving the undesired LED's out would result in the desired effect, maintaining the viability of the circuit.

Please help me understand if this project is feasible, or if its value does not exceed its cost (both monetarily and time-wise). I have attached the schematics I would like to use and a visual guide for the plan. Thanks,

~Nate

 

[captainate]

Anybody? I'd like to build this soon... thanks,

~Nate

 

[PCBWING]

Design is very attractive, I think is feasible.

 

[3lectrokid]

For the VU meter i would take out every other LED so you would leave only LED 1,3,5,7,9.

 

[Nigel Goodwin]

For the VU meter i would take out every other LED so you would leave only LED 1,3,5,7,9.

As long as you use bar and not dot!

 

[3lectrokid]

Oh yeh. I forgot about that

To put the VU display into bar mode you must connect pin 9 to pin 3. Otherwise the led will no show on half the levels

 

[Nigel Goodwin]

You could still do it, a couple of diodes for each LED to make an OR gate.

 

[audioguru]

The circuit does not have a -9V supply, it has only a +9V supply and 0V.
The resistors at pin8 of the LM3916 set its 10th LED to light when its input is +9.1V peak and when its supply is at least +10.6V. The resistors ratio should be reduced so that the LEDs light at lower levels.

Without a peak detector circuit the LEDs will appear as a dim blur.
I use this peak detector circuit on my Sound Level Indicator project:

 

[captainate]

The resistors ratio should be reduced so that the LEDs light at lower levels.

How does that math work?

 

[audioguru]

The resistor between pin 7 and pin 8 has 1.25V across it and its current sets the current in the LEDs according to the graph about the LED Current vs Reference Loading .
Then the resistor from pin 8 to ground has the same current in it that is in the first resistor.
Then simply use Ohm's Law and simple arithmatic to see the voltage at pin 7.

For example, if the first resistor is 560 ohms and the second resistor is also 560 ohms then the 10th LED lights when the input is 2.5V. The current in each LED is 20mA.

Another example is the first resistor is 560 ohms and the second resistor is a piece of wire.
Then the 10th LED lights when the input is 1.25V and the current in each LED is also 20mA.