LED – Simple VU Meter with LM3915

Russell mentioned probably more ways to solve the problem than you care about, but that's Russell: always gives extensive answers :-)

The Greinacher multiplier he mentioned at the end of his answer may work, but the diodes may mean that you already need a rather high input voltage, I don't know if it will work well on the line output. (The Greinacher is typically used to multiply power supply voltages, not signals. It can be used to get kilovolts from a 110V or 230V line voltage, for instance.)

Russell also mentioned this opamp amplifier:

This is one of the basic opamp circuits, and is easy to understand if you know some basic things about opamps:

• The ideal opamp has infinite amplification
• The output signal is the amplified difference between the inputs
• The ideal opamp has zero input current

The first two items imply that you'll get a finite output signal if both inputs are almost at the same level. And that's what an opamp in a negative feedback system will do: it will try to make the level on the - input (called the inverting input) the same as on the + input (the non-inverting input). So that will be \$V_{IN}\$, and the current through R1 is

\$I_{R1} = \dfrac{V_{IN}}{R1}\$

Since there doesn't flow current in the - input the current through R1 is the same as the current through R2, so the voltage over R2 is

\$V_{R2} = I_{R1} \times R2\$

and therefore

\$V_{OUT} = V_{IN} + \dfrac{V_{IN}}{R1} \times R2\$

or

\$V_{OUT} = (1 + \dfrac{R2}{R1}) \times V_{IN}\$

So, if we choose R1 = 1k\$\Omega\$ and for R2 a 100k\$\Omega\$ potentiometer, we can vary the amplification between 1x and 101x. This will be a good amplification for a line out signal, and you can set the amplification to the LM3915's full scale.