Lm3915 Calculator Apr 2026

[ V_\textRLO = V_\textLO - \text(offset) \quad \textand \quad V_\textRHI = V_\textRLO + \fracV_\textHI - V_\textLO10^(9/10) ]

( V_\textRHI = 1.5 ) V. Check: 1.5 V peak corresponds to ~1.06 Vrms → ~0.5 dBV (close to 0 dBV).

RLO = 0 V (ground). RHI = 5.0 V (to reference). But now the highest LED triggers at ( V_\textin \approx 5.0 ) V peak? That’s far above 1.414 V. So we must attenuate input.

0 dBV = 1 Vrms → peak = 1.414 V. -30 dBV = 0.0316 Vrms → peak = 0.0447 V. LM3915 Calculator

| Problem | Consequence | |---------|--------------| | Choosing R1/R2 for a specific full scale | Incorrect clipping level | | Converting dBu or dBV to required input voltage | Mismatch with line-level audio | | Setting RLO/RHI for offset display (e.g., -20 dB to +10 dB) | First LED never lights | | Resistor selection for precise 1 mA/LED | Burnout or dim display |

[ V_\textin,peak = \sqrt2 \times V_\textrms ]

Then choose ( R_\textin1, R_\textin2 ) as a voltage divider. [ R_\textset = \frac12.5I_\textLED ] [ V_\textRLO = V_\textLO - \text(offset) \quad \textand

From Vref = 5V to RHI = 1.5V: Use voltage divider between pin 7 and ground, middle to pin 4. Choose Rtop = 10 kΩ, Rbottom = 4.285 kΩ (approx 4.3k).

For a desired max LED at ( V_\textin,peak = V_\textmax ):

But for simplicity, designers often set ( V_\textRLO = V_\textLO ) and ( V_\textRHI = V_\textref ) (if ( V_\textref ) is scaled to match highest LED threshold). More practically: The LM3915’s internal divider has a ratio of ~1.25 dB per step in voltage terms, so the voltage at step n is: RHI = 5

( R_\textset = 12.5 / 0.015 = 833.3 \ \Omega ) → use 820 Ω.

Typically ( R1 = 1.2 \textk\Omega ) (recommended min). Example: To set ( V_\textref = 2.5 \textV ), ( R2 = 1200 \times (2.5/1.25 - 1) = 1200 \times 1 = 1.2 \textk\Omega ). If the lowest LED lights at ( V_\textin = V_\textLO ) and the highest at ( V_\textin = V_\textHI ), then:

[ \textAttenuation factor = \fracV_\textref,desiredV_\textmax ]

Choose R1 = 1.2 kΩ. ( R2 = 1200 \times (5.0 / 1.25 - 1) = 1200 \times (4 - 1) = 3600 \ \Omega ) (3.6 kΩ).

[ V_\textth,n = V_\textRLO \times 10^(n-1)/10 \times \fracV_\textRHIV_\textRLO \times 10^9/10 ]