Lm3915 Calculator May 2026

[ R2 = R1 \times \left( \fracV_\textref1.25 - 1 \right) ]

| Parameter | Formula | Standard value example | |-----------|---------|------------------------| | ( R_\textset ) | 12.5 / I_LED | 620 Ω for 20 mA | | ( V_\textref ) | 1.25 × (1+R2/R1) | 5.0 V: R1=1.2k, R2=3.6k | | LED step voltage (n from 1 to 10) | ( V_\textRLO \times 10^(n-1)/10 ) (if RHI/RLO = 1:0 ratio) | Step 6: ×3.16 from step 1 | | Power (bar mode) | ( 10 \times V_\textLED \times I_\textLED ) | 10×2V×0.02A = 0.4W |

[ V_\textth,n = V_\textRLO \times 10^(n-1)/10 \times \fracV_\textRHIV_\textRLO \times 10^9/10 ] LM3915 Calculator

A dedicated calculator solves these with direct equations. 4.1 Reference Voltage Divider (R1, R2) Given desired ( V_\textref ):

Desired input at pin 5 for LED10 = 5.0 V (peak). Actual peak input = 1.414 V. Thus, we need gain , not attenuation. Instead, set RHI lower: Use a voltage divider from Vref to set RHI = 1.5 V (peak). Then: [ R2 = R1 \times \left( \fracV_\textref1

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

Then choose ( R_\textin1, R_\textin2 ) as a voltage divider. [ R_\textset = \frac12.5I_\textLED ] Thus, we need gain , not attenuation

However, the standard application simplifies by setting ( V_\textRHI = V_\textref ) and ( V_\textRLO = 0 ) for ground-referenced input. For line-level audio (e.g., 1.228 Vrms = +4 dBu), an input voltage divider is needed before pin 5: