Lm3915 Calculator May 2026

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

Choose R1 = 1.2 kΩ. ( R2 = 1200 \times (5.0 / 1.25 - 1) = 1200 \times (4 - 1) = 3600 \ \Omega ) (3.6 kΩ). LM3915 Calculator

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: Example: To set ( V_\textref = 2

| 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 | Thus, we need gain , not attenuation

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: