Lm3915 Calculator 95%

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:

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

Example: For 20 mA (typical bright LED), ( R_\textset = 12.5 / 0.02 = 625 \ \Omega ). Use 620 Ω standard. Design goal: Audio level meter for -30 dBV to +6 dBV (36 dB range, but LM3915 only does 30 dB, so compress or shift). Desired: LED1 = -30 dBV, LED10 = 0 dBV (30 dB span). Reference voltage = 5.0 V (from 12V supply). LED current = 15 mA.

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 |

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

| 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 | Typically ( R1 = 1

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

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.

[ \textAttenuation factor = \fracV_\textref,desiredV_\textmax ] Use 620 Ω standard

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

Then choose ( R_\textin1, R_\textin2 ) as a voltage divider. [ R_\textset = \frac12.5I_\textLED ]

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