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:
[ V_\textRLO = V_\textLO - \text(offset) \quad \textand \quad V_\textRHI = V_\textRLO + \fracV_\textHI - V_\textLO10^(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
[ R2 = R1 \times \left( \fracV_\textref1.25 - 1 \right) ] Instead, set RHI lower: Use a voltage divider
Then choose ( R_\textin1, R_\textin2 ) as a voltage divider. [ R_\textset = \frac12.5I_\textLED ]
( V_\textRHI = 1.5 ) V. Check: 1.5 V peak corresponds to ~1.06 Vrms → ~0.5 dBV (close to 0 dBV).
| 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 |