LED阵列电源参数计算求助：确定最优供电电压方法

Hey Dylan, great question—calculating the right power supply for your LED array is way more reliable than just tinkering with an adjustable supply, so let’s break this down clearly, step by step.

Your 1W LEDs have a rated current of 700mA (0.7A), but you’ll also need their forward voltage (Vf)—this is non-negotiable for accurate calculations. For most 1W white LEDs, Vf sits between 3.2V and 3.4V at 700mA; if you’re using colored LEDs (red, blue, etc.), this number will shift (e.g., ~2V for red). Pro tip: Measure the Vf of a few of your actual LEDs at 700mA with a multimeter—nominal specs can have small tolerances that impact performance.

You mentioned "普通并联LED串" (regular parallel LED strings), so I’ll break down the two most likely setups you’re working with:

Scenario 1: Series Strings in Parallel (Most Reliable Array Setup) #

This is the standard for LED arrays: each "column" is a string of N LEDs wired in series, and all 14 columns are wired in parallel. Here’s how to calculate:

1. Calculate total forward voltage per column:
   V_column = N × Vf
   (Example: If each column has 2 LEDs, V_column = 2 × 3.3V = 6.6V, using a middle-of-the-road Vf of 3.3V)

2. Choose your optimal supply voltage:
   You need a voltage slightly higher than V_column to leave room for a current-limiting resistor (required to prevent overcurrent from LED Vf tolerances). Aim for 1-2V above V_column—this balances efficiency (minimizing resistor power loss) and reliability (ensuring consistent current across columns).
   (Example: For V_column=6.6V, pick an 8V or 9V constant-voltage supply)

3. Calculate the current-limiting resistor for each column:
   R = (V_supply - V_column) ÷ If
Power rating of R = (V_supply - V_column) × If
   (Example: 8V supply, 6.6V column, 0.7A current: R = (8-6.6)/0.7 = 2Ω; Power = 1.4×0.7 = 0.98W → use a 2W rated resistor to be safe)

4. Total power supply requirements:
   Total current = 14 columns × 0.7A = 9.8A
   Minimum supply power = 9.8A × V_supply = 9.8×8 = 78.4W → pick a 100W supply to leave a safety margin

Scenario 2: All LEDs in Direct Parallel (Regular Parallel Strings) #

This is when all 14 LEDs are wired directly in parallel (no series strings). Note: This is less reliable because LED Vf tolerances will cause current to be uneven (LEDs with lower Vf will draw more current and burn out faster). But if you’re using this setup:

1. Choose supply voltage:
   Pick a voltage 0.2-0.3V above your LED’s Vf (e.g., 3.5V for LEDs with Vf=3.2-3.4V)

2. Add a current-limiting resistor per LED:
   (Do NOT use a single shared resistor—this will make current imbalance worse!)
   R = (V_supply - Vf) ÷ If
Power rating of R = (V_supply - Vf) × If
   (Example: 3.5V supply, 3.3V Vf, 0.7A current: R = (3.5-3.3)/0.7 ≈ 0.286Ω → use a 0.3Ω resistor; Power = 0.2×0.7 = 0.14W → a 0.25W rated resistor works)

3. Total power supply requirements:
   Total current = 14 × 0.7A = 9.8A
   Minimum supply power = 9.8×3.5 ≈ 34.3W → pick a 40W+ supply

LEDs are current-driven devices, so a constant-current supply is more efficient and reliable than constant-voltage + resistors. For your 14-column series-parallel array:

• Total current = 14 × 0.7A = 9.8A
• Supply voltage range needs to cover the minimum and maximum V_column (e.g., 6.4V to 6.8V for 2-LED columns with Vf=3.2-3.4V)
• Pick a 9.8A constant-current supply with a voltage range of 6-10V—no resistors needed, and current stays consistent even with LED Vf variations.

内容的提问来源于stack exchange，提问作者Dylan144GT