One light emitting diode display circuit

Light-emitting diodes (LEDs) can be powered using direct current (DC), alternating current (AC), and pulse drivers. A typical buck LED display circuit is illustrated in Figure 13-1. The current-limiting resistor (R) for LED tubes can be calculated using the following formulas. For DC drive applications (refer to Figure 13-1 (a) and (b)), the formula for R is R = (E - UF) / IF, where E is the power supply voltage in volts (V), UF is the forward voltage drop of the LED (typically between 1.7V and 2V), and IF is the LED current, generally set to a maximum operating current (IFm) of 1/3 to 1/5 of the rated current in milliamperes (mA). The power rating (Pr) of the resistor should exceed the actual power consumption to prevent overheating. For AC drive applications (see Figure 13-1 (c) and (d)), the formula is R = 0.45 * (U - UF) / IF, where U is the RMS AC voltage in volts (V). A diode (VD), such as the 1N4007 rated at 1A and 1000V, is utilized to protect the LED from negative half-cycles of the AC signal. For pulse drive applications (refer to Figure 13-1 (e)), the formula is R = E / (2 * (UCE + UF) * IF), where UCE is the voltage drop across the transistor in saturation (approximately 0.3V).

The described circuit utilizes light-emitting diodes (LEDs) in various configurations depending on the type of power supply. In DC applications, the circuit design incorporates a current-limiting resistor to ensure that the LED operates within its specified current range, preventing damage due to excessive current. The resistor value is derived from the difference between the supply voltage and the LED's forward voltage drop, divided by the desired forward current.

In AC applications, the circuit must accommodate the alternating nature of the power supply. The current-limiting resistor is adjusted to account for the effective (RMS) voltage of the AC source. The inclusion of a protective diode is critical, as it safeguards the LED from reverse polarity conditions that occur during the negative half-cycle of the AC waveform.

For pulse-driven configurations, the circuit is designed to handle short bursts of current, necessitating a different approach to calculating the resistor value. The transient response of the circuit must be considered, particularly the voltage drops across the transistor in saturation and the LED itself, to ensure proper operation during the pulse.

Overall, the design of the LED display circuit requires careful consideration of the power supply type and the characteristics of the components used to maintain the longevity and reliability of the LEDs. Proper calculations for resistor values and the selection of protective components are essential to achieve optimal performance in each application scenario.Light-emitting diodes can be used DC, AC and pulse and other drivers, resistance typical buck LED display circuit as shown in Figure 13-1. LED tubes reported the string loop current limiting resistor R can calculate the following formula; O DC drive [see Figure 13-1 (a), (b)] when R = EU, P 21FR formula R- limiting resistor , KN; E power supply voltage, V; UF-emitting diode forward voltage drop (as 1.7-2V), V; IF-emitting diode current, generally the maximum operating current IFm for 1 / 3-1 / 5, mA; resistor power Pr selected, the actual power consumption is Wei R, in order to prevent overheating of the resistance, to be more than sufficient quantity, w. AC drives [see Figure 13-1 (c), (d)] when R = 0 45U-UF IF formula U- RMS AC voltage, V. Diode VD is used to protect the light emitting diode AC negative half weeks when the breakdown is not, the choice 1N4007,1A / 1000V.

pulse drive when there [see (e)] R = E two (UCEs + UF) fF formula B a power supply voltage, V; U. . . - Tube voltage drop transistor saturated conduction time (about 0. 3V), V.