constant-current Archives

AL5809 constant current driver

The AL5809 series of devices are a very convenient two-terminal constant current driver and ideal for many LED applications.

The AL5809 turns on when the voltage between IN and OUT swings from 2.5V up to 60V enabling it drive long LED chains. The floating ground, 60V Voltage rating between Input and Output pins designed to withstand the high peak voltage incurred in offline applications.

15mA, 20mA, 25mA, 30mA, 40mA, 50mA, 60mA, 90mA,
100mA, 120mA, and 150mA versions are available in PowerDI123 package.

AL5809 constant current LED driver — The AL5809 can be used as a low-side or high-side driver.

While the device looks rather like a diode, the internals are far more complex.

AL5809 internal schematic diagram — The AL5809 contains a factory-set constant current regulator with associated control circuitry.

A low drop-out (LDO) voltage regulator provides power for the internal circuitry. This will determine the minimum operating voltage of the device which is 2.5 V. This means that the minimum supply voltage required is the V^F of the diodes + 2.5 V. ( V^IN(min) = V^LED_CHAIN + 2.5 V.) A fixed preset LED current setting resistor sets the reference current of the Current regulation block. The LED current setting resistor varies with each variant of the AL5809. An accurate current mirror within the Current regulation control block increases the reference current to the preset LED current of the AL5809.

Power considerations

While the devices have a wide operating voltage care must be taken to avoid excess power dissipation. Maximum power is given in the Package Thermal Data table.

Table 1. Maximum power dissipation.

Package	P^DIS
PowerDI123	0.68 W
Power DI123 on GETEK PCB	1.24 W
SOD-123	0.36 W

Read the datasheet notes carefully to understand how these specifications were developed and, in particular, the PCB type, vias and pad size.

If we take the 0.68 W specification of the first line of the power table we can calculate the maximum voltage for each AL5809 variation.

Table 2. V^MAX across the device as a function of current assuming 0.68 W maximum dissipation.

Current	V^MAX
15 mA	45 V
20 mA	34 V
25 mA	27 V
50 mA	13.6 V
100 mA	6.8 V
150 mA	4.5 V

Devices may be connected in parallel to increase current or power.

AL5809 parallel configuration allows higher power dissipation.

PWM dimming

PWM can be used with these devices subject to the t^ON_MIN and t^OFF_MIN of 0.5 ms. At 100 Hz this would limit PWM control to 5 to 95%.

AL5809 datasheet.

LM317 constant-current power supply

The LM317 adjustable voltage regulator can be used to make a simple constant-current power supply. The device is over forty years old but is still very popular with beginners due to it’s low cost, availability and thousands of practical applications. LM317 datasheet.

Constant current

The LM317 regulates by adjusting it’s output until it is 1.25 V higher than the voltage on the adjust pin. For a constant current supply we just need to add in a series resistor to drop 1.25 V at the required current.

The LM317 can handle currents up to 1.5 A but be careful to so some power dissipation calculations and heatsink adequately if the power goes over a watt or two. (See ‘Temperature rise’ below.)

Voltage drop and headroom

For the LM317 to regulate properly it must have adequate supply voltage to allow for the sum of the voltage drops in the circuit. These are:

The minimum voltage drop across the regulator itself. This is quoted in the datasheet as the difference between input and output voltage, \( V_I – V_O \) = 3 V.
The voltage drop across R1. This is always 1.25 V.
The voltage drop across the load. For LEDs it will be \( V_f \times n \) where \( V_f \) is the forward voltage drop of each LED and \( n \) is the number of series LEDs.

Explaining the ‘headroom’ voltage

The \( I_{adj} \) reference current generator feeds 50 to 100 µA through the 1.25 V voltage reference.
The built-in Zener means that the op-amp inputs won’t equalise until the voltage on the output is 1.25 V above the adjust pin.
If the output voltage is low then the op-amp inverting input voltage will fall below that of the non-inverting input and the op-amp output will rise.
As (3) rises the Darlington transistor will turn on …
… turning on the second transistor. The Darlington arrangement will cause about 2 × 0.7 = 1.4 V of the voltage drop between the input and output due to the forward voltage of the two base-emitter junctions.
Finally the internal current sense resistor will account for most of the remainder of the voltage drop. (The op-amp may require a little more than 4, 5 and 6.)

Calculation example

Calculate the value of R1 to supply 100 mA to 5 series-connected blue LEDs with \( V_f \) = 3.1 V. The circuit will be powered from a 24 V supply.

First the resistor: \(R = \frac {V_{REF}}{I} = \frac {1.25}{0.1} = 12.5 \ \Omega \).

Now check the required input voltage:

\( V_{IN\ min} = 3 + 1.25 + 3.1 \times 5 = 19.75 \ \mathrm V \) minimum. Our 24 V supply is above this so it is fine.

We need to do one more thing: calculate the power dissipated in the LM317. This will be the voltage across the LM317 times the current:

\( P = ( V_{IN} – V_{OUT}) I = (24 – 19.75) \times 0.1 = 4.25 \times 0.1 = 0.425 \ \mathrm{W} \)

Temperature rise

We’ll take a simple approach and use the LM317’s \( R_{\theta (JA)} \), the Junction to Ambient thermal resistance parameter (and abuse it as complained about in TI’s Application Report SPRA953C). Frot the KCT TO-220 package this is 37.9°C/W. This results in a temperature rise of \(\) 37.9 \times 0.425 = 16.1°C. Even with reasonably high ambient surroundings the junction temperature will come nowhere near its 125°C maximum.

Simple constant-current driver

How it works

On power-up Q1 and Q2 are off. There is no collector current so L1 is off.
If the digital control input on the left is brought high (5 V) Q1 will turn on. Current will flow through L1, Q1 and R2.
As the voltage drop across R2 increases to about 0.6 V Q2 will start to turn on and shunt some of the base current away from Q1.
The result is that the circuit will settle at whatever Q1 emitter current will drop 0.6 V across R2.
When the input signal drops to zero Q1 and L1 turn off.
If an unswitched version is required then just connect R1 to Vbb.

Note that the usual series resistor is not present with the LED. Q1 acts as a variable resistor in this case adjusting to maintain the required current.

Example

Calculate R2 value to set L1 current at 30 mA.

\( R = \frac [V][I] = \frac [600m][30m] = 20\ \Omega\).

(We’re using ‘m’ as shorthand for ‘milli’.)