Switching circuit for driving LED at different current values

Question

^{simulate this circuit – Schematic created using CircuitLab} Above schematic is for limiting current to 4A when battery is connected to LEDs.

Datasheet of LED(BXRA-56C5300H 50W) are 23.5 Vf and 2.1 A typical current. Led max rated current is 4 A @ 28.5 V only for pulsing at 10% duty cycle. My actual project has 64 such LEDs

Current design

Powering 32 such LED with DC power supply of 24V 67A. Limiting current to each parallel connected LED to 2.1 A using current limiting resistor of 1 ohm 5W rating. LEDs are mounted on heat sink

New design

Requirement is to drive each LED string at Max rated current of 4A when required and then switch back to regular operation which is described in current design above.This switching should be digitally controllable.I understand my current DC power supply cannot supply so much of high current.So I was thinking to use another power source.I have in mind to use SLA(car batteries).

Since LEDs are connected in parallel each LED will required its own current limiting circuit so equal amount current is supplied to each LED.

Can anyone suggest me circuit for achieving this digitally switching between current limiter circuit of 4A(shown above) and 2.1A?

I simulated this circuit in proteus and I was getting 3.8 A flowing through LED string. But I am still figuring out how can I switch between this limited current to my current setup of 2.1 A . I need to run entire array of LEDs @2.1A most of the time and only on requirement need to switch to 4A . So I need to switch between two these two power supplies.

Answer 1

LEDs are 23.5 Vf and 2.1 A typical current.

Typical isn't good enough in this case. A look at the I/V curve shows that \$ V_f \$ varies at 2.1 A from about 20.8 V to > 26 V.

Figure 1. \$ V_f \$ for the 5300 series lamps.

There are four curves given from MAX Vf through "typical" to MIN Vf all at 25°C. Then there is the red line at 70°C.

The green load-line represents your 1 Ω resistor. It will drop 1 V/A so at 4 A the voltage will have dropped to 20 V. The voltage and current for each LED curve can be estimated by the intersection of the curve with the load-line.

1. With worst-case maximum Vf you can expect 0.4 A.
2. The typical case would draw 1.3 A.
3. The minimum Vf at 25°C would draw 2.1 A.
4. As the LEDs heat up to 70°C the current increases to 2.6 A. This, of course, will make the LEDs hotter, the Vf will drop and the current increase until something gives up - the PSU, the resistor or the LEDs.

The problem is that you haven't enough headroom for a simple resistive current limiter. If you had several more volts available the load-line wouldn't be so steep and the current wouldn't vary so much.

Note also that you are assuming your batteries will give constant voltage. They won't!

You need dimmable constant current power supplies.

Figure 2. A typical mains-powered constant current power supply.

Figure 3. The dimmable constant-current power-supply control circuit. Source: Dimmable mains PSU control | LEDnique.com.

Many of the LED power supplies such as those by Mean Well, etc., offer three modes of control: output constant current level can be adjusted through the control input

• by connecting a variable or fixed resistance to the control terminals,
• using a 0 ~ 10 V DC control signal,
• using a 10 V PWM signal between.

In your case you would purchase a PSU model that would supply the required peak current and run it normally in dimmed mode (reduced current).

I haven't checked all the details for you so you have some engineering to do. In particular, make sure that the current can be controlled over the range of Vf voltages you are interested in.

I've written more on the subject in the article lined above. Load-line resistance graphic tool may also be of interest in explaining in more detail how to use the load-line - although the article deals with lower voltage circuits.

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^{simulate this circuit – Schematic created using CircuitLab}

Figure 4. Op-amp control.

Try something like Figure 4. The 2.1 A input can be connected to 3.3 V permanently. The 4.0 A input can be connected to a GPIO. Select the resistors to give you about 300 mV at the non-inverting input per amp of LED current.

Answer 2

I see some problems with your approach. With a 24V power supply, 23.5 Vf and 1 ohm resistor you are only going to get 0.5 Amps, not 2.1. Also Vf is a typical value and slight changes in Vf could result in big changes in current and brightness because Vf is so close to yout 24 V power supply. You really need a an LED driver, Linear Tech is a good place to start. FET switches could be used to switch between the two power sources, but you may find an LED driver that can achieve what you want with a single power supply.

Answer 3

You can't put 50W Leds in parallel because there is no current regulating device for such powerful Leds. Your resistor will not make it: it's rated 1/10 of the Led power and 1 Ohm regulates almost nothing. Transistor explained that more scientificaly with the chart. Each Led must have its own AC/DC constant current supply, and even two if you want to switch different currents as there is no other way to control current at these power levels. Each supply being monitored with a relay. Also I wouldn't use the LEDs at or near maximum current. 3A should be the maximum in your case, and for a very short time, one minute or so and with an X-large heatsink. 3A is 75W. Also consider that you don't put 64 such LEDs next to each other, so add to this cables with monstruous diameter. Were your LEDs be 10 times smaller, it would be another story.