Buck convertor calculation on Vout close to Vin

Question

I'm currently stuck in calculating the right inductor for driving very bright LED's with The TPS92518 led driver. (http://www.ti.com/product/TPS92518-Q1)

This is a 2 channel LED driver and on every channel I want to drive two of the following LEDs in series. http://www.farnell.com/datasheets/573490.pdf

One LED has 10.2V voltage drop and are rated 640mA max. So two LEDs in series need at least 20.4V

Since my supply voltage is 24V my Vin en Vout of the bucket convertor are very close to each other.

So I end up with a duty cycle of: D = 22 / (24*0.9) = 1.02 (0.9 for efficiency factor)

Basically 102% duty cycle.

I want to be able to analog dim the LEDs and since the LEDs are used in an image recognition setup the switching frequency needs to be >300kHz

How does this work? Am I working with the wrong chip here? Because Vin and Vout are almost equal this is not really a buck convertor right? Or is this called a constant voltage buck convertor?

Does it control the current by changing the switching frequency?

Answer 1

The LED datasheet mentions: Forward voltage @ 640mA, 25°C: minimum 9V, maximum 11.5V. Therefore, two LED modules in series will have a Vf between 18V and 22V.

Let's suppose we have a 24V +/- 5% supply, its minimum voltage is 22.8V.

The Buck driver supports high duty cycles and long ON-times for the FET, but it is a bootstrapped high side driver, so it can't be ON continuously. It has to switch to recharge the bootstrap cap. Let's assume a max duty cycle of 99.5% (datasheet paragraph 8.3.7.2).

Min supply voltage & Max LED voltage: dropout voltage is 24 * 0.95 * 0.995 - 22 = 0.69V.

Max supply voltage & Min LED voltage: dropout voltage is 24 * 1.05 * 0.995 - 18 = 7V.

The buck driver's input and output voltage are close enough that we can assume input current equal to output current. Thus, in this 0.69V margin we must fit all the resistive losses: wires from the supply, MOSFET RdsON, inductor resistance, wires to LED, plus extras. At 640mA, it is doable: the MOSFET's RdsON and inductor resistance should easily be below 0.1R. So, it should work.

Switching frequency is calculated from \$ V = L \frac{di}{dt} \$. Pick an inductor ripple current value. The inductor current goes up when the FET is ON, and goes down by the same amount when the FET is OFF. This allows to calculate inductor value and frequency.

Switching frequency will be low. What is the OFF time?

With a 640mA current, let's go with a 320mA ripple. When the FET is OFF, LED voltage is across the inductor, so 18-22V. With The chip can run at 2 MHz, and I can't find minimum off time information in the datasheet, so... I'll pick dt = 500ns. Thus L = 20V * 500nv / 320mA = 34µH, we could go with 33µH or 47µH, at this current a 47µH inductor is easy to get, so let's pick that. So, with an OFF-time of 500ns, we have a current ripple of 234mA.

What is the ON time? That depends on the voltage across the inductor when the FET is ON, which depends quite a bit on resistive losses since the dropout voltage can be quite low.

Worst case: suppose we have 0.2V across the various resistive losses, thus 0.49V remains to crank up inductor current when the FET is ON. In this cas ON-time is 22.5µs, so frequency is about 45kHz.

Best case, with max power supply voltage and min LED voltage, gives about 480 kHz.

Conclusion:

If you need very stable LED output for your imaging applications (to avoid stripes on your images for example) then, if you're unlucky and get LEDs with high Vf, or a supply that is a bit low, you won't get the frequency you mention in the question.

This might need a capacitor across the LED to smooth out the current, or a power supply that can be adjusted to make sure there is enough voltage headroom, or gettin glocky with the LEDs.

Why not use a dimmable AC to DC LED driver instead? Although these don't have fast ON/OFF capability, so if you need to blink the LED quickly, that won't work. But if you just need to light it, it's cheap and pretty simple.