Tip / Sign in to post questions, reply, level up, and achieve exciting badges. Know more

## TLD5045EJ - Inductor selection and min drop voltage

Level 1
Level 1

Two Questions for the Litix TLD5045EJ that I could not find in the datasheet:

a) how do I select the inductor value (the 220uH mentioned in some sections seems quite high especially for the higher frequencies)

b) what is the min dropout voltage, or the max LED voltage at say 12V input and 700mA LED current?

Thanks

KPN

1 Solution

## Re: TLD5045EJ - Inductor selection and min drop voltage

Moderator
Moderator

Hello Peter,

with 4red  LED (2.5 V @ 700 mA) you have still room for this buck (about 2 V if the minimum Vin is 12 V).

here the steps I suggest you to design your LED driver:

1. set your peak current by choosing RSET (9.5k fits well)
2. Select your inductor to fulfill the ripple requirements (220 uH at 100 kHz generates +/-40 mA ripple over 700 mA average output)
3. Adjust Rfreq, Rcomp and Cfreq fo achieve 100 kHz switchng frequency (Rfreq = 7.3k, Rcomp = 100k, Ccomp = 100n fit 100 kHz switching)

You can try our evaluation board and play with it to see how powerful TLD5045EJ is!! 😁

One remark: TLD5045EJ is an high integrated LED driver (Power elements are embedded). With this load configuration the output power is 7W. Considering 85% efficiency the IC has to dissipate about 1.2 W: please consider the cooling area below the device.

Regards,

Fausto

4 Replies

## Re: TLD5045EJ - Inductor selection and min drop voltage

Moderator
Moderator

Hi KPN,

Thank you for contributing to the Infineon community.

The inductance value can be calculated by using the below formula,

Vout - Output voltage

Vin-Input voltage

Fs-Switching frequency

∆IL- Inductor current ripple

The inductance value will vary with respect to the frequency.

There will be a minimum drop of 1.4V at 700mA load across the power transistor and freewheeling diode as noted in the datasheet below.

Regards,

Nishanth

## Re: TLD5045EJ - Inductor selection and min drop voltage

Level 1
Level 1

Thanks Nishanth,

a) thanks for the formula - might be worth adding that to the datasheet 😉

b) yes I saw the dropout voltage in the datasheet, but sometimes that is not the whole story. Again it would have been easier to have a graph showing e.g. the efficiency of the buck converter with regards to the input and output voltage. That is a more meaningful way of determining the suitability for a particular application, especially when evaluating different chips.

The real question was can I drive 4 red LEDs (approx 10V) from 12-14V vehicle power with 700mA?

Answer: looks marginal, so probably not.

Regards,

Peter

## Re: TLD5045EJ - Inductor selection and min drop voltage

Moderator
Moderator

Hello Peter,

with 4red  LED (2.5 V @ 700 mA) you have still room for this buck (about 2 V if the minimum Vin is 12 V).

here the steps I suggest you to design your LED driver:

1. set your peak current by choosing RSET (9.5k fits well)
2. Select your inductor to fulfill the ripple requirements (220 uH at 100 kHz generates +/-40 mA ripple over 700 mA average output)
3. Adjust Rfreq, Rcomp and Cfreq fo achieve 100 kHz switchng frequency (Rfreq = 7.3k, Rcomp = 100k, Ccomp = 100n fit 100 kHz switching)

You can try our evaluation board and play with it to see how powerful TLD5045EJ is!! 😁

One remark: TLD5045EJ is an high integrated LED driver (Power elements are embedded). With this load configuration the output power is 7W. Considering 85% efficiency the IC has to dissipate about 1.2 W: please consider the cooling area below the device.

Regards,

Fausto

## Re: TLD5045EJ - Inductor selection and min drop voltage

Level 1
Level 1

Thanks Fausto,

I will set up a quick prototype with the chip to test the performance based on the suggestion.

Much appreciated.

KPN