Powering TFT LED backlight - Page 1
Forum section: electronics / beginners

author: Kaisha
author_rank: Contributor
timestamp: August 22, 2021, 12:28:19 am
content: I need to power a TFT LED backlight (datasheet:https://www.buydisplay.com/download/manual/ER-TFT032A3-3_Datasheet.pdf). I currently have a buck regulator supplying 3.3V to my microcontroller (ESP32) which will interface with the LCD. The problem is the LED backlight has a forward voltage of 3.2V - 3.4V, so while my power supply can supply the voltage necessary, I'm not sure how I should current limit the backlight.Normally a simple resistor is used, calculated R = (Vs - Vf) / I. Problem is Vs = Vf in this case, which leaves me with a resistor size of zero. This (of course) doesn't work.Another option would be to use a dedicated constant current led power supply like the tps54200 (datasheet:https://www.ti.com/lit/ds/symlink/tps54200.pdf). Problem with that is that it seems overly complex for something simple like a 100mA backlight, I'm running out of room on my PCB, and the suggested components (in particular Rsense) doesn't make any sense.Any simple/easy solutions that come to mind?

author: langwadt
author_rank: Super Contributor
timestamp: August 22, 2021, 12:44:16 am
content: what is your input voltage?

author: Kaisha
author_rank: Contributor
timestamp: August 22, 2021, 02:24:49 am
content: The input to the buck regulator is either usb (5V) or two Li Ion batteries in series (9.4V to 6.6V). The output of the regulator is 3.3V.

author: amyk
author_rank: Super Contributor
timestamp: August 22, 2021, 03:41:46 am
content: 100mA LDO wired for CC operation, if you don't mind the power dissipation...

author: Kaisha
author_rank: Contributor
timestamp: August 22, 2021, 03:46:00 am
content: That would be a lot of power wasted. Given that this is a battery operated (primarily) and the LED backlight is most likely going to be the highest power draw, I'd prefer not to.

author: Zero999
author_rank: Super Contributor
timestamp: August 22, 2021, 09:19:43 am
content: Quote from: Kaisha on August 22, 2021, 03:46:00 amThen you have two options:Power the backlight off 3.3V, perhaps via a low value resistor and hope the current remains within the rating.Use a separate, current regulated buck converter for the backlight, like the tps54200 mentioned in your first post. What doesn't make sense, regarding the resistor values? You only need 100mA, so the inductor can be much smaller.There's no other way.

author: Kaisha
author_rank: Contributor
timestamp: August 22, 2021, 11:52:50 am
content: QuoteOk, I wasn't sure. I thought maybe I could drive it from an OpAmp, or perhaps a microprocessor DAC and transistor or something like that...QuoteAccording to the datasheet, calculating the value for Rsense gives me 2.2 Ohm. Not 2.2k... just 2.2. This seems ridiculously small, and I'm worried it'll overheat. I don't really know how much power is going through it though... If it says in the datasheet I missed it somewhere. And unlike the buck regulator I used, the tps54200 isn't supported in Webench, and I don't see any examples online, so I can't double check my math.QuoteFor the TPS54331 (regulator I'm using) you need a larger inductor for smaller currents, I assumed it would be similar for the tps54200.

author: Zero999
author_rank: Super Contributor
timestamp: August 22, 2021, 12:13:31 pm
content: Quote from: Kaisha on August 22, 2021, 11:52:50 amThe problem is, 3.3V isn't guaranteed to power a single LED properly and powering it from a higher voltage either requires the energy to be dissipated as heat, or an inductor to temporarily store it, to dump it into the LED, i,e. a switched mode power supply.QuoteP = I2R, so the power dissipation will be tiny at 100mA.QuoteA higher value inductor doesn't always mean physically bigger. The size is proportion to the stored energy, which depends on the current as well as the inductance. E =1/2×L×I2and you only need one tenth of the current, so even if the inductance value is higher, the inductor is quite likely much smaller.

author: Kaisha
author_rank: Contributor
timestamp: August 22, 2021, 07:33:04 pm
content: I appreciate the answers, that really clear up a lot.

author: Zero999
author_rank: Super Contributor
timestamp: August 22, 2021, 10:29:44 pm
content: Quote from: Kaisha on August 22, 2021, 07:33:04 pmI've had a couple of other ideas:Use a buck regulator to get 4.5V and run the backlight using a plain old resistor and the MCU with a 3.3V low dropout regulator.Run the backlight using a current mode buck regulator and connect the MCU in parallel with it, with the LED forming a shunt regulator.#1 is the saner option, but involves a bit of power loss in the resistor and low dropout regulator.#2 is a little more dodgy, since the LED current will vary, with the MCU current and the MCU will be destroyed, if the LED fails open circuit. It also means the backlight has to be run continuously, as the MCU is powered.

author: Kaisha
author_rank: Contributor
timestamp: August 23, 2021, 12:35:21 am
content: #1 sounds interesting... I like it. I wouldn't know how to do #2.

author: grrmachine
author_rank: Contributor
timestamp: August 23, 2021, 07:44:48 am
content: Quote from: Kaisha on August 22, 2021, 11:52:50 amI'd probably do this!Build a cheap current sink with an opamp, small NMOS, and a shunt resistor(s). Power off the circuit via +5V USB or battery, current would still be regulated.Using a dual opamp, one to amplify the shunt current feedback with sub 100mshunt (Rds of the small Mosfet could bottleneck) and minimize the power dissipation at 100mA. And spare a DAC to set the comparator's reference to dim/turn off the backlight.

author: Zero999
author_rank: Super Contributor
timestamp: August 23, 2021, 09:22:11 am
content: Quote from: grrmachine on August 23, 2021, 07:44:48 amThat will work, but he wants to save power, especially when it's run off batteries. If backlight is run off a linear regulator, over 60% of the power will be wasted, when the battery voltage is 8.4V.

author: grrmachine
author_rank: Contributor
timestamp: August 23, 2021, 12:28:56 pm
content: Yep, True! Especially when the input is 8.4V and battery operated.

author: langwadt
author_rank: Super Contributor
timestamp: August 23, 2021, 01:02:56 pm
content: Quote from: Zero999 on August 22, 2021, 10:29:44 pmor for the sketchier version of #1, up the buck output voltage a bit to and use a couple of diodes to the drop the mcu suppy to ~3.3V. make it a bit safer by putting the diodes before the buck sense divider

author: Zero999
author_rank: Super Contributor
timestamp: August 23, 2021, 02:37:45 pm
content: Quote from: langwadt on August 23, 2021, 01:02:56 pmThat's a good idea. One of the diodes could be Schottky to lower the voltage drop and save power.

author: langwadt
author_rank: Super Contributor
timestamp: August 23, 2021, 05:05:33 pm
content: Quote from: Zero999 on August 23, 2021, 02:37:45 pmneed enough voltage for the transistor Vce and a Vbe over the resistor, so it might get marginal

author: Zero999
author_rank: Super Contributor
timestamp: August 23, 2021, 06:56:51 pm
content: Quote from: langwadt on August 23, 2021, 05:05:33 pmReducing VBwill proportionally decrease the voltage across the sense resistor, so it won't make much difference to VCE. Most small transistors also have quite a low saturation voltage at 100mA. The BC327 has a saturation voltage of 100mV when IB= >4mA and IC= 100mA.https://www.onsemi.com/pdf/datasheet/bc327-d.pdf

author: langwadt
author_rank: Super Contributor
timestamp: August 23, 2021, 07:34:34 pm
content: Quote from: Zero999 on August 23, 2021, 06:56:51 pmyeh it would work, could also skip the transistor and just use a single diode and the resistor

author: Kaisha
author_rank: Contributor
timestamp: August 23, 2021, 08:51:07 pm
content: Well for the record I went with a dedicated power supply (tps54200), but I'm finding this discussion fascinating.I'm unsure as to the purpose of the transistor in the schematic that langwadt serves.

author: Ian.M
author_rank: Super Contributor
timestamp: August 23, 2021, 09:06:33 pm
content: Quote from: Kaisha on August 23, 2021, 08:51:07 pmConstant current source. Its base voltage is reasonably well defined by the two diodes (assuming the load on the 3.3V rail is reasonably stable), which sets the voltage across the emitter resistor, and thus its collector current.

author: Zero999
author_rank: Super Contributor
timestamp: August 23, 2021, 09:35:39 pm
content: Quote from: Kaisha on August 23, 2021, 08:51:07 pmI would also use a dedicated LED drive IC./The transistor is a constant current source. The diodes roughly drop 0.7V each, giving a voltage drop of 1.4V, at the base and the transistor's base-emitter junction also drops 0.7V. This gives a fairly constant 0.7V across the 7 Ohm resistor, so roughly 100mA will flow through it.