I'd like to design a circuit that utilizes a small 1.3" LCD display with maximum LED backlight forward current of 40 mA. Since I'd like to use brightness control, I am hesitating between using a standard PWM transistor circuit or a dedicated LED driver for the backlight. I've seen both applications used in dedicated products, however I don't know which one would be appropriate in my case.

Therefore the question: What are the pros and cons of a transistor PWM circuit vs LED driver circuit for a display backlight?

Here are some examples of the two circuits:

  1. Transistor PWM circuit:


simulate this circuit – Schematic created using CircuitLab

  1. Using a dedicated dimming LED driver e.g. aw9364 (cheaper or more appropriate may exist, if you agree then please recommend one): enter image description here

  2. Using an LED driver e.g. ZXSC310 (cheaper or more appropriate may exist, if you agree then please recommend one): enter image description here

From my perspective, I find the transistor circuit much cheaper and easier to implement. I am hesitant about any EMC issues that may be caused by the PWM, however the transistor circuit and the display will be placed within max of 1 cm of each other. The MCU will also be within same distance and definitely has a dedicated PWM output.

My application will not use a battery but a constant power supply source.

  • 1
    \$\begingroup\$ look for "reasons for individual series resistors", and try to (roughly) estimate the power efficiency. \$\endgroup\$ Commented Aug 14, 2021 at 14:01
  • \$\begingroup\$ The series resistor will be 10ohms. Considering a 40mA of max current gives 400mW. Since we are talking about only one LCD backlight, I can't see the relevance of power efficiency in this application? \$\endgroup\$
    – Phill Donn
    Commented Aug 14, 2021 at 14:07
  • \$\begingroup\$ 400 mW is a lot. I mean, my whole smartphone can be used for ca 8h actively, i.e. with backlight on, with a 2400 mAh 4V battery, that's 9.6 Wh; if 3.2 Wh went to backlight alone, that would severely limit the CPU power I could use. Plus, you're forgetting about your transistor. \$\endgroup\$ Commented Aug 14, 2021 at 14:11
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    \$\begingroup\$ did you just, on the fly, completely change your second option thereby invalidating my answer? not really cool... \$\endgroup\$ Commented Aug 14, 2021 at 14:13
  • \$\begingroup\$ @MarcusMüller I realized that I made a mistake in the option since it was not viable. The edited LED driver was not dimmable hence I had to replace it to avoid unnecessary discussions unrelated to the topic. Even though I changed it, the same question still applies. \$\endgroup\$
    – Phill Donn
    Commented Aug 14, 2021 at 14:25

3 Answers 3


The LEDs should always be driven with a constant current unless used for indication purposes (e.g. 1 to 5mA, unimportant illumination). Full stop.

The LEDs in your application are used as a backlight. So, to me, the best practice is to use a dedicated LED driver.

The typical drive current is 30mA (The need for an overdrive to 40mA depends on the brightness requirements). The boost converter that is supplied from 3.3V as shown in the OP will not work, because the input is 3.3V and the output is 3V (Vf of the LED). A buck converter would be overkill. A CV boost to get something around 5V and using a series resistor could be an option. But...

If you use a series resistor: The required series resistor will be RLED=10R and its dissipation will be less than 10mW. One issue with this technique is that the drive current will slightly increase as the circuit runs: The LED (actually, the junction) will get warm as the current flows through it. As the junction temperature increases, its forward voltage, VF, decreases. Thus (VCC-VF) difference increases and this results in a slightly increased drive current. It may not be a problem, but still is a thing to consider.

So, either design your own current source with a transistor with VCC > 3.3V or use a low-dropout linear LED driver. Again, using a buck converter would be overkill.


From original question, the second option was:


which is fundamentally different from what you're showing now.

  1. You need a series resistor per LED in your parallel-sunk transistor-based circuit. So, to compare fairly: for 4 LEDs, you'd need 4 series resistors¹, the transistor and the biasing for the transistor; that's N_LED + 3 components in addition to the LEDs. Your boost-converter based circuit is always only 6 additional components – no matter how many LEDs you put in series.

  2. Your transistor circuit has to convert the voltage difference between your supply (3.3V) and your LED's desirable forward voltage (e.g. 2V for a red LED) to heat – that means that (in this 2V Vf example), 1.3 V of 2V that 1/3 of your energy just heats up resistors and a transistor. That's stupid for battery-operated devices.

  3. your single-transistor-based circuit only works when the supply voltage is higher than Vf of the LED plus the forward voltage of the conducting collector-emitter junction of the transistor, typically 0.7 V. So, if, for example, you're using a 3.3V supply, and your LED is a blue one with Vf=2.8V, then you simply can't operate that LED at full brightness, because 3.3V - 0.7 V < 2.8 V.

¹ If you're not aware why you should need a series resistor per LED: this has been answered a hundred times on here, look for "thermal runaway parallel LED" or so.


The backlight in your LCD has this characteristic: -

enter image description here

And, wiring two of them in parallel as per the diagram in your question is not good practice unless, they are very well matched (and thermally also) because you will get illumination differences between the two and one may actually hog all the current and burn, rapidly followed by the 2nd one burning so, it's bad practice: -

enter image description here

They should be wired in series but in doing so, the terminal voltage needed is significantly greater that the 3.3 volt supply rail hence, you have no option other than to use circuit 2 (the booster).

What are the pros and cons of a transistor PWM circuit vs LED driver circuit for a display backlight

Quite simply, one stands a chance of destroying the LEDs and one looks like it will work just fine.

  • \$\begingroup\$ The LEDs are in the same package (i.e. 2-chip LED), so they should have nearly the same electrical characteristics. \$\endgroup\$ Commented Aug 14, 2021 at 14:12
  • \$\begingroup\$ @RohatKılıç - there is only one LED in the package (pin 2 and pin 3 of the connector) so, the OP has to be considering two displays. \$\endgroup\$
    – Andy aka
    Commented Aug 14, 2021 at 14:24
  • \$\begingroup\$ I don't have control over the connection of the backlight LEDs. If you check page 6 in the datasheet, they are internally connected in parallel inside the display. It's only one display. \$\endgroup\$
    – Phill Donn
    Commented Aug 14, 2021 at 14:27
  • \$\begingroup\$ @Andyaka - p.6 shows the LED connection. There's most likely a 2-chip LED in the same package. It's also indicated in the table you copied into your answer. \$\endgroup\$ Commented Aug 14, 2021 at 14:29
  • \$\begingroup\$ Then why does your circuit in your question imply that you are connecting two LEDs in parallel (I assumed you were using two separate displays of course). You are making quite a mess here @PhillDonn \$\endgroup\$
    – Andy aka
    Commented Aug 14, 2021 at 14:29

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