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I'm trying to turn on/off the backlight of this LCD for the power saving when my ESP8266 goes to deep sleep. Because the ESP8266 has a maximum output current of 12mA and the LCD needs a current about 60mA, I decided to use a PNP transistor;(BC177.) According to the datasheet, it can handle currents about 100mA, so it seems to me that it's safe and good to use the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

This page includes the state of the GPIOs when the ESP8266 goes to the deep sleep. I've tested this circuit and it works but I'm not sure if the circuit is appropriate and safe. Does turning off the backlight save power? The mentioned LCD has two other power supplirs which are unaffected when the ESP8266 goes to deep sleep.

Edit: We have the following information from the mentioned page:

  • Backlight type: LED*4 current 60mA
  • Voltage: working voltage VDDI=VDD= 2.8V-3.3V
  • Total power consumption: 0.22W

With the above circuit, I'm getting about 2.96V collector voltage measured by oscilloscope.

Edit2: I found the datasheet for the LCD and uploaded it here.

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2 Answers 2

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What you probably need is for the transistor to turn on so that it drops very little voltage between collector and emitter. This in turn means that your LED circuit receives the full 3.3 volts across it. Of course, this assumes that the LEDs are voltage driven and not current driven. However, your LCD is "defined" by an Aliexpress page and, they don't usually link data sheets so I'm not going to try and figure that out.

So, assuming you want the smallest volt drop across the BJT in order to maximize drive level to the backlight, you need a much lower base resistor. This is because the current gain of the BC177 might be very low when driving the BJT into saturation.

enter image description here

Page 2 of the data sheet tells you this; \$V_{CE(SAT)}\$ is 200 mV when the collector current is 100 mA and the base is driven with 5 mA. So, with a bit of hand-waving, your base needs 5 mA (or maybe a tad more) and this means R1 should have about 2.6 volts volts across it when 5 mA flows.

That's a resistance of 520 Ω and not 10 kΩ.

There'll be 0.2 volts dropped between collector and emitter so, be aware of this and double check that the LEDs can be voltage driven (hint you need to locate the data sheet and read it).

Also does turning off the backlight save power?

Yes, it should save power.

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  • \$\begingroup\$ Thanks for the answer. I couldn't find the datasheet but in the mentioned page we have following information: Backlight type: LED*4 current 60mA Voltage: working voltage VDDI=VDD= 2.8V-3.3V Total power consumption: 0.22W. I think this means we should struggle to have 60mA collector current, right? Also do you think it's safe to switch off/on VDDI and VDD using the same circuit as backlight? In this way we can save more power. \$\endgroup\$
    – S.H.W
    Commented Sep 4, 2021 at 20:21
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    \$\begingroup\$ Be aware that most backlights (in my experience) don’t include current-limiting resistors, so you’ll need an external one, or preferably a formal current source. This makes sense because it allows you (as designer of the circuit) to determine how much backlight current you want and how to control it. From a power-saving point of view, 60mA is far more than the rest of the LCD will draw, and probably more than the power budget for the CPU. \$\endgroup\$
    – Frog
    Commented Sep 4, 2021 at 20:58
  • \$\begingroup\$ Basically @S.H.W find a data sheet. If I were you I wouldn’t use an LCD that wasn’t supported by a data sheet and produced by a manufacturer with a solid demonstrable quality system. \$\endgroup\$
    – Andy aka
    Commented Sep 4, 2021 at 23:35
  • \$\begingroup\$ @Andyaka Thanks. I added the datasheet. Please take a look. \$\endgroup\$
    – S.H.W
    Commented Sep 5, 2021 at 11:00
  • \$\begingroup\$ @S.H.W you need to apply a constant current into the LED backlight. 90 mA will be too much so, if you are OK with 60 mA, that sounds more respectable. You don't want to apply a constant voltage because, as the LED warms up, the current could rise above the upper limit i.e. you need to control the current and not apply a hard voltage to the LEDs. That's a different question if you don't know how to do that. \$\endgroup\$
    – Andy aka
    Commented Sep 5, 2021 at 11:27
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Based on the Hfe of the PNP, even if the backlight draws up to 100mA, you should be under 1mA draw from the ESP. With that said, it won't draw current, but with a PNP, it should sink current. And 1mA is very, very low. With that said, you might also just want to check if the backlight has a built-in current limit feature, such as say a current limiting resistor.

But yes, the LED's will probably draw a bit of power, and powering them down will save some power. But why don't you just switch off the whole screen instead?

Also, R1 is 10k, with a typical Vbe of 0.7V, the base will sit at about 2.6V, and if you want to make sure you can power down the backlight, allow at least 80mA - giving Ib about 0.5mA (depends on the transistor - seems they vary). So at 0.5mA and 2.6V drop, do the math to get the resistor value. Then it should all be safe and good :)

Best Jono

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  • \$\begingroup\$ A P-channel MOSFET would work good and have almost no voltage drop. \$\endgroup\$
    – Gil
    Commented Sep 4, 2021 at 20:48
  • \$\begingroup\$ Please provide additional details in your answer. As it's currently written, it's hard to understand your solution. \$\endgroup\$
    – Community Bot
    Commented Sep 5, 2021 at 0:47

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