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I'm sketching out a design for an easily and cheaply built handheld computer based on an ARM microcontroller, and thinking of using a small LCD display. The display has an LED backlight rated for up to 80mA with a typical \$V_f\$ of 3.2V, though a smaller current will probably be enough. I want to power it from a couple of AA cells, so some kind of voltage boost is needed. The microcontroller will be a convenient source of a clock signal, but the I/O pins do not have anything like the required current drive. So I am thinking of the following circuit, with push-pull drive for the voltage doubler. A prototype seems to work reasonably well, though the diode losses are significant.

schematic

simulate this circuit – Schematic created using CircuitLab

The idea is that when the clock input is low, C1 will charge through D1 and the PNP emitter follower Q2. When the clock is high, Q1 conducts and lifts the left end of the capacitor to over 2V, and the capacitor can then discharge into the LED (generic type shown). For the backlight application, no smoothing of the output is needed, and the LED can act as its own rectifier.

I expect the 100kHz signal to be close to 0V low and over 3V high, since the microcontroller (like the display logic) will be on its own power supply, with a 3.3V commercial boost converter module. Potentially, it will be able to drive the bases of the transistors a little beyond the battery voltage when the clock is high, reducing the voltage loss owing to the \$V_{BE}\$ of Q1. I've included R1 to prevent the microcontroller from providing significant charging current for C1 through the BE junction of Q1.

Can you suggest a better, simple solution? Is there an easy addition to the circuit that would permit measuring the average LED current with a microcontroller pin? Should I just give up and add another IC just to power the backlight?

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    \$\begingroup\$ Have you estimated the power efficiency? \$\endgroup\$
    – Andy aka
    Dec 13, 2020 at 18:09
  • \$\begingroup\$ Difficult to marry theory and practice in assessing the efficiency. From one point of view, every coulomb of charge that flows through the LED has been through one, two, three diode drops elsewhere in the circuit, so we would do well to beat 50%. From another angle, the current shown on my bench supply is about twice the average LED current as measured with a scope across a 10 ohm resistor in series with it. I've not yet had chance to measure the supply current more accurately than that. \$\endgroup\$ Dec 13, 2020 at 19:13
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    \$\begingroup\$ Use a simulator. I think you’ll find efficiency to be not as good as a proper boost circuit and that’s important for a battery supply. \$\endgroup\$
    – Andy aka
    Dec 13, 2020 at 20:02
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    \$\begingroup\$ You could check white LED drivers for smartphone flash. These take a LiIon input voltage which is a good match for 3 AAs and have good efficiency, and due to economies of scale they're pretty cheap. \$\endgroup\$
    – bobflux
    Dec 13, 2020 at 21:33

2 Answers 2

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One suggestion is to use a n chan mosfet and an inductor in boost configuration. Use a pwm signal from the microcontroller to vary brightness and set the maximum led current.

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  • \$\begingroup\$ I will have to get some inductors and try it out. Fewer components too! \$\endgroup\$ Dec 13, 2020 at 22:24
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    \$\begingroup\$ The voltage doubler has the perhaps significant advantage that it won't be damaged if a software crash leaves the control signal stuck in either state. \$\endgroup\$ Dec 13, 2020 at 22:48
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    \$\begingroup\$ @MikeSpivey You can use a version of the Joule Thief, by adding a series resistor to the LED and another BJT as a comparator. It will boost to any required voltage necessary to provide the design current in the LED. I don't know if that is of any interest, but I could write up an example. If it matters to you. \$\endgroup\$
    – jonk
    Dec 13, 2020 at 22:54
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Just because it's fun, I'll toss out an efficient version of a Joule Thief that will self-correct for about \$20\:\text{mA}\$ through the LED with a \$V_f\approx 3.2\:\text{V}\$. (It will work fine with larger and smaller forward voltages, too.)

Normally, I'd want to include a diode for reversed base-emitter voltages that may damage the switching BJT. But a snubber works more efficiently and achieves similar results. Something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Some tweaking will be required to get optimal efficiency. But somewhere in the vicinity of 75% should be achievable. It should be set to run at about \$50\:\text{kHz}\$ given the non-RF type of small signal BJTs. Playing with \$C_1\$ and \$R_2\$ should be considered for this purpose and for tweaking the efficiency. It should work well as the batteries decline in voltage, as well, so it doesn't require fresh batteries in order to operate.

I've not built this particular circuit and I don't have your LEDs, either. So this is merely a suggestion on the assumption that you can adjust it per your actual situation. (\$L_1\$ and \$L_2\$ are the primary and secondary of a transformer capable of operating at \$100\:\text{kHz}\$ or better, in case it's not already obvious. If a ferrite core and ungapped, it will likely require a cross-section of \$2.5 \:\text{cm}^2\$ to stay under \$100\:\text{mT}\$. I'd probably look at gapping.)

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  • \$\begingroup\$ D1 and maybe C2 could be removed. Removing the diode drop will help the numbers \$\endgroup\$
    – Kartman
    Dec 14, 2020 at 6:04
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    \$\begingroup\$ @Kartman I'd already tried that idea and simulation showed it reduced the efficiency by about a percent or two. Obviously, the behavior is also different (the LED pulses rather than exhibiting a more continuous current.) I'd need to delve into more details to find out exactly why Spice says what it says. It's likely "in the details" below where I'd like to bother, right now. (Likely due to \$Q_2\$ behavior differences.) But your point remains. Two less parts may be worth having. A concern may be operation at a fully charged \$3\:\text{V}\$, though. I was depending on \$D_1\$ for a small drop. \$\endgroup\$
    – jonk
    Dec 14, 2020 at 6:15
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    \$\begingroup\$ @Kartman It appears some inefficiency from removing those two parts arrives due to a much higher peak LED currents and losses, thereby, in the current sense resistor as well as differences in driving/over-driving \$Q_2\$. The smoothing helps that as well as providing more predictable behavior across part variations. Also, the LED dissipates more due to bulk resistance as well as somewhat lower efficiency in converting electrical energy into light (accounting for these in spice is 'difficult' but they exist just the same.) \$\endgroup\$
    – jonk
    Dec 14, 2020 at 7:20

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