Electronic – LEDs inside of solar-powered calculators

calculatorledsolar cell

Why do a lot of solar-powered calculators contain red or infrared LEDs that are hidden from the user? So far I have only seen this in Texas Instruments calculators.

Here are some examples of this in Texas Instruments calculators, taken from user-contributed guides on iFixit:

Older TI calculators, from datamath.org:

Here's a newer TI-106II, with what appears to be an infrared LED:

I think I first saw this in a Youtube video from either EEVBlog or bigclivedotcom and I remember whoever doing the video was confused as to why there was an LED inside. If anyone has seen this video and could provide the URL that would be good.

Any ideas?

Best Answer

I remember a label printer I dissected as a kid having the same. From the position of the LEDs, it looked like they were pretty close to the EEPROMs on the device. My best guess hence is: This might actually be a way for the manufacturer to, at low cost, add a programming header to the devices that allows for different functionalities being programmed in a manual labor assembly line, where you just want to decide at last minute, before you put the back cover on, into which market you'll sell that product.

Why solar calculators? If they ran QA on those, they'd want a way to test them, so one would assume you just want to put your calculator in a light box, and run a few tests. Now "run a few tests" is a bit complicated with a device that runs at wildly varying voltages, and probably also wildly varying clock rates: you either end up somehow electrically coupling the device with sensitive connectors (which are expensive) to something that does the level shifting in your test stand, or you do the level shifting in your calculator and end up having an unnecessary level shifter in every calculator. Or, you just add some wireless link, like, in the simplest case, two LEDs, one used as a photodiode (receiver), one as a Light emitting diode (transmitter). You place the half-assembled calculator manually on a frame that is being lit from a below, and communicate with the LEDs in the shadow of the calculator itself.

My other best guess is: LEDs have become a mass product, and they come with a relatively well-defined band gap. So if you need a 2.15 V voltage reference for whatever, why not go and buy the cheapest yellow LED you can find?

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Electronic – Driving a LED with an open drain digital output

This circuit will work, but not be efficient, power-wise. If the output is high the LED is on and you'll have 20 mA current through LED and resistor.

But with the output low, the LED is short-circuited and therefore off, but the resistor current will be even higher: 50 mA!, which is not only bad for your wallet and the environment, but also for the output pin; most I/Os will only allow 25 mA.

Place the LED in series with the resistor. Then a high output won't light the LED because the output FET is switched off. A low output will draw the 20 mA through LED, resistor and open-drain output.

Note that this will invert the logic: in your first schematic the LED will be off when the output is active, in the second circuit it will be on with active output.

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If the output can't sink enough current you'll need an external transistor to increase that.

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If the output is active (low) there will flow a 1.5 mA from +3.3 V through Q2's emitter-base junction, R2 and Q1. That 1.5 mA will allow Q2 to source more than 100 mA for a typical general-purpose transistor. You will only get about 22 mA, though, because R1 won't allow more.

If Q1 is off there won't be any base current through Q1, so the LED will remain off. Q1 will have a small leakage current, and to avoid that this would get amplified by Q2 I added R3. As long as the leakage current is less than 0.7 V / 15 kΩ = 50 µA all the leakage current will flow through R3, so that will ensure Q2 will be completely off.

24VAC LED flood with 12VAC driver works on 12VDC—how

What is going on?

 Why does something sold as an AC24V device contain an AC12V driver?
 Why did applying a DC voltage to an AC device not blow it up?

 Why could I only get red out of the LED and not any other colours,
 and why did the  IR/control circuitry seem to not work?

If it doesn't use a transformer for lowering the supply voltage there's a very good chance is does work from DC. Most wall-warts will run from a DC input because (like your LED panel) they have a bridge rectifier that converts ac into dc. Bridge rectifiers pass ac or dc exactly the same - they are just steering diodes that always push the most positive input voltage to the +Vout and the most negative input voltage to -Vout

It did not blow up (or not-work or get too warm or short the supply out) because it doesn't use an input voltage transformer (as per the answer in the paragraph above).

When you powered it from 12V dc I suspect that there isn't enough voltage to allow all the circuits to work correctly. AC 24V will produce a peak voltage of about 34V and this will likely feed through a bridge rectifier to give about 32Vdc - a lot higher than 12V dc you fed in and this I suspect is the main reason.

Ditto for AC 12V - it has a peak voltage of 17V and after a couple of diode drops in the bridge will settle at 15.6VDC - if you had a bench supply and fed it with 17Vdc I'd bet it would work.

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Related Solutions

Electronic – Driving a LED with an open drain digital output

24VAC LED flood with 12VAC driver works on 12VDC—how

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