Re: TI-H: EL Backlight installed!

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Re: TI-H: EL Backlight installed!


On Tue, 14 Apr 1998, David Knaack wrote:

> The LCD requires AC to run, correct?  Where is the AC power for the
> LCD produced?  (I don't have schematics).

I *think* so. It should be produced inside the LCD driver chip.

> >Try putting an ammeter on the positive power lead to the LCD board
> and see
> >if the current draw increases when the calc gets turned on.
> 
> I don't think I want to disconnect that ribbon cable, they are a bit
> too hard to work with.  At this point I'd prefer to just use the
> switch.

True.. I wasn't sure if you had it on or off. I would suspect that there
are probably some pins which go high (or low) when the LCD gets turned on
and then switch back when it goes off. Maybe if you probe around for
something like that (or get pinouts for the LCD drivers..?) it might be
possible to find something.

>> If it does, you can put a resistor there and put the base of the
>> transistor on one side and the emmiter (?) on the other side. Calculate
>> the resistor value such that the change in the current being drawn
>> (between ON state and OFF state) will produce a voltage big enough to
>> turn the transistor on and off. > > Where exactly would the resistor
>> go?


I learned this one from a lead-acid battery charging circuit. :)

               R1
+5v ---+----/\/\/\/---+-----------> some circuit
       |              |
       |              |
       |    Q1        |
       +---> |        |
            >|--------+
       +___/ |
       |
       +--> the switched circuit, I think

Then you choose R1 to turn Q1 on at the desired current through R1:
I(R1)=R1/0.6

Here's the theory of the circuit (as I understand it):

The supply voltage here is 5v dc. The exact value doesn't matter much, but
that's just what number I picked. So there is a circuit on the top right
side which pulls some amount of current in it on state. Let's say it pulls
20 mA. When turned off, that circuit pulls 5 mA.

We want the transistor to come on when the circuit is on, or in other
words, when the circuit is pulling greater than something like 15 mA. That
is high enough that the transistor won't come on if the circuit pulls too
much OFF current, but it will come on if the ON current isn't quite 20 mA.

In this circuit, there is a resistor in series with whatever is drawing
current. When the gizmo draws a low amount of current, the voltage drop
across the resistor will also be low. At a higher current, the voltage
drop will be higher. Since Q1 will turn on when it gets 0.6v across those
two pins, we want to choose a resistor that will have a 0.6v drop across
it when the circuit is turned on, or when it is at whatever ON current we
might be using for the circuit.

So the switch should happen at 15 mA. Ohm's law says to pick R1 like this:
.015A = R1/0.6 so we want R1 to be .009 ohms. 

Okay, so a 9 milliohm resistor is hard to find.. A coil of wire might
work. The battery circuit here draws like 1A at 14v, so its resistor would
be about 1.7 ohms, which is a little easier to come up with.. Maybe this
circuit would be best saved for high-current projects. Or maybe we can
tweak the circuit so it'll work here anyway. Any suggestions?


-- 
Greg Hill
greg-hill@bigfoot.com
www.comports.com/link
References: