Tristated output to two common Base transistor issue

[flotulopex]

Luciano,

Je dois avoir les yeux dans mes poches, c'est pas possible autrement!

Nevertheless, it looks not possible to control the LEDs the way I thought I could...

[Luciano]

Hi Roger,

Try that:

If you use standard LEDs, replace the 22 Ohm resistors with 100 Ohm resistors.

Best regards,

Luciano

EDIT: The note about the resistors is only if you use this circuit.
To drive two standard LEDs (max 20 mA) you don't need transistors.

[Darrel Taylor]

20 LED's,... 20 Transistors, 20 Resistors. Start adding diodes, half-watt resistors.... Yuck!

I think I'd just go with 3 of these

<table><tr><td><a href="http://www.allegromicro.com/en/Products/Part_Numbers/6277/6277.pdf"><img src="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1974&stc=1&d=118911366 1"></a></td><td>

Don't even need resistors on each LED.

Up to 24V for LED drive.

Set it for 50ma constant current.
Shiftout the LED states

And that's it.
</td></tr></table>

http://www.allegromicro.com/en/Produ.../6277/6277.pdf
About $1.50 each in single quantities.
<br><br>

[RussMartin]

One problem I see with Luciano's suggested circuit is that the two 22-ohm resistors form a voltage divider (!) pulling a maximum of 114 mA (5 volts / 44 ohms = 0.114 amps) when the LEDs are doing nothing. 10 channels of this is 10 x 114 mA = 1.14 amps.

[Luciano]

Hi Russ,

The circuit is for the brake light of a car, so the circuit gets the power
only when the brake pedal is pressed. Also consider that power is generated
by the alternator, so this current will also decelerate the car when the driver
activates the brakes ...

Best regards,

Luciano

[RussMartin]

The circuit is for the brake light of a car, so the circuit gets the power only when the brake pedal is pressed.

Aha! Thanks! In that case, Roger should go ahead and give your circuit a try!

[Luciano]

Hi,

There are ony two 22 ohm resistors for the 20 LEDs.


Best regards,

Luciano

[RussMartin]

BrianT advised me by private message that I had oversimplified my earlier analysis, and he is right. I looked only at the voltage divider created by the resisitors built into the transistors and neglected to consider the diode drop across the base-emitter junctions of the transistors. The circuit, redrawn, looks like this:

<IMG SRC="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1980&stc=1&d=118913663 8">

With the diode junctions in place (a nominal 0.7 VDC drop across each), two voltages change: At A, 4.3 volts; at C, 0.7 volt. At B, it remains 2.5 volts. Both transistors are still "turned on" if there is no voltage present (a high-impedance state) at B.