Tristated output to two common Base transistor issue

[BrianT]

With no PIC or any other drive, the PNP and NPN transistors are both turned ON by each other and the two series resistors (not shown) which are inside the cases. If you used regular PNP and NPN without intrinsic series resistors the circuit would instantly blow up two transistors.

IF the LED supply is the 12 volt auto battery things get worse. The PNP is always conducting because its base is at either 0 or 5 volts while the emitter is at 12 volts. The PNP base sees either 7 or 12 volts drive so it is always ON.

A rethink is needed.

HTH
Brian

[chips123]

I would think some diode isolation with a pull-up on the PNP base and a pull-down on the NPN base. Something along those lines anyway.

[RussMartin]

I assumed on the basis of information given that the LED supply is 5 VDC:

50 mA (current through LED) x 47 ohms (current-limiting resistor) gives a voltage drop of 2.35 VDC.

Here is what Brian is trying to tell you:

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R1a and R1b are built into Q1; R2b and R2a are built into Q2.

As wired, the built-in 22K resistors create a voltage divider, shown in the circuit above. Brian is not quite correct in saying that "the PNP and NPN transistors are both turned ON by each other" (they are not), but he is right about the effect of the intrinsic resistors. When the input (at B) is in a high-impedance state (Z), the voltage at A is 3/4 of +VDC, at B, it is 1/2 of +VDC, and at C, it is 1/4 of +VDC.

In other words, if +VDC is 5 volts, and the pin connected to B is high-impedance (Z), the voltage at A will be 3.75 volts, at B 2.50 volts, and at C 1.25 volts. The voltages present at A and C will keep the transistors turned partly on, which sounds like the problem you described.

The effects of a logic high (nominal +5 VDC) or a logic low (nominal 0 VDC) at B I will leave as an exercise for those who are curious.

[Acetronics2]

Hi, Russ

Look at old audio C class fully transistorized power amplifiers output stage ... the scheme is quite the same ... and see how the quiescent current is adjusted ...

There is THE solution !!!

Alain

[flotulopex]

Okay, some additional infos here.

First, I thought to use transistors for the high-power LEDs because I have to drain more than the acceptable 20-25mA a PIC can hold.

Yes, the circuit is +5V powered.

Acetronics, I will try with diodes but, if this works, I will have to add 30 diodes to the circuit!!!! (c'est pas une question de poids.... mais ça fait du monde sur le PCB ).

BrianT, you may be right: I'm an electrician, not an electronician so I don't "master" those kind of Z-state subtilities and don't really know how to handle them a correct way...

RussMArtin, I need to light the LEDs separately, never simultaneously.

If my way of doing (thinking) is completely wrong, how would this kind of problem be resolved in an industrial environnement? Would they just use a 20 I/O ports PIC?

Finally, I ask myself if this can be done ... or not?

[Acetronics2]

Hi, Roger

the simplest (component count ) is to use "classic" transistors and use resistors for 3.5v Hi state switching, and 1.5v Lo state switching ...

That gives "R1" = 10k and "R2" = 2.2 k ... and class "C" transistors ( gain > 400 ).
The original Logic transistors can't work this way due to the too high b>e resistor

AND do not forget the +V/2 pullup 2 resistors ...

BUT ... what do you want to realize ??? other solutions possible !!!

Alain

[RussMartin]

If my way of doing (thinking) is completely wrong, how would this kind of problem be resolved in an industrial environnement? Would they just use a 20 I/O ports PIC?

Finally, I ask myself if this can be done ... or not?

This is why engineering is an art, not a science!

Let's sum up what we know so far:

1. It is a 5 VDC circuit.

2. The LEDs require more current than a PIC pin can source or sink.

3. There are three permitted states for any pair of LEDs--one or the other is on, or they are both off. They are never both on at the same time.

4. There are 10 output channels.

Beyond this, we don't know what you are attempting to do, or why. In other words, we don't know your intended application.

And there are questions: Do you have to use such current-hungry LEDs? Why? Of 10 channels, how many LEDs are likely to be on at the same time? Do you have size and/or board space limitations?

Alain suggests using "classic" transistors (I think he means ones without the built-in resistors). This sounds like one good approach. I would go further and suggest generic switching transistors driven well into saturation. Going to a 20-port PIC would also work, with the advantage of allowing you to use transistors that are all the same.