Tristated output to two common Base transistor issue

[flotulopex]

Hello,

I have to power automotive LEDs that drain 50mA.

I use twenty LEDS and have a 16F630, so my circuit connects two LEDs to 1 I/O port and I use the Z state to switch them OFF.

To make this without destroying the PIC, I use transistors like this (one port example):
<img src="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1965&stc=1&d=118901631 5">

Problem: when I have set the port direction to Input (=Z state), both LEDs still light; less than normal, but still enough to make this look not good.

The SMD transistors have built-in 22k resistors like this (the PNP one has same connections):
<img src="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1964&stc=1&d=118901608 9">

I tried to add 1k resistors at each Tr's Base, but... still the same.

Is there a way to sort out this issue?

NB: how do I put text beside the images???

[Acetronics2]

Hi, Roger

Supposing your leds are 5v powered ...

inserting 3x "1N4148" in series in EACH "base" connexion, AND two resistors ( equal high value ) to set "third state" output to V+/2 ...

that should work !!!

Alain

PS: dis, tes montages tordus ..., tu vas les pècher où ??? ... LOL !!!

[RussMartin]

Can the circuit be simplified?

Is there a special reason that each LED must have its own transistor driver? And that one is PNP while the other is NPN?

Have you considered using one transistor to drive two LEDs in series? (Two LEDs in series use the same 50 mA of current!) It would reduce your parts count a lot.

Is there a special reason for setting the output pin to high-Z (input) to turn the LEDs off? (Are you then using the pin as an input?) Have you considered leaving the output pin as output, adding a pull-up (or pull-down) resistor, and simply toggling it high and low?

[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:

<IMG SRC="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1971&stc=1&d=118905553 3">

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.