Tristated output to two common Base transistor issue

[RussMartin]

Scratch this post.

[Luciano]

Hi Russ,

These are the test results. I have used the old transistors
and also new transitors but the results are the same. The
picture below is with the new transistors.

The picture shows the two LEDs ON with about 25% luminous intensity.




Best regards,

Luciano

[Luciano]

Hi,

With the circuits we are exploring here, the two LEDs are turned OFF by setting
the I/O of the PIC as input. (Input tristated). When the I/O is set as input,
the voltage present on the I/O pin must be recognized as logic LOW or as a
logic HIGH. Undefined voltage levels are dangerous for the input buffer of the I/O.
Floating inputs on digital pins are a no-no.

Best regards,

Luciano

[RussMartin]

Thanks for an excellent result! (And the photo!) And thanks for bearing with me . . . I'm doing this on paper, not on a breadboard. I don't know what your LEDs are, so I'm having to guess at the Vf (I'm using 1.6 V) and assuming an If of 20 mA.

If you please, try increasing both Rcl's each by the same amount until the LEDs just go out, then test high and low at Rb again.

I appreciate your concern about floating inputs, but we can deal with that next. (I have something in mind.)

[Darrel Taylor]

Just to throw a little confusion in the mix, .... how bout ....

<table border=1><tr><td><img src="http://picbasic.co.uk/forum/attachment.php?attachmentid=1994&stc=1&d=118930280 7"></td><td>
Or, it may need a normal NPN
<img src="http://picbasic.co.uk/forum/attachment.php?attachmentid=1995&stc=1&d=118930343 1"></td></tr></table>
Also just on paper
<br>

[Bruce]

How about something like this?

[Luciano]

Hi Bruce,

Here are the results:


When Tab is open the whole circuit uses 1mA.
I have used 3 transistors BC558.

Thank you!

Best regards,

Luciano

P.S.

Do not connect LEDs in parallel like that with just one series resistor!
If you need to connect LEDs in parallel, use one resistor for each LED.

This was just a quick test to see if 50mA were available.