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enter image description here

I tried to find an appropriate answer for the question but couldn't find what I needed exactly. I found out that "Initially the output from the PIC is floating and we use the pull up resistor R7 to turn OFF the transistor". My specific questions are:

  1. How will the floating output from the PIC affect the base of the transistor? Will it make it fluctuate between ON/OFF state?
  2. What would happen if we don't have the pull up resistor?
  3. Is there any risk of the transistor being destroyed because of the floating output form the PIC?
  4. How will the pull up resistor put the transistor in OFF state when there is floating output from the PIC?
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  • \$\begingroup\$ Why do you think the output would float? \$\endgroup\$ May 16 '16 at 12:13
  • \$\begingroup\$ Because that is the default state in which the microcontroller starts. @IgnacioVazquez-Abrams \$\endgroup\$ May 16 '16 at 12:14
  • \$\begingroup\$ The default state of many embedded processors is "input". But you can turn that around very quickly to an output at boot up. Using a pull up resistor in your design can ensure the state of the transistor is always known. \$\endgroup\$
    – st2000
    May 16 '16 at 12:32
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A floating output can be due to the pin being configured as an input (quite common) or it is truly tri-state

1.How will the floating output from the PIC affect the base of the transistor? Will it make it fluctuate between ON/OFF state?

With the pullup in place, the transistor is solidly pulled to the OFF state.

2.What would happen if we don't have the pull up resistor?

Without the pull up, the state of the transistor will be indeterminate; there will only be leakage (in the order of μA) in or out of the controller pin, so the transistor might be either on or off, or possibly fluctuate as you mention.

3.Is there any risk of the transistor being destroyed because of the floating output form the PIC?

Highly unlikely, but it might depend on the transistor load.

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  • \$\begingroup\$ is it possible to show me the diagrammatic representation of the question 1? It would be really helpful. \$\endgroup\$ May 16 '16 at 12:26
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    \$\begingroup\$ That may be difficult. We are discussing a situation that should not happen (not be designed) and if it does will be dependent on many factors as the values are very small. That is to say, we are concerned with such small values that we need to look at every circuit detail including noise. For instance, not all PIC processors are the same and not all pins on a PIC have exactly the same circuit inside. So already the problem of describing what might happen is complex. \$\endgroup\$
    – st2000
    May 16 '16 at 12:37
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Floating outputs are used in special cases. For example, when tying multiple signals together in a wired OR arrangement. Normal digital outputs are push-pull :

enter image description here

...and are driven both high and low. Such outputs do not need a pull up resistor. You should not operate circuits designed for digital purposes outside of saturation or cutoff. So you should not leave the input to transistors in a floating state. In this state, electrical noise can influence the transistor. Also, in this state, the transistor could be partially conducting and generate more heat. Heat that could damage the transistor as it was likely designed (chosen) "small". That is, it was expected not to have to operated outside of (lower power demanding) cutoff and saturation states.

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  • \$\begingroup\$ Could you also explain me, (how) will the pull up resistor pull the transistor to a OFF state? Especially in this case. @st2000 \$\endgroup\$ May 16 '16 at 12:34
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    \$\begingroup\$ A silicon PNP transistor is turned on when the base is lower by about 0.6 volts relative to the positive power supply. The beta or amplification of most transistor's is large (usually ~100) so it only takes a little current to produce large results. It follows that almost any pull up resistor (for example 1K ohms, 2K ohms or even 10K ohms) will be enough to raise the base to the positive power supply level and turn the transistor off. \$\endgroup\$
    – st2000
    May 16 '16 at 12:54
  • \$\begingroup\$ ok i understood that very clearly, but now i am confused with another question (what will the pull up resistor do when we want the transistor to be turned on? OR why does it only work during the floating of PIC?, wont it still raise the voltage at the base towards positive power supply level?) @st2000 \$\endgroup\$ May 16 '16 at 13:08
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    \$\begingroup\$ Yes, so we pick a pull up resistor large enough so the processor's output can still pull the voltage down without using too much current. For example for 5 volt logic a 10k ohm resistor would only take 0.5 mA That is, if the processor pin can sink 0.5 mA the transistor should turn on. \$\endgroup\$
    – st2000
    May 16 '16 at 13:16
  • \$\begingroup\$ ok, that was clear. One last question, Why does the PULL UP only work during FLOATING of the PIC output? because even though the PIC is floating it is still giving some output right? @st2000 \$\endgroup\$ May 16 '16 at 13:19

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