Help With 555 Timer Circuit Transistor Swap

Discussion in 'The Projects Forum' started by jimburns, Sep 29, 2011.

  1. jimburns

    Thread Starter New Member

    Jan 24, 2011
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    Hi folks, have a question I'm hoping someone can help with, my electronic skills are terribly rusty and I'm outside my comfort zone on this.

    I have a 555 timer circuit I borrowed from a site and got it put together on a breadboad and it works just as I need and expected it to.

    But given my stock of parts, I set out to explore how to swap the NPN BC547 transistor with a PNP, in this case I've been trying to use a 2907A.

    I've just had no luck. Attached is the original circuit and some of my scrawl.

    Ok, again, rusty remember... I think my biggest problem is the right polarity on the base lead. But since this ties directly into the output of the 555 I've gotta fully understand the 555, which I don't. I haven't done this sort of thing in years and I'm flying by the seat of my pants.

    I sorta understand that the output is being fed back into the lower comparitor, pin 2 I think and a logic probe on pin 3 with the NPN transistor does show the output in constant hi/lo pulsing. But I'm unable to keep the correct operation of the 555 part of the circuit while I try to modify it for the PNP....

    In my case I'm using a piezo motherboard buzzer... and in general it works pretty good.

    I'm not using the same caps, I'm using two .47uf caps instead.

    They have an effect on tone pitch but unlike in other basic 555 circuits without the transistors where they provide reverse current to keep the buzzer sounding during switching of the timer circuit (that I thought I at least understood fairly well), this one, I think uses the caps to cause the 555 to pulse which drives and undrives the transistor to the buzzer leg. Any confirmation and clarification on this is appreciated.

    At this point, I'm just dying to understand the entire thing better. Anyone in the mood for some explaining? :)

    Thanks in advance,

    /jim
     
    Last edited: Sep 29, 2011
  2. DickCappels

    Moderator

    Aug 21, 2008
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    I do not understand what you are trying to do, but generally, PNP's cannot be used to replace NPN's, no matter how much you swap leads. If you need some BC546's (very similar to BC547) send a PM with you snail mail address.
     
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  3. iONic

    AAC Fanatic!

    Nov 16, 2007
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    When looking at the 2907(PNP) - the flat side, the pinout is EBC. The emitter in this case should be tied directly to the source voltage and your components tied to the collector to ground. But the base is biased by a low signal from the 555.

    [​IMG]
     
    Last edited: Sep 29, 2011
  4. jimburns

    Thread Starter New Member

    Jan 24, 2011
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    iONic, thanks for the thoughts... You are right, I flipped the E and C, as you can see in my sketch, I added the PNP transistor in sharpie out to the right of the original circuit.

    So, are you saying it matters where my load goes? Because you're right, I flipped the E & C but I left the buzzer and small resistor between the new Emitter and source. I didn't think exactly where the load was placed in the E-C side of the circuit mattered as long as we had charge flowing.

    About the base bias... forgive me as I try to understand... in the original circuit with the NPN, with the Base being positive, yes, I agree it's being biased low...

    then when I move to PNP, shouldn't I reverse this? Everything I've read over the last couple of days suggests that, in terms of what's actually happening inside BJT's, there's no reason we can't alternative NPN or PNP but we have to reverse our bias voltages appropriately. Not sure what I'm actually saying here..... :) D'oh!

    Thanks....

    /jim
     
    Last edited: Sep 29, 2011
  5. jimburns

    Thread Starter New Member

    Jan 24, 2011
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    Thanks Dick, I appreciate the thought.

    I'm not sure I agree... of course, I don't have the electronics experience to argue the matter... I do know, of course, that given any circuit, you can't just swap NPN/PNP and expect it to work as designed. I understand that.

    I'm trying to modify the original 555 circuit so I can get a PNP transistor to drive my buzzer based on the 555's output.

    I do have other NPN's.... and they work.... it may have started out because I didn't have many NPN options but it's turned into an interesting exercise now as I try to knock some dust off....

    In terms of how the circuit works, it's a great conductivity circuit, which, with a few modifications and additions, I've made into a small matchbox sized water alarm for under sinks, in water closets, etc... anywere you want to monitor for leaks.

    I just think it would be nice to have a circuit design for different types of transistors....

    Many thanks,

    /jim
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    One of the problems with trying to use a PNP from a 555 output, is that the 555's output is a Darlington voltage follower. It can go nearly to 0v, but when the output is high, it won't go to Vcc.

    So, if you are powering the 555 with 6v, the output (pin 3) will only be ~ 4.7 to 5.3v, give or take. The problem here is that the PNP transistor will start conducting when the base voltage is ~0.63 less than the emitter voltage.

    The PNP transistor will never really get turned completely off.

    So, along with your 1k resistor from pin 3 to the base of your PNP transistor, connect another 1k resistor from the base to your +6v supply. This 2nd resistor will turn the transistor off when the 555 output is trying to go high.

    You also need to move the load (the buzzer and the 39 Ohm resistor) to between the 2N2907's collector and ground.
     
  7. jimburns

    Thread Starter New Member

    Jan 24, 2011
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    Ok, are V measurements like this all relative to ground?


    If you look closely, the line between pin 3 and the 33k resistor on the left of the 555 at Pin 8 is cut.... I took that to mean it doesn't connect to...

    You'll notice, and I scratched it out, but I tried to disconnect the path from Pin 3 to the 33k resistor and at the top right corner notice I pulled a resistor up to the source rail. It didn't help me but...

    See my last note, I'm confused on this. Figuring that if charge was flowing, well then... charge was flowing, I'm not seeing how it matters what side the load is on. Can you say more on this?


    Thanks for the discussion, I know it's probably dopey to a lot of you, but you'd be surprised how much every little clue helps me pull more and more "ahhha's" out of my skull.

    Best,

    /jim
     
  8. jimburns

    Thread Starter New Member

    Jan 24, 2011
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    0
    Are are correct. You can see it in my notes in the sketch, but an ammeter confirms that not only does the PNP not close, it pulls more current with the probes open or off than with them closed. Of course, that's mostly a lot of nonsense since I understand that the entire circuit needs to be modified... but I was trying to see what I could see.....

    How did you know the PNP trans will conduct at .63 less than emitter....

    I AM looking at the datasheets.... just bear in mind, they don't make a lot of sense just yet! :)


    Best,

    /jim
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    When I suggest a voltage without adding some point of reference, the voltage is using ground as the 0v reference; which is the default.

    Sometimes I'll use a term like "5v Vce", which for a transistorized circuit means that "there is 5v on the collector, as referenced to the emitter" - the emitter is the 0v point in that case.

    Oops! No, it's supposed to connect.

    That's an older way of showing that wires are crossing each other rather than being connected. Another way you'll see is a wire jumping over another, sort of like an Omega symbol but with straight sides.

    Nowadays, wires are drawn simply crossing each other without the breaks and the jumps. Connected wires (a junction) are shown by a dot.

    You need to have that 33k Ohm resistor connected from pin 3 to pins 2 & 6, and the 10nF cap from pins 2 & 6 to ground - otherwise, the 555 won't run.

    When you're using a transistor as a saturated switch, the load goes on the collector, and the emitter gets connected to ground (if NPN) or +V (if PNP)


    It's not dopey. :) It's a stage of learning or re-learning electronics.

    I have re-drawn your circuit using the PNP transistor and the resistor that I suggested, along with moving the load to the PNP's collector.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    I've been at this a week or two longer than you have. ;)

    Seriously though, 0.63v is about where you start getting some usable current from the collector. If you look in the datasheets, they'll specify "cutoff" voltage somewhere around 0.5v/500mV; that's where the collector current is at some minuscule amount; basically [pun intended] considered turned off.

    It'll take time. Motorola/OnSemi and Fairchild have pretty good datasheets on their transistors; lots of graphs for you to look at. The saturation curves are very helpful to look at. You'll almost always see "Ib=Ic/10" somewhere on the plot.

    For example, pull up OnSemi's datasheet for a 2N3903, 2N3904 NPN transistor:
    http://www.onsemi.com/pub_link/Collateral/2N3903-D.PDF
    These are pretty common, and carry a maximum Ic rating of 200mA. However, look at page 7, Figure 16, the collector saturation region plot. Notice that the highest the plot goes is Ic=100mA with Ib=10mA; beyond that, Vce(sat) gets too high and power dissipation in the transistor increases dramatically.
    Take a look at Figure 17, "ON voltages", just down and to the left. There you can see that when Ic=1mA, Vbe is that 0.63v I mentioned earlier. As Ic goes up, Vbe needs to increase. You can see that after 100mA, the Vce is rapidly getting higher.

    A basic "rule of thumb" for transistors is to look at the maximum rated collector current - and don't go over 1/2 of that. If you need more current, then find a transistor with a higher rating.
     
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