problem with 3904 transistors in a 555&4107 circuit

Discussion in 'The Projects Forum' started by beginnersluke, Oct 11, 2015.

  1. beginnersluke

    Thread Starter New Member

    Aug 26, 2015
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    Hi,

    I'm prototyping a circuit and the 3904 transistors are not doing what I'd expect, and I'm not sure why. I was hoping maybe someone here could help me track down the problem, or just point out the problem with my thinking.

    Here are the circuit basics. I'm basically building a countdown clock, which uses a 555 & 4107 to sequentially light up 10 LEDs, then flash the last LED a few times.

    I'll add a basic schematic, but basically it's an astable 555, with a 1 minute cycle, with the 555 output providing the clock input for the first 4107. The first 9 outputs feed the first 9 LEDs. The 10th output feeds power to the next 4107 (and disabled the 4017), which uses the same clock output as the first. This lights up the last LED. The second output powers another astable 555 and 4107 (and disables further counting on the 4107).

    (I know the datasheet recommends a different circuit to cascade 4107s, but since I don't ever need it to come back around to the first output, my implementation above seemed to be all I needed).

    Now, I was worried about drawing too much current through the 4107s to power the later subcircuits, so I though, I'll run that output and use it to switch on a transistor, and that will be a better way of powering the further components. (Maybe this wasn't a problem, I'd be curious about this too. I could probably struggle through the calculations of what was necessary, but just figured this way, I'd just be safe.)

    HERE IS THE PROBLEM and QUESTION.

    I connect the 4017 to the base of the 3904 (though a base resistor), connect the collector to the power rail, and the emitter to power the subcircuit.

    However, the voltage I get at the emitter is LOWER than the voltage from the 4107. This seems wrong to me.

    For example, on the first transistor, the voltage out of the 4107 is 5.2v (this is the v on the power rail for the circuit). After the base resistor, I read 4.9v. At the emitter, I read 4.6v.

    Shouldn't I be getting 5.2v, the rail voltage, at the emitter?

    This problem gets worse at the next 4107 output/3904 transistor (i.e the voltages get even lower).

    (For now, I've rebuilt this with a series of inverters built with these same 3904s -- inverting the 4107 output twice, giving me the rail voltage.)

    The circuit works well (even with the basic transistors, it works, the LED is just dimmer than it could be), but I can't figure out why the transistors aren't acting as I'd expect. (For the record, I figured when the base gets powered, the transistor will "switch on" and give me the voltage of the rail at the emitter.) I also tried connecting the first transistor in a darlington pair (i.e. added a second transistor), but the voltage dropped even more (again, to my surprise).

    Can anyone help me figure out what's going on here?

    Thanks so much,

    Luke
     
  2. #12

    Expert

    Nov 30, 2010
    16,261
    6,770
    Basic beginner mistake. The voltage at the emitter is ALWAYS lower than the base. If you want full voltage, put the loads on the collector side and ground the emitter. And don't be stingy with the drive current. The transistor base needs a tenth of the load current in order to switch it on good and hard.
     
  3. KJ6EAD

    Senior Member

    Apr 30, 2011
    1,425
    363
    Luke, stop calling the 4017 a 4107.
     
    beginnersluke likes this.
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    As pointed out, it looks like you want to use a common emitter saturated switch, but are using an emitter follower instead. One way to fix that is to use a PNP 2N3906 with the emitter tied to Vcc and the collector going to the chip. But that is not the preferred way to control signals.

    Both the 555 and the 4017 have enable inputs. A better way is to power all chips continuously from the circuit power source, and use the enable inputs to control the outputs. Take a swing at that arrangement and post another schematic.

    Also, the first 4017 will continue to cycle after it reaches the 9th output; you do not show it being disabled. Also, the Q9 output can enable the next 4017 directly, without a transistor in the way. Also, the 2nd 4017 can enable a 555 directly without a transistor in the way.

    ak
     
  5. beginnersluke

    Thread Starter New Member

    Aug 26, 2015
    10
    0
    Ha!

    I'll try. If only you knew how often in life I do this. The thing I hate most are the 57 mph speed limits on the interstate. Why'd they make it so precise? :)

    Thanks,

    Luke
     
  6. beginnersluke

    Thread Starter New Member

    Aug 26, 2015
    10
    0

    Thanks, yeah. I'll play around with the enable controls.

    I didn't note it in the schematic, but I use those to keep the 4017s from cycling back around. So using them for both purposes will take a bit of thinking and maybe some logic gates, but it may be a better way. I will certainly play around with it and post an update.

    As far as just skipping the transistors and powering the next phase directly from the previous 4017 output, I just wasn't sure how to go about calculating the current drain and whether this would work. I could have spent the time to figure that out, but just figured I'd use the transistor switch in a 'better safe than sorry' way of thinking. (And I figured I'd learn something from trying it that way which I did.

    So long story short, I'll try using the enable pin and some logic gates, because that does seem like the proper way to do what I want to do. On the other hand, if I can just run it all through the chips and skip the logic gates (since I'm already using the enable pins to stop the counter), then that would yield the simplest circuit.

    Thanks a lot for the pointers everyone. I'm mostly doing this to have fun and learn, so I'm succeeding on both counts.

    (The real success will be if my little 10 minute countdown clock actually prompts my daughter to eat her breakfast in a reasonable amount of time, but that's another story!)

    Thanks again,

    Luke
     
  7. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    You misunderstood my response. My point was *not* to power any chip from anything except the Vcc power rail. Powering one chip from the output of another, might work, but it is a terrible practice to adopt. Power pins are for power. Enable pins are for enabling. Never the twain shall meet.

    ak
     
  8. beginnersluke

    Thread Starter New Member

    Aug 26, 2015
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    0
    Ah yes. Sorry. I'll work that up. Thanks again.
     
  9. beginnersluke

    Thread Starter New Member

    Aug 26, 2015
    10
    0
    Oh, I know why I did not use the enable pins to control the 2nd and 3rd 4017s. Because until I want that second 4017 to start counting, I need it to have no output at all. If it starts on zero and doesn't start counting (isn't enabled) until it's time, it will still be lighting it's LED, which should be off during that part of the sequence.

    I've now tried to google for a good way around this (have enable not only enable counting, but any output at all), but haven't found anything good.

    I suppose an AND gate would do it. Input 1 would be the inverted enable signal, and input 2 being Q0 on the 4017. The LED would only light when both the enable is pulled low (to start the Counter) and that output is on.

    That was the problem though, that originally stopped me from pursuing that design. Are there any better tricks to accomplishing that?

    Thanks again; this forum is great.

    Luke
     
  10. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,528
    1,248
    Based on that additional information, nothing will be a simple as what you started with, but better-implemented. As mentioned above, use a PNP as a saturated switch. Emitter to +5V, collector to the 4017 Vdd, and the base through 3.3K to the first 4017. Add a 0.1 uF cap from the collector to GND. A 10 K resistor from the collector to GND will help the part turn off more quickly.
     
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