TS555 (cmos) not working properly in NE555 circuit

Discussion in 'General Electronics Chat' started by Standisher, May 29, 2015.

  1. Standisher

    Thread Starter Member

    Jan 16, 2015
    47
    3
    Hi, I'm new to this forum and relatively new to electronics so hope you'll be gentle with me :) I built a very effective signal generator using the NE555 IC in astable mode. It generates sound at a good volume though a mini speaker and simultaneously illuminates a LED showing the output. The circuit operates on a 9V battery and is still functioning well after 3 months on the same battery. I Have just purchased some (10) CMOS 555's (believing that they would be compatible with all NE555 circuits) and thought I'd pop one into the IC socket of the signal generator to compare output. I was shocked to find that the volume of the signal through the speaker was considerably lower (almost inaudible) and the LED extremely dim. It would appear therefore that the output voltage of the TS555 is a lot lower than that of the NE555? I'm not very good at interpreting datasheets yet but a quick glance at the relative datasheets seems to confirm this (if High-Level output voltage is what I should be looking at?). Would be grateful if you could confirm or refute my suspicion that output voltage of TS555 is lower than NE555.
     
  2. ian field

    Distinguished Member

    Oct 27, 2012
    4,415
    784
    The CMOS part can't source or sink as much current as the bipolar one.

    You could try adding a complementary pair of emitter followers between its output and load.

    With a square wave you might get away with just connecting the 2 transistor bases together, if you get a crossover glitch, you'll need a pair of bias diodes between the bases - then it gets more complicated, you can feed the square wave in between the 2 diodes put you need a couple of bias resistors to pull each base toward its nearest supply rail.
     
    Standisher likes this.
  3. Standisher

    Thread Starter Member

    Jan 16, 2015
    47
    3
    Thanks for the speedy response Ian. I'll experiment with your suggested possible solution (on breadboard as the current 'boxed-up' generator is working perfectly well with the old NE555 IC) so that I will know what is required in future projects.
     
  4. ian field

    Distinguished Member

    Oct 27, 2012
    4,415
    784
    The CMOS 555 can work to about 2MHz, while the bipolar type runs out of puff at around 200kHz. Emitter followers that I suggested can be a bit slow - so you wouldn't get the full 2MHz the CMOS part is capable of.
     
    Standisher likes this.
  5. crutschow

    Expert

    Mar 14, 2008
    13,050
    3,244
    At audio frequencies a complementary emitter follower driver should work well to amplify the output current, even higher than what the standard NE555 can deliver.
    Use the common 2N2222 and 2N2905 or 2N2907 for example.
    It needs no additional resistors, just the two transistors.
     
    Standisher likes this.
  6. Standisher

    Thread Starter Member

    Jan 16, 2015
    47
    3
    That's interesting, if I have understood it properly. Do you mean something like this:

    [​IMG]

    If so, would the input Voltage to the transistors be taken directly from the 9V supply?
     
  7. ian field

    Distinguished Member

    Oct 27, 2012
    4,415
    784
    Complementary pair emitter followers were used to buffer the video output on the better resolution CRT PC monitors, but they required peaking inductance in the cascode collector circuit that was driving them. It was still a compromise, emitter followers need careful design to get much speed out of them, but they did a better job of driving the cathode capacitance than the high output impedance cascode stage.
     
  8. ian field

    Distinguished Member

    Oct 27, 2012
    4,415
    784
    In an audio amplifier, that as drawn will introduce crossover distortion - each transistor needs 0.7V between base and emitter to make it conduct at all, so you get a small dead zone on the zero crossing point. Normally you'd have a series pair of diodes between the bases to provide them with the 0.7V each.
     
  9. crutschow

    Expert

    Mar 14, 2008
    13,050
    3,244
    Yes.
    But upon further thought, you don't need the PNP when driving a low impedance speaker connected to ground for a 0-9V signal, you just need the NPN emitter follower.
    Connect the collector to 9V, the base to the 555 output, and the emitter to the speaker.
    That will more than fast enough to amplify the audio square-wave from the 555.
     
    Standisher likes this.
  10. Standisher

    Thread Starter Member

    Jan 16, 2015
    47
    3
    That's excellent. Many thanks crutschow.
     
  11. crutschow

    Expert

    Mar 14, 2008
    13,050
    3,244
    I should amend that to say that you don't need the PNP if you are not driving the speaker through a capacitor in series (to block the DC component in the signal).
     
    Standisher likes this.
  12. Standisher

    Thread Starter Member

    Jan 16, 2015
    47
    3
    That's absolutely fine. The speaker is not being driven through a capacitor for this application.
     
  13. flat5

    Active Member

    Nov 13, 2008
    403
    17
    crutschow, I think this is what you suggest.
    Emitter follower1.png
    I would normally do it this way.
    Emitter follower2.png
    Aside from not using a resistor, is there a good reason to put the load in the emitter path instead of the collector path?
    (I could not think of a better word than path.)
     
  14. crutschow

    Expert

    Mar 14, 2008
    13,050
    3,244
    One advantage of the emitter follower circuit (besides no resistor) is that it does not saturate and thus has no saturation delay as the second circuit would.
    That being said, the saturation delay will likely not be enough at audio frequencies to be noticeable.
     
  15. flat5

    Active Member

    Nov 13, 2008
    403
    17
    Could the saturation be controlled by that resistor? The resistor could make sure that the transistor does not fully conduct, right?
    Is it also true that the speaker receives all the current if the transistor is switched, so providing a little more audio output in the first circuit?
     
  16. crutschow

    Expert

    Mar 14, 2008
    13,050
    3,244
    You can't reliably control saturation by the input resistor since current gain varies over a wide range between different transistors of the same part number (look at a data sheet) and also varies with temperature.

    Yes, that configuration gives a slightly higher voltage across the speaker (the supply voltage minus the saturation voltage of perhaps 0.1-0.2V versus an emitter follower which can only go to within about 0.7V of the supply).
     
    Standisher likes this.
  17. Standisher

    Thread Starter Member

    Jan 16, 2015
    47
    3
    Just tried your suggestion crutschow, very effective and very simple. Many thanks for your input.
     
  18. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,147
    204
    Note: The design equations are slightly different too.
     
Loading...