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A square wave test signal generator for testing electronic projects (2)

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  • MultiMedia SRL
    SO WHAT GOOD IS THAT? A lot. A unique feature of square wave signals is that they are RICH in what we call harmonics . You not only get 5 Khz, you also get
    Message 1 of 2 , Oct 30, 2002
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      SO WHAT GOOD IS THAT?

      A lot. A unique feature of square wave signals is that they are "RICH" in what we call "harmonics". You not only get 5 Khz, you also get 10, 15, 20, 25 ........450, 455, 460 Khz etc. forever etc. until the signal becomes too weak.

      Now here is the beauty of the device, if we build a multivibrator running at say around 2 Khz we're going to get those rich harmonics all the way up into the r.f. region. If we are trouble shooting an audio amplifier we have harmonics at 2 Khz. 4, 6, 8 etc. which our audio amplifier can easily reproduce. If we are trouble shooting an i.f. amplifier at 455 Khz then I would expect the harmonics to still be quite stong enough to be heard up there as well as being quite usable much higher in frequency.

      It's cheap to construct, it's relatively easy to construct - with some practice for some novices and, it's a pretty versatile tool to own.

      HOLY MOSES WHERE'S THE MULTIVIBRATOR CIRCUIT?

      Here it is in fig 2.

      multivbrator / square wave signal generator schematic diagram


      Fig 2 multivbrator / square wave signal generator schematic diagram

      Firstly you will notice it is powered by 9V which of course can easily be provided by a convenient size battery (so it's portable). There are only two general purpose NPN silicon transistors used (types to consider 2N2222A, BC107 - nothing sacred here just any convenient NPN silicon you can easily and cheaply get hold of). Find out for absolutely certain the pin outs! That is which are EBC. It needs a gain (Hfe) of at least 120 with these values - see addendum later.

      HOW DOES IT WORK?

      This circuit consists of two cross coupled amplifiers comprising Q1 and Q2 ( NOTE: where the diagonal lines cross over in the schematic in fig 2 (and later fig 4) there is NO connection.

      REPEAT AFTER ME: "where the diagonal lines cross over in the schematic in fig 2 (and later fig 4) there is NO connection".

      Neither of these transistors are EXACTLY identical so when we switch on, one of them is going to take control first. Let us assume this is Q1. It will turn on hard (just like a light switch) but C1 will now begin to charge up through R1 (timing is R1C1) and because the base of Q2 is connected to this point it will rise toward the 9V with the same timing. At about 600 mV Q2 will start to conduct (switch on) and the voltage at the point of connection between the collector of Q2, C2 and RL2 will begin to drop.

      Because this decrease in voltage is coupled back to the base of Q1 through C2 so Q1 begins to switch off with Q2 now switching on even harder but C2 will now begin to charge up through R2 (timing is R2C2)and the whole process just repeats itself over and over. All the while the voltage at the collector of Q2 switches from about 9V (when Q2 is OFF) to 0V (when Q2 is ON) - BTW in case you're wondering, the 9V disappears as a voltage drop across RL2. Now in the real world you won't get a switching action of exactly between 0 and 9 volts AND it won't be an exact square wave as depicted in fig 1. What you will get is a rough approximation to suit our end purposes, i.e. an harmonic generator to use for trouble shooting.

      Now for the purists I could have given a more detailed and scientific explanation BUT is that what we're on about here with this beginner (and even not so beginner) electronic project?

      The ON / OFF period (called "mark/space" ratio for the purists) - look back at fig 1 that's what it's for, is determined by R1C1 and R2C2. If we make R1 = R2 and C1 = C2 it can be shown mathematically, (and I won't bore you) that Fo = 0.7 / (R * C).

      Using this formula and assuming we have a resistor value of 39K for R and a capacitor value of 0.01 uF (same as 10 NF) we get a frequency of approximately:

      0.7 / [ 10 X 10-9 X 39 X 10+3]

      Using my calculator I get an answer of:

      0.7 / 0.00039 = 1795 Hz

      Did you get that answer? NO? If not, why not?

      In case your confused we are using the exp function on your calculator. 10 X 10-9 means I entered 10, then hit exp, then entered 9, then hit the +/- key (not the subtract key). Do the same for the 39 X 10+3 figures (having hit the "X" or multiply button first) but this time not using the +/- key. I put that "sum" in memory with M+ then enter .7 and divide with RM (recall memory). Your milage might vary with the calculator you use. Learn how to use it!

      Now 1795 Hz is ideal for our purposes. NO it doesn't have to be exactly 2,000 Hz or 2 Khz.

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