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Microcontrollers and digital systems

How to connect frequency divider 7474

jafsuso1
hace 5 años
hace 5 años
xtal-tester.gif [4.89kb] Hello everyone, I have a crystal tester and a 20 mhz oscilloscope, I would like to make a frequency divider to be able to check crystals of more range because I am limited to those 20 mhz of the oscilloscope but if I make a divider I could do what I want . Well, the problem arises where to connect each leg of the 7474 splitter and how to do it taking into account that: 1 = NC 2 + 6 + 11 = inverter output of splitter x4 3 = frequency input 4 = NC 5 = splitter output x4 7 = mass (0V.) 8 + 12 = inverter inverter output x2 9 = splitter output x2 10 = NC 13 = NC 14 = Vcc (+ 5V) where I connect 2 + 6 + 11 = inverted output of the splitter x4 8 + 12 = inverter inverter output x2 and the frequency input from where I take it from the tester? I think I'm getting a mess, but an orientation would appreciate it. Thank you very much in advance to the community
ricbevi
ricbevi
17.055
hace 5 años
hace 5 años
The crystals in fundamentally rarely exceed 20MHz. Those that are marked in higher frequencies are third or fifth harmonic over-tone and with that circuit that samples will be unable to oscillate in over-tone. Ex: a crystal marked at 26,965MHz (Channel 1 of BC) actually oscillates at approximately 9,321.6MHz in fundamental and will do so in the third overtone on the other frequency. I don't know why you want to see the oscillation if you want to measure it with the visual indication (led in your scheme that started the oscillator) and a frequency meter is enough. If you do not have a frequency meter and you are going to work with crystals, it would be good to think about building one or buying. Personally I have frequency meters built for more than 30 years (when the thing was not so easy to order and buy) and they work without problems every day. On the Web it is flooded with frequency meters to build based on microcontrollers and a few more elements. I doubt that with that scheme you can attack the digital input of a 7474 and make it divide stably. If you enter 3 the output of the first splitter [b] x2 [/b] it is 5 connected to 11 (input) and the splitter x4 comes out 9 or 8 Regards. Ric.
juite
juite
372
hace 5 años
hace 5 años
Remembering that you are in TTL, do not forget the signal levels. This is the basics for a divisor by four:
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