Clock Signal using an Oscillator

Discussion in 'The Projects Forum' started by MrL, Mar 24, 2012.

  1. MrL

    Thread Starter Member

    Oct 21, 2009
    46
    0
    Hi,

    I made a thread a couple of weeks ago about the generation of a clock signal, and i have now purchased an 8.192 MHz oscillator in order to generate this clocking signal. The oscillator i have purchased is shown below:

    http://uk.farnell.com/jsp/search/productdetail.jsp?SKU=1842156
    http://www.txccrystal.com/images/pdf/7w.pdf

    I have constructed the following circuit, using this oscillator:

    http://img849.imageshack.us/img849/1351/circuite.png

    I'm using a 9V supply along with a potential divider in order to obtain the 5V input.

    After testing with an oscilloscope, i am getting the correct frequency from the output. However, my high output voltage is only 1V, were as the datasheet states this should be 0.9 Vdd, which in this case would be 4.5V. Would anyone please be able to help me out as to why i'm only get 1V? Any help would be appreciated.

    Thanks.
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    The datasheet you posted says it is a 3.3V device.
    A potential divider will be loaded down by the current that the oscillator draws. You need a voltage regulator that regulates to the voltage required by your oscillator. For example, if you really do have a 5 volt oscillator, use a 78L05 or something similar.
     
  3. MrL

    Thread Starter Member

    Oct 21, 2009
    46
    0
    Ah yeah, my mistake, thanks. However, my potential divider was providing the intended 5V when connected to the oscillator, or would you still recommend a voltage regulator? Thanks.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Post a schematic of your oscillator with the potential divider connected to it, showing resistor values. The schematic you posted implies that the package has 14 pins. The datasheet you posted has only 4 pins. What's up with that?
    How are you measuring the output voltage?
    Why do you have 22pF connected from the output to ground?
     
  5. MrL

    Thread Starter Member

    Oct 21, 2009
    46
    0
    Here is my full clock circuit schematic:

    http://img220.imageshack.us/img220/957/clockcircuit.png

    I'm measuring the output voltage (from the oscillator) using an oscilloscope, and i have a 22pF capacitor connected from the output to ground as most example test circuits on various different datasheets had a load capacitor. Is this not required?

    Thanks.
     
  6. MrChips

    Moderator

    Oct 2, 2009
    12,446
    3,361
    I would omit C2 and add 10uF electrolytic in parallel with C1. I would change C1 to 0.1uF.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    You said you measured 5V on the oscillator power supply pin. This is not possible with the circuit you posted. Your voltage divider puts out 5V unloaded. The Thevenin resistance of this divider is 5.56kΩ. The oscillator draws a maximum of 20mA (see the datasheet) If we assume this is resistive, this is equivalent to a resistance of 5V/20mA=250Ω. This is loading your 5.56kΩ 5V source, so your loaded voltage is 215mV. It is probably higher than this, but it still won't work. You need a voltage regulator.
     
  8. MrL

    Thread Starter Member

    Oct 21, 2009
    46
    0
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Your oscillator output voltage, and maybe frequency, will be affected by the load you put on the oscillator. The voltage divider also wastes more power than a regulator, which will shorten your battery life.
     
    Last edited: Mar 27, 2012
Loading...