# questions about crystal oscillators

Discussion in 'General Electronics Chat' started by msoof, Sep 11, 2008.

1. ### msoof Thread Starter New Member

May 28, 2008
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0
hi all,

I was reading about crystal oscillators (consisting of invertors and series/parallel resonant quartz) and had these queries in my mind. Hopes u guys can help...

1. What starts the oscillations? What's the very initial input that is fed into the invertor input? An AC or a DC? Will either one work?

2. Assuming we have a uController which utilizes the pierce oscillator mode. When the uController is powered up initial, I guess the very first 'task' is to establish the clock frequency. What is being fed into the invertor input? If AC, what is the frequency of this AC? How does the uController establish this AC's frequency when the crystal oscillator haven't even start oscillating?

2. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
218
Here is an example link to a tutorial on the subject. There are many others that can be accessed through google.

hgmjr

3. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
Have a look at this Wikipedia entry:
http://en.wikipedia.org/wiki/Pierce_oscillator

When power is first applied, it's most likely that the input to U1 will be at a low level due to C1, and that will make the output of U1 high. If that isn't enough to start it oscillating, current will flow from the output of U1 through R1 to charge C1, eventually raising the input to a logic high level, which will toggle the output of U1 low. The sudden transition will likely trigger the xtal into oscillating. If not, C1 will discharge via R1 to the output of U1 until it again toggles the state of U1. The RC time is generally significantly lower than the xtal frequency. It takes a number of oscillations for it to become stable.

Apr 5, 2008
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5. ### msoof Thread Starter New Member

May 28, 2008
6
0
hi there,

from what u said that 'When power is first applied, it's most likely that the input to U1 will be at a low level due to C1...',

- is the very initial input to U1 at a low level due to the charging of C1? i.e. the DC fed into U1 will go to C1 to charge it up...and C1's voltage will ramp up..

in other words... we can just feed in a DC to get the oscillator started, making use of the capacitance of the C1, or the internal crystal shunt capacitance C0 (if we are not loading any external capacitors).. true?

The reason for my asking is because, my IC requires a 24mhz crystal oscillator and upon power up, probing the crystal shows a flat DC on both ends of the crystal. i am using a 24mhz crystal with an external 1Mohm bias resistor and a pair of 10pF to ground on each side of the crystal.

i am wondering if its a dead crystal? or maybe its an IC problem?

thanks very much for your inputs...

6. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
That's not really what I was implying. Actually, on that side of the inverter, C1 has no charge path except through the resistor from the output of U1. This is assuming that the input of U1 has nearly infinite impedance.

I tried to explain it in very simple terms. You're going off base with this talk.

You can't probe a crystal oscillator circuit successfully with a meter, O-scope, or whatever. If you're looking at the output and it's not oscillating, you have a problem.

Try creating a simple oscillator circuit with the crystal.
Perhaps use a 4093 NAND gate as a driver, a 20pF cap to ground on the intput, the XTAL between the input and output, and a 20pF to 120pF cap to ground on the output. Use another stage on the 4093 to square up the waveform. Make sure to ground the unused inputs.

Crystals break; they're very fragile. If you drop them from more than a couple of feet to a hard floor, it's likely that they are fractured.