How can I turn ON-OFF this oscillator circuit?

Question

On my prototype board I used an integrated oscillator which gave me the desired frequency without any extra component. Then it's being divided by a divider ic. The clock pulse going from the oscillator to the 8 stage divider 74AHC1G4208 is turned on and off by an AND gate. The ON-OFF signal comes from a D Flip-Flop CD4013. On the prototype I use a 10 stage divider. But in the final version I will use a 8 stage divider, which is the same. So I linked to the 8 stage version of the chip.

This circuit works perfectly. Now that I plan production in some quantity, I would like to replace the oscillator ic by a much cheaper and also more available, crystal and its passive components, as shown on this schematic from the divider ic 74AHC1G4210 datasheet. This chip has an built-in output X2 for oscillator circuit. X1 is the clock input being divided. I plan to copy this circuit.

The crystal is an IQD 4.9152 MHz.

Could you tell me how I can turn this circuit on and off as I did with the AND gate. If I put an AND gate between the crystal and X1, I imagine it won't work because the AND gate will prevent resonance with the crystal.

Maybe I could sink the current coming from X2 to the crystal, limited by R1 (in this case 4.7K or 10K) with a N-MOSFET? But wouldn't that perturb the crystal?

The first 5 or 6 pulses can be irregular, as long as the next ones are at the good frequency.

I also have no idea what value should be for Rbias, C1 and C2, except from the indication on this schematic. It reads: "R1 is the power limiting resistor, its value depends on the frequency and required stability against changes in VCC or average ICC. For starting and maintaining oscillation a minimum transconductance is necessary, so R1 should not be too large. A practical value for R1 is 2.2 kΩ."

Answer 1

Could you tell me how I can turn this circuit on and off as I did with the AND gate.

You could feed the enable signal into the X1 pin through a diode (eg. 1N4148) with Anode to X1 so it pulls down to ground and turns the oscillator off when low, but doesn't affect it when high.

The first 5 or 6 pulses can be irregular, as long as the next ones are at the good frequency.

Without knowing the exact requirements of your circuit I can only say that crystal oscillators take a long time to the start up, so if you try to turn the oscillator on and off directly there will be a lot more disturbance than just 5 or 6 'irregular' pulses. There could be a delay of tens or hundreds of milliseconds before the oscillator level builds up enough to clock the divider reliably.

Since your existing circuit works, I suggest using a separate Gate IC (eg. 74LVC1GU04) for the oscillator.

Answer 2

This type of oscillator is called a Pierce Oscillator.

Could you gate the oscillator signal after the eight stage divider, if you could live with the first pulse after the oscillator is enabled being an odd length? Failing that you could use your own NOT gate to build the oscillator. Gating then possible. This enable/disable gate should be a schmitt Nand to square up the oscillator's output because Rbias biases the NOT gate into the linear region.

C1 & C2 would be chosen so that CL (the total load capacitance) on the crystal is 18pF, the data sheet recommended value. The total load capacitance on the crystal is made up from C1, C2, the input and output capacitances of the inverter + strays. Using a value for CL away from the recommended load capacitance will pull the crystal off frequency.

Using the standard equation for calculating a CL of 18pF requires C1 = C2 = 27pF. A ball park figure for the input and output capacitances of the inverter would be 5pF and stray capacitance could be taken to be 2pF. Then using this equation.

As a rule of thumb R1 should have a value equal to the reactance of C2 at the oscillator's frequency.

R1 = 1/(2 * pi * 4.9152M * 27pF) = 1k2

Once built you would need to check that it functions correctly over a range of temperatures.