Crystal load capacitance?

samy555 · Aug 6, 2015

Hi
From page 1.17 of:

I read:

The author says: "Our crystal had marked frequency of 7045 KHz. This was the specified frequency for operation with a 32-pF load capacitance."
My question is: If I had a 20MHz crystal, how will I know the exact value of its load capacitance?
Thank you

betwixt · Aug 6, 2015

The load capacitance is what you specify when you get the crystal made. Ready made crystals will have the load capacitance specified for their frequency and you would design your circuit to match that capacitance. The frequency moves slightly as the capacitance changes, that's why its necessary to know it.

Brian.

KlausST · Aug 6, 2015

Hi,

Usually you find the capacitor value in the crystal datasheet.

Klaus

Audioguru · Aug 6, 2015

I think students do not look at the datasheet for anything because the electronic parts are cheap junk bought on ebay with no manufacturer's name and no datasheet.

samy555 · Aug 7, 2015

Audioguru said:
....... because the electronic parts are cheap junk bought on ebay with no manufacturer's name and no datasheet.

Yes, you're right
What do you do if you were this student?

Q2: For the RC LPF shown above, is it important to have a frequency of 159KHz, or it is not critical?

Q3: Is that oscillator a common collector colpitts oscillator?

thank you alot

betwixt · Aug 7, 2015

Q1. Buy parts from a reputable supplier. If that isn't possible, the best you can do is make a guess, most Quartz crystals are designed around 15 - 30 pF load capacitance.
Q2. Almost certainly it isn't critical, the values only determine the impedance at the collector. The capacitor is essentially keeping the collector at ground signal level and the resistor is providing a degree of signal isolation from supply line noise.
Q3. Correct.

Brian.

samy555 · Aug 7, 2015

betwixt said:
Q1. Buy parts from a reputable supplier. If that isn't possible, the best you can do is make a guess, most Quartz crystals are designed around 15 - 30 pF load capacitance.

Q3. Correct.

Brian.

Excellent answers, I benefited a lot of them, Thank you Brian.

Q2. Almost certainly it isn't critical, the values only determine the impedance at the collector.

OK

The capacitor is essentially keeping the collector at ground signal level

Xc of that 10n capacitor @ 7MHz = 2.27Ω ,, yes short circuit

and the resistor is providing a degree of signal isolation from supply line noise.

How???? it is very small value?
thanks

samy555 · Aug 8, 2015

Q4: (Repeated): For the RC Low Pass Filter shown below, is it important to have a frequency of 159KHz, or it is not critical?
broken link removed

Q5: the base voltage (VB) = 6 volt, VE=5.4 volt (if VBE=0.6 V),,, IC = VE/RE= 2.45 mA
I think that IC is relatively large and the circuit designer could accomplish the same performance if he work with 1mmA or even 0.5mA. Do you agree with me in this opinion?

thank you very much

FvM · Aug 8, 2015

The RC combination is an optional power supply filter, intended to remove switching noise or RF signals generated by other parts of the circuit. Dimensioning isn't critical at all, in case of doubt, you'll use a larger capcitor to imoprove the filter effect.

erikl · Aug 8, 2015

samy555 said:
Q5: the base voltage (VB) = 6 volt, VE=5.4 volt (if VBE=0.6 V),,, IC = VE/RE= 2.45 mA
I think that IC is relatively large and the circuit designer could accomplish the same performance if he work with 1mmA or even 0.5mA. Do you agree with me in this opinion?

Sure. But it depends on the input capacitance of the stage which receives the oscillation power from the emitter output. With too large input cap, oscillation might not start. Should be worst case, corner and - if possible - MC simulated.

BradtheRad · Aug 8, 2015

samy555 said:
Q4: (Repeated): For the RC Low Pass Filter shown below, is it important to have a frequency of 159KHz, or it is not critical?
broken link removed

Re the 100 ohm resistor:

Playing with simulations I have seen values work from 100 to 1000. It provides some degree of isolation between the oscillating loop and the power supply, so the oscillating loop can exhibit wider voltage swings.

Q5: the base voltage (VB) = 6 volt, VE=5.4 volt (if VBE=0.6 V),,, IC = VE/RE= 2.45 mA
I think that IC is relatively large and the circuit designer could accomplish the same performance if he work with 1mmA or even 0.5mA. Do you agree with me in this opinion?

Those are in the range of reasonable levels. It is usually possible to adjust a value or two somewhere, and get the circuit to oscillate. You might adjust the bias current, or supply voltage, or emitter resistor, collector resistor, etc.

Some of it depends on the characteristics of the real transistor you use.

samy555 · Aug 10, 2015

FvM said:
The RC combination is an optional power supply filter, intended to remove switching noise or RF signals generated by other parts of the circuit. Dimensioning isn't critical at all, in case of doubt, you'll use a larger capcitor to imoprove the filter effect.

I mean from the very beginning that the frequency of 159 KHz is far away from the operating frequency (7MHz) and also far away from the electricity humm (50/100 Hz).
Is the 100Ω resistance and 10 nF cap are a low pass filter or not?

thank you very much

- - - Updated - - -

erikl said:
Sure. But it depends on the input capacitance of the stage which receives the oscillation power from the emitter output. With too large input cap, oscillation might not start. Should be worst case, corner and - if possible - MC simulated.

Sorry do not tell me about things that can not be calculated! can you calculate the input capacitance of the stage?

Then what is the relationship between current and input capacitance?
thanks alot

- - - Updated - - -

BradtheRad said:
Re the 100 ohm resistor:

Playing with simulations I have seen values work from 100 to 1000. It provides some degree of isolation between the oscillating loop and the power supply, so the oscillating loop can exhibit wider voltage swings.

Those are in the range of reasonable levels. It is usually possible to adjust a value or two somewhere, and get the circuit to oscillate. You might adjust the bias current, or supply voltage, or emitter resistor, collector resistor, etc.

Some of it depends on the characteristics of the real transistor you use.

I'd expect better answers than those you have given me like tell me that the filter must have frequency ranges between X & Y.

Oscillator can not work with weak currents like 0.5mA, or there is no difference between the use of 1mA or 3mA.
Thank you very much.

betwixt · Aug 10, 2015

Is the 100Ω resistance and 10 nF cap are a low pass filter or not?

Yes, as we repeatedly stated, it is to let clean DC reach the oscillator from the supply line so it doesn't risk being modulated by supply line noise. The capacitor also keeps the collector at (very close to) ground from the oscillation perspective.

Do not read too much into the 159KHz figure, it is not relevant to the oscillators function, the values were chosen as 'mid-range' suitable ones rather than being calculated for a specific cut-off frequency. Personally, I would probably add an additional capacitor of 1uF to 10uF across the 100nF for even better supply rejection.

Brian.

erikl · Aug 10, 2015

samy555 said:
Sorry do not tell me about things that can not be calculated! can you calculate the input capacitance of the stage?

Not exactly calculate, but at least I can estimate the input capacitance of the next stage. Or even better: analyze it by simulation!

samy555 said:
Then what is the relationship between current and input capacitance?

The current determines the emitter output resistor, by this the output capacitance of the oscillator stage to the next stage. Lower current needs higher resistance, so generates higher output impedance. This - on the other hand - determines, if the output impedance is low enough to drive the input impedance of the next stage. If this isn't true, the oscillator won't start oscillating. Try it by simulation!

samy555 · Aug 10, 2015

betwixt said:
Yes, as we repeatedly stated, it is to let clean DC reach the oscillator from the supply line so it doesn't risk being modulated by supply line noise. The capacitor also keeps the collector at (very close to) ground from the oscillation perspective.

Do not read too much into the 159KHz figure, it is not relevant to the oscillators function, the values were chosen as 'mid-range' suitable ones rather than being calculated for a specific cut-off frequency. Personally, I would probably add an additional capacitor of 1uF to 10uF across the 100nF for even better supply rejection.

Brian.

Thank Brian,, the following question directed to you personally:
If you want to re-design this osillator for working at a frequency of 100 MHz, What are the new valuse you choose for the 100Ω and 10nF?
Greetings

- - - Updated - - -

erikl said:
Not exactly calculate, but at least I can estimate the input capacitance of the next stage. Or even better: analyze it by simulation!

You and I know that simulation programs are often far from reality.

Would you recommend fitter program that is close to some extent to the real world.

The current determines the emitter output resistor,

yes

by this the output capacitance of the oscillator stage to the next stage.

I could not understand this sentence

Lower current needs higher resistance, so generates higher output impedance. This - on the other hand - determines, if the output impedance is low enough to drive the input impedance of the next stage. If this isn't true, the oscillator won't start oscillating. Try it by simulation!

That's OK
Thanks very much

Audioguru · Aug 10, 2015

Nobody makes a 100MHz crystal so you need a lower frequency crystal in a frequency synthesizer circuit or operate a lower frequency crystal oscillator at a harmonic of the crystal frequency.

erikl · Aug 10, 2015

samy555 said:
You and I know that simulation programs are often far from reality.

If you have the right models, simulation results are much closer to reality than simple estimations.

samy555 said:
Would you recommend fitter program that is close to some extent to the real world.

You can do that, but I think it's not necessary.

samy555 said:
by this the output capacitance of the oscillator stage to the next stage.

Click to expand...

could not understand this sentence

Should read

by this the output impedance of the oscillator stage to the next stage.

betwixt · Aug 10, 2015

Thank Brian,, the following question directed to you personally:
If you want to re-design this osillator for working at a frequency of 100 MHz, What are the new valuse you choose for the 100Ω and 10nF?

As already pointed out, you probably wouldn't use that oscillator configuration at 100MHz but answering your specific point, I would use the same values while accepting they are not critical to the oscillator functions. You could short out the 100 Ohm resistor and omit the capacitor completely if you were sure the supply impedance was low enough and it would make almost no difference to operation. If I really did build that circuit at 100MHz my main concern would be the properties of the 100nF capacitor, I would ensure it was a type that worked well at high frequencies or alternatively connect a good HF one in parallel with the existing 100nF. When you work at frequencies like 100MHZ the deficiencies of some capacitor constructions becomes significant, some types start to show higher than expected impedance and thus may not be good at keeping the collector at ground signal potential.

Brian.

samy555 · Aug 12, 2015

I continue my reading in the book, and I will ask some questions…..

Q6: Now we have a 2.3KΩ at the collector of Q2, and the 50 Ω load is reflected to be 50*16 =800 ohm, My question is why the designer did not make the ratio 12:2, so we have
50*36 = 1800 ohm? I think it's the best and we can transfer greater power?

Q7: To calculate the input impedance (Zin) of the buffer (Q2 stage)

Is Zin (=3.4K) big enough to not load the oscillator?
Thank you very much

erikl · Aug 12, 2015

samy555 said:
Q6: Now we have a 2.3KΩ at the collector of Q2, and the 50 Ω load is reflected to be 50*16 =800 ohm, My question is why the designer did not make the ratio 12:2, so we have 50*36 = 1800 ohm? I think it's the best and we can transfer greater power?

The answer is given in your fig.8: blue underlined. Means the primary inductive reactance of the transformer shouldn't have much impact, if the output load impedance changes (from 50Ω). In terms of greater power transfer, however, I think you're right.

samy555 said:
Q7: To calculate the input impedance (Zin) of the buffer (Q2 stage)
Is Zin (=3.4K) big enough to not load the oscillator?

No, I think it's too low for loading from the base (in parallel to the crystal). The input of the second stage should be connected from the emitter (via 100pF), which has a much lower output impedance than the base.

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