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walid

Crystal examiner

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Hi
often have some friends with their remote control to to repair it. This requires that you have the examiner Cristal to ensure the validity of this piece. In the past, I was immediately replaced them with new, but I know I need to examine the crystal. Searched in the internet and I found this circuit and structure but did not work, therefore decided to understand it very well that to succeed when structure the next time.
I deleted the 56p cap because I do not need to watch the freq. I know that the first transistor worked as oscillator and the o/p sig rectified by the two diodes to the final stage which will be ON if the first stage oscillate.
My questions:
1) is the 1st stage is colpitts osc.
2) How the value of frequency depends on crystal?
3)Why 680p and 150p caps, What if others these values? 

post-2833-14279143277696_thumb.jpg

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Hi
often have some friends with their remote control to to repair it. This requires that you have the examiner Cristal to ensure the validity of this piece. In the past, I was immediately replaced them with new, but I know I need to examine the crystal. Searched in the internet and I found this circuit and structure but did not work, therefore decided to understand it very well that to succeed when structure the next time.
I deleted the 56p cap because I do not need to watch the freq. I know that the first transistor worked as oscillator and the o/p sig rectified by the two diodes to the final stage which will be ON if the first stage oscillate.
My questions:
1) is the 1st stage is colpitts osc.
2) How the value of frequency depends on crystal?
3)Why 680p and 150p caps, What if others these values?   


The caps will set the feedback ratio. If this ratio is too small (as in this case) the circuit will not oscillate as the loop-gain is less than 1. Try swapping the 2 caps around then it should work. As you go higher in frequency the cap values should be reduced. 20 MHz crystal would require values of about 47pF and 68pF. Emitter resistance also have some influence.

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Plots shows happy and more happy circuit values. First plot shows osc struggling to start because of low loop gain.

This circuit will not work with very low value crystals like 32kHz watch crystals as the crystal will be loaded too much. You need an oscillator circuit with a FET input for this.

If you do want to try other values try and keep C2 slightly higher in value than C1. C1=C2 is also ok in most cases.

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Hi AN920
Thank you for the wonderful and important answer.
Questions:
(a)

{The caps will set the feedback ratio. If this ratio is too small (as in this case) the circuit will not oscillate as the loop-gain is less than 1.}

The feedback ratio is the loop-gain = 150/680 = 0.2 < 1, are u mean that?
(b)
{As you go higher in frequency the cap values should be reduced. 20 MHz crystal would require values of about 47pF and 68pF.}

The remot control's crystal is 455KHz, can u please tell me the appropriate values for those caps, What the role?
©
{Emitter resistance also have some influence.}

How can influence?
(d)
{This circuit will not work with very low value crystals like 32kHz watch crystals as the crystal will be loaded too much. You need an oscillator circuit with a FET input for this.}

I know that the FET have a very large i/p impedance so not loading the 32KHz crystal. But the thing that I do not know is why a high freq crystal can work fine with BJT While a low freq crystal can't?
Thank you very much.

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I will respond more complete when I have time, but here is some idea of minimum parrallel (load resistances) across crystal that will still allow circuit to oscillate.

Freq MHz       Res value
20                  3k
10                  4k
1                    27k
0.5                 370k
(5kHz)            6M

The loop gain is a bit  more complicated as just the ratio of C1,2 and include data from the crystal, gm of transistor etc. Here we don't know the crystal data unless we measure it or get it from vendor data and have to make some educated guesses.

The series internal resistance of crystals go up as the frequency goes down. That is the reason why if we load the crystal too much the losses over the internal resistance will kill the Q of the crystal and the prevent oscillation from happening. At higher frequencies the internal resistance is much lower (~20 Ohm at 20MHz) and can tolerate much lower loads across it.

You can see this from the model of 32k crystal

LS= 0.007 CS= 3.5e-015 RS= 18000 CO= 1.7e-012

and 15M crystal

LS= 0.0075 CS= 1.5e-014 RS= 25 CO= 4.5e-012

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Thank you AN920,

(1)

Step 2: Get crystal model data from vendor or measure with network analyzer.
Our case: LS = 0.08H, CS = 4.5pF, RS = 50

(a)It was difficult to obtain any information from the seller in my country. How can measure these parameters by network analyzer?
(b) Do you not see that the value of Ls = 0.08H is very large value?

(2)
Higher Q values will give better stability crystal.

Can you shed more light on this point and the relationship between Q and stability?

(3)
Step 4: Make sure transistor is biased at collector current that will give highest Ft. From the datasheet of BC182 we see that we can get this with a collector current of about 10mA. We will use a DC gain val of 100 with this transistor.

From datasheets i found:
ft ic good.jpg
Yes we have more ft (200 MHz at 10mA) than at 0.5mA, and this is more clear with the following graph:
fticgoodgraph.jpg
Now, At Ic =10mA, DC hfe = 100 as shown in the following graph. Why you take the DC hfe not AC hfe?
ftdcbeta.jpg
(4)
Select R1 to have about 50% of V1 on emitter.

In other words, you want to say that: Make VCE =  50% of V1?

(5)
Calculate R2 to give about 10mA collector current. Check that this value is at least (200 x RS, the higher, the better) for minimal loading on crystal.

Lets calculate it:
Vcc=9V, hfe = 100, VE = 4.5V and Ic = 10mA ==> IB = 10m/100 = 100uA
Vcc - RB*IB - VBE - VE =0
VBE at IC = 10mA (from one of the graphs in the datasheets) = approx. 0.68 V, so:
9 - 100u * RB - 0.68 - 4.5 = 0
RB = (9 - 0.68 - 4.5)/100u = 38.2 K ohm (far from 82 K)???
AND note that if IC = 10 mA then VE = 5.6 V not 4.5V. Please I want further clarification.
Correction : Rs * 200 = 50 * 200 = 10 K , yes you are right 38.2 K > 10 K

(6)
Check that this value is at least (200 x RS, the higher, the better) for minimal loading on crystal.

How a low value of R2 can load the crystal, please Explain this point.

I will complete the remaining questions tomorrow. Thank you very much. 

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(a)It was difficult to obtain any information from the seller in my country. How can measure these parameters by network analyzer?
I will provide a method for you later. Your network analyzer should be able to accurately measure Fs and Fp

(b) Do you not see that the value of Ls = 0.08H is very large value?
Yes, this is one of the characteristics of crystals, Large LS and small CS

(2)
Higher Q values will give better stability crystal.
Can you shed more light on this point and the relationship between Q and stability?
The higher the Q the more rapid the rate of phase shift vs frequency, meaning that the frequency have narrow margin for drift in short term. Long term drift may still occur due to temperature, aging and power dissipation in the crystal.

(3)
Step 4: Make sure transistor is biased at collector current that will give highest Ft. From the datasheet of BC182 we see that we can get this with a collector current of about 10mA. We will use a DC gain val of 100 with this transistor.
From datasheets i found:
ft ic good.jpg
Yes we have more ft (200 MHz at 10mA) than at 0.5mA, and this is more clear with the following graph:
fticgoodgraph.jpg
Now, At Ic =10mA, DC hfe = 100 as shown in the following graph. Why you take the DC hfe not AC hfe?
The equation takes into account the calculation of amount negative resistance generated by the transistor by using DC gain, which is needed for oscillation to occur.
ftdcbeta.jpg
(4)

Select R1 to have about 50% of V1 on emitter.
In other words, you want to say that: Make VCE =  50% of V1?

Yes more or less, not very critical.

(5)Calculate R2 to give about 10mA collector current. Check that this value is at least (200 x RS, the higher, the better) for minimal loading on crystal.

Lets calculate it:
Vcc=9V, hfe = 100, VE = 4.5V and Ic = 10mA ==> IB = 10m/100 = 100uA
Vcc - RB*IB - VBE - VE =0
VBE at IC = 10mA (from one of the graphs in the datasheets) = approx. 0.68 V, so:
9 - 100u * RB - 0.68 - 4.5 = 0
RB = (9 - 0.68 - 4.5)/100u = 38.2 K ohm (far from 82 K)???
AND note that if IC = 10 mA then VE = 5.6 V not 4.5V. Please I want further clarification.
Correction : Rs * 200 = 50 * 200 = 10 K , yes you are right 38.2 K > 10 K

Yes, that is the calculated value but depends on the param or transistor self. Transistors from Philips will be different to say Motorola's. It is wise to verify the Ic finally and readjust Rb to get correct Ic. In my case I just simulated the circuit and with 39k resistor and BC182 model Ic was about 13mA. I increased Rb and Re to bring it down to 10mA. You don't want Re too low as it will reflect back to the base terminal x gain and load the crystal too much.

(6)
Check that this value is at least (200 x RS, the higher, the better) for minimal loading on crystal.
How a low value of R2 can load the crystal, please Explain this point.
R2 will appear in parallel over the crystal with the transistor input impedance reducing the Q of the crystal.

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Hi AN920
I was very happy when I read your reply, I wish you a long and happy life, thank you very much.

I will complete the remaining questions tomorrow.

(1)
Step 5: Calculate SQRT(RS/(1/DC gain));  = 70

(a) That is the same as: SQRT(RS* DC gain) = SQRT(50 * 100) = about 70 (correct me)
(b) What this value? I mean what does that mean?

(2)
Step 6: Calculate (C1 + C2); C12 = 1/(70*omega) = 454 pF

omega = 2 * pi * freq (yes it is)

The rest is very clear, but I can use laws, although I do not understand how it came, and what it means.
This by no means diminishes the value of the rare information you provided. Thank you very much.
Now lets discuss about some relevant points :
(1)The Crystal examiner that we are discuss about it. I built it to check for one type of crystal which is 455KHz. Here is its image:
6810293.jpg
from: http://www.ersatzteile-onlineshop.de/gruppen/3853000000-1.htm
All the remote controls that They were supposed to do repair containing this type.
When i get its model data LS, CS and RS, I'll redesign the original circuit according to the method you provided me and put it here for comments and corrections.

(2)Some sources have labeled them Resonator and others said crystals, Is there any difference?

(3) I understand easily what resistor do in a circuit, also the cap and coils, but i can't imagine How crystal functioning in electronic circuit? 

(4) When I asked my first question on this topic, the circuit did not operate because of the error in the wiring but after correcting the error it worked well.
I kept some of the damaged crystal because I noted that some of the damaged  crystals back to work after a period. I noted that by coincidence.
What I want to say here that I also noted that the testing device glows strongly with the good crystal quality and lighting weak with bad crystal.
From all of the foregoing Can I say that the crystal does not destroy just like transistors (0 or 1) it weaken or its condition worsened.     

(5) The last question: I have heard repeatedly that the "remote control" pattern working with the carrier wave of 40 kHz, why use Crystal written on it 455 kHz?


Thank you very much

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(a) That is the same as: SQRT(RS* DC gain) = SQRT(50 * 100) = about 70 (correct me) yes
(b) What this value? I mean what does that mean? Get ratio for loop gain vs C1, C2 values needed to be able to oscillate with given RS

Now lets discuss about some relevant points :
(1)The Crystal examiner that we are discuss about it. I built it to check for one type of crystal which is 455KHz. Here is its image:
6810293.jpg
from: http://www.ersatzteile-onlineshop.de/gruppen/3853000000-1.htm
All the remote controls that They were supposed to do repair containing this type.
When i get its model data LS, CS and RS, I'll redesign the original circuit according to the method you provided me and put it here for comments and corrections.

(2)Some sources have labeled them Resonator and others said crystals, Is there any difference? Uses ceramic materials instead of quartz. Q values not nearly as high as quartz but much higher than a normal LC tuned circuit. Model is the same


(3) I understand easily what resistor do in a circuit, also the cap and coils, but i can't imagine How crystal functioning in electronic circuit?  It is a tuned circuit with L and C values, but with very high Q

(4) When I asked my first question on this topic, the circuit did not operate because of the error in the wiring but after correcting the error it worked well.
I kept some of the damaged crystal because I noted that some of the damaged  crystals back to work after a period. I noted that by coincidence.
What I want to say here that I also noted that the testing device glows strongly with the good crystal quality and lighting weak with bad crystal.
From all of the foregoing Can I say that the crystal does not destroy just like transistors (0 or 1) it weaken or its condition worsened. Crystals and resonators can become lazy or damaged. Dropping quartz crystals on a floor is not a good idea, because the quarts wafer becomes very thin and  fragile at high frequencies. They can also become damaged by very high drive level current through them which increases the internal dissipation     

(5) The last question: I have heard repeatedly that the "remote control" pattern working with the carrier wave of 40 kHz, why use Crystal written on it 455 kHz? 455KHz resonators is used in radio IF circuits as well and cheap. They divide this frequency to get the 40kHz. The controlling micro can also be clocked by the 455kHz


Thank you very much

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To get the values of LS,CS, RS and Co

Measure the points Fs and Fp very accurately. Use narrow span on NWA

Measure holder capacitance Co with LCR meter at low test frequency (1-5kHz)

Then:
CS = Co{(Fp/Fs)^2 -1}
LS = 1/{[(2*pi*Fs)^2]*CS}
RS = {2*(NWA impedance)*[10^(insertion losses/20)-1]}; NWA impedance normally 50 or 75 Ohm

Insertion loss will be what you read for Fs on the MAG display in dB below 0dB. This loss will increase with increasing crystal Q values.

From this you can complete your model. It is handy to write a mathcad program to calculate all his if you are going to use it a lot.

It is possible to do this with a signal generator and a scope as well (will take much more time and not so accurate) as long as they have the same input and output impedances. Many better scopes have a 50 Ohm input selector and most RF generators are near 50 Ohm over their full frequency range. The scope must have a bandwidth much higher (5-10 times) than the Fs and Fp levels you are trying to measure. You will have to convert your relative display readings to dB.

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Hi AN920
Thanks to the rapid and excellent reaction.
I have only a digital multimeter Equipped with the function of measuring capacitance.
i have not an oscilloscope or function generator.
Therefore, I will not be able to know the parameters of any crystal using the methode mentioned above.

The discussion has been more than impressive, and the information provided by AN920 was excellentand new, at least for me. I will stop discussion on this subject in the hope that I met with Mr. AN920 in another new subject. Thank you very much and goodbye. 

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I will stop discussion on this subject in the hope that I met with Mr. AN920 in another new subject. Thank you very much and goodbye. 


Thank you for your kind remarks, but actually I am O+ who loves EE in a mainly <-O world.  ;)

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Just to explain the concept of negative resistance. If we take the transistor circuit without the crystal connected and use a NWA to look at the real part of the input impedance (red) and available gain (blue), we can see that our values generate the plots as shown.

We can see that the red curve indicates that the circuit will have a negative resistance of about -58 Ohm (this more than cancels our +50 Ohm from RS) at 5MHz. We also see that the gain is near maximum at 5 MHZ confirming potential for oscillation as soon as we add the tuned circuit (crystal).

It is also clear that if we had a crystal with a RS of around 10 Ohm and frequency at 9 MHz, the circuit will not have enough gain (gain < 1 at this point) with our present capacitor values to oscillate.

The last plot show the result of increasing the capacitor values. We still have enough gain, but now we have only -40 Ohm negative resistance and not enough to cancel RS.
 

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Hi An920
While I am looking into the Internet, I encountered the following firure from the Web site :
http://www.k8iqy.com/qrprigs/2n215/2n215page.htm
CRYCOLPITTSOSC.jpg
As you see in the picture C21 is less ten times than the upper C20, and this violates what we agreed upon in previous commentary.
I think that this design done by senior scientists.

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It is difficult to comment without the whole picture. The missing parts are the crystal data. It is possible to use other ratios as long as you can get enough gain and enough negative resistance to satisfy oscillation. The article explains in detail most of the other circuit design values but none on the crystal oscillator? Many people get crystal oscillators to work by trial and error, not that I am suggesting that that was the case here.

By making C20 much larger the loop gain will drop off more rapidly with frequency which will attenuate harmonics. The designer may have experimented with values that gave the lowest harmonics with just enough gain to oscillate at 14.3MHz

Another crystal of the same value in series on the output will produce a very good sine wave output with low harmonics.

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Hi
TO audioguru:

The 'scope pic of the oscillator's output shows severe distortion that is cleaned up with the 3rd-order lowpass filter.

(a) Clear from your talk that you agree with AN920 that this is not good design.
(b)
3rd-order lowpass filter
I thought that it is banpass filter. i think it is 3rd order because it is 3 elements filter. Let me look at the internet for further information on this filter.

TO AN920
Many people get crystal oscillators to work by trial and error, not that I am suggesting that that was the case here.

I have long believed that many of the designs are in a trial and error because of the different specifications of some electronic components from each other.
Another crystal of the same value in series on the output will produce a very good sine wave output with low harmonics.

(a) That is a very good and new idea, are you sure that Another crystal of the same value in series on the output work as lowpass filter?
(b) Can shed further light on the use of crystal as a LPF.

thanks GURU
thanks AN920

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