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Posts posted by walid

  1. Hi AN920

    I thank you for your wonderful to deliver some of the concepts that are new, at least for me.
    I want to know from a practicle perspective, I want to know the importance of this subject. Where we need to calculate the -ve resistance?
    I did some search and get the following:
    I'll discuss with you:

    Considering the amount of attention given to superconducting materials which have an effective electrical resistance of zero it is surprising that the property usually called Negative Resistance is so little known.

    From this i understand that Negative Resistance = superconducting.

    In general, the current tends to rise with increasing voltage, but there is a region between the peak voltage and valley voltage, where the the current falls as the voltage is increased. This is called the Negative Resistance Region because in this voltage range the dynamic resistance, r<0 .

    I think that this is the best simplified  definition of the concept of the Negative Resistance. Did agrees with me?

    (3)If you read the rest of the page and saw the pictures following, you'll know that there were three cases :

    a- R be greater than 0 . This is the factual situation that we know.
    b- R be equal to 0, a situation virtually impossible.
    c- Or to be less than 0, and here I ask how it can be made so?

    After receiving reply I'll discuss with u the contents of the next page :

    thank u very much
  2. 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. 

  3. 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.

    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?

    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:
    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
  4. Thank you AN920,


    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?

    Higher Q values will give better stability crystal.

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

    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:
    Now, At Ic =10mA, DC hfe = 100 as shown in the following graph. Why you take the DC hfe not AC hfe?
    Select R1 to have about 50% of V1 on emitter.

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

    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

    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. 
  5. Hi AN920
    Thank you for the wonderful and important answer.

    {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?
    {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?
    {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.
  6. Thank you AN920, It is very clear
    (a) After considering all of the above:
        Can we consider that C1 play a impedance matching between the antenna and the tuned circuit?

    (b) Another point is still unclear to me is : if we consider that the length of one meter of the wire antenna is of 50 Ohm resistance.
        So 10 meters have a 500 ohm and 2m have 100 ohm resistance.
        Now, with 10 meter (500 ohm) the recommended C1 is about 5pF, please look at the total Z:
    zwith 10meters.jpg
    note 31.8 Kohm
    with 2 meter (100 ohm) the recommended C1 is about 50pF, please look at the total Z 
    note, 3.18 Kohm
    Big difference ten times??!!

    © As seen above, the writer said:

    {Every reception antenna behaves as a voltage generator, having its own internal resistance and capacitance.}

        Why did not say internal inductance also?
    (d) Regarding the last point you mentioned about a switchable narrow-wide FM detector in a measuring instrument, I think you mean C16 (100nF) that big value cause a wide BW.
        I noteced some strange thing, the collector of Q1 is not connected to any supply VCC.
  7. Hi AN920
    I have read with interest all that came in your last reply, so I read it more than five times, I want to further shed light on the following points :

    {The capacitor increase the series impedance Xc to minimize loading.}

       Are u mean the series impedance of the antenna? And if so, What can the antenna be considered from this point?
    {Connecting the antenna directly will be like connecting a low value resistance in parallel with the tuned circuit, killing the selectivity. }

       Are u mean that with C1 it is will be like connecting a high value resistance in parallel with the tuned circuit.
    {Connecting a low value resistance in parallel with the tuned circuit, killing the selectivity}

       I want to offer my interpretation of this phrase and you should correct me if I was wrong:
       Connecting this low value resistance in parallel with the tuned circuit make the total i/p impedance is always low for every signal of any freq so there is no selectivity at all.
    (4) Why u consider the antenna as alow resistance? I have heard that it is 50, 75 300 ohm, why antennas are low value resistance?
    (5) Why antenna is considered to be // to the tuned circuit, why not inseries?
    [glow=red,2,300]NOTE: [/glow] Please reply to questions one after another, not all at once so I understand very well not have to repeat the question in different ways.
    [glow=red,2,300]I forgot to tell you that your answer above, consisting of 4 lines better than 2 pages in the book. [/glow] thank u very much
  8. I am sorry if continued reading found the answer which is:

    Every reception antenna behaves as a voltage generator, having its own internal resistance and capacitance. Antenna's resistance damps the oscillatory circuit and reduces its selectivity (which manifests as the "mixing" of stations) and sensitivity (which exerts as signal strength reduction), and antenna's capacitance reduces the reception bandwidth. More precise, antenna's capacitance reduces the upper bound frequency of the reception bandwith (Pic.3.2), making reception of the stations laying close to this frequency impossible. Both these features are undesirable and manifest themselves as less as the capacitance C1 is smaller. On the other hand, the smaller the capacitance C1, the weaker the signal that goes through it from the antenna, the reception therefore getting weaker. As you can see, the compromise solution is a thing to go for, i.e. one must find the capacitance at which the signals from the antenna won't be much weakened while simultaneously keeping the selectivity and the bandwidth big enough. You can start with C1 being about 30 pF. Then,

    using C, tune yourself to some radio stations you can receive. If all the stations that interest you are there, and the strongest one of them still does not jam the reception of other stations all's well. Try then with some bigger capacitance for the capacitor C1. The reception will be getting louder, so do continue increasing C1 as long as it is still possible, by changing C, to receive all the stations of your interest that can be heard in your place, without the interference of some strong or local station. If, however, reception of some nearby station isn't possible, smaller C1 should be tried out. In this manner the biggest capacitance for C1 should be found, that allows optimal reception both regarding selectivity and bandwidth. The simplest solution is using variable capacitor for C1, its capacitance ranging from few picofarads to few dozens pF, adjusting it to obtain optimal reception for each station individually. During this, whenever C1 is being changed, the receiver must be re-tuned to the station using C.

    But if you have any useful addition, welcome
  9. Hi all
    These days I read a book on radio receivers from the site :
    During the reading, sometimes I find it difficult to understand some of the points, so I ask you to help me understand. Thank you very much.
    Question #001 (from sec: 3.1.4. Other Components)
    The writer said:
    this is the figure he refered to:
    as you see, few pF (3 - 7 pF)for 10 meters antenna and few dozens (30-70 pF) for 2 meter antenna.
    note C1 value is In inverse proportion with the antenna length.
    (a) Why this inverse proportion?
    (b) What would happen if we removed this C1?
    [glow=red,2,300]Thanks alot.[/glow]

  10. Dear AN920, I do not find the right words to express to you my deepest thanks and respect. thank you very much.
    (1)The transformer Comprising L1, L2 and L3 have DOT at the bottom of each coil.
      I remember I studied something of this kind before, but now I forgot its purpose, can u please tell me, and what happen if we reverse one of them?
    (2)I see a command line: K1 L1 L2 L31, i know K1 is the mutual inductance and      L31 is the inductance affect L3 due to current pass through L1. but how to write this command and how to choose those values? 

    NOTE: There is a small problem in the program which is the nodes numbers that appear in the graph didnot shown in the circuit, and one can't know some graph belong any node. 

  11. I am not expert like Mr. guru, I will try to answer using my simple information :
    Oscillators can be built using opamps or transistors or may be some special IC's.
    opamps can not handle high frequency but transistors can
    if u want to build a function generator try to search one in google or tell me to find one for u
    That is all I know regarding your question, and I hope that others help me in the answer
    Do not hesitate to ask about everything that comes to you

  12. Hello AN920, in fact, I thank you from the bottom of my heart to respond quickly. It is clear you made an effort to explain the idea but unfortunately I did not understand much, lets see:
    In your circuit, L = 4.7+4.7 = 9.4 uH
    C = 560p
    So the resonance freq fo is about 2.2 MHz.
    At fo i see from your graph that gain = 0 dB (Because of the absence of amplifier, as u said) and phase = 0 degree.
    when using amplifier, we get mor gain (5 dB as u said) and 360 degree phase angle and all these coditions are needed to maintain oscillation==> OK 
    My questions:
    (1)First, before I forget, Can you please give me the name of the program which is used in the simulation to help me in analyzing some vague points.
    (2)360 deg is the same as 0 deg, Why you considered it as 360 and not 0?

    {Oscillators with more than 6-8dB of loop gain is normally not a good thing, as it produces many other unwanted products. } 

        (a) why you said "loop gain" and not simply "gain"?
        (b) Why Oscillators with more than 6-8dB of loop gain is normally not a good thing?
    (4)If I want to calculate the phase angle to the folowing circuit:
    using jwL and 1/jwC, can I look at caps and coils as: [(L1 + L2)//C1] +C2 or what?
    (5) Phase angle before using amplifier = 0 and after using amplifier = 360. What is the role of amplifier in this change. Can you explain this point.
    thank you AN920
  13. Hi

    The statements which is not true is:

    d. An oscillator must have a negative feedback circuit.

    because  An oscillator must have a positive feedback circuit to maintain oscillation.
    negative feedback is good for amplifiers to get more stability.

  14. Hi AN920,
    First, thank you for clarification and assistance, there are still some questions:

    Before that we have to agree on some things following:
    this is an AM radio reciever antenna, local osc and mixer
    AM band is from 530 KHz up to 1600 KHz
    the local osc' freq is always above that value of 455 KHz, so it can be set from (530+455) to (1600+455), that is from 985 to 2055 KHz.

    {it has low enough reactance (Xc ~ 10 Ohm) at the frequency of interest.}

    Xc = 16 Kohm at  1000 KHz, and Xc = 8 Kohm at 2000 KHz

    {Oscillators with more than 6-8dB of loop gain is normally not a good thing, as it produces many other unwanted products. }

    Please I want further explanation on that point.
    {Think of an oscillator as an amplifier with A>1 with in-phase feedback.}

    A>1 To what extent? 
    {in-phase feedback}

    I want to discuss this point in some details,
    Lets assume that there is some harmonic sig at the collector of 1000 KHz and 1mV rms.
    and Lets assume that the local osc transformer have the following turns numbers:
    the coil connected to the collector is of 10 turns (Let us call the primary coil).   
    the other coil (secondary) is of 30 turns and tapped at 10 turns from the bottom.
    NOW: if the sig at the primary is 1000 KHz and 1mV rms, it will be of 1000 KHz and 3mV rms at the secondary also at phase with the original signal.
    At the tap we get the same sig with reduced amplitude to 1mV rms, when this sig pass through the 10n cap it will be 90 degree out of phase, so the it is not in-phase feedback??!!
    What is your reply to this talk?
  15. Thank you AN920, I saw this link previously, commentary good, but not enough.
    The most important thing in that link :
    my question:
    in my drawing:
    Can I consider the 10n cap connected to the emitter as a bypass cap and thus get a high voltage gain:
       I know that one purpose of using that cap is to block the DC from going to ground through the transformer coil.
    that cap have an reactance of about 16k at 1 MHz and about 8 k at 2 MHz.
    is the 10n value is choosen carefully?

    I thank everyone help me to understand the idea.

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