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another FM transmitter project

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by hannobisschoff@gmail.com, Apr 14, 2011.

  1. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Deleted duplicated post.
     
    Last edited: May 31, 2013
  2. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The power dissipated in a high value resistor (27k) with a low voltage across it or a low current in it is almost nothing.
     
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  3. Nepaliman

    Nepaliman Member

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    Dear guru,
    I know each basic electrical formulas but I couldn't apply it :( . Your few calculation inpired me lot. Can you calculate the value and its watt of this resistor?

    - and how to calculate output voltage if I ised 220 ohms resistor there? Idea needed please!
     

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  4. dave

    Dave New Member

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  5. audioguru

    audioguru Well-Known Member Most Helpful Member

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    A resistor does not regulate the output voltage. If the current changes then the output voltage also changes. You should use a 7805 voltage regulator IC instead of the resistor.

    Ohm's Law says that the resistance is the voltage across it divided by the current.
    12V - 5V= 7V across the resistor. The resistance is 7V/0.5A= 14 ohms which is not a standard resistor value.

    The power dissipated by the resistor is simply the voltage across it times the current. 7V x 0.5A= 3.5W. A 5W resistor will get fairly hot.

    If you use a 220 ohm resistor and it has a current of 0.5A then it must have a high voltage across it.
     
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  6. Nepaliman

    Nepaliman Member

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    Hahaha laughing at me because I am being puzzeled :| ! Look my attach and check my silly Voltage calculation:

    V= RxI= 220Ω X 0.5A= 110V :p

    I think I invented step-up transformer using Ohm's Law!
     

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  7. audioguru

    audioguru Well-Known Member Most Helpful Member

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    If your load is SHORTED then Ohm's Law says that the current is 12V/220 ohms= only 54.6mA which is MUCH LESS that the 0.5A that you need.
    If the load voltage is 5V then Ohm's Law says that the current is (12V - 7V)/220 ohms= only 31.8mA.

    See, calculations are simple!
     
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  8. Willen

    Willen Well-Known Member

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    One thing amazed me!

    Tweeeter speaker of powerful audio amplifier has a 1uF or 4.7uF capacitor of its series which passes only higher audio frequencies but a small value capacitor- 0.33uF series on Mod-4 passes all audio frequencies, How?
     
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  9. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Another basic calculation rule similar to Ohm's Law.

    A series capacitor and its load resistance to ground is a simple highpass filter. The capacitor has reactance that cuts low frequencies.

    The formula for the cutoff frequency (the level is cut -3dB) is "1 divided by (2 pi times the capacitance in Farads times the resistance in Ohms)".

    A 4.7uF capacitor feeding an 8 ohm tweeter has a cutoff frequency of "1 divided by (2 x pi x 4.7uF x 8 ohms)"= 4255Hz. Higher frequencies will be passed and lower frequencies will be cut.

    A 0.33uF capacitor feeding a 10k resistor to ground= 48.2Hz. Frequencies above 48.2Hz are passed and frequencies below 48.2Hz are reduced.
     
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  10. Willen

    Willen Well-Known Member

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    And may I know what is 'pi' and its value while calculating? Please
     
  11. audioguru

    audioguru Well-Known Member Most Helpful Member

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    You can find the value of pi in Google.
    Pi is always used to calculate the cutoff frequency of a highpass or lowpass RC filter.

    I have not used the accurate number for pi for many years because the formula is always 1 divided by (2 x pi x RC) so 1 divided by (2 x pi) is almost simply 0.16.

    Accurately, 1 divided by (2 x 3.14159 x 330nF x 10k)= 48.2Hz.
    But 0.16 divided by (330nF x 10k)= 48.5Hz which is very close (less than 1% higher).

    I corrected the error I made in my previous post.
     

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  12. Willen

    Willen Well-Known Member

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    Wow! Yes, I learnt about more about cutoff frequency (fc) of capacitor. Found a nice calculator too: http://www.2pif.com/high-low-pass-filter.php It is showing 48.23Hz same as your. BUT...where is the 10K load resistor to ground (series with 330nF) on Mod 4 to form High Pass Filter?

    Your audio attenuator (27K+1K) taught me lot actually! From 2 days I am modifying my ancient :) VCD Player. It has lost its remote control functionality so sound of its internal amplifier was horriable high. I found amplifier's input point and set 2 Potentiometer as a gain controller for L and R. Thus I made volume controller. There I found two type of attenuator stage for FM (attenuator B) and for MP3 (attenuator A), look at schematic. What is the difference and its difference application among them?

    Another, I found a sentence- "our ears perceive sound in a logarithmic fashion." Also you said about this once. What they mean? Can you say about it in simple please! Can't we compare our hearing with Microphone's hearing?

    Sorry for asking lot more at once!
     

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  13. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Every coupling capacitor has a resistor load to ground and forms a highpass filter.
    1) My Mod4 circuit has a 330nF coupling capacitor from the electret mic to the input of the preamp transistor:
    The preamp transistor operates at a collector current of 0.28mA and the datasheet for the 2N3904 has a graph showing its typical current gain at that current is 100 and another graph shows its base-emitter input impedance is 11k ohms. Then the transistor's current gain times the emitter resistor value of 470 ohms is 47k. The total input resistance of the preamp transistor is 11k + 47k= 58k ohms.
    The 160k and 30k biasing resistors for the preamp transistor are effectively in parallel with a total of 25.3k which is parallel to the 58k of the transistor for a total of 17.6k.
    The electret mic and the 10k resistor powering it is about 2.7k in series with the coupling capacitor so the total resistance is 17.6k + 2.7k= 20.3k.
    Then the cutoff frequency is 24Hz.
    2) The output of the preamp transistor has an impedance of about 10k ohms and its 330nF coupling capacitor feeds the oscillator transistor that also has an input impedance. This forms another highpass filter that combines with the first filter causing the output to be -6dB at the combined frequency. Then the -3dB cutoff frequency of the entire circuit is higher than 24Hz and might be 35Hz.

    Attenuator A shorts the signal which might overload the signal source. Attenuator B is a volume control.

    Yes, our hearing's sensitivity to loudness is logarithmic then we can hear a very wide range of loudness from a whisper to a nearby jet airplane.
    3dB is double or half the power. But it sounds only a little louder or a little less loud.
    10 times the power sounds twice as loud and 1/10th the power sounds half as loud. That is why a volume control has an "audio taper" (it is logarithmic).
     
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  14. Willen

    Willen Well-Known Member

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    Wow! Nice calculation! Now I knew that why 4.7uF capacitor passes only high freq. on 8 ohms tweeter speaker, bacause it forms a high pass filter and its cutoff is 4232Hz.

    If I made a low pass filter using 56 ohms resistor with 25pf capacitor, it will pass only below than 113MHz. It will block harmonics on FM but I think 'RC filter' will not work on RF and VHF band. I need to use LC Filter instead, isn't it?

    Thank you for your long chain of calculation :)
     
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  15. audioguru

    audioguru Well-Known Member Most Helpful Member

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    A series resistor feeding a capacitor to ground as a lowpass filter or a series capacitor feeding a resistor to ground as a highpass filter are very simple "first-order" filters that have a very gradual slope of attenuation which is 6dB (1/2 the signal level) per octave (double or half the frequency). Most harmonics are barely attenuated.
    But an LC tuned circuit with a high Q passes the frequency you want and attenuates harmonics very well.
     
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  16. Nepaliman

    Nepaliman Member

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    Hi,
    a simple thing still confusing me :)

    I know the main current flow (as a signal) through transistor is EC or CE (totally at saturated mood) but how this happens, look my attach please!

    I would be so happy if you clear me about such operation of transistor and its current flow! I searched lot and found also lot but could not able to make concept about this by reading their deep explanation. :| So, in simple please!

    (on attachment, mistakenly I wrote EBC for diode view of PNP. Please read it correctly- CBE)
     

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    Last edited: Jun 8, 2013
  17. audioguru

    audioguru Well-Known Member Most Helpful Member

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    You cannot make a transistor with two diodes.
    When a transistor has some forward base-emitter current then there is much more collector-emitter current.
     
  18. JoeJester

    JoeJester Active Member

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    This doesn't look like two diodes. :)

    [​IMG]
     
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  19. Nepaliman

    Nepaliman Member

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    :( I know I cannot make transistor with two diodes. Because middle part of two diodes (junction) has been combined very critically on transistor.

    ( It is little hard to visualize like- existence of 'impedance' on non inductor circuit :( )
     
  20. Willen

    Willen Well-Known Member

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    There is 150uA and 450uA peak to peak wave form for I(c) and there is 300uA is mid point, so you called Collector current of preamp has 300uA (which is mid-point) for simple.

    I got a wave form which has 0.5uA to 1.5uA peak to peak and I also called "it has 1uA" (mid-point) for simple.

    But look at one of my attachment, it has -3.5uA and +3.5uA peak to peak waveform and its mid-point is Zero! I am being fool, is it 0.0uA (no current conducting)? :)
     

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  21. audioguru

    audioguru Well-Known Member Most Helpful Member

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    A coupling capacitor passes AC but blocks DC then the output of the coupling capacitor with no load is 0VDC and its voltage swings positive and negative.
     

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