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There seems to be very little literature to explain the practical construction of a radio and how a super regenerative circuit actually act as a selector, amplifier, how a antenna is actually to be connected etc. I have a circuit that I have built which goes like this

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When my circuit did not work (I used a BF199), I did a check on the LC tank circuit that is supposed to select the frequency. An important part of that was the coil. My calculations looked like this

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As per my calculations, if I want to tune in to a 1 Mhz frequency, my capacitor will need to be of 3.38 x 10^-21 F. The circuit diagram has a Trimmer Capacitor 3-22pF. I tried calculations for different frequencies and it seems that such a circuit will only be able to tune into about 50Hz or so. I would like to know what is happening here. How will this tuner circuit work as according to me it needs to oscillate at desired frequency we want to tune it to? I checked my calculations a million times I can't see what is wrong. Please help!

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    \$\begingroup\$ You've done something strange in your calculation. That capacitance value is waaaaaay off. \$\endgroup\$ – brhans Jan 14 '16 at 20:55
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    \$\begingroup\$ First rearrange that formula to calculate C in terms of L and F. In the process you'll probably get a more sensible answer - if not, post your new formula (edit the question). \$\endgroup\$ – Brian Drummond Jan 14 '16 at 21:01
  • \$\begingroup\$ 1.0/(2*3.14*(1.3e-7*3.4e-21)^0.5) = 7.6 TeraHertz. My goodness! \$\endgroup\$ – jpcgt Jan 14 '16 at 21:59
  • \$\begingroup\$ The actual value of C is about 1900 pF. You need to check your calculation a million plus one times. \$\endgroup\$ – Barry Jan 14 '16 at 22:24
  • \$\begingroup\$ Inductance is miles off for 1 MHz too. \$\endgroup\$ – Andy aka Jan 14 '16 at 22:58
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Try 100 uH and 100 pF just as an example: -

F = \$\dfrac{1}{2\pi\sqrt{LC}}\$ = 1.592 MHz i.e. not bad for a guess.

You need to be approximately in this ball-park areas to tune 1 MHz.

134 nH is absolutely way out and 10^-21 farads is preposterous.

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  • \$\begingroup\$ @KunalVerma Did you get your circuit working with the new value range I proposed? \$\endgroup\$ – Andy aka Feb 8 '16 at 17:47

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