pls help guys about capacitors?

3 Comments

1. Ron R says:

   July 7, 2009 at 5:13 pm

   I agree with paulie2shoes about the formula and resonant frequency, although in the real world the load on the resonant circuit will usually shift the frequency somewhat. There are commercially available 300pF capacitors, but not 500pH inductors.

   Since you asked about availability of parts, you must be planning on building this circuit. Let me start by saying I think the values for the inductor and capacitor are unrealistic. I don’t know the purpose of this circuit, but an inductor of 500pH is so small it is nearly impossible to make from wire. For example if you use number 20 wire, it only takes about 1.6 mm of wire to get 500pH! (source: ARRL Handbook for Radio Amateurs 1994). This makes it impractical to use a discrete inductor for your circuit. The inductance of the lead wires is larger than the desired inductance. Also, the formula for the impedance of an inductor is Z = 2 pi f L. For your inductor this is 2 * pi * 410 * 10^6 * 500 * 10^-12 = 1.28 ohms. This is an unusually low impedance.

   300pF capacitors are widely available (see http://www.digikey.com). There are many other specifications you must make besides the capacitance. At frequencies as high as 410 MHz you need a low-loss dielectric material in the capacitor. For low-power a ceramic or polystyrene dielectric would probably work well. For high-power you will probably need a capacitor with an air dielectric. Note that the amount of RF power the circuit will handle will determine the voltage breakdown requirement for the capacitor and the current capability of the inductor.

   There is a catalog of air-dielectric variable capacitors on http://www.orenelliottproducts.com/capacito.htm. As you can see these capacitors are fairly large when they have a capacitance around 300pF. The Inductance of the lead wires to these capacitors is larger than 500nH, so they won’t work for you. You need a very small capacitor, which probably rules out high-power.

   Without knowing the purpose of your circuit, it is hard to suggest what would be the best approach. However, note that if you increase the inductor by a factor of 10 to 5nH and reduce the capacitor by a factor of 10 to 30pF, the circuit will still resonate at the same frequency. The inductor can then be made from 16 mm of number 20 wire, which is much more easily handled and the capacitor will be much smaller in size. The impedance will, of course, be 10 times higher; 12.8 ohms.

   I hope this helps.
2. paulie2shoes says:

   July 7, 2009 at 5:13 pm

   The value of the Capacitor is 300 pf, that’s pico farads (300 x 10 -12)’
   The Inductor is 500 pico henries (500 x 10 -12)
   The answer is 410.936296 Mhz

   The formula for resonant frequency is : f= 1 over 2 pi x square root of LC.
3. Mr. Un-couth says:

   July 7, 2009 at 5:13 pm

   Paulie2S… gave the correct resonant frequency. But here is a few pointers for what ever they are worth.

   1. A tank circuit does not generate a resonant frequency or any other frequency. It must be activated and sustained from an external Voltage source in order to start and maintain oscillations. A tank circuit has no internal energy source of it`s own and will therefore dissipate all of it`s stored starting energy from the external source after a few cycles of oscillations and cease to oscillate if not sustained by an outside source.
   2. A tank circuit resonates at the specific single frequency at which it`s capacitive reactance (Xc) equals it`s inductive reactance (XL). Setting the equations for these two equal to each other and solving for f in terms of L and C enables you to derive the formula for the resonant frequency.
   3. I think you can find a capacitor close to 300 pf in value but I don`t think you will be able to find a 500 ph inductor.