2   Tuned Circuit Formula Part 1

What is a resonant tuned circuit? A very good question with a not too easy answer. Let’s see. The reactance of a capacitor, C, reduces when the applied frequency increases but, in an inductance, L, the reverse happens, with the reactance increasing as the frequency increases.

Importantly, at the resonant frequency(fres) the reactances of C and L are equal. (Xc = XL)

In any capacitor, when a voltage is applied there is a large current flow initially and the voltage takes time to build up.  So the current leads the voltage (by 90 degrees for AC). In any inductance, due to the back EMF produced when a voltage is applied, the current takes time to reach a maximum, so the voltage leads the current (also by 90 degrees). A simple rule is:

In a C, I comes before V. (CIV) and V comes before I in an L, (VIL) CIV VIL = CIVIL

What has all this got to do with resonance you ask ? Read on McDuff. Learn and digest.

Using vector diagrams we can actually SEE what happens if a signal (AC) is applied at fres. By convention, voltages and currents are represented by lines whose lengths represent their amplitude.(size, voltage, height!) Phase is indicated by their direction. Also by convention, vectors rotate round the axis in an anticlockwise direction at whatever the frequency is; i.e. at so many revolutions a second. A vector diagram freezes the action at one instant in time.

Plainly, in a parallel tuned circuit the voltages across both L & C must be in phase, but you can see that the current in the capacitor is equal and opposite to the current in the inductor so the currents cancel out. If we have voltage without little or no current flow, this must mean that the impedance must be very high, so there is little attenuation of the applied signal. But,  signals NOT at fres will produce currents that do not cancel out, so the impedance will therefore be lower for these, so smaller amounts of these off-frequency signals will be passed on.

The further from fres a signal is, then less of it is passed on to the next (amplifier) stage.

We have a different set of condition if C and L are in series, as the current is common to both and must be in the same phase. So, if we apply CIVIL again, the voltages across the C and across the L are now equal and opposite at resonance, and so they cancel out. Having current flow with very little voltage must mean that the impedance is VERY LOW. In this case, signals at fres pass through easily with little loss. However, signals which are higher or lower in frequency find the impedance increasing as the voltage cancellation deteriorates, so less of them gets through to the next stage. Once again, the further from fres the signals are, progressively smaller amounts are passed on to the next stage. In all this, the inevitable DC resistances have not been mentioned. R must be present in the coil wire and the interconnecting leads in any circuit, which is one reason to use fat wire and keep all the wiring as short possible. This also reduces the lead inductance, which is increasingly a problem as we go up in frequency as even the smallest straight wire has some inductance! Oh, another thing, the R affects the Q of a resonant  tuned circuit too! What’s that? Find out in a later article.

That’s it for this time. Mike G3JKX