Variable Capacitor - capacitance vs. rotation angle

Question

I have an Elenco AM / FM radio kit. For the AM section I am trying to figure out the relationship between the frequency shown on the calibrated dial and the angular position of the dial.

For a capacitor, the capacitance is defined as C = epsilon * A / d, epsilon is the permittivity of the dielectric material between the plates, A is the plate area, and d is the plate separation.

The capacitance seems to be a straightforward linear function of rotation angle. For a variable capacitor like this,

which is representative only, with maximum capacitance being when the rotating blades are all interleaved with the stator blades, I defined the capacitance as C = -k1 * theta + k2, theta being 0 at the maximum capacitance value.

The kit does not specify the inductor value of the ferrite-core antenna and coil, but I estimated it to be about 0.2 mH.

The resonant frequency for an LC circuit is f = 1 / (2 * pi * sqrt(L * C)).

I plotted two lines.

The "Plotted dial data" curve resulted from reading 6 frequencies from the dial and the angles for which they would tune. The "Theoretical Equation data" is the plot of the frequency equation with the linear C equation substituted into it with k1 and k2 being determined so that the curves meet at 540 kHz and 1600 kHz, the lower and upper ends of the AM band. The respective theta is 10 and 170 degrees.

You can see that the two have a superficially similar shape, but I cannot determine why the theoretical curve is not more like the plotted dial curve.

I am missing something.

Thanks for any help.

Answer 1

Rotary tuning capacitors are not necessarily simple sectors in their plate geometry. Rather, more complex shapes are sometimes used.

If one wants linear change of frequency with respect to dial rotation, this cannot be achieved by linear change in capacitance such as caused in an angle-proportionate change in area. Across a narrow frequency band it may be close enough, and in some applications the dial markings do not really need to be linear. But for something like the local oscillator of a short wave communications radio, it was really quite important than dial markings correspond closely to actual frequency across the sub-band.

I've personally seen, and little web searching confirms, that more complex plate shapes with a slight teardrop character, plates offset from the axis, etc, were not uncommon. A little care could yield a plate design to suit any reasonable desired capacitance curve calculated back from a desired change in resonant frequency. Some examples can be seen at http://www.vk6fh.com/vk6fh/tuning%20capacitors.htm

(please don't edit to "linkify" that - as an external resource, it's meant to be only a supplemental example, and should it change the details of the URL such as the creator's amateur radio callsign may be key to finding it again)

As a practical matter, it's also critically important to note that an "open" design as shown in the question statement must sit behind a shielded panel, otherwise "hand capacitance" may distort the tuning. More compact "pocket radio" designs such as actually used in the asker's kit presumably have built-in shielding; additionally broadcast channels are often wide enough that exact tuning is less critical than in some other applications.

Answer 2

You are on the right track, in that the angle of the variable capacitor is linearly related to the capacitance, and that the formula for resonant frequency is

\$f_0=\frac{1}{2\pi\sqrt{LC}}\$

However, the frequencies we need to look at are 455kHz away from what you read on the dial of the AM receiver.

By far the most popular type of AM radio receiver is a super-heterodyne receiver. There is a local oscillator that is controlled by a variable capacitor, that creates a signal that is mixed with the received radio signal. This mixing causes a new signal (the IF or intermediate frequency) signal at around 455kHz. This IF signal is amplified and then detected (converted to audio).

So, when your AM receiver says 1MHz on the dial, I believe the local oscillator is generating a frequency of either 1.455MHz or 1.000-0.455 = 545kHz. Both have apparently been in use. See Article on Superheterodyne receivers

Knowing that, re-do your math to see what you come up with.