Varactor ?

They give the mean DC voltage, VR, as well. In operation, the diodes are subjected to a DC reverse bias with a (hopefully) smaller AC signal component superimposed. You have to apply an AC signal to measure capacitance! 1 MHz is a good test frequency for an AM radio tuning diode.

Hmmm. Well, thanks for the reply... Now I just have to try and understand why an AC signal is necessary for measuring capacitance. Not as smart as I thought I was. I don't remember covering varactors in school (30 years ago!) but I thought I understood the principle. If you would be so kind, why *is* an AC signal necessary for the measurement of capacitance?

Thanks again,

Dave

"Dave

** This will be worth waiting for.....

..... Phil

Yes, using the formula Q=CV you can measure capacitance using DC

Q = Charge = time integral of current

So you measure the voltage change across the capacitor over a time interval and integrate the current flowing during the same time.

Alternatively, you can measure capacitance using AC using the fact that reactance Xc = 1/wC e.g. by placing it in a tuned circuit, or as part of an RC network.

Varactor capacitance measured at DC is not quite the same as at RF frequency, so the datasheet quotes the capacitance at the frequency of interest, which, for an AM tuning varactor is around 1 MHz.

In principle it isn't, but in practice it is. If you had an ideal large-valued capacitor you could apply a known DC voltage through a known resistance and measure the time the capacitor took to charge to a given fraction of the source voltage. That gets pretty tricky when the capcitance is as small as we're talking about in a varactor... it can't handle a huge voltage (even if you had one handy) so you'd need to measure a ridiculously small time interval (and final voltage). And that measurement would assume the capacitor was ideal... which it isn't.

So it's much easier to measure the *reactance* of the capacitor, which is essentially its resistance to AC. It's proportional to the applied frequency as well: Reactance = 1/ (2 * pi * f * C).

You could use a simple RC voltage divider to measure the capacitance. Apply a known AC voltage at a known frequency, and adjust R until the AC voltage at the center is exactly half. Measure R, which will be equal to the reactance of the capacitor, and rearrange the above formula to solve for the capacitance.

Best regards,

Bob Masta DAQARTA v4.51 Data AcQuisition And Real-Time Analysis

Scope, Spectrum, Spectrogram, Sound Level Meter FREE Signal Generator Science with your sound card!

"Andrew Holme"

** Huh ??

A " voltage change" defies the notion of DC voltage.

DC voltages do not change -

else they acquire an AC component.

.... Phil

First, because when you're dealing with a device whose capacitance varies with voltage, the exact definition of capacitance can get a bit fuzzy. Second, because the varactor doesn't act like a perfect capacitor, and it's apparent capacitance can vary by frequency.

The "DC" definition would be C = q/V, so (assuming you could measure charge accurately) that implies that you would measure the charge and voltage, divide the measured charge by the voltage, and get the capacitance value.

That would be nearly useless, because you don't care about the total capacitance of the diode -- what you care about is it's incremental capacitance about the given bias voltage:

C(Vb) = dq/dV | | V = Vb

One very reasonable way to measure this is to impose a sinusoidal voltage (or current) at some frequency, and measure the current (or voltage) that results. If you do this, then you'll also take care of those parasitic effects (mostly resistive losses at 1MHz, although at higher frequencies the package inductance would become important). Because the parasitic effects change the apparent capacitance, you want to measure the thing at the frequency it's going to be used -- hence 1MHz for a diode that's being marketed for AM radio use.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

Also worthy of consideration, with a varying voltage on a varactor there will result also varying capacitance.

Varactors are diodes that are reverse biased with a DC voltage to modulate the thickness of the depletion region between the P and N doped regions. The DC bias is listed on the datasheet as the value of Vr.

By modulating the depletion thickness, the small signal capacitance is changed. The capacitance is measured with an applied small signal at a frequency of 1MHz.

See:

(or:

for some more details.

Regards,

Bob.

"Bob Masta"

** The condition for equal impedance is when the AC voltages appearing across both R and C are the same - but it will not be "half" each.

When the R value equals the reactance of the capacitor, the AC voltage across R or C is 0.7071 (1/ sq.rt.2 ) times the applied voltage.

..... Phil

Many, many thanks, to all for the replies. I have a lot of info to assimilate and digest, and I do appreciate the information. Please forgive this one-size-fits-all note of appreciation, I just don't know yet what to say to each of you other than "many thanks. It is appreciated."

Dave

Absolutely correct! (Sorry, brain fart on my part.)

To the OP: This is a simple RC filter, and when the reactance equals the resistance the output is at half *power*, not half voltage. Since power is proportional to the square of voltage, it is reduced to 0.7071^2 = 0.5.

Apologies for the confusion, and best regards,

Bob Masta DAQARTA v4.51 Data AcQuisition And Real-Time Analysis

Scope, Spectrum, Spectrogram, Sound Level Meter FREE Signal Generator Science with your sound card!