Varactor ?

D

Dave

Guest
Looking at the datasheet for the NTE618, and wondering why it lists a
frequency (1MHz) after each entry for the interterminal capacitance. Anyone
have any ideas on this? I figured the applied voltage would be DC. No?

Thanks,

Dave
 
"Dave" <db5151@hotmail.com> wrote in message
news:q92dnUSDJI9JzojUnZ2dnUVZ_sHinZ2d@posted.internetamerica...
Looking at the datasheet for the NTE618, and wondering why it lists a
frequency (1MHz) after each entry for the interterminal capacitance.
Anyone have any ideas on this? I figured the applied voltage would be DC.
No?
Click to expand...

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.
 
"Andrew Holme" <ah@nospam.co.uk> wrote in message
news:VjeRk.17592$uq3.6030@newsfe14.ams2...
"Dave" <db5151@hotmail.com> wrote in message
news:q92dnUSDJI9JzojUnZ2dnUVZ_sHinZ2d@posted.internetamerica...
Looking at the datasheet for the NTE618, and wondering why it lists a
frequency (1MHz) after each entry for the interterminal capacitance.
Anyone have any ideas on this? I figured the applied voltage would be
DC. No?


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.
Click to expand...
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
"Andrew Holme"
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?
Click to expand...

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




...... Phil
 
"Phil Allison" <philallison@tpg.com.au> wrote in message
news:6nlfnlFm0rduU1@mid.individual.net...
"Dave
"Andrew Holme"
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?



** This will be worth waiting for.....
Click to expand...
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.
 
On Sat, 8 Nov 2008 06:37:18 -0600, "Dave"
<db5151@hotmail.com> wrote:

"Andrew Holme" <ah@nospam.co.uk> wrote in message
news:VjeRk.17592$uq3.6030@newsfe14.ams2...

"Dave" <db5151@hotmail.com> wrote in message
news:q92dnUSDJI9JzojUnZ2dnUVZ_sHinZ2d@posted.internetamerica...
Looking at the datasheet for the NTE618, and wondering why it lists a
frequency (1MHz) after each entry for the interterminal capacitance.
Anyone have any ideas on this? I figured the applied voltage would be
DC. No?


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
Click to expand...
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
www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
FREE Signal Generator
Science with your sound card!
 
"Andrew Holme"
"Phil Allison"
"Dave
"Andrew Holme"
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?



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


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.
Click to expand...

** Huh ??

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

DC voltages do not change -

else they acquire an AC component.




..... Phil
 
On Sat, 08 Nov 2008 06:37:18 -0600, Dave wrote:

"Andrew Holme" <ah@nospam.co.uk> wrote in message
news:VjeRk.17592$uq3.6030@newsfe14.ams2...

"Dave" <db5151@hotmail.com> wrote in message
news:q92dnUSDJI9JzojUnZ2dnUVZ_sHinZ2d@posted.internetamerica...
Looking at the datasheet for the NTE618, and wondering why it lists a
frequency (1MHz) after each entry for the interterminal capacitance.
Anyone have any ideas on this? I figured the applied voltage would be
DC. No?


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
Click to expand...
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
 
"Bob Masta" <N0Spam@daqarta.com> wrote in message
news:4915956e.2914154@news.sysmatrix.net...
On Sat, 8 Nov 2008 06:37:18 -0600, "Dave"
db5151@hotmail.com> wrote:


"Andrew Holme" <ah@nospam.co.uk> wrote in message
news:VjeRk.17592$uq3.6030@newsfe14.ams2...

"Dave" <db5151@hotmail.com> wrote in message
news:q92dnUSDJI9JzojUnZ2dnUVZ_sHinZ2d@posted.internetamerica...
Looking at the datasheet for the NTE618, and wondering why it lists a
frequency (1MHz) after each entry for the interterminal capacitance.
Anyone have any ideas on this? I figured the applied voltage would be
DC. No?


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


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.
Click to expand...
Also worthy of consideration, with a varying voltage on a varactor there
will result also varying capacitance.
 
On Sat, 08 Nov 2008 02:54:35 -0600, Dave wrote:

Looking at the datasheet for the NTE618, and wondering why it lists a
frequency (1MHz) after each entry for the interterminal capacitance.
Anyone have any ideas on this? I figured the applied voltage would be
DC. No?
Click to expand...
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:
http://www.electronics-radio.com/articles/electronic_components/diode/
varactor-varicap-diode.php

(or: http://tinyurl.com/59k3m8)

for some more details.

Regards,

Bob.
 
"Bob Masta"

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.
Click to expand...

** 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

"Dave" <db5151@hotmail.com> wrote in message
news:q92dnUSDJI9JzojUnZ2dnUVZ_sHinZ2d@posted.internetamerica...
Looking at the datasheet for the NTE618, and wondering why it lists a
frequency (1MHz) after each entry for the interterminal capacitance.
Anyone have any ideas on this? I figured the applied voltage would be DC.
No?

Thanks,

Dave
Click to expand...
 
On Sun, 9 Nov 2008 14:27:58 +1100, "Phil Allison"
<philallison@tpg.com.au> wrote:

"Bob Masta"


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.


** 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.
Click to expand...
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
www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
FREE Signal Generator
Science with your sound card!
 
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