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For my own amazement, measured for comparison two diodes, MVAM109 and BB910.
A schematic of the N5IB design?test fixture** is attached. All stabilized at room temperature for measurement.
?????????????? Capacitance, pF
V??????? MVAM 109????? BB910
0????????? 750????????????????? 52
1????????? 481????????????????? 35
2????????? 340????????????????? 27
3????????? 234??????????????? 22
4????????? 152????????????????? 17
5????????? 100????????????????? 13
6????????? 66??????????????????? 9
7???????? 45??????????????????? 7
8???????? 33??????????????????? 6
9???????? 27????????????????????5.9
10?????? 23???????????????? 5.73
11?????? 21?????????????????? 5.59
12?????? 20?????????????????? 5.45
Highest voltage available from the battery.
The capacitance range of the BB910 is similar to the 60 pF section of a polycon often used with HF?VXOs. The MV-209 has a comparable C range.
** N5IB produced a kit of the fixture, since retired.? Fairly easy to build one from?parts on-hand.
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
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Hey Chuck..... what does this mean.... "All
stabilized at room temperature for measurement."
On 12/16/2019 09:27 AM, Chuck Carpenter
wrote:
toggle quoted message
Show quoted text
For my own
amazement, measured for comparison two diodes, MVAM109 and BB910.
A schematic of the N5IB design?test fixture** is attached. All
stabilized at room temperature for measurement.
?????????????? Capacitance, pF
V??????? MVAM
109????? BB910
0?????????
750????????????????? 52
1?????????
481????????????????? 35
2?????????
340????????????????? 27
3????????? 234???????????????
22
4?????????
152????????????????? 17
5?????????
100????????????????? 13
6?????????
66??????????????????? 9
7????????
45??????????????????? 7
8????????
33??????????????????? 6
9????????
27????????????????????5.9
10?????? 23????????????????
5.73
11??????
21?????????????????? 5.59
12??????
20?????????????????? 5.45
Highest voltage available from the battery.
The capacitance range of the BB910 is similar to the 60 pF section
of a polycon often used with HF?VXOs. The MV-209 has a comparable
C range.
** N5IB produced a kit of the fixture, since retired.? Fairly easy
to build one from?parts on-hand.
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
_._,_._,_
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On 12/16/19 9:27 AM, Chuck Carpenter
wrote:
toggle quoted message
Show quoted text
For my own amazement, measured for comparison two diodes, MVAM109
and BB910.
A schematic of the N5IB design?test fixture** is attached. All
stabilized at room temperature for measurement.
?????????????? Capacitance, pF
V??????? MVAM
109????? BB910
0?????????
750????????????????? 52
1?????????
481????????????????? 35
2?????????
340????????????????? 27
3????????? 234???????????????
22
4?????????
152????????????????? 17
5?????????
100????????????????? 13
6?????????
66??????????????????? 9
7????????
45??????????????????? 7
8????????
33??????????????????? 6
9????????
27????????????????????5.9
10?????? 23????????????????
5.73
11??????
21?????????????????? 5.59
12??????
20?????????????????? 5.45
Highest voltage available from the battery.
The capacitance range of the BB910 is similar to the 60 pF section
of a polycon often used with HF?VXOs. The MV-209 has a comparable
C range.
** N5IB produced a kit of the fixture, since retired.? Fairly easy
to build one from?parts on-hand.
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
|
I'm curious how the 1n4001 compares with purpose built varactor diodes.
Figure 3 here shows it going from about 20 to 5 pf as the reverse voltage goes from 0 to 10 volts.
? ??
Would be interesting to verify that with measurements, trying several different manufacturers.
|
Virtually ANY diode will work, some better than
others. I have a number of projects where I used a red LED as a
varactor. I usually got from low-teens pf to ~50 pf. Smaller
ones tended to be better than larger ones. Red was better than
other colors which have a much more limited range. The only
problem I found was light sensitivity. You have to block the
lens (tape, dipped in paint, whatever) or it will pick up 60
cycle variations from lighting and other weird effects.
Didn't have much luck with gp diodes. They work.
Just not as much range as red LEDs. I've used them since about
2005. A couple are still plugging away and work fine.
Eric KE6US
On 12/16/2019 11:42 AM, Jerry Gaffke
via Groups.Io wrote:
toggle quoted message
Show quoted text
I'm curious how the
1n4001 compares with purpose built varactor diodes.
Figure 3 here shows it going from about 20 to 5 pf as the reverse
voltage goes from 0 to 10 volts.
? ??
Would be interesting to verify that with measurements, trying
several different manufacturers.
|
I used a 1N4007 in my NC-40 build. More range than the specified varactor. The PIN structure helps. 73, Gary? WB6OGD?
toggle quoted message
Show quoted text
On Dec 16, 2019, at 1:20 PM, Eric KE6US <eric.csuf@...> wrote:
?
Virtually ANY diode will work, some better than
others. I have a number of projects where I used a red LED as a
varactor. I usually got from low-teens pf to ~50 pf. Smaller
ones tended to be better than larger ones. Red was better than
other colors which have a much more limited range. The only
problem I found was light sensitivity. You have to block the
lens (tape, dipped in paint, whatever) or it will pick up 60
cycle variations from lighting and other weird effects.
Didn't have much luck with gp diodes. They work.
Just not as much range as red LEDs. I've used them since about
2005. A couple are still plugging away and work fine.
Eric KE6US
On 12/16/2019 11:42 AM, Jerry Gaffke
via Groups.Io wrote:
I'm curious how the
1n4001 compares with purpose built varactor diodes.
Figure 3 here shows it going from about 20 to 5 pf as the reverse
voltage goes from 0 to 10 volts.
? ??
Would be interesting to verify that with measurements, trying
several different manufacturers.
|
I have uploaded a couple of pdfs for some diode junction capacitance comparisons that I did several years ago. They are in the K5DW directory. I used whatever I had in the junque box at the time. I only used reverse voltages up to 9 volts because I noticed that above that the capacitance flattened out considerably.
Anyways, use them or ignore them as you?see fit!
Don Wines, K5DW
On Mon, Dec 16, 2019 at 1:42 PM Jerry Gaffke via Groups.Io <jgaffke= [email protected]> wrote: I'm curious how the 1n4001 compares with purpose built varactor diodes.
Figure 3 here shows it going from about 20 to 5 pf as the reverse voltage goes from 0 to 10 volts.
? ??
Would be interesting to verify that with measurements, trying several different manufacturers.
|
Hi all
See a number of diode measurements here centering around a lot of LED measurements.
73 Hans G0UPL? ?
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Jerry, the shape of the curve in fig 3 is typical of?junctions of non?"purpose built" diodes. (like that term).
I've attached a set of curves from K7QO.? The curve at the bottom is a power diode.? Note that there is not much change from about 5 V so higher voltages don't help much.
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
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Gary,? What was the specified part?
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
|
Chuck,
On page 16 of? ?
there's an equation for the capacitance of a reverse biased diode
? Cj =? ?A/2 * ((2*q*e/(V0-V))*(Nd*Na/(Nd+Na)))**(1/2)
From that, it appears that the junction capacitance is inversely proportional
to the sqrt of the voltage applied (voltage measured relative to V0)
Which to my eye, agrees with those curves.
If you happen to be exceptionally curious, the derivation is in the pages prior.
And if you're lucky, he defines some of the variables and constants in lecture 26.?
Extra credit:?
So, as the voltage across the diode decreases, the change in capacitance due to some delta-V (voltage noise) increases.
But since we are using a resistive divider to create the voltage across the diode,
any noise in the power supply voltage is reduced proportional to that resistive divider.
Which one wins?
Jerry
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Show quoted text
On Thu, Dec 19, 2019 at 12:12 PM, Chuck Carpenter wrote:
Jerry, the shape of the curve in fig 3 is typical of?junctions of non?"purpose built" diodes. (like that term).
I've attached a set of curves from K7QO.? The curve at the bottom is a power diode.? Note that there is not much change from about 5 V so higher voltages don't help much.
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
|
Jerry,? That's a fun equation!
The half-power Sqrt sez is not linear. I can't quite visualize** how the parallel capacitor calculation works out.? I may just have to get really curious and check out the constants and variables and work some examples from the back of the book.
In the meantime, I'll just use my measuring gadget...8^)
Fun Stuff
**I did this stuff about 60 years ago.? Kinda hazy now.
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
|
Chuck, It took me awhile to figure out how to make sense of it. Could be errors, but this is what I am currently thinking:
Here's that equation again, a power of (1/2) means we take the square root:
Cj = A/2 * ((2*q*e/(V0-V))*(Nd*Na/(Nd+Na)))**(1/2)
V0 is very close to zero volts so we can ignore it, and V is a negative voltage.
A is the cross sectional area of the cap, Nd and Na have to do with doping levels.
Those all get determined when the diode is fabricated, and won't change for us.
e (epsilon) is the permitivity, should be about the same for all silicon diodes.
q is the charge of an electron, determined back when the universe was born.
If we lump all those constants into a single value I'll call K, we get:
Cj = K/sqrt(V)
Assume we have a varactor diode that measures 100pf when we put 1 volt across it.
What is the capacitance when we put 10 volts across it?
At one volt: K = Cj*sqrt(v) = 100*sqrt(1) = 100
At 10 volts: Cj = 100/sqrt(10) = 31.62 pf
Getting back to our original conversation, assume our 10v power supply
has 0.1v of pk-pk ripple on it. How much variation in capacitance do we get
with this particular diode for a bias at 1v versus a bias at 10v?
Here's the difference between max and min capacitance at a nominal 10v of bias:
delta-C = 100/sqrt(10-0.1) - 100/sqrt(10+0.1) = 0.316 pf
When we adjust the pot to divide down the power supply to give 1v of bias,
we also divide down the noise by a factor of ten.
Here's the difference between max and min capacitance at a nominal 1v of bias:
delta-C = 100/sqrt(1-0.01) - 100/sqrt(1+0.01) = 1.000 pf
Conclusion:
Our oscillator will be significantly more sensitive to power supply noise
at low bias voltages across the varactor diode. This is in spite of the fact
that the voltage fluctuation on a bias of 1 volt is 1/10 what it would be for a 10 volt bias.
?
On Fri, Dec 20, 2019 at 02:12 AM, Chuck Carpenter wrote:
Jerry,? That's a fun equation!
The half-power Sqrt sez is not linear. I can't quite visualize** how the parallel capacitor calculation works out.? I may just have to get really curious and check out the constants and variables and work some examples from the back of the book.
In the meantime, I'll just use my measuring gadget...8^)
Fun Stuff
**I did this stuff about 60 years ago.? Kinda hazy now.
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
|
A minor error in my previous post where I said:? "assume our 10v power supply has 0.1v of pk-pk ripple on it" My subsequent calculations for the variation in capacitance assume noise on the power supply of 0.2v pk-pk, not 0.1v pk-pk. Chuck wrote: > **I did this stuff about 60 years ago.? Kinda hazy now. My college days were back in the 1970's. So I've got a 15 year advantage on you. It's scary to revisit this stuff and realize how much I forgot. For those who forgot high school algebra, this might be confusing:
If we lump all those constants into a single value I'll call K, we get:
Cj = K/sqrt(V)
Assume we have a varactor diode that measures 100pf when we put 1 volt across it.
What is the capacitance when we put 10 volts across it?
At one volt: K = Cj*sqrt(v) = 100*sqrt(1) = 100
At 10 volts: Cj = 100/sqrt(10) = 31.62 pf
We could break it down a bit further:
Cj = K/sqrt(V)? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?# Forget the nasty formula.? All we care about is that capacitance is proportional to? 1/sqrt(V)
Cj * sqrt(V) = K/sqrt(V) * sqrt(V)? ? ? ? ? ? ?# multiply both sides by sqrt(V)
Cj * sqrt(V) = K * (1/sqrt(V)? * sqrt(V)? ? ?# regroup Cj * sqrt(V) = K? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?#? sqrt(V) divided by itself is equal to 1 K = Cj * sqrt(V)? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?#? switch sides So if we know the capacitance Cj is 100pf at a bias of 1v, we can calculate the value of K: K = Cj * sqrt(V)? = 100 * sqrt(1) =? 100? The two different instances of "100" are an unfortunate coincidence, and thus a source of confusion. The first 100 is 100pf, the second 100 is the computed value of the constant K for this particular diode. Now that we know the value of K for this diode, we can calculate the capacitance Cj for a bias of 10v: Cj = K/sqrt(V)? = 100/sqrt(10) = 31.62 pf Jerry,? ?KE7ER
On Fri, Dec 20, 2019 at 07:55 AM, Jerry Gaffke wrote:
Getting back to our original conversation, assume our 10v power supply has 0.1v of pk-pk ripple on it. How much variation in capacitance do we get with this particular diode for a bias at 1v versus a bias at 10v?
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The 1n4001 seems to follow the? ?Cj = K/sqrt(V)? ?formula rather closely.
From the graph of fig 4 on page 2 of the 1n4001 datasheet at? ?
? ??
I read the following sample of values:??1v:30pf? 9v:10pf? 30v:5pf? 80v:3pf
Using the midrange 9v:30pf pair to compute? ?K = 10*sqrt(9) = 30,
we get the following results from Cj = K/sqrt(V) = 30/sqrt(V):
? ? 1v: 30.0pf? ? 9v 10.0pf? ? 30v: 5.48pf? ?80v: 3.35pf
The figures computed from? Cj=K/sqrt(V)? follow the graph quite well.
Some diodes don't follow our formula so closely.
Take a look at the figure on the left side of page 2 for the 1SV322:
? ??
Using that figure, I read:? ? 6v:5pf 4.5v:6pf 2.8v:10pf 1.6v:20pf? 0.5v:38pf
Using the 2.8v:10pf center value pair, I find that K=16.73? and Cj=K/sqrt(V) gives:
? ?6v: 6.83pf? ? 4.5v: 7.88pf? ? 2.8v: 10.0pf? ? 1.6v 13.23pf? ? 0.5v: 23.66pf
Note that values from the figure have higher capacitance at low voltages than our computed values.
So this specially designed 1SV322 varactor diode is even more non-linear than the 1n4001.
Do I have something wrong here?
Why would we not use a 1n4001 instead?
Perhaps performance at high frequencies, where the 1n4001 might show a bit of PIN behavior?
Curious.
Jerry, KE7ER
?
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Correction, at 9v the 1n4001 graph shows 10pf, not 30pf:
? wrong:? ?Using the midrange 9v:30pf pair to compute? ?K = 10*sqrt(9) = 30,
? right:? ? ?Using the midrange 9v:10pf pair to compute? ?K = 10*sqrt(9) = 30,
Just in case anybody was trying to make sense of this stuff.
> Perhaps performance at high frequencies, where the 1n4001 might show a bit of PIN behavior?
I usually blame poor performance at high frequencies on capacitance, but not the case here.
So how would a 1SV322 be better than a 1n4001?
To show "PIN behavior" it would have to occasionally be forward biased, that's not the case here.
But is there a similar behavior that might cause trouble at high frequencies?
No diode used as a variable capacitor will have a linear response to voltage, but some are better than others.
This is significant when trying to tune a VFO, don't want all the action to be happening at the bottom end of pot rotation.
How close to linear is not a characteristic of varactor diodes I've ever seen discussed.
Most analysis is a quick look at min and max capacitance.
I think I'm done here, will go back to fiddling with si5351's.
We now return you to your regularly scheduled programming.
Jerry, KE7ER
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Chuck Carpenter
Don K5DW