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my varactor extension

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I was wondering if the transistor schematic below would also qualify for a varicap, just because transistors can output some capacitance, just because they are two diodes with anodes or cathodes tied together.

and because 1/2 the manufacturers do NOT provide a chart displaying the capacitance of a diode, is there a formula I can use to calculate capacitance based on voltage?

I would think it would involve using log, ln, sin or cos, because capacitance isn't linear when dealing with diodes.
 

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Hi MStechca,
Look again (and again and again). The datasheet for a 2N3904 doesn't say it has 8pF. That's its max and only a few in 100 of them will be max. The curves show that its typical capacitance is only 4pF. Some might be only 2pF.

Its typical tuning ratio is next to nothing: 2.8pF at 5V, 4pF at 0.5V.
Its emitter-base is a low voltage zener diode whose leakage current at 5V will messup your DAC.

You nave millions of other transistors to look at but the 2N4401 that is paralleled in the circuit I posted looks to be much better. Notice that the circuit uses its low-leakage-current collector-base junction as its varactor diodes, and it tunes only a small part of the FM band.
 

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2N3904 to 2N4401 comparison:
 

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Look at the capacitance ratio on a real varactor diode.
This BB134 has a low capacitance for UHF. The BB132 and BB133 have double the capacitance for VHF tuning.
 

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The perfect varactor for me is one that can produce an output capacitance of 0 to 10 pF based on an input voltage from 0 to 9.
 
mstechca said:
The perfect varactor for me is one that can produce an output capacitance of 0 to 10 pF based on an input voltage from 0 to 9.
That's not nearly enough capacitance.
Add about 10pF of stray capacitance and then the range becomes only from 10pF to 20pF. It won't tune the entire FM band.

The voltage of your 9V battery will quickly drop to 7,2V then slowly drop to 6V over its life. The tuning will automatically scan up in frequency as the battery runs down.
 
It won't tune the entire FM band.

I don't need it to go that far. I'm making a transceiver, and consistently jamming the FM band will bring on the RF cops. :lol:

The voltage of your 9V battery will quickly drop to 7,2V then slowly drop to 6V over its life. The tuning will automatically scan up in frequency as the battery runs down.
The good news is that I made an outsmarting feature.
Since I can make an LED to detect the signals, I can ram the LED signal into the clock of the counter, and make the receiver continuously scan until it finds a station that transmits nothing.
because the LED detects amplitude, I can set a threshold on how loud the audio must be before it would count as interference.

so your tuning theory in my books is now outdated

hee hee heeeee :lol: :lol: :lol: :twisted:
 
1) Why use an LED as a detector diode instead of an ordinary detector diode?
2) You're making a scanning radio in reverse. It stops scanning when it detects nothing. Then your transmitter starts modulating and the scanning resumes. Good-bye communication.
3) Don't you think that AM is outdated?
 
audioguru said:
1) Why use an LED as a detector diode instead of an ordinary detector diode?
WRONG again. The LED shows the output. I can have it autoscan without the LED.

2) You're making a scanning radio in reverse. It stops scanning when it detects nothing. Then your transmitter starts modulating and the scanning resumes. Good-bye communication.
thats ok, because when it gets to the point that it stops, I'll rip out the connection to the clock that restarts the scanning, and it wont rescan any more :wink:

3) Don't you think that AM is outdated?
why should the obsolete factor of a technology have to do with its availability or productivity?
in fact, thinking that way SUCKS!
because the landfill now contains millions of old PC's 8086's, 286's etc, because what, windows 3.1 is outdated, and Microsoft steers users away from it?

here is the interesting part. I'm saving my machine from landfill :lol:

and no, AM isn't outdated for me.
 
Cmos ICs have a higher output resistance at lower supply voltages. Using a 5V regulator at its minimum of 4.5V, a typical 74HCxx output can provide 35mA at 2.25V which calculates to be a 64 ohm resistor. Its minimum current is 12mA so its max resistance is 188 ohms.
All the 74HC4017 ICs I've used in my 6V Ultra-bright Chaser project measure typical or better, that's why they're so darn bright!

If you use 10k resistors as the minimum value in your resistor ladder at the outputs of the counter then the circuit's max error would be only 188/10k= 1.88%. If you use 100k resistors then the error is 0.2%.

If you operate the counter with a 6V supply (its absolute max is 7V) then the error is halved.

ok, now that I know the minimum resistors to use at the output of my counter, I can't remember the absolute maximum resistances I can use,and why are there maximums when the error level is lower?

and does your error level stated above determine the chances of a short circuit?
 
mstechca said:
ok, now that I know the minimum resistors to use at the output of my counter, I can't remember the absolute maximum resistances I can use,and why are there maximums when the error level is lower?
If you use 5% resistors in your ladder then you don't need an extremely low error with very high resistor values. You also get error with extremely high resistor values due to leakage current of the varactor diode and you won't be able to measure the tuning voltage with a multimeter without a high error.

does your error level stated above determine the chances of a short circuit?
Of course not! A 74HCxx IC on a 5V or 6V supply can provide much more than its absolute max continuous output current rating of only 25mA to a short and will go "poof". It can momentarily provide an output current of 60mA to quickly charge stray capacitance for its high speed. An ordinary CD4xxx Cmos IC can provide an output current of only a couple of mA to a short with a 5V or 6V supply.
 

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Here is another good "real" varactor diode:
 

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And here are Zetex's hyperabrupt varactors. They have wider tuning range, but they also have fairly high temperature coefficients. This is not a problem if you are using them in a VCO that is within a PLL, but it might be when run open-loop, as mstecha is doing.
There are other mfr's of hyperabrupt varactors.
 

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If you use 5% resistors in your ladder...
I always use 5% tolerance resistors.

You also get error with extremely high resistor values due to leakage current of the varactor diode and you won't be able to measure the tuning voltage with a multimeter without a high error.
If you remember, I'm not using a varactor diode.
I have been playing, and so far, I am using a transistor as a varactor diode. I connected emitter and base together and grounded them, and I connected collector to the coupling series capacitor and the resistors.

So do I have to make the resistances low enough to provide current greater than the diode's leakage current to override it or am I off here?
 
mstechca said:
So do I have to make the resistances low enough to provide current greater than the diode's leakage current to override it or am I off here?

No, as you've been told many times already THERE SHOULD BE NO CURRENT, or at least too small to measure or make any difference. I would suggest using resistors in the 100's or 10's of kilo-ohms - although compared to the current demands of the rest of the circuit it's pointless getting carried away reducing the D2A current excessively.
 
mstechca said:
If you use 5% resistors in your ladder...
I always use 5% tolerance resistors.

You also get error with extremely high resistor values due to leakage current of the varactor diode and you won't be able to measure the tuning voltage with a multimeter without a high error.
If you remember, I'm not using a varactor diode.
I have been playing, and so far, I am using a transistor as a varactor diode. I connected emitter and base together and grounded them, and I connected collector to the coupling series capacitor and the resistors.

So do I have to make the resistances low enough to provide current greater than the diode's leakage current to override it or am I off here?
DAC resistance affects 2 things in your circuit:

1. The Q of your tank.
The reactance XL of your 100nH inductor at 100MHz is 63 ohms. The resistance of your DAC is about half the resistance of the MSB (lowest value) resistor (0.533 for a 4-bit DAC, to be more precise). Q=R*XL. To avoid killing the Q, you should probably keep the resistance at least 100*XL, or greater than about 6k. Nigel's advice is spot on, as usual.

2. The settling time after a tuning change.
The time constant is DAC resistance times the coupling cap (1nF in your schematic). This will only be an issue if you are doing really fast scanning.
 
Ron H said:
DAC resistance affects 2 things in your circuit:

1. The Q of your tank.
The reactance XL of your 100nH inductor at 100MHz is 63 ohms. The resistance of your DAC is about half the resistance of the MSB (lowest value) resistor (0.533 for a 4-bit DAC, to be more precise). Q=R*XL. To avoid killing the Q, you should probably keep the resistance at least 100*XL, or greater than about 6k. Nigel's advice is spot on, as usual.
I would hope that the impedance of the resonating tank would be much higher than just the XL of its inductor. Therefore DAC resistors should be in the hundreds of k ohms. :lol:
 
audioguru said:
I would hope that the impedance of the resonating tank would be much higher than just the XL of its inductor. Therefore DAC resistors should be in the hundreds of k ohms. :lol:

This is all ignoring the series resistor feeding from the D2A to the varactor, which Audioguru has rightly included on various diagrams he's posted.
 
Nigel Goodwin said:
This is all ignoring the series resistor feeding from the D2A to the varactor, which Audioguru has rightly included on various diagrams he's posted.
Leaving the series resistor out would save a whopping 10 cents. :lol:
 
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