This has become my biggest doubt of my life...........?

[aj_silverthunder]

guys please please solve this thing........
here i have made two different circuits.....but i have a big big doubt that which is correct and which is wrong...
1)








2)

now guys please please tell me which will configuration is correct..
in the first schematic the D1 anode is DIRECTLY grounded,in order to get biased by 3V....
IN 2 schematic,as i think the D1 anode is grounded by the ground connection in between the 2 inductors...
so people .please tell me which configuration is correct..................its my biggest confusion of life......
plz plz help me :-(

 

[betwixt]

Both will work. There is no correct and incorrect in this schematic, all you are doing is applying the variable capacitance to a different part of the tuned circuit.

To understand the varicap, forget that it's in an oscillator circuit for now and just look at the DC voltage path across it. As long as there is a DC connection its capacitance will vary according to the voltage you apply. For simulation purposes, change the diode to a fixed capacitor of the same value the varicap would appear to be and measure the output frequency.

Note that the varicap is used in non-conducting mode (reverse biased) so although there has to be a DC path, it doesn't actually draw any current.

Brian.

 

[aj_silverthunder]

thank u friend.that was my biggest doubt ......

Added after 1 minutes:

will u tell me which one is more reliable..the schematic 1 or schematic 2.....
i like to know it from..cause i feel u r very good in RF...plz.

 

[betwixt]

Thanks for the compliment, I have been qualified in RF communications since age 14 but I would never say I was an expert, even though many years have passed since then :|

Personally, if I had to choose between the two I would use the one with the varicap grounded at one end but I would change the resistor values quite a lot. This is my reasoning:

The tuning voltage source and the audio input (V2 & V3) can be assumed to have almost no impedance. This means the two resistors R4 and R5 are effectively in parallel across the varicap and the oscillation. Placing 500Ω across the signal will quite likely stop the oscillator running or make its output very low. As no current from the tuning voltage passes through the varicap you can make R4 much bigger, try 100K instead. Now you are no longer conecting a low resistance across the signal you can also increase the value of R5 without losing modulation voltage. If you do this you can safely also reduce the value of C6 which will improve the response of changes to the tuning voltage. Remember that current from the two voltage sources are also providing current to charge or discharge C6 so keeping its value as small as possible will allow charging and discharging to be quicker. The effect of dropping the value on the audio is only that lower values give a poorer low frequency response but unless you are using a high quality receiver, amplifier and loudspeakers you probably wouldn't notice much difference. To see how the response changes, try simulating with a frequency sweep at V3, the difference will only be at low frequencies so simulate from say 10Hz to 10KHz.

If I was designing from scratch I would still use the oscillator circuit I gave you ages ago. The reason being that the varicap is grounded and it uses a single tuning coil which is easier to make. It also has fewer components !

Brian.

 

[E-design]

What frequency is this supposed to work on? 40MHz?

Design 1 may have a lot of amplitude instability

Design 2 may have this as well as bad harmonics

 

[aj_silverthunder]

k so higher value R4 and R5 will not damped the amplitude of the frequency...i have altered it as in this schematic..


Added after 14 minutes:

and can u tell me am i right in calculation of the frequency of this oscillator...
L=120nH(60nH+60nH)
C=i think the C4,C5 and D1 are in series..
C4=100pF;
C5=100pF;
D1=40pF@3VDC..
AS C4,C5 and D1 are series(need confirmation on this)so 100pF,100pF and 40Pf will make C=22.22pF

So L=120nH and C=22pF..
do u think my calculation on finding L and C is right or Wrong..
and i need to know are D1,C5 and C4 are in series or D1 AND C5 ARE IN SERIES AND ARE(D1,C5) PARALLEL TO C4..AM I RIGHT..?
please tell me is it wrong or right..plz B'coz i need to know which are in parellel and which are in series to calculate the C value...
Added after 2 minutes:

 

[E-design]

D1 and C5 will have very little influence on setting the resonant frequency. Your frequency will be determined by L2,3 & C4. C1,2 affect the amount of coupling and hence loading on the resonator. With current values resonance will occur around 40~41MHz. Simulation also shows amplitude instability of your design. You should be able to see this in your LT CAD simulations as well.

 

[aj_silverthunder]

SO then accorcding to ur simulator...u r getting the frequency as 40Mhz..
check this if its correct..
L=120nH..
and C=100pF
so LC ffrequency will be 45Mhz..
now lest take D1 and C5..
D1 40pF@3V
C5=100pF
they are in series so it will be 28pF
..
now D1 and C5 are in parellel to C4..
so 28+100=128pF
so L=120nH and C=128pF..so F=40.609Mhz..
am i right...?

Added after 1 minutes:

By the way,E-design which simulator are u using..i hope it would help me a lot if u named it..

Added after 2 minutes:

So my Final doubt is ....

i need to know are D1,C5 and C4 are in series or ""D1 AND C5 ARE IN SERIES AND ARE(D1,C5) PARALLEL TO C4..AM I RIGHT..?""
B'coz i need to know which are in parellel and which are in series to calculate the C value... to know the resonant frequency

 

[E-design]

D1 and C5 will be in series and in parallel to L2. Plots done on 2 different simulators Multisim and Genesys. BTW LTspice shows the same instability

 

[aj_silverthunder]

D1 and C5 will be in series and in parallel to L2. Plots done on 2 different simulators Multisim and Genesys. BTW LTspice shows the same instability

so ur saying D1 and C5 will be in series and in parallel to L2......aren't they parallel to C4..?

Added after 1 minutes:

can u tell me which simulator u have used to check the gain vs load graph..in which frequency is also mentioned..which is that software..

 

[FvM]

To overcome the instability, simply adjust the L3:L2 ratio (L3 < L2) or reduce C1 to about 10 pF.

 

[aj_silverthunder]

i feel that C5 and d1 are in series with C4 parallel..wat do u think u can see it in this below sch..

 

[FvM]

It would be parallel, if the bottom node of C4 is grounded. But it isn't.

In my opinion, the question of interest is to know df0/dC(D1) (how much does the resonance frequency change
with D1 capacitance variation). It can be exactly calculated by considering all circuit capacitances, not only C5 and C4.

 

[aj_silverthunder]

do u think now its D1 and C1 are in series and are parallel to the C2 and L1..
and my doubts in this is:
1)is it better to use resistor at the collector than a inductor..






Added after 24 minutes:

SO calculating the frequency..
L=100nH
C=25pf+4pF(D1=30pF@5VDC and C1=5pF are in series so it apporx..4pF)+5pF(from C5 and C4)
so C=25+4+5=34pF
so the output frequency of the oscillator would be 86Mhz...
AM i right in calulation please please confirm me ..plz..brian..im waiting for ur reply and all eda members

 

[betwixt]

Sorry, my internet connection hasn't been working properly for the past few days, it's very slow at the best of times and sometimes it just isn't there at all. I can still access it with a mobile phone but it's very expensive and tedious when there are attachments.

Your calculation is correct but bear in mind that when such low capacitor values are being used the effects of 'stray' capacitance can be significant. All oscillator designs will suffer from this and the only way to keep it under control is to use very short component wires and keep the tuned circuit components close to each other.

The advantages of this oscillator are:

1. more of the tuned circuit components are grounded at one end, including the varicap so it's easier to put the DC and modulation on it.

2. One component, C2, can be used to alter the center frequency, you could make this a variable capacitor. Note that because one side of C2 is ground, you could make it variable with the screwdriver slot on the ground side to minimize the de-tuning you would get when the screwdriver shaft touched it.

3. No center tapped components! - easier to construct.

4. The output is taken from a point that is not directly connected to the tuned circuit. This makes it less likely that the load will pull the oscillator off frequency.

Brian.

 

[aj_silverthunder]

u r really good in RF...
one thing i noticed is,in 2nd point..i also thought of using a variable capacitor...but it has three pins..like this



as u can see there are 3 pins A,B,C;
"B"the middle one will be connected to ground..
so with reference to the above circuit...one end of the cap "A" pin will be connected and the other end that's pin "C" will be grounded..so now "B" and "C"are grounded..
do u think i am right ?

Added after 23 minutes:

and as every oscillator has feedback..
do u think the feedback to the transistor is given to emitter by the capacitive voltage divider ...and is it called negative feedback or positive feedback..?

 

[FvM]

Where did you get your knowledge about the varcap pin assignment? If it's the same type, that I
have in my drawer, then the two opposite pins (A + C) are connected to the rotor - and should be
grounded. It looks very much like a part that's now made by Vishay:
https://www.vishay.com/doc?28527 (Originally produced by Roederstein or Philips, I think).

Also without a datasheet, a quick view from the downside reveals the internal connection.

P.S.: It's been Philips Components series 2222 808

P.P.S.: I linked the wrong document, corrected now.

 

[aj_silverthunder]

And also don't u think C3(56pF) is series with C1 and C2..i have red marked those 2,plz check it out



Added after 7 minutes:

Where did you get your knowledge about the varcap pin assignment? If it's the same type, that I
have in my drawer, then the two opposite pins (A + C) are connected to the rotor - and should be
grounded. It looks very much like a part that's now made by Vishay:
https://www.vishay.com/doc?28526 (Originally produced by Roederstein or Philips, I think).

Also without a datasheet, a quick view from the downside reveals the internal connection.

P.S.: It's been Philips Components series 2222 808

should A and C be grounded..? im really confused

Added after 10 minutes:

please tell me which pins will be inputs and which pin should be connected to ground in trimmers..plz dont confuse me..

 

[FvM]

C3 is rather is series with C4 and C5, at least it's also influencing the resonance frequency. The problem
arises by introducing more frequency determining capacitors then necessary. The basic circuit (a common
collector colpitts oscillator) doesn't have that much capacitors.

You can simplify the calculation by making the C3 capacitance considerably higher, e.g. > 1 nF. But's not an
actual problem, I think.

P.S.:

please tell me which pins will be inputs and which pin should be connected to ground in trimmers..plz dont confuse me..

Calm down, read the above linked datasheet. A capacitor hasn't in- or output, just two terminals. In
case of a varcap, it's useful to ground the rotor, if ever possible, for the reason betwixt explained.
Check which pins connect to the rotor and use them accordingly.

 

[aj_silverthunder]

do u think the trimmer should be used like this



Added after 2 minutes:

C3 is rather is series with C4 and C5, at least it's also influencing the resonance frequency. The problem
arises by introducing more frequency determining capacitors then necessary. The basic circuit (a common
collector colpitts oscillator) doesn't have that much capacitors.

You can simplify the calculation by making the C3 capacitance considerably higher, e.g. > 1 nF. But's not an
actual problem, I think.

P.S.:

please tell me which pins will be inputs and which pin should be connected to ground in trimmers..plz dont confuse me..

Calm down, read the above linked datasheet. A capacitor hasn't in- or output, just two terminals. In
case of a varcap, it's useful to ground the rotor, if ever possible, for the reason betwixt explained.
Check which pins connect to the rotor and use them accordingly.

Are A and C the two terminals of the trimmer and wat is B pin given..?

Added after 3 minutes:

And as u said about increasing the C3 to >1nf is rights,if C3 is increased to 1nF,it wont have much influence in change of frequency,may change very very little..