Discussing Stagger tuning in VHF receivers....

[Averagesupernova]

20 uF seems quite large for radio frequencies.

 

[Baluncore]

It will also need a resistor in series with the 100pF ceramic replacement for C1, to linearise the VHF gain.

 

[Baluncore]

I have fixed it so it works, now you only need to neaten it up and fix the unrealistic gain.

 

[tech99]

I think R3 and R4 need to be bypassed to ground, otherwise they place a resistance in series with Q2 base.
Also not sure how you are setting the Q of the two circuits. Q2 low input impedance seems to place an unwanted low impedance across L2/C4.

 

[Baluncore]

I think R3 and R4 need to be bypassed to ground,

The R3, R4 resistance there sets the Q of resonator 2, and the base bias voltage, but the base is still somewhat undefined.

Notice that when the resonators have identical component values, there are two resonant peaks.
The transistors will need to be “realised” with a part number.

I will answer any meaningful question, or pick it up if the simulation falls over, but I'm going to let brainbaby sort his circuit out as a learning exercise in simplification.

 

[brainbaby]

In order to stagger tuned the circuit the centre frequency of 1st stage and 2nd stage will be varied by a factor of B/2√2.

So the centre frequency of first stage becomes w1 = w0 - B/2√2

and for second stage it is w2 = w0 + B/2√2

where B is the bandwidth of the circuit.

and w0 is the centre frequency after the circuit is stagger tuned.

First to estimate B we have to find the 0.707 point in the figure.
Initially for a start I have taken the value of C3 and C4 to be 5.6 pF.
So I ve also assumed figure to be acc. to scale.

The 0.707 point is 50.4 db

B = 225 - 201 = 24 mhz.

w0 is found out to be approx. 213 Mhz

Now,

w1 = 213 - 24/2√2

= 204 mhz

w2 = 213 + 24/2√2

= 221.4 mhz

Initially the circuit is synchronously tuned and in order to stagger tune it we require suitable value for C4.

w2 ^2 = 1 / L2 C4

(221.4) ^2 = 1/ 100nH X C4
...C4 = 2.04 nF

But as I simulate the circuit with a new value of C4 I get a steep notched response curve which do not agree with the output response when the circuit is stagger tuned..
since the overall effect of stagger tuning is to produce a narrow band with a maximum flat response with steeper fall offs.

where seems to be the mistake….??

Fig1..when C4 =5.6pF

Fig2.. when C4 = 2.04nF

 

[Baluncore]

where seems to be the mistake….??

Please post the full circuit and or the circuit.asc.txt
Don't forget that the BW is a function of Q and that in your circuit, Q is set by external resistance or load.
What is your RF transformer coupling coefficient?

 

[Tom.G]

Fig2.. when C4 = 2.04nF

It appears your algebra and the resonant frequency formula are incorrect. Please look up the formula for resonant frequency and re-calculate. I get a resonant frequency of ≅11MHz for the second tuned circuit. (Start the simulation frequency sweep around 5MHz to see it.)

 

[brainbaby]

Actually the circuit is same which you have posted in post#33..the only change which I made was that I have removed the function .param and simulated the circuit with value of C3 and C4 to be 5.6pF as I was unsure about the values to have a start..

My objective was to visualise the effect of stagger tuning of flattening the output response so I thought to stagger the stages to a constant value of B/2√2...
Also I was wrong in framing out the value of Bandwidth B, which I took just as a mere difference of two mean and extreme values 201/225 Mhz...
I earlier had a thought about taking the quality factor into consideration but then still I needed the value of R which was still unknown to me..

 

[Baluncore]

When staggering resonator tuning it is very important that you have a well controlled and stable Q.

An ideal collector has an infinite impedance and is a perfect current sink. You are driving the first resonator without any defined resistance and so it has an unspecified Q. You need to add a resistor in parallel with L and C to control the Q.

An ideal base is also an infinite impedance so you must do the same to the second resonator. Real components external to the resonators will add collector and base reactance plus resistive loading the tuned circuits. That will stagger the tuning without you trying. I added a 2k2 resistor across each resonator and set the coupling to 0.065

The response I posted in #37 shows what is possible with a few circuit changes. I take my output from the emitter of Q2, that way my circuit represents a one transistor stage of an amplified double tuned bandpass filter. It reduces the gain to that expected from a single transistor stage and provides a realistic input and output impedance to the coupled resonators. There are a couple of other changes needed.

Note that at 200 MHz, VHF BPFs are significantly more sensitive to stray circuit parameters than are 35 MHz TV, 10.7 MHz FM or 455 kHz IF amplifiers. You can expect the LTspice model to be sensitive to component values.

 

[jim hardy]

I earlier had a thought about taking the quality factor into consideration but then still I needed the value of R which was still unknown to me..

Using the rules of thumb that
Bandwidth = Resonant Frequency / Q
and Q = X / R

you could guesstimate some values for R to plug into your simulation and observe its effect.

 

[brainbaby]

Ok by joining resistor nearly approx 2K to the base of the transistor of two satges will give the feasibility to calculate the quality factor..
Since
Q = X/R...i)
where X = 1/2 pi. fc. C
= 1 / 2 pi X 213 X 1000 X 5.6 pF
= 1.33 X 10 ^5 ohm.

substituting the value of X in equation i)
Q = 1.33 x 10^5 / 2k
= 66
So the quality factor equals to 66.

now finding bandwidth
B.W = fc / Q
= 213 x 1000 / 66
= 3227.27 Hz i.e 3.22 Mhz

w1 = 213 - 3.22/2sqrt 2
= 211.85 Mhz

w2 = 213 + 3.22/2 sqrt 2
=214.13 Mhz

(w2)^2 = 1 / L2 x C4
(214.13)^2 = 1/ 100nH X C4

So...C4= 2.18 X 10 ^ 8 pF...
C4 = 0.000218 pF

Fig_1 when C4 = 5.6pF synchronous

Fig_2 when C4 = 0.000218 pF staggered

Conclusion is absolutely reverse of my expectation
When C3 = C4 = 5.6pF (synchronously tuned) the curve is more flat and steeper as expected in stagger tuning
but when C3= 5.6pF and C4 = 0.000218 pF (stagger tuned) the response curve seems to be more like as expected in synchronous tuning...

 

[Baluncore]

The base is high impedance, so putting 2k0 in series with it will have no effect on the Q of resonator 2.

Make two identical RLC parallel tuned circuits, with a high impedance collector input, or base output.
To do that move the 2k0 to parallel the second resonator.

Then sweep the capacitance parameter to find the best capacitance tuning.
Then play with the coupling coefficient over the range 1% to 10% to see if it improves the sweep flatness.

Hints;
1. Avoid using the F for farad. It will catch you out with super capacitors when it is interpreted as femto. 10F = 10^-14 farad.
2. Avoid decimal points, it is more readable when you can use an SI multiplier 6.8p = 6p8

 

[brainbaby]

sweep the capacitance parameter to find the best capacitance tuning.

In post 42 Fig_1 seems to be over saturated which I overlooked earlier as amplitude value of 700V makes no sense...

Could you just tell me the permissible values of the capacitance C4 to be sweeped which agrees to my calculated values..

 

[Baluncore]

Your numbers are not for modified circuit. Move R2 to stabilise Q of resonator 2. Add C6 to give resonator 2 something to work against while driving the base. Base bias through inductor, so minimum load on the base. Take the output from emitter to avoid high ringing voltages. Plot Mag(V(out)). I would expect gain per stage to be between 12dB and 24dB for one transistor at 200 MHz. Play with the values of R2, Kst, notice R5,6,7,8.