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For what value of L, will this circuit resonate at 332kHz?

CDRIVE

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May 8, 2012
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When you replaced the inductor (bad move) did you destroy it? If not do you still have it?
I would very much like to see a some photos of this receiver, including the front and rear panels.

On another note I agree with duke regarding the high Q of this circuit. Inductance was kept low to keep coil resistance low.

Chris
 

Laplace

Apr 4, 2010
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This was an interesting circuit to analyze, forcing one to learn new techniques for dealing with complicated equations. It was necessary to assume that the 0.1 uF capacitors provided a direct path to RF signal ground so that the C11, C66, & R11 components could be removed from the circuit. As it was, MathCAD complained that the results were too large to format for display, otherwise with those 3 components included the computer algebra system was unable to produce a result at all.

Given the component values shown, L1 and L2 should both be 82 uH for 332 KHz, but it is L2 that primarily determines the frequency peak. A change of 1 uH in L2 will shift the center frequency ~2 KHz whereas a change in L1 has very little effect. The attachment shows frequency-magnitude plots for L1 values of 82, 20, and 300 uH for comparison.
 

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  • Mathcad - MathCad_EP-23_3.pdf
    40.4 KB · Views: 97

ubercool

Nov 6, 2012
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Here are the answers to your questions:

1. Yep but the receiver was not receiving at these connections. It was receiving at different positions of switches which I adjusted randomly.

2. Yep but there was no change

3. Yep...isnt this the repetition of question 1?

4. Around 1uH...these are variable inductors and the variation is from 0.8uH to 1.3uH

5. What is a bodge?

6.Yep I have a radiocommunication service tester



I agree Chris but he has not told us if he has b******d it.

The original circuit uses large capacitors and very small inductors. The circuit Q is very high and losses in the inductors will be important.
The bandwidth will be very small and a variable inductor will be necessary to tune the filter 'on the nose'.
If the 15pF capacitor is actually 23pF, this will drop the tuned frequency and adversly affect the response curve.

1. Did you have the correct switch connections?
2. Did you clean the switches?
3. Did you bridge the appropriate switch contacts?
4. What is the inductance of the inductors? You have a meter.

5. Do you have room to fit in a bodge?
6. Do you have a 332kHz accurate signal generator?
 

ubercool

Nov 6, 2012
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In the tools option there is:

Diagram Window
Netlist editor
Find cimponent
Re-read symbol database
Re-compile library
Re-build library
New Macro Wizard
Edit macro properties (transparent)
Export macros (transparent)
Renumber components
Lock Schematic editor (ticked)
Dock netlist editor (ticked)

But there isnt any option for filter design...? :(


It's 6 in the drop down list under "Tools".

Chris
 

ubercool

Nov 6, 2012
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No I havent replaced the inductors yet...I am just trying to redesign through software at this time. Once I get the right results on that, I will replace the variable inductors with fixed inductors

OK I will try to post some photos


When you replaced the inductor (bad move) did you destroy it? If not do you still have it?
I would very much like to see a some photos of this receiver, including the front and rear panels.

On another note I agree with duke regarding the high Q of this circuit. Inductance was kept low to keep coil resistance low.

Chris
 

duke37

Jan 9, 2011
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Hi Laplace, you made my mind boggle 55 years ago. You seem to be able to do it still.

I would have checked all that the switches were all correct and working properly and, if only one frequency was involved, then the contacts could be bridged so there would be no likelyhood for poor contact. If there was some reception with the switches in the wrong positions, either the switches were wrong or a component had changed value. the capacitors are as likely to be faulty as the inductors.

A bodge is a crude remedy for a problem. It is not normally recommended.

The original circuit was presumably designed to a high standard but could be simplified if switched frequencies are not needed.

You will not be able to use fixed inductors, variation will be necessary to tune it 'on the nose'.
 

CDRIVE

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Tina-TI is free but it's a stripped down version. It's not surprising that it doesn't include Filter Design.

Chris
 

ubercool

Nov 6, 2012
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Hmmm and if I download TI filterpro, it will be useless, as it is for active filters only right?
 

CDRIVE

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Well, it's not useless but yes, it's for active filter design.

Does the TI version give you AC Transfer Characteristic?

Chris
 

CDRIVE

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yep it does give AC transfer characteristic

Well, this option will plot the bandpass of a filter you design but it won't design it for you.

Chris
 

ubercool

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I have used AC analysis to plot the resonant plot of the circuit. It is posted here alongwith the TINA schematic.There is one problem, this circuit is showing resonance with inductor values of 100uH whereas the inductors in the original circuit are variable from 0.66uH to 1uH only
 

Attachments

  • Resonance plot.jpg
    Resonance plot.jpg
    38.1 KB · Views: 107
  • TINA ckt.jpg
    TINA ckt.jpg
    94.5 KB · Views: 105
Last edited:

duke37

Jan 9, 2011
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The original circuit is two parallel tuned circuits, top and bottom coupled so very complex.
You have simulated a series circuit top coupled to a parallel circuit.
The source impedance is clearly 39k, you have not included any damping whatever so I do not know what has determined the peak voltage.

In the Williams design, I chose a bandwidth of 5kHz and a load of 39k. (see #17). The peak is at -6dB which is the maximum you could get with equal source and termination impedances. Your peak of +75dB is absurd and is due to the series circuit.

I do not know the effect of top and bottom coupling, whether they add or subtract and whether they affect the inductances.
 

CDRIVE

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I have to repeat much of what duke said. A totally lossless filter (impossible) would output 0dB! Your circuit is not the same as the OEM. There are also a number of things that I don't understand about your circuit. For one thing; what is C1? Your print indicates it's a 35000uF cap across the source. I don't get the dashed leads of this cap either. Tina doesn't provide this symbol. I don't have your voltage test points either.

When you redraw this please don't let labels overlap another label or component. You can click the labels to position them independent of the component.

Chris
 
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