Selective amplifier resonance frequency is less than computed with basic equation

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
I have selective amplifier with LC elements of 10 uH and 10 pF .
This equation returns the resonant frequency of 15.915 MHz : var freq_computed = 1 / (2*3.1434*Math.sqrt(L*C));
But inside simulator, the resonant frequency appears to be moved to smaller number, the value of 12.88 MHz.
Is this normal behavior? Is this happening because of transistor?
I am including screenshot with bode plotter inside multisim.

I tried also test using 3 uH and 3 pF, here the resonance freq. is higher, but inside simulator still moved to smaller number.
 

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Thread Starter

georgefrenk

Joined Dec 24, 2023
54
The 2N2222 has an output capacitance of 8pF, which will be in parallel with your load capacitor.
// This is LC tank with C of 10 pF and L of 10 uH
var C = 10 / 1000 / 1000 / 1000 / 1000;
var L = 10 / 1000 / 1000;

var C_output_2N2222 = 8 / 1000 / 1000 / 1000 / 1000;

var Cfinal = C + C_output_2N2222; // here I sum the LC tank's capacitance and transistors output capacitance of 8 pF.

var freq_computed = 1 / (2*Math.PI*Math.sqrt(L*Cfinal));
freq_computed = freq_computed / 1000 / 1000;
console.log('freq_computed (MHz) = ' + freq_computed.toString());

Now I get value of 11.862 MHz via equation, which is much closer to measured resonant freq.12.88 MHz inside simulator's bode plotter.

Am I doing this the right way?
 
Last edited:

Papabravo

Joined Feb 24, 2006
21,225
The uncertainty in pi is rather less the uncertainty in Cob of the 2N2222.
Multiple mistakes don't improve the situation at all. If one is trying to understand what a problem is all about you have to find all of the mistakes. Sloppy writing hardly ever brings credit to the author.
 

crutschow

Joined Mar 14, 2008
34,392
So the discussion has degenerated due to a .058% difference between two calculations. :rolleyes:
For most engineering calculations, using 22/7 for the value of Pi is more than sufficient (within 0.04%).
 

Thread Starter

georgefrenk

Joined Dec 24, 2023
54
To conclude, Ian0's suggestion to use 2N2222's transistor's 8pF capacitance in my equation has helped me very much. I tried some L and C values which have higher resonance frequencies and now difference between simulated peak frequency inside bode plotter and calculated one is much smaller than without taking into account transistor's 8 pF of capacitance. And of course from now on I will use Math.PI constant always inside my javascript code.
 

crutschow

Joined Mar 14, 2008
34,392
10pF is a very small capacitance, that will be significantly altered by real world parasitic capacitances.
You should use values at least 10 times that to minimize that problem if possible.
If not, then you will have to tweak the value after the circuit is built to compensate for the parasitics.
 
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