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# Harmonic currents calculation with LTSpice

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#### kahlenberg

##### Junior Member level 3
Hi,
I can simulate following circuit in LTSpice and I can also generate FFT for current.
The circuit has a square wave generator with 5V amplitude, 1ns rise-time, 1ns fall-time, 4ns on time, 10ns period (100MHz).
I want to calculate effective value of currents at 100MHz, 300MHz, 500MHz ...
When I calculate, I get different results than LTSpice.
What is the correct way to calculate the effective currents?

My calculations:
Z1=sqrt(R2^2+(2*pi*f1*L)^2); (R2=100, f1=100e6, L=10e-6), then I1=U/Z, Ieff1=I1/sqrt(2)=5.55mA.
Z3=sqrt(R2^2+(2*pi*f3*L)^2); (R2=100, f3=300e6, L=10e-6), then I3=U/Z, Ieff3=I3/sqrt(2)=1.87mA.
They are different than LTspice.

Your schematic resembles a buck converter with duty cycle 40 percent. The standard formula says average voltage at the load is duty cycle times supply V.
That is, 2V (40% of 5V). Resulting average current is 20mA at any switching frequency.

I could be wrong. This doesn't factor in the rise-fall times.

Hi / Servas.

I guess there are several issues:

You calculate Z using R and X_L.
For R you use R2 only.
--> You miss the 100 R of the source. R = R_source + R2 = 200 Ohms.

In opposite to Brad I see a 0V / 5V pulse with 50% duty cycle. 1ns rise, 4ns HIGH, 1ns fall, 4ns LOW = 1-4-1-4

I don´t see what you used as "U" (Volts) in your calculation.
You may see your generator signal as 2.5V DC + a square wave with 2.5V amplitude (-2.5V ... + 2.5V)
it also equals to 2.5V RMS.
For a -2.5V/2.5V square wave (t_r = t_f = 0) you get a fundamental sine with (4/Pi) * 2.5V amplitude
Now due to rise and fall time your fundamental amplitude may be a little smaller than 3.18V

The 3rd overtone has a 1/3 of the fundamental amplitude. (so less than 1.06 V amplitude)
The 5th overtone has a 1/5 of the fundamental amplitude (so less than 0.637V amplitude)
.. and so on.

RMS of a sinewave is: amplitude / 1.414

Hope this helps.

Klaus

corrected above, because I missed the factor of "4/Pi" in amplitude.

Last edited:
Hi,

I think that's a simple math problem. You have not stated what you used for your voltage amplitude here
Z1=sqrt(R2^2+(2*pi*f1*L)^2); (R2=100, f1=100e6, L=10e-6), then I1=U/Z, Ieff1=I1/sqrt(2)=5.55mA.
Z3=sqrt(R2^2+(2*pi*f3*L)^2); (R2=100, f3=300e6, L=10e-6), then I3=U/Z, Ieff3=I3/sqrt(2)=1.87mA.

Have a look here [1], to calculate the individual amplitudes. For your example with a duty cycle of 50%, a DC offset of 2.5 V and a HIGH level voltage of 5 V you are getting the following amplitudes

V(0 Hz) = 2.5 V
V_pk (100 MHz) = 2 • 5 / pi • sin(pi • 0.5) = 3.1831 Vpk --> I_pk(100 MHz) = 3.41349 mA
V_pk (300 MHz) = 2 • 5 / (3 • pi= • sin(3• pi • 0.5) = -1.06103 Vpk --> I_pk(300 MHz) = 395.8 µA

Here, I have considered the series impedance (100 Ohm) of the voltage source as well, to calculate Z_1 & Z_3.
The calculations are done for a square wave and are already in good agreement with your SPICE results. Note your excitation source has a kind of trapezoid shape.

[1] https://www.dspguide.com/ch13/4.htm

BR
--- Updated ---

... Sorry for the "double" information, I have just now seen @KlausST 's reply.

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