# Spice Simulation of Parallel LRC Circuit

Discussion in 'CAD' started by Thomas D. Dean, Nov 11, 2003.

1. ### Thomas D. DeanGuest

I am using spice3f5. I don't get what I expect when I simulate a parallel LRC
circuit.

* parallel RLC Circuit
V1 1 0 0 ac 1
V2 3 0 0
R1 1 2 10
L 2 3 10uh
C 1 3 1000pf
R2 1 3 100meg
..end

run
ac lin 10k 1meg 3meg
plot v2#branch

Does not display what I expect, v2#branch approaching zero at approx 1.6Mhz.
What am I doing wrong?

The series LRC circuit behaves as expected.

* Series RLC Circuit
V1 1 0 dc 0 ac 1
V2 4 0 0
R 1 2 10
L 2 3 10uh
C 3 4 1000pf
..end

run
ac lin 10k 1meg 3meg
plot v2#branch

v2#branch peaks at approx 1.6Mhz

tomdean

2. ### Thomas D. DeanGuest

Sorry to answer my own post. v2#branch is a complex current. So, I should plot
the magnitude.

* parallel RLC Circuit
V1 1 0 0 ac 1
V2 3 0 0
R1 1 2 10
L 2 3 10uh
C 1 3 1000pf
R2 1 3 100meg
..end

run
ac lin 10k 1meg 3meg
plot mag(v2#branch)
plot phase(v2#branch)

tomdean

3. ### qrkGuest

Are you sure you want R1 = 10 Ohms? That gives you a bandwidth of 15.9
MHz. Kind of silly for a 1.59 MHz center freq.

Fun with filters: to calculate the bandwidth of a simple LRC filter:
BW = 1/(2*pi*R*C)
In real life, you need to figure in the effective parallel resistance
across the inductor which is generally frequency dependent.

Mark

4. ### Paul BurridgeGuest

I've simulated your circuit in LT spice and it does appear to show a
clear peak at 1.6Mhz. I'll post a screen shot of the result to a
binary group if that'll be any help to you.
Alt.binaries.schematics.electronic is the group we usually use for
this purpose. Can you find that okay if I do?

p.

5. ### Helmut SennewaldGuest

Hello Thomas,
not the voltage v(2) has to be zero at the resonance frequency.
It is the current i(V2) who is zero at this frequency if you
wouldn't have any resistance in your circuit. Give it a try.

The only purpose of the zero Volt source V2 is to measure current,
because there is no other way to do it in standard SPICE.

Best Regards
Helmut

6. ### Winfield HillGuest

Paul Burridge wrote...
Thomas, I don't understand why you're "grounding" both
ends of the resonant circuit. Keep in mind that voltage
sources have zero impedance by definition, so you'll not
be able to see energy flowing from the inductor to the
capacitor and back, right?

Thanks,
- Win

whill_at_picovolt-dot-com

7. ### Helmut SennewaldGuest

Hello Mark,

BW = f0 / Q = f0 / ( sqrt(L/C) / R ) = f0 / 12 = 132kHz

Best Regards
Helmut

8. ### qrkGuest

Sorry for taking so long to reply, but I was on holiday. My equation
stands for a standard RLC bandpass. I assumed that the OP wanted a
bandpass filter since his comment card stated "parallel RLC Circuit".
Your equation is correct for a band-reject filter which he was working
on previous to this circuit.

Upon further examination of his deck, I don't know what he is after
since the capacitor is directly across the AC source (i.e., he has the
equivalent of an RL circuit).

Aloha, Mark

9. ### Helmut SennewaldGuest

Hello Mark,
have you seen my other posting?
I think that Thomas has tried to plot the wrong value.
If we plot I(V2), we will get a current peak for the series RLC
circuit and we will get a notch for the parallel RLC circuit.
So both cases show a resonance when looking for the current I(V2).

As usual, we will never know what the correct question/answer is,
because we have no feedback from the questioner.

Best Regards,
Helmut