0
\$\begingroup\$

I've created a circuit in LTSpice to model a fan assembly I'm looking to build. In the below circuit the current sources represent the fan and the voltage sources represent the voltage source I'll be connecting to the fan. Filtering occurs between the two. I'm attempting to prevent the noise from the fans getting on the voltage lines.

enter image description here

Below is the graph for the voltage between the two inductors, L1 and L3. There are multiple lines because of the step function on C3.

enter image description here

According to the graph, between 1Hz and 100Hz the voltage is -40dB. If dB is referenced to 1V this would mean at low frequencies the voltage between L1 and L3 is 10^-2V. Shouldn't it be 12V? Is there a problem in how I am interpreting the graph or am I not understanding how my circuit works?

FOLLOW UP QUESTION

If the AC analysis does not take into account the +12V DC, what exactly is the AC analysis doing? Is it replacing the the +12V source with a sinusoidal waveform? I also have a sinusoidal currant draw from AC1. Is this varied at all with the AC analysis?

\$\endgroup\$
7
  • \$\begingroup\$ 12v don't exist in the AC analysis..They are only used to get the operation point..Hover any schematic wire after doing the AC analysis and the DC operation point (i.e. dc voltage, current, power) will be displayed at the bottom. \$\endgroup\$
    – tobalt
    Commented Jan 7, 2022 at 19:18
  • \$\begingroup\$ @tobalt after running the analysis hovering over schematic nets only says "Click to plot V(N###)" \$\endgroup\$
    – timsweeney7
    Commented Jan 7, 2022 at 19:21
  • \$\begingroup\$ Yeah I noticed it too. Hmm strange. I am sure, that it is usually displayed there. But I don't know right now, how to get it back there. \$\endgroup\$
    – tobalt
    Commented Jan 7, 2022 at 19:28
  • \$\begingroup\$ The .AC analysis first calculates the operating point for the whole linearized circuit. Any source that is not DC, such as SIN() or PULSE() will be replaced with their DC value at time=0 (the first value): I1 will be 1 A, I2 will be 0 A, I3 will be 0.2 A. That means any DC voltages may influence the behaviour of the circuit if there are active elements. Since you only have passive elements, any DC will be useless (plus C1 is completely useless across a voltage source, unless there is some series resistance or you use the current through it). \$\endgroup\$
    – a concerned citizen
    Commented Jan 7, 2022 at 20:41
  • \$\begingroup\$ @aconcernedcitizen ok then after everything is linearized what happens? Which of the 6 sources would be used as the source of the AC sweep? \$\endgroup\$
    – timsweeney7
    Commented Jan 7, 2022 at 21:42

1 Answer 1

Reset to default
2
\$\begingroup\$

12 V is the DC voltage (0 Hz); between 1 and 100 Hz, the -40 dB means the signal amplitude is 10 mVAC (40 dB below 1 V). This is the AC component of the signal -- which only makes sense for an AC (i.e. not 0 Hz) signal.

\$\endgroup\$
2
  • \$\begingroup\$ -40dB is 10mVAC? And so if I wanted to get the true voltage at any given frequency do I need to add 12V to the AC value? \$\endgroup\$
    – timsweeney7
    Commented Jan 7, 2022 at 19:13
  • \$\begingroup\$ yes. -40 dB represents 10^(-40/20) = 10^-2 = 0.01. This is w.r.t 1 V, so 10 mV \$\endgroup\$
    – jp314
    Commented Jan 7, 2022 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.