0
\$\begingroup\$

I have a simple question about a low-pass filter. Without a load I have a resonance effect.

Low pass filter

/!\ I forgot to delete a modification on the previous screenshot the 0.5 Ohms resistor after V1 that's my bad...

With a load of 1 Ω the result is what I expect:

With a load of 1 Ω

And with 100 kΩ:

With a load of 100 kΩ

Load is a motor, Faulhaber 2342S024CR, driven by 28V PWM. And is not represented on the simulation.

Thanks you guys for the informations and your time. Next time I will try to be clearer I'm really sorry for the mistakes.

\$\endgroup\$
12
  • 2
    \$\begingroup\$ You haven’t asked a question. \$\endgroup\$
    – winny
    Commented Feb 20 at 9:20
  • 1
    \$\begingroup\$ Node numbers mean nothing here. \$\endgroup\$
    – Andy aka
    Commented Feb 20 at 9:51
  • \$\begingroup\$ What are you measuring ? Voltage accross R1 ? Or one R1 node ... \$\endgroup\$
    – Antonio51
    Commented Feb 20 at 10:17
  • \$\begingroup\$ Oh excuse me I measure the output between R1. And my question is why there is a gain when the output is an "open circuit" and not when the load is very low. \$\endgroup\$
    – SteveninJ
    Commented Feb 20 at 10:25
  • 1
    \$\begingroup\$ Something is screwy. With a 0.5ohm -> 1.0ohm divider you should get a passband gain of something around -3.5dB. Not 0dB. \$\endgroup\$
    – Ste Kulov
    Commented Feb 20 at 14:43

2 Answers 2

Reset to default
2
\$\begingroup\$

If the load is absent the high output voltage comes from the resonance of the inductors and capacitors in your circuit. You can perhaps use a time-domain SPICE simulation to see how the signal in that case will swing up after it is started.

If the load is present (and has a low resistor value) then you are to some extent shorting the output signal. That is especially imprtant around the resonance frequency because there the output impedance is high, so the load has much influence.

\$\endgroup\$
1
\$\begingroup\$

Here is a working simulation of your circuit for loads of 1Ω, 5Ω and almost open-circuit. I've moved the ground to the output side so that it's not necessary to subtract two voltages.

You'll notice that the behaviour near resonance appears different (peaking but no notch) and the passband gain is correct with the heavy load.

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ Hi, "I've moved the ground to the output side so that it's not necessary to subtract two voltages." what do you mean by substract I just probed the positiv side of R1. What do you mean by passband gain is correct, this filter is supposed to be a lowpass filter ?.. :) \$\endgroup\$
    – SteveninJ
    Commented Feb 23 at 12:33
  • 1
    \$\begingroup\$ I am taking the output voltage as the voltage across your R1, not the voltage from top of R1 to the (-) of the source voltage. That's a fundamental difference. It's how I would expect a common mode filter to work. As to the second question, raised in a comment above, you have (well, 'had', you edited it out, which is rather bad etiquette because it invalidates answers people have spent time composing) a series resistance of 0.5Ω and a load of 1Ω so 0dB is wrong. \$\endgroup\$
    – Spehro Pefhany
    Commented Feb 23 at 12:41

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