LPF behavior

[BorisRap]

Hi guys,

I'm trying to simulate a circuit I have to build. I have a pulsing current source of amplitude 1mA and duty cycle 50%. The frequency of this source will vary from DC to 1Mhz.
Just to test I put together this circuit in LTSpice to check the behavior.
I set a low pass filter to around 33Hz, the rest of the filtering can be done digitally.
I simulated the frequency at 1Khz and 1Mhz and I found that the output voltage of the opamp is the same (500mV).

1Khz:


1Mhz:


 Why is that? I thought the output at 1Mhz would be way lower due to the LPF.
That is actually the behavior that I want for my circuit, but I just can't understand why isn't being attenuated at higher frequencies

Thanks!

[Zero999]

Your low pass filter is working. It is attenuating the high frequency AC components of the signal and allowing through the DC.

The input signal is 1V square wave, with a 50% duty cycle, which is equivalent to 0.5VDC plus a 0.5V bipolar squarewave (i.e. going from +0.5V to -0.5V) superimposed on top of it.

Refer to the simulation below.


V3 = V1+V2

Therefore:
squarewave1 = squarewave2

If you add a low pass filter, it will remove V2, which is entirely AC, leaving only V1. A high pass filter will have the opposite effect and you'll be left with a bipolar 0.5V squarewave.

[BorisRap]

Your low pass filter is working. It is attenuating the high frequency AC components of the signal and allowing through the DC.

The input signal is 1V square wave, with a 50% duty cycle, which is equivalent to 0.5VDC plus a 0.5V bipolar squarewave (i.e. going from +0.5V to -0.5V) superimposed on top of it.

Refer to the simulation below.

(Attachment Link)
V3 = V1+V2

Therefore:
squarewave1 = squarewave2

If you add a low pass filter, it will remove V2, which is entirely AC, leaving only V1. A high pass filter will have the opposite effect and you'll be left with a bipolar 0.5V squarewave.

Thanks for the explanation!

[radiolistener]

Why is that? I thought the output at 1Mhz would be way lower due to the LPF.

Because DC voltage (500mV in your case) is 0 Hz frequency which is below 33 Hz cut-off of your LPF, so it is passed through filter with no attenuation