Why does the passive highpass filter give such results in spice analysis?

[electronx]

I am designing a passive high pass filter, here are the values: 2M ohms and 1uf cut-off frequency is 0.07Hz +-20V and when I apply a 1 khz sinus signal, I get such a result. What is the reason for this? The voltage level looks like this ım using pspice

[iMo]

Because your resistor is 2 Milliohm?
Try with 2Meg or 2000k..

[Chalcogenide]

There is a chance that you placed a 2 mOhm resistor instead of a 2MOhm; check the current across the resistor.
SPICE is typically not case sensitive, so the 1e6 multiplier is usually written as "MEG" (or just use scientific notation to be always sure).

[electronx]

Thank you  all .I feel super noob now

[MarkT]

I'd say Spice is just dumb here, its a real gotcha...  I suppose the justification is Spice originally was written in Fortran using punched cards...  I doubt it understands p v. P either (pico / peta) - not that petaohm resistors are a thing!

[TimFox]

SPICE is backwards-compatible for its notation.
The original SPICE, in compiled Fortran on Hollerith cards, was not case-sensitive, so "m" or "M" always means "milli" and for 10⁶ one needs to write "Meg" or "MEG".
Similarly, "p" always means "pico" and "u" always means "micro".

[MrAl]

Hi,

Just to verify that it is a 0.002 Ohm resistor and not a 2M resistor (2M as we usually write 2 megohms) the theoretical calculation comes out to:
Vout=251.327 nV

which matches your plot result closely.

If you had an actual 2 megohm resistor here, it would have came out very very close to 1v.

For reference, the theoretical calculation comes out to:
Vout=(2*pi*f*C*R)/sqrt(4*pi^2*f^2*C^2*R^2+1)

with f, C, and R positive, and pi=3.14159265...