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Phase shifts (lag or lead)???

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diarmuid

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As an example:

A common-source amplifier is an inverting amplifier because it adds 180deg phase shift.

Is the output signal phase shifted by +180 (i.e. it leads the input signal by 180deg) or
is it phase shifted by -180 (i.e. it lags the input signal by 180deg)?

I think it is +180 but intuitively why is that?

Does this mean the output signal comes after the input which is why it incurs a +ve
phase shift?

Thanks,

Diarmuid
 

The output signal is phase shifted by 0 .. -180° (i.e. it lags the input signal by this amount). This phase shift refers to the positive, i.e. non-inverting input of the amplifier. See this snippet from Johns & Martin EE/CS 5720/6720 lecture on Op-Amp Simulation: View attachment phase.pdf
 
If you consider a periodical input signal, you can't distinguish between + or -180 degree.
But if you associate phase shift somehow with time advance/delay, you recognize that the output isn't phase shifted, just inverted.
 
FvM said:
If you consider a periodical input signal, you can't distinguish between + or -180 degree.
But if you associate phase shift somehow with time advance/delay, you recognize that the output isn't phase shifted, just inverted.
Click to expand...

Yes - and I think, this becomes more clear if you consider as a signal input not a sinusoidal but - for example - a squarewave signal with a dury cycle other than 50%.
In this case, you immediately can recognize that there is no phase shift but simply a signal inversion.
 
Thanks for the responses guys.

However, im still not totally clear.

Attached is the mag/phase response of the common-source amp output where you can see (phi(vout) = +180deg)).

As per you arguments I would expect the output signal to just be inverted, not phase shifted and by -180deg. Is
the "+" real therefore, or just some numerical artifact?

Also, I have always said that in a -ve feedback loop, the loop intrinsically adds -180deg phase shift.
This is not technically correct I think, should it be: in a -ve feedback loop, the loop inverts the output signal thereby adding the "-" part.
Is this correct?

Thanks,
 

Attachments

  • +ve_phi_shift.pdf
    204.5 KB · Views: 61

The AC simulation plot displays +180 degree phase at 1Hz to fit the principal value range of -180..+180 degree. If you add a small positive phase shift, the plot would probably start near -180 degree and jump to +180 degree somewhere.

As explained, +180 (or -180) degree phase shift is identical to signal inversion for pure sine (and similar symmetrical periodic) signals. In AC analysis, inversion is displayed as additional 180 degree phase shift, either with positive or negative sign.
 
Thanks. Im clearer now. Basically mathematically -180/+180 are the same thing (graphically too for a periodic signal).

Cheers.
 

diarmuid said:
Also, I have always said that in a -ve feedback loop, the loop intrinsically adds -180deg phase shift.
This is not technically correct I think, should it be: in a -ve feedback loop, the loop inverts the output signal thereby adding the "-" part.
Is this correct?
,
Click to expand...

Yes - I think so.

If - in addition - within the loop there is phase shift of -140 deg we arrive at a total phase shift of -180-140=-320 deg. (+40 deg above -360 deg)
Thus, we have a phase margin of +40 deg.
We find the same margin using the expression +180-140=+40 deg. (+40 deg above 0 deg).
 
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