Question: 90 degree phase shift

Hi guys :wink:,
I am trying to do a 90 degree phase shift of my signal. My lecturer said I can use comparator and integrator to do so. I search it in the book and in the internet, but I can't find it. Anyone know how can I use comparator and integrator to do a 90 degree phase shift?

When you talk 90 degree shift the first concept that comes to mind is to use capacitors or inductors.

Is your signal AC?
Run it through a capacitor and current will lead by up to 90 degrees.
Run it through a coil and current will lag by 90 degrees.
That's the principle anyway. How to implement it is another matter.

Or is your signal pulsed DC?
You can introduce a delay by attaching a capacitor to a pulse output. (This is an integrator.)
When the pulse changes state, it then charges (or discharges) the capacitor.
The new output voltage catches up in a fraction of a second. Automatic delay.
It's not the same as a phase change but it may suit.

You may not even need a comparator. Just feed the delayed signal to the following stage.
You must tailor the capacitor value to your particular circuit. You may need to add a resistor somewhere as well.

I am not sure what he meant by "comparator". Maybe he meant to say op amp. But an op amp set up as an integrator has a transfer function of 1/s. A pole at zero (1/s) means a 90 degree phase lag.

In practice you have to keep any dc content on your signal from pegging the op amp, though, since it has very high dc gain.

Thanks BradtheRad and biff44 for your reply:smile: . Yes, maybe I heard wrong. I think what he said about comparator should be to compare between the original signal and the output of the integrator.

I've another question. If I am using an OP-Amp as the integrator, what value R and C should be? Can just use any values?
By the way my signal is a square wave signal(low-level: 1mV and high-level: 3.5V).

capital_zach said:

..............
I've another question. If I am using an OP-Amp as the integrator, what value R and C should be? Can just use any values?
By the way my signal is a square wave signal(low-level: 1mV and high-level: 3.5V).

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I don`t know how exact the phase shift of 90 deg must be achieved in your design.
But you should know that an opamp integrator - strictly spoken - introduces a signal phase shift of 90 deg at one single frequency only! The reason for this deviation from the ideal response is the frequency dependency of the opamp itself (at least two poles to be considered). Due to the finite low frequency gain the phase shift goes up to 0 deg for lower frequencies and goes beyond -90 deg due to the second opamp pole.
The values for R and C of course are determined by the time constant required by your application.

You can use a comparator and adjust it to change state at any desired point on the capacitor charge/discharge slope.

It can be a comparator IC. Or it can be an op amp set up to behave the same.

You have two inputs. The output changes state when one input V crosses the V level of the other input.

You would send your signal to one input. The signal is falling and rising because the capacitor takes time to charge and discharge.

At the other input you'd attach a reference V. This can come from a potentiometer. You select a voltage ranging between output V and zero.

The comparator changes state where the rise/fall slope meets the reference V.

The time delay will not be the same when going from hi to lo compared with going from lo to hi. You get equal time delays only when you set at the halfway V. It's up to you to decide how you want to use it.

Again this concept gives a fixed time delay. It's not the same as a phase shift.

What value should the capacitor be? It depends on the time between your square waves.
Say your signal runs at 1khz. Then you have .0005 second between transitions.

A .01 uF capacitor in series with a 10K ohm resistor has a time constant of .0001. It should have a steep enough rise/fall slope between transitions.

Or if your signal runs at 100 hz, then use a .1 uF capacitor and 10k resistor. Or .01uF and 1k ohm. Etc.

By the way some comparator IC's sink current at the output but do not source current. You would have to use a pullup resistor to obtain a positive high. For that reason it's easier to use an op amp.

if I use ua741 as my OP-Amp. Can I increase the slew rate of the output signal, so that the output signal can be a fine square wave?

Can I increase the slew rate of the output signal

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Yes, if you use a different OP type. E.g. uncompensated ua709. By the way, you neither told about the involved signal waveform, nor the intended circuit. A lot of guessing is required to answer your questions.

FvM said:

Yes, if you use a different OP type. E.g. uncompensated ua709. By the way, you neither told about the involved signal waveform, nor the intended circuit. A lot of guessing is required to answer your questions.

Click to expand...

My signal wave is a square wave and the intended circuit is a normal OP-Amp used as comparator. So the slew rate for ua741 is fix right? There is no way to increase it except by using other type of OP-Amp? Because in my p-spice there is only ua741.

To get faster slew rate, run an op amp at maximum gain. You may not need a feedback loop.

However running at high gain means you get lower bandwidth. A few kHz in a 741. Check specs to find out what your situation allows.

The 741 has a quirk. Its output can only go down to 2V above its ground pin. You may need to use a bipolar supply.

A comparator may be more suitable. It makes an internal connection from output pin to ground. It doesn't need a feedback loop. It has a faster slew rate at high gain. You get 2 Mhz bandwidth.