# Phase comparator gain calculation

Discussion in 'Electronic Design' started by John Wilkinson, Apr 14, 2005.

1. ### John WilkinsonGuest

Hi,
I am trying to design a PLL using the 74HCT7046 chip.
I am using the Phase frequency detector and feeding this into a single
supply op-amp active PI filter. This has its non-inv input tied at
vdd/2=2.5V whilst the amp is on +15V to give the VCO tuning voltage
+2-+15V for a 5MHz change in frequency.

Now what is the gain of the PD. I have reckoned it as 3*vdd/2pi, is this
correct?
Do I need to take into account that the non-inv input is at half VDD?

Has anyone got a schematic for a TTL PFD feeding a fast totem pole type
discrete output at +15V, to up the voltage swing?

Thanks,
John.

2. ### Andrew HolmeGuest

Yes.

When the pull-up transistor is on, the phase detector output voltage w.r.t.
op-amp virtual-earth is +2.5V
When the pull-down transistor is on, it's -2.5V.

If you want to think of it as a voltage-output phase detector, the gain is
2.5/2pi

Alternatively, treat it as a current output: Let's say the resistor between
the PD output and the summing node is R, then the gain is (2.5 / R) / 2pi.
You now have a current-input active loop filter, of which R is not a part.

BTW You probably want to split R in two and put a small capacitor to ground
from the middle. This takes the sharp edge off the pulses before they reach
the op-amp.

3. ### John WilkinsonGuest

Thanks for the reply Andrew, and sorry for being obtuse, but firstly I
had the gain for the PFD wrong, it is VDD/4pi.
Are you saying that now I have the split supply to the op-amp this is
now doubled? ie 2*vdd/4pi.

Many thanks,
John.

4. ### Andrew HolmeGuest

No. The PFD gain is not affected by the op-amp power supply.

5. ### John LarkinGuest

The charge-pump pd's are generally loaded into a capacitor, and the
chip output is usually considered to be a current source. So the
transfer function from phase to pin voltage is of the form k/S, an
integral.

John