# phase shifter circuit

Discussion in 'Electronic Design' started by [email protected], Oct 18, 2007.

1. ### Guest

anyone know how to design a phase shifter circuit with standard
components?
input signal: 6.1v ac rms, 488hz
phase shift: 23 ±5 deg lead
output signal: must be capable of driving a DDC DRC-10520 (Ref
signal).
±15V on the board

2. ### Jan PanteltjeGuest

488Hz = 2.049180328 ms period time.
23 degrees is 23/360 x that = 130.9198543 us.
5 % of that = 6.54992714 us
So say we are safe with 131 us
That is 131 cycles of a 1 MHz clock.

So,
comparator - 131 cycle delay -
|
1MHz
clock

Solution: PIC.
Use 8 pin PIC. 12F629 it has the comparator.
pins: +, -, in, xtal in, xtal out, out: makes 7.
One spare pin to break off.

But actually, if you want to use the DCR-10520 to convert to sine,
then a PWM on the PIC can do that too.

3. ### Phil AllisonGuest

"Jan Panteltje"

Seems he wants to have his info before it arrives.

........ Phil

4. ### Jan PanteltjeGuest

My wrong, delay one cycle - lead = 2049.180328 - 130.918542 = 1918.261786
say 1918 us.
So delay 1918 clock ticks of 1MHz.

5. ### Phil AllisonGuest

"Jan Panteltje"

** That is 337 degrees of lag - according to the OP.

Betcha.

...... Phil

6. ### Jan PanteltjeGuest

Well, we will have to wait and see, I have hopes.

7. ### RST Engineering \(jw\)Guest

Not enough information. You **can** use a simple RC phase shifter with an
amplitude follower and about four or five stages to give you 337 degrees of
phase LAG, which is the same thing as 23 degrees of phase LEAD for the
following cycle.

Jim

--
"If you think you can, or think you can't, you're right."
--Henry Ford

anyone know how to design a phase shifter circuit with standard
components?
input signal: 6.1v ac rms, 488hz
phase shift: 23 ±5 deg lead
output signal: must be capable of driving a DDC DRC-10520 (Ref
signal).
±15V on the board

8. ### Phil HobbsGuest

For a fixed frequency, you can get small amounts of phase lead easily
with an RC highpass network. The phase of such a network is

tan(23 degrees)
phi = atan(1/(2*pi*f*RC)), so RC = --------------- = 138 microseconds.
2*pi*(488 Hz)

You could use a 140k resistor and a 1nF capacitor. You'll want to use a
decent cap, so that your phase doesn't vary everyplace, but your spec
corresponds to about +- 25% capacitor tolerance, so something like a 10%
film cap will do fine.

You can clean up the edges with a comparator.

Cheers,

Phil Hobbs

9. ### Tom BruhnsGuest

Seems like other posters so far haven't quite groked what a DRC-10520
is. You need to drive the differential reference input with a
sinewave (not digital) at 3.4V RMS, and the input impedance is
guaranteed within 0.5% of 26 kohms. If your input 6.1V is isolated
(not ground referenced), it should be easy to use series capacitors
and resistors to scale and phase shift. Phil Hobbs' posting gives
you the basic info you need for the phase shift part...you just need
to understand how much attenuation there is in the high pass filter so
you know the voltage out of it, and then pick the load resistance such
that the added resistors and the internal DRC-10520 resistance form a
voltage divider to get you the rest of the way to 3.4 volts.

If your 6.1V is ground referenced, then I'm unsure whether you will be
able to use that directly or not. The DRC-10520 data sheet is very
vague on the common mode range of the reference input, and whether it
must be driven differentially or if instead RL may be grounded. I'd
call DDC about that one, to find out. If it can be grounded, then you
can just use a single series RC from your 6.1V to get the phase shift
and the attenuation. But DDC's data sheet for this part only shows
the reference being driven from an isolated transformer winding, so
it's not really clear how you're supposed to drive it single-ended.
It's also unclear to me just what they mean by listing different
single-ended and differential reference input impedances; perhaps you
can drive it single-ended with RL floating. Again, get ahold of DDC
and ask them! They should be happy to provide you with some
applications info about the part. It's to their advantage to have you
use the part successfully.

Cheers,
Tom