Robert,
Here is the info on the power direction indicator as I first recieved it.
As shown, some current limiting in the Voltage senseing loop is lacking.
Function is as follows.:
An image of the Voltage transformer neutral is established between the 
active diode pair as voltage rises to the diode conduction threshold 
in each half wave cycle.  The current transformer current circulates 
'hung off' the neutral, developing a voltage, + or - relative to the
voltage transformer center tap, depending on the current direction in 
the monitored line. The location of the neutral image swaps places in
the other half cycle.
 
The output is half wave rectified ac clipped at a 1 diode forward 
conduction voltage.
 
Output voltage can be increased by replacing each diode with multiple 
diodes in series, but at the expense of turn-on time each cycle.
 
I built the circuit (with trippled diodes), but there is not much drive 
to the output.  I think the idea will work for control if I put the 
output to the gates of a pair of oppositely connected UJT's. 
 
That's to try next.  Thanks for your suggestion earlier. 
 
--------------------------------------------
Msg#:49379 *ELECTRONICS*
05/05/96 23:01:52
From: Dave Lyle
  To: Jim Mcandrew
Subj: Reply to Msg# 32171 (POWER DIRECTION INDICATOR)
(( On 05-03-1996, Jim Mcandrew wrote to All: ))
 
JM>Anyone have an idea how to make a simple AC power flow direction
JM>indicator?  By which I mean, a device which could energize a relay,
JM>depending on which end of an ac conductor pair was the source of power
 
I -think- this will work for you.  It's basically a double balanced
mixer configured as a phase detector.  Give it a try......
 
 
 
         T1              +------------------+     T2
                         |                  |
        ||(----+---------)------+           |     ||(---
        ||(    |         |      |  diodes   |     ||(
  >---\ ||(   100 Ohms   +-->|--+-->|--+    +----)||(
  I    )||(    |___      |             |     ____)||( Voltage
  in   )||(    |   |Gnd  |             |    |    )||(
  >---/ ||(   100 Ohms   +--|<--+--|<--+----)----)||(
        ||(    |                |           |     ||(
        ||(----+----------------+           |     ||(---
                                            |
                                            |
                                          Output
 
T1 is the current transformer, suggest a 6.3 volt filament transformer
with a secondary capable of carrying whatever the full load current is.
The current passes through the 6.3 volt winding, and the 117 volt
winding is wired to the diodes.
 
T2 is a 12.6 volt center tapped filament transformer, but a 100 ma
rating would be sufficient. The 117 volt side is connected to the sensed
line, and the 6.3 volt side is connected to the diodes.
 
The two resistors will see a maximum of about 2 volts across them.
Half watt-ers should be sufficient.
 
The 4 diode will need to be able to handle the full load current divided
by T1's turns ratio.  So if you need to pass 10 amps through T1, then
the diodes need to be rated at at least .5 amps.  1N4004's would be a
possibility in that case.
 
The output will be DC and the polarity an indication of which way
the power is flowing.  A filter cap might be advantageous here, just
make sure it is non-polar.  I'll leave it to you to figure out how it
works. Hint: Yes, T1's secondary -is- shorted out by the diodes. One
pair when the current is positive, and the other when it is negative.
 
Dave
davel@execpc.com
------------------------------------------
 
We exchanged some messages on theory and fine points later.  
I'm using 5 amp rectifier diodes in the prototype I built.  Also,  
current limiting in the Voltage circuit is a must.
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