Suggestion for Driving Ternary Inputs from  3.5V logic output.

1) We measure at the TERNP and TERNN inputs .980V and 1.05V
   respectively.   For DSM  the design is different and may change.
   but not by a lot. 
   
2) Objective is to pull current in nearly equal units in the range
   180uA < unit <  230uA.  With standard Ternary count of 0, -1, -2
   units..
   
 
 DC coupled resistors seem to cause problems so lets build a simple
 diode switch that switches 1 unit of ternary current.  Two coupled
 in parallel will pull two units which is good for one ternary input. 
 
   Each ternary input  (two inputs per channel) has two independent
  diode switches that steer 1 or 2 units of current  independently
  into the ternary receiver. Four switches total per ternary receiver
  
  
  The diode is a standard silicon diode 0.7V typical. 
  
  To keep the capacitance low we can use a three pin dual diode
  common anode type like a ??BAW56?? or other low capacitance (2pF) 
  device.  The anode goes to the  Ternary input. The two cathodes 
  each connect to a two resistor series combination as shown below.
  
 1 UNIT current control  ( one of two that attach to a common anode dual
  diode per ternary input.)
  
  
   Assume FBGA output (0,3.5V)    negative voltage -3V supply
   
   ....Schematic....
   
   FBGA out>--R2---cathode---R1---<-3V		R1=20K  R2=8.2K
   
   CASE 1 'zero' state FBGA out = 0   1 unit of current flows into 
    ternary input. 
  
   Diode is 'on'  so V(cathode) = 1V-.7 = .3V
   
   I unit = .3V/8.2K + 3.3V/20K = 202uA 
   
   CASE 2 'true' state FBGA = 3.5V 
   
   No current thru diode if Vcathode > 1V --Drop across R1 > 4V
   
   V(R1) = 6.5V(R1/(R1+R2)) =  4.6V 
   
  Possible problems include a slow response due to capacitance at the
  junction of the two resistors and diode. 
  
  I recommend that  the resistor divider network be located near
  the dual diode pack and that the dual diode pack be located near
  the ternary input. Decouple the -3V supply near the ternary receiver(s)
  to gnd.
  
  If this doesn't work we may need to replace the diodes with
  transistor current sources. 
  
  					hope this helps,
  					       		    Mitch