Method of Determining Secondary Current Ratings in D.C. Circuits

Half Wave Rectifier (HWR)
Full Wave Center Tap (FWCT)
Full Wave Bridge (FWB)
Dual Complementary Rectifies (DCR)
Adding A Regulator

Half Wave Rectifier (HWR)

  • Inefficient, DC flux component.
  • Use generally for 0.5 watt or less.

    WITHOUT CAPACITOR: 

    VAC = 2.2 x ( VDC + 1)    
IAC = 1.6 x IDC           
WATTS = VDC x IDC (output) 
VA = 3.5 x ( WATTS + loc* )

    Example: Specify a transformer 5V,
    100 ma DC output. 

    VAC = 2.2 x ( 5 + 1) = 13.2 V    
IAC = 1.6 x 0.1 = .16 A          
WATTS = 5 x 0.1 = 0.5 WDC        
VA = 3.5 x ( 0.5 + 0.1 ) = 2.1 VA
    		WITH CAPACITOR: 

    VAC = .88 x ( VDC + 1 )    
IAC = 2.6 x IDC            
WATTS = VDC x IDC (output) 
VA = 2.3 x ( WATTS + loc* )

    Example: Specify a transformer 5V,
    100 ma DC output. 

    VAC = .88 x ( 5 + 1 ) = 5.3 V    
IAC = 2.6 x 0.1 = 0.26 A         
WATTS = 5 x 0.1 = 0.5 WDC        
VA = 2.3 x ( 0.5 + 0.1 ) = 1.4 VA

    Full Wave Center Tap (FWCT)

  • More Efficient, although secondary not fully utilized
  • Preferred in ver low voltage supplies where extra diode voltage drops unwanted

    WITHOUT CAPACITOR: 

       VAC = 2.2 x ( VDC + 1 )     
   IAC = .8 x IDC              
   WATTS = VDC x IDC ( output )
** VA = 1.4 x ( WATTS + IDC* ) 
core

    Example: Specify a transformer 5V,
    2 A DC output. 

       VAC = 2.2 x ( 5 + 1 ) = 13.2 V      
   IAC = .8 x 2 = 1.6 A                
   WATTS = 5 x 2 = 10 WDC              
** VA = 1.4 x ( 10 + 2 ) = 16.8 VA 
core
    		WITH CAPACITOR: 

       VAC = 1.7 x ( VDC + 1 )     
   IAC = 1.2 x IDC             
   WATTS = VDC x IDC ( output )
** VA = 1.7 x ( WATTS + IDC* ) 
core

    Example: Specify a transformer 5V,
    2 A DC output. 

       VAC = 1.7 x ( 5 + 1 ) = 10.2 V      
   IAC = 1.2 x 2 = 2.4 A               
   WATTS = 5 x 2 = 10 WDC              
** VA = 1.7 x ( 10 + 2 ) = 20.4 VA 
core

    Full Wave Bridge (FWB)

  • Most Efficient, secondary is fully utilitzed
  • Preferred for smallest transformer size and cost at higher voltages

    WITHOUT CAPACITOR: 

    VAC = 1.1 x ( VDC + 2)     
IAC = 1.1 x IDC            
WATTS = VDC x IDC (output)  
VA = 1.2 x ( WATTS + 2 IDC* )

    Example: Specify a transformer 5V,
    2 A DC output. 

    VAC = 1.1 x ( 5 + 2) = 7.7 V   
IAC = 1.1 x 2 = 2.2 A          
WATTS = 5 x 2 = 10 WDC         
VA = 1.2 x ( 10 + 4 ) = 16.8 VA
    		WITH CAPACITOR: 

    VAC = 0.8 x ( VDC + 2 )    
IAC = 1.8 x IDC            
WATTS = VDC x IDC (output) 
VA = 1.4 x ( WATTS + 2 IDC* )

    Example: Specify a transformer 5V,
    2 A DC output. 

    VAC = 0.8 x ( 5 + 2 ) = 5.6 V  
IAC = 1.8 x 2 = 3.6 A          
WATTS = 5 x 2 = 10 WDC         
VA = 1.4 x ( 10 + 4 ) = 19.6 VA

    Dual Complementary Rectifies (DCR)

  • Most Efficient, when load is balanced.
  • Provides a (+) and (-) suppl with a common ground.

    WITHOUT CAPACITOR: 

    VAC = 1.1 x ( VDC + 2)     
IAC = 1.1 x IDC            
WATTS = VDC x IDC (output) 
VA = 1.2 x ( WATTS + 2 IDC* )

    Example: Specify a transformer for +/- 15V,
    (30V total) 2 A DC output. 

    VAC = 1.1 x ( 30 + 2 ) = 35.2 V
IAC = 1.1 x 2 = 2.2 A           
WATTS = 30 x 2 = 60 WDC        
VA = 1.2 x ( 60 + 4 ) = 76.8 VA
    		WITH CAPACITOR: 

    VAC = 0.8 x ( VDC + 2 )    
IAC = 1.8 x IDC            
WATTS = VDC x IDC (output) 
VA = 1.4 x ( WATTS + 2 IDC* )

    Example: Specify a transformer +/- 15V,
    (30V total), 2 A DC output. 

    VAC = 0.8 x ( 30 + 2 ) = 25.6 V
IAC = 1.8 x 2 = 3.6 A          
WATTS = 30 x 2 = 60 WDC        
VA = 1.4 x ( 60 + 4 ) = 89.6 VA

    Adding A Regulator To The Circuit "With Capacitor" Above:

  • Adding a regulator will require higher VDC after the rectifier to, keep the ripple above the regulator voltage, provide for a regulator voltage drop and provide for low line conditions.
  • The higher VDC will translate back to a higher VAC and higher transformer VA rating.
  • Note that VDC referred to is after the rectifier and not final Regulator output VOUT
  • The power supply designer should determine value of VDC after the rectifier suitable to his Regulator Power Rating, Capacitor Size and Low Line Limit.
  • Typicals are: VREG >= 3 VRIPPLE = 10% x VDC peak.

    - Formula Factors have been rounded off for convenience.
    * Formula assume 1 drop per diode.
    ** In FWCT circuits, core VA can be less than secondary V x A if specially designed.
    Typical resistive load. (Batteries are considered as a capacitor.)