PATENT ABSTRACT
A system and method are disclosed for providing a low voltage bandgap reference circuit that provides a substantially constant output voltage over a range of values of temperature. For example, the bandgap reference circuit could be capable of providing output voltages that are as low as one hundred millivolts. Also, no special start-up circuitry may be required to initiate the operation of the bandgap reference circuit. The bandgap reference circuit could further require fewer transistors and fewer resistors than prior art bandgap reference circuits.

PATENT DESCRIPTION
TECHNICAL FIELD OF THE INVENTION 
     The present invention is generally directed to the manufacture of bandgap reference circuits and, in particular, to a system and method for providing an improved low voltage bandgap reference circuit. 
     BACKGROUND OF THE INVENTION 
     A bandgap reference circuit is commonly used to provide a reference voltage in electronic circuits. A reference voltage must provide the same voltage every time the electronic circuit is powered up. In addition, the reference voltage must remain constant and independent of variations in temperature, fabrication process, and supply voltage. 
     A bandgap reference circuit relies on the predictable variation with temperature of the bandgap energy of an underlying semiconductor material (usually silicon). The energy bandgap of silicon is on the order of one and two tenths volt (1.2 V). Some types of prior art bandgap reference circuits use the bandgap energy of silicon in bipolar junction transistors to compensate for temperature effects. 
     As the design dimensions of electronic circuit elements decrease, the magnitude of the power supply voltages have also decreased. Lower power supply voltages reduce the total power requirements of an electronic circuit. This is especially important in electronic circuits that operate on battery power. Electronic circuits that use lower supply voltages also require bandgap reference circuits that provide lower reference voltages. 
     Therefore, there is a need in the art for a bandgap reference circuit that is capable of providing a low reference voltage. Specifically, there is a need in the art for an improved low voltage bandgap reference circuit that can provide a reference voltage having a magnitude less than one and two tenths volts (1.2 V). 
     Before undertaking the Detailed Description of the Invention below, it may be advantageous to set forth definitions of certain words and phrases used throughout this patent document: the terms “include” and “comprise,” as well as derivatives thereof, mean inclusion without limitation; the term “or,” is inclusive, meaning and/or; the phrases “associated with” and “associated therewith,” as well as derivatives thereof, may mean to include, be included within, interconnect with, contain, be contained within, connect to or with, couple to or with, be communicable with, cooperate with, interleave, juxtapose, be proximate to, be bound to or with, have, have a property of, or the like. 
     Definitions for certain words and phrases are provided throughout this patent document, those of ordinary skill in the art should understand that in many, if not most instances, such definitions apply to prior uses, as well as to future uses, of such defined words and phrases. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
       For a more complete understanding of the present invention and its advantages, reference is now made to the following description taken in conjunction with the accompanying drawings, in which like reference numerals represent like parts: 
         FIG. 1  illustrates a schematic representation of a first embodiment of a low voltage bandgap reference circuit of the present invention; and 
         FIG. 2  illustrates a schematic representation of a second embodiment of a low voltage bandgap reference circuit of the present invention. 
     
    
    
     DETAILED DESCRIPTION OF THE INVENTION 
       FIGS. 1 and 2 , discussed below, and the various embodiments used to describe the principles of the present invention in this patent document are by way of illustration only and should not be construed in any way to limit the scope of the invention. Those skilled in the art will understand that the principles of the present invention may be implemented with any type of suitably arranged bandgap reference circuit. 
       FIG. 1  illustrates a schematic representation of a first embodiment of a low voltage bandgap reference circuit  100  constructed in accordance with the principles of the present invention. The input voltage V IN  is connected to a first current source  110  that produces a current having a value of I 1  and to a second current source  120  that also produces a current having a value of I 1 . As shown in  FIG. 1 , the input voltage V IN  is also connected to the collector of bipolar junction transistor Q 3  and to the collector of bipolar junction transistor Q 4 . 
     The output of first current source  110  is connected to the collector of bipolar junction transistor Q 1 . The output of first current source  110  is also connected to the base of bipolar junction transistor Q 4 . The output of second current source  120  is connected to the collector of bipolar junction transistor Q 2 . The output of second current source  120  is also connected to the base of bipolar junction transistor Q 3 . The emitter of bipolar junction transistor Q 3  is connected to the base of bipolar junction transistor Q 2 . The emitter of bipolar junction transistor Q 3  is also connected through resistor R 2  to the base of bipolar junction transistor Q 1 . 
     The emitter of bipolar junction transistor Q 1  is connected to ground. A first end of resistor R 1  is connected to the base of bipolar junction transistor Q 1  and a second end of resistor R 1  is connected to ground. The current that flows through resistor R 1  is designated as I 2 . 
     The emitter of bipolar junction transistor Q 2  is connected to the voltage output terminal V OUT . The emitter of bipolar junction transistor Q 2  is also connected through resistor R 3  to ground. The current that flows through resistor R 3  is designated as I 3 . 
     The emitter of bipolar junction transistor Q 4  is connected to the collector of bipolar junction transistor Q 5 . The base of bipolar junction transistor Q 5  is connected to a node between the emitter of bipolar junction transistor Q 4  and the collector of bipolar junction transistor Q 5 . The emitter of bipolar junction transistor Q 5  is connected to the voltage output terminal V OUT . 
     The output voltage V OUT  is the sum of the voltage across resistor R 2  and the difference between the base-emitter voltage V BE  of transistor Q 1  and transistor Q 2 . The current through transistor Q 1  is equal to I 1  and the current through transistor Q 2  is also equal to I 1 . 
     The area of transistor Q 1  is equal to a unit value of area. That is, the transistor Q 1  has a value of area equal to one square unit (designated “ 1 x” in  FIG. 1 ). The area of transistor Q 2  is equal to “A” times the area of transistor Q 1 . That is, transistor Q 2  has a value of area equal to A square units of area (designated “Ax” in  FIG. 1 ). 
     With equal currents (I 1 ) through transistor Q 1  and through transistor Q 2  and with an area ratio of “one” to “A” (1:A), the difference voltage (ΔV BE ) is given by the expression:
 
Δ V   BE   =V   T  ln( A )  (Eq. 1)
 
     where the term V T  represents the thermal voltage of the transistor at the absolute temperature T. 
     The current I 2  flows through resistor R 1 . Ignoring the base currents in transistor Q 1  and in transistor Q 2 , the value of current flowing through transistor R 2  is also I 2 . Transistor Q 3  supplies the I 2  current and the value of the current I 2  is given by the expression: 
     
       
         
           
             
               
                 
                   
                     I 
                     2 
                   
                   = 
                   
                     
                       V 
                       
                         BEQ 
                         1 
                       
                     
                     
                       R 
                       1 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     2 
                   
                   ) 
                 
               
             
           
         
       
     
     where the term V BEQ     1    represents the base-emitter voltage of transistor Q 1 . This means that the voltage V R     2    across resistor R 2  is given by the expression: 
     
       
         
           
             
               
                 
                   
                     V 
                     
                       R 
                       2 
                     
                   
                   = 
                   
                     
                       
                         R 
                         2 
                       
                       
                         R 
                         1 
                       
                     
                     ⁢ 
                     
                       V 
                       
                         BEQ 
                         1 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     3 
                   
                   ) 
                 
               
             
           
         
       
     
     Adding the PTAT (Proportional to Absolute Temperature) difference voltage (ΔV BE ) to the voltage V R     2    across resistor R 2  provides a first order temperature independent output voltage V OUT .
 
 V   OUT   =ΔV   BE   +V   R     2     (Eq. 4)
 
     
       
         
           
             
               
                 
                   
                     V 
                     OUT 
                   
                   = 
                   
                     
                       
                         V 
                         T 
                       
                       ⁢ 
                       
                         ln 
                         ⁡ 
                         
                           ( 
                           A 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         ( 
                         
                           
                             R 
                             2 
                           
                           
                             R 
                             1 
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         V 
                         
                           BEQ 
                           1 
                         
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     
                       Eq 
                       . 
                       
                           
                       
                       ⁢ 
                       5 
                     
                     ) 
                   
                   ⁢ 
                   
                       
                   
                 
               
             
           
         
       
     
     Transistor Q 3  supplies the current I 2  and controls the bases of transistor Q 1  and transistor Q 2  to keep the collector of transistor Q 2  at a voltage value of 2V BE +V OUT . Transistor Q 4  and transistor Q 5  control the output voltage V OUT  to keep the collector of transistor Q 1  at a voltage value of 2V BE +V OUT . Transistor Q 5  is only used to balance the collector voltages of transistor Q 1  and transistor Q 2 . 
     The current I 3  flows through resistor R 3 . The value of resistance of resistor R 3  should be selected to provide a current value of approximately I 1  through transistor Q 4  and transistor Q 5 . The absolute value of the current I 3  is not critical. 
     The value of the resistance of resistor R 3  is approximately equal to the output voltage V OUT  divided by the sum of the current I 1  plus the current through transistor Q 4 . Because the value of the current through transistor Q 4  is approximately equal to the current I 1 , the approximate value of the resistance of resistor R 3  is given by the expression: 
     
       
         
           
             
               
                 
                   
                     R 
                     3 
                   
                   ≅ 
                   
                     
                       V 
                       OUT 
                     
                     
                       ( 
                       
                         
                           I 
                           1 
                         
                         + 
                         
                           I 
                           Q4 
                         
                       
                       ) 
                     
                   
                   ≅ 
                   
                     
                       V 
                       OUT 
                     
                     
                       2 
                       ⁢ 
                       
                         I 
                         1 
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     
                       Eq 
                       . 
                       
                           
                       
                       ⁢ 
                       6 
                     
                     ) 
                   
                   ⁢ 
                   
                       
                   
                 
               
             
           
         
       
     
     The minimum value of the input voltage V IN  for bandgap reference circuit  100  is given by the expression:
 
 V   IN (minimum)=2 V   BE   +V   SAT   +V   OUT   (Eq. 7)
 
     The term V BE  represents a value of base to emitter voltage of said first bipolar junction transistor Q 1 . The term V SAT  represents a minimum voltage required for the current sources ( 110 ,  120 ). The term V OUT  represents the output voltage. The currents I 1  in the current sources ( 110 ,  120 ) may be constant or they may be proportional to absolute temperature (PTAT). Typical values of V IN  (minimum) are in the range of one and eight tenths volt (1.8 V) to two volts (2.0 V). 
     The low voltage bandgap reference circuit  100  of the present invention provides a low value of output voltage V OUT  that is constant with temperature over a pre-selected range of temperature values. The value of output voltage V OUT  can be significantly less than one and two tenths volt (1.2 V). The value of output voltage V OUT  can be as low as approximately one hundred millivolts (100 mV). The lowest value of output voltage V OUT  achievable by prior art devices is approximately two hundred millivolts (200 mV). 
     The value of output voltage V OUT  that is provided by the low voltage bandgap reference circuit  100  of the present invention depends on the ratio of the value of the resistance of the R 1  resistor to the value of the resistance of the R 2  resistor (R 1 /R 2 ). The value of the resistance of the R 3  resistor is not critical. No special start-up circuitry is required to operate the low voltage bandgap reference circuit  100  of the present invention. Start-up is initiated simply by supplying the I 1  currents. 
     The optimal values of the resistances of the resistors (R 1 , R 2  and R 3 ) may be selected using the analysis set forth below. The basic equation for the base-emitter voltage V BE  for the bipolar junction transistor Q 1  is: 
     
       
         
           
             
               
                 
                   
                     V 
                     
                       BEQ 
                       1 
                     
                   
                   = 
                   
                     
                       E 
                       GE 
                     
                     - 
                     
                       H 
                       ⁡ 
                       
                         ( 
                         
                           
                             E 
                             GE 
                           
                           - 
                           
                             V 
                             
                               BE 
                               o 
                             
                           
                         
                         ) 
                       
                     
                     + 
                     
                       
                         V 
                         To 
                       
                       ⁢ 
                       H 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         ln 
                         ⁡ 
                         
                           ( 
                           
                             
                               I 
                               1 
                             
                             
                               I 
                               0 
                             
                           
                           ) 
                         
                       
                     
                     - 
                     
                       η 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         V 
                         To 
                       
                       ⁢ 
                       H 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         ln 
                         ⁡ 
                         
                           ( 
                           H 
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     
                       Eq 
                       . 
                       
                           
                       
                       ⁢ 
                       8 
                     
                     ) 
                   
                   ⁢ 
                   
                       
                   
                 
               
             
           
         
       
     
     The expression E GE  represents the silicon bandgap voltage. A typical value for the silicon bandgap voltage is approximately one and two tenths volt (1.2 V). The letter H represents the ratio of the absolute temperature T to the room temperature T 0 . 
     
       
         
           
             
               
                 
                   H 
                   = 
                   
                     T 
                     To 
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     9 
                   
                   ) 
                 
               
             
           
         
       
     
     The room temperature T 0  is equal to twenty seven degrees Celsius (27° C.) and equal to three hundred degrees Kelvin (300° K.). The expression I 1  represents the current through transistor Q 1  at the temperature T. The expression I 0  represents the current through transistor Q 1  at room temperature T 0 . 
     The expression V BE     0    represents the value of base-emitter voltage V BE  of transistor Q 1  when the temperature is room temperature T 0  (and the current through transistor Q 1  is I 0 ). The expression V T     0    represents the thermal voltage at room temperature T 0 . 
     
       
         
           
             
               
                 
                   
                     V 
                     
                       T 
                       0 
                     
                   
                   = 
                   
                     
                       
                         kT 
                         0 
                       
                       q 
                     
                     ≅ 
                     
                       26 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       millivolts 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     10 
                   
                   ) 
                 
               
             
           
         
       
     
     The letter k represents Boltzmann&#39;s constant and the letter q represents the electron charge. The Greek letter η in Equation 8 represents the exponent of T in the saturation current of transistor Q 1 . The expression η is referred to as XTI in the SPICE™ circuit simulation program and has a value of approximately four (4) for diffused silicon junctions. 
     We use the expression for V BE Q     1    of Equation 8 in Equation 5 (reproduced below): 
     
       
         
           
             
               
                 
                   
                     V 
                     OUT 
                   
                   = 
                   
                     
                       
                         V 
                         T 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         ln 
                         ⁡ 
                         
                           ( 
                           A 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         ( 
                         
                           
                             R 
                             2 
                           
                           
                             R 
                             1 
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         V 
                         
                           BEQ 
                           1 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     5 
                   
                   ) 
                 
               
             
           
         
       
     
     For convenience, ratio R 2 /R 1  will be represented by the Greek letter α. The letter H also represents the ratio of the thermal voltage V T  at the absolute temperature T to the thermal voltage V T     0    at room temperature T 0 . 
                   H   =       V   T       V     T   0                 (     Eq   .           ⁢   11     )               
Using these expressions, Equation 5 becomes:
   V   OUT   =V   T     0     H  ln( A )+α V   BE Q     1     (Eq. 12) 
     The goal is to find a value for the ratio α and a value for the area A such that the partial derivative of V OUT  with respect to H is zero. 
     
       
         
           
             
               
                 
                   
                     
                       ∂ 
                       
                         V 
                         OUT 
                       
                     
                     
                       ∂ 
                       H 
                     
                   
                   = 
                   0 
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     13 
                   
                   ) 
                 
               
             
           
         
       
     
     For a current I 1  that is proportional to absolute temperature (PTAT), the letter H also represents the ratio of the current I 1  at the absolute temperature T to the current I 0  at room temperature T 0 . 
     
       
         
           
             
               
                 
                   H 
                   = 
                   
                     
                       I 
                       1 
                     
                     
                       I 
                       0 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     14 
                   
                   ) 
                 
               
             
           
         
       
     
     Using Equation 8 and Equation 14 one may express Equation 12 as follows:
 
 V   OUT =α└E GE   −H ( E   GE   −V   BE     0   )+ V   T     0     H  ln( H )−η V   T   H  ln( H )┘+ V   T     0     H  ln  (Eq. 15)
 
     Taking the derivative with respect to H gives: 
     
       
         
           
             
               
                 
                   
                     
                       ∂ 
                       
                         V 
                         OUT 
                       
                     
                     
                       ∂ 
                       H 
                     
                   
                   = 
                   
                     
                       α 
                       ⁡ 
                       
                         [ 
                         
                           
                             - 
                             
                               ( 
                               
                                 
                                   E 
                                   GE 
                                 
                                 - 
                                 
                                   V 
                                   
                                     BE 
                                     0 
                                   
                                 
                               
                               ) 
                             
                           
                           + 
                           
                             
                               
                                 V 
                                 
                                   T 
                                   0 
                                 
                               
                               ⁡ 
                               
                                 ( 
                                 
                                   1 
                                   + 
                                   
                                     ln 
                                     ⁡ 
                                     
                                       ( 
                                       H 
                                       ) 
                                     
                                   
                                 
                                 ) 
                               
                             
                             ⁢ 
                             
                               ( 
                               
                                 
                                   - 
                                   η 
                                 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                         
                         ] 
                       
                     
                     ⁢ 
                     
                       V 
                       
                         T 
                         0 
                       
                     
                     ⁢ 
                     
                       ln 
                       ⁡ 
                       
                         ( 
                         A 
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     16 
                   
                   ) 
                 
               
             
           
         
       
     
     Setting the derivative in Equation 16 equal to zero and evaluating at H=1 gives:
 
α└−( E   GE   −V   BE     0   )− V   T     0   (η−1)┘+ V   T     0    ln( A )=0  (Eq. 17)
 
     This gives an expression for α as follows: 
     
       
         
           
             
               
                 
                   α 
                   = 
                   
                     
                       
                         V 
                         
                           T 
                           0 
                         
                       
                       ⁢ 
                       
                         ln 
                         ⁡ 
                         
                           ( 
                           A 
                           ) 
                         
                       
                     
                     
                       
                         ( 
                         
                           
                             E 
                             GE 
                           
                           - 
                           
                             V 
                             
                               BE 
                               0 
                             
                           
                         
                         ) 
                       
                       + 
                       
                         
                           V 
                           
                             T 
                             0 
                           
                         
                         ⁡ 
                         
                           ( 
                           
                             η 
                             - 
                             1 
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     18 
                   
                   ) 
                 
               
             
           
         
       
     
     This result for α is placed into Equation 12 in order to find the value of V OUT  where H equals one. The value of V OUT  when the value of H equals one will be referred to as the “magic” voltage. When the value of H equals one, then Equation 12 reduces to:
 
 V   OUT   =V   magic   =V   T     0    ln( A )+α V   BE     0     (Eq. 19)
 
     Substituting the value of α from Equation 18 gives: 
     
       
         
           
             
               
                 
                   
                     V 
                     magic 
                   
                   = 
                   
                     
                       
                         V 
                         
                           T 
                           0 
                         
                       
                       ⁢ 
                       
                         ln 
                         ⁡ 
                         
                           ( 
                           A 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         
                           V 
                           
                             BE 
                             0 
                           
                         
                         ⁢ 
                         
                           V 
                           
                             T 
                             0 
                           
                         
                         ⁢ 
                         
                           ln 
                           ⁡ 
                           
                             ( 
                             A 
                             ) 
                           
                         
                       
                       
                         
                           ( 
                           
                             
                               E 
                               GE 
                             
                             - 
                             
                               V 
                               
                                 BE 
                                 0 
                               
                             
                           
                           ) 
                         
                         + 
                         
                           
                             V 
                             
                               T 
                               0 
                             
                           
                           ⁡ 
                           
                             ( 
                             
                               η 
                               - 
                               1 
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     20 
                   
                   ) 
                 
               
             
           
         
       
     
     Factoring out the expression V T     0    ln(A) and rewriting the result gives: 
     
       
         
           
             
               
                 
                   
                     V 
                     magic 
                   
                   = 
                   
                     
                       V 
                       
                         T 
                         0 
                       
                     
                     ⁢ 
                     
                       ln 
                       ⁡ 
                       
                         ( 
                         A 
                         ) 
                       
                     
                     ⁢ 
                     
                       ( 
                       
                         
                           
                             E 
                             GE 
                           
                           + 
                           
                             
                               V 
                               
                                 T 
                                 0 
                               
                             
                             ⁡ 
                             
                               ( 
                               
                                 η 
                                 - 
                                 1 
                               
                               ) 
                             
                           
                         
                         
                           
                             ( 
                             
                               
                                 E 
                                 GE 
                               
                               - 
                               
                                 V 
                                 
                                   BE 
                                   0 
                                 
                               
                             
                             ) 
                           
                           + 
                           
                             
                               V 
                               
                                 T 
                                 0 
                               
                             
                             ⁡ 
                             
                               ( 
                               
                                 η 
                                 - 
                                 1 
                               
                               ) 
                             
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     21 
                   
                   ) 
                 
               
             
           
         
       
     
     For a constant value of current I 1  the expression (η−1) may be replaced with the expression η. For resistor R 1  and resistor R 2  where the thermal conductivity (TC) is non-zero, the expression (η−1) may be replaced by the expression (η−1+σ) where the Greek letter σ is equal to the thermal conductivity (expressed as a reciprocal of degrees Celsius) times the room temperature T 0  (expressed in degrees Celsius).
 
σ=( TC )( T   0 )  (Eq. 22)
 
     The selection of the design parameters using the analysis set forth above proceeds as follows. First, the value of resistance of resistor R 1  is set to be approximately equal to the base-emitter voltage V BE Q1  of transistor Q 1  divided by the current I 1 . 
     
       
         
           
             
               
                 
                   
                     R 
                     1 
                   
                   ≅ 
                   
                     
                       V 
                       
                         BE 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         Q1 
                       
                     
                     
                       I 
                       1 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     23 
                   
                   ) 
                 
               
             
           
         
       
     
     Then Equation 21 is used to find the area A from the desired value of output voltage V OUT . Alternatively, Equation 21 can be used to find the value of output voltage V OUT  from the desired value of area A. 
     Then Equation 18 is used to find the value of α. Then the value of resistance of resistor R 2  is determined from:
 
R 2 =αR 1   (Eq. 24)
 
     Then the value of resistance of resistor R 3  is determined from Equation 6: 
     
       
         
           
             
               
                 
                   
                     R 
                     3 
                   
                   ≅ 
                   
                     
                       V 
                       OUT 
                     
                     
                       2 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         I 
                         1 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     25 
                   
                   ) 
                 
               
             
           
         
       
     
     To illustrate the process of finding the design parameters as set forth above consider the following numerical example. Assume that the following values have been determined: 
     E GE =1.17 volt 
     V BE     0   =0.65 volt 
     I 1 =10.0 microamperes (μA) 
     A=10.0 square units of area 
     ρ=2 
     V T     0   =26 millivolts 
     The value of resistance of resistor R 1  is determined by Equation 23 as follows: 
     
       
         
           
             
               
                 
                   
                     R 
                     1 
                   
                   = 
                   
                     
                       
                         0.65 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         volt 
                       
                       
                         10 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         μ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         amps 
                       
                     
                     = 
                     
                       65 
                       ⁢ 
                       k 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       Ω 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     26 
                   
                   ) 
                 
               
             
           
         
       
     
     Then the given values are used in Equation 21 to determine the V magic  value for the output voltage V OUT .
 
V magic =V OUT =0.131 volt  (Eq. 27)
 
     Equation 18 gives the following value for α:
 
α=0.1099  (Eq. 28)
 
     Then Equation 24 gives:
 
 R   2   =αR   1 =(0.1099)(65 kΩ)=7.14 kΩ  (Eq. 29)
 
     Then Equation 25 gives: 
     
       
         
           
             
               
                 
                   
                     
                       R 
                       3 
                     
                     ⁢ 
                     
                       
                         V 
                         OUT 
                       
                       
                         2 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           I 
                           1 
                         
                       
                     
                   
                   = 
                   
                     
                       
                         ( 
                         
                           0.131 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           volt 
                         
                         ) 
                       
                       
                         2 
                         ⁢ 
                         
                           ( 
                           
                             10.0 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             μ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             amps 
                           
                           ) 
                         
                       
                     
                     = 
                     
                       6.55 
                       ⁢ 
                       k 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       Ω 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     30 
                   
                   ) 
                 
               
             
           
         
       
     
     Table One below illustrates the variation of the value of output voltage V magic  as a function of the area A of transistor Q 2 . 
                                                 TABLE ONE                           Area A in   3.0   4.0   5.0   10.0   20.0           square           units           V magic  in   62.5   78.9   91.6   131.0   171.0           millivolts                        
The residual curvature in the output voltage V OUT  is given by the equation:
   V   CURVE   =V   OUT   −V   magic   (Eq. 31) 
     Equation 31 can also be expressed as:
 
 V   CURVE   =V   T     0   α(η−1)[( H− 1)− H  ln( H )]  (Eq. 32)
 
     This expression for V CURVE  is similar to that for a prior art bandgap reference circuit except that the value of V CURVE  is reduced by the factor of α. The percent of curvature to output voltage V magic  is the same as the prior art. 
     Increasing the value of V OUT  by increasing the ratio α will cause a negative temperature coefficient and vice versa. This result is opposite to that obtained from a prior art bandgap reference circuit. In a prior art bandgap reference circuit, the PTAT (Proportional to Absolute Temperature) voltage is scaled. In the bandgap reference circuit of the present invention, the base-emitter voltage (V BE ) is scaled. If one adds more PTAT voltage to the value of V OUT  (by increasing the ratio α) then one obtains a higher value of V OUT  and a positive temperature coefficient. If one adds more base-emitter voltage (V BE ) to the value of V OUT , then one obtains a higher value of V OUT  and a negative temperature coefficient. 
       FIG. 2  illustrates a schematic representation of a second embodiment of a low voltage bandgap reference circuit  200  constructed in accordance with the principles of the present invention. The input voltage V IN  is connected to a first current source  210  that produces a current having a value of I 1  and to a second current source  220  that also produces a current having a value of I 1  and to a third current source  230  that produces a current having a value of I 2 . The input voltage V IN  is also connected to the collector of bipolar junction transistor Q 3  and to the collector of bipolar junction transistor Q 4 . 
     The output of first current source  210  is connected to the collector of bipolar junction transistor Q 1 . The output of first current source  210  is also connected to the base of bipolar junction transistor Q 4 . The emitter of bipolar junction transistor Q 4  is connected to the output voltage terminal V OUT . 
     The output of second current source  220  is connected to the collector of bipolar junction transistor Q 2 . The output of second current source  220  is also connected to the base of bipolar junction transistor Q 3 . The emitter of bipolar junction transistor Q 3  is connected to a fourth current source  240  that produces a current having a value of I 3 . The output of fourth current source  240  is connected to ground. 
     The base of bipolar junction transistor Q 2  is connected through resistor R 2  to the base of bipolar junction transistor Q 1 . The output of third current source  230  is connected to the base of bipolar junction transistor Q 2 . 
     The emitter of bipolar junction transistor Q 1  is connected to ground. A first end of resistor R 1  is connected to the base of bipolar junction transistor Q 1  and a second end of resistor R 1  is connected to ground. 
     The emitter of bipolar junction transistor Q 2  is connected to the voltage output terminal V OUT . The emitter of bipolar junction transistor Q 2  is also connected through resistor R 3  to ground. 
     The emitter of bipolar junction transistor Q 5  is connected to the base of bipolar junction transistor Q 2 . The collector of bipolar junction transistor Q 5  is connected to ground. The base of bipolar junction transistor Q 5  is connected to a node between the emitter of bipolar junction transistor Q 3  and the fourth current source  240 . 
     The area of transistor Q 1  is equal to a unit value of area. That is, the transistor Q 1  has a value of area equal to one square unit (designated “ 1 x” in  FIG. 2 ). The area of transistor Q 2  is equal to “A” times the area of transistor Q 1 . That is, transistor Q 2  has a value of area equal to A square units of area (designated “Ax” in  FIG. 2 ). 
     The second embodiment of the invention in the low power bandgap reference circuit  200  replaces the “diode” equivalent around the transistor Q 2  of bandgap reference circuit  100  with a “folded buffer” arrangement that comprises transistor Q 3  and transistor Q 5 . This puts a value of voltage that is equal to (V BE +V OUT ) on the collector of transistor Q 1  and on the collector of transistor Q 2 . 
     Therefore, the minimum input voltage V IN  in bandgap reference circuit  200  is less than the minimum input voltage V IN  in bandgap reference circuit  100 .
 
 V   IN (min)= V   BE   +V   SAT   +V   OUT   (Eq. 33)
 
     The term V BE  represents a value of base to emitter voltage of said first bipolar junction transistor Q 1 . The term V SAT  represents a minimum voltage required for the four current sources ( 210 ,  220 ,  230 ,  240 ). The term V OUT  represents the output voltage. 
     Equation 7 gives the minimum input voltage V IN  for the bandgap reference circuit  100 .
 
 V   IN (min)=2 V   BE   +V   SAT   +V   OUT   (Eq. 7)
 
     In Equation 33 the output voltage V OUT  can be as low as approximately one hundred millivolts (100 mV). A low value of V OUT  in Equation 33 provides headroom for the fourth current source  240  that provides the 13 current. 
     The third current source  230  provides the I 2  current for resistor R 1  and transistor Q 5 . In one advantageous embodiment the value of the I 2  current is given by: 
     
       
         
           
             
               
                 
                   
                     I 
                     2 
                   
                   = 
                   
                     
                       
                         V 
                         
                           BE 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           Q1 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           MAX 
                         
                       
                       
                         R 
                         
                           1 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           MIN 
                         
                       
                     
                     + 
                     
                       I 
                       1 
                     
                   
                 
               
               
                 
                   ( 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     34 
                   
                   ) 
                 
               
             
           
         
       
     
     This value of current for I 2  provides transistor Q 5  with a current that has a value of current that is equal to I 1 . It is noted that compensation capacitors may be required in low voltage bandgap reference circuit  200 . 
     The low voltage bandgap reference circuits of the present invention ( 100  and  200 ) have several advantages over prior art bandgap reference circuits. First, no start-up circuitry is required. Second, the error amplification function is carried out by NPN bipolar junction transistors. Third, the bandgap reference circuits of the present invention require fewer transistors than prior art circuits. Fourth, the bandgap reference circuits of the present invention require fewer resistors than prior art circuits. 
     The foregoing description has outlined in detail the features and technical advantages of the present invention so that persons who are skilled in the art may understand the advantages of the invention. Persons who are skilled in the art should appreciate that they may readily use the conception and the specific embodiment of the invention that is disclosed as a basis for modifying or designing other structures for carrying out the same purposes of the present invention. Persons who are skilled in the art should also realize that such equivalent constructions do not depart from the spirit and scope of the invention in its broadest form. 
     Although the present invention has been described with an exemplary embodiment, various changes and modifications may be suggested to one skilled in the art. It is intended that the present invention encompass such changes and modifications as fall within the scope of the appended claims.