Abstract:
A driver device may include a driver transistor providing a regulated current; and a stabilization circuit to produce a stabilized reference voltage to be applied to said driver transistor. The stabilization circuit may include: first and second bipolar stabilization transistors; a voltage divider including a first resistance and a second resistance, said voltage divider being interposed between the bases of said first and said second transistors, with the first resistor connected between the base of said second transistor and said partition point of said voltage divider and the partition point acting on the base of the first transistor; and a polarization network to determine the base-emitter voltages of said first and said second stabilization transistors, wherein: said first resistance has a value lower than the value of said second resistance, and the base-emitter voltage of said first transistor is higher than the base-emitter voltage of said second transistor.

Description:
CROSS-REFERENCE TO RELATED APPLICATION 
     This application claims priority to Italian Patent Application Serial No. TO2009A000358, which was filed May 4, 2009, and is incorporated herein by reference in its entirety. 
     TECHNICAL FIELD 
     This disclosure relates to current regulation devices. 
     This disclosure was devised with specific attention paid to its possible application to the thermal stabilization of linear devices that regulate the driving current of light sources such as, for example, light emitting diodes (LEDs). 
     BACKGROUND 
     In backlighting applications, discharge lamps (or fluorescent lamps) have recently been increasingly substituted with LED lighting modules. Such lighting modules may comprise several printed circuit boards (PCBs), each of them with one or more LEDs, the PCBs being connected through wires or cables in order to be flexible and fit into channel letter light signs, that normally require custom light sources, or adaptable systems. The high number of units (or modules) connected in parallel requires a low cost driving solution. 
     The linear current regulation device according to the diagram of  FIG. 1  is an example of a solution presently used for the purpose. 
     In this example, that refers to the driving of two LEDs connected in series and respectively indicated by LED 1  and LED 2 , the device includes two bipolar transistors (BJT), in this case of a pnp type, indicated by the reference numbers T 1  and T 2 , and two resistances R 1  and R 2 . 
     Referring to  FIG. 1 , the positive terminal of a direct voltage input generator VDC 1  is connected to the emitter E 1  of the first transistor T 1  (node A). The resistance R 1  is connected between the emitter E 1  and the base B 1  of the first transistor T 1 . The emitter E 2  of the second transistor T 2  is connected (node C) to the base B 1 . The collector C 1  of the first transistor T 1  is connected, via the node D, to the base B 2  of the second transistor T 2 . The resistance R 2  is connected between the node D and the node B, i.e. the negative terminal of the voltage generator VDC 1 . The collector C 2  of the second transistor T 2  feeds the series of both LED 1  and LED 2 . Finally, the cathode of the second diode LED 2  is connected to the above mentioned node B on the second supply line  20 . 
     In operation, the high-impedance resistance R 2  polarizes the first transistor T 1  with a very low collector current Ic (μA), but the base-emitter voltage V BE1  of the first transistor T 1  is set to the value V BEon . 
     The current I LED  that flows through and drives the LEDs is the same as the collector current I C2  of the transistor T 2  (which approximately equals the emitter current I E2  of the same transistor) and therefore it is dependent, through the resistance R 1 , on the value of the voltage dropping between the emitter and the base (reference voltage V BE1 ) of the transistor T 1 , according to the following relation: 
     
       
         
           
             
               I 
               LED 
             
             = 
             
               
                 
                   I 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     2 
                   
                 
                 ≈ 
                 
                   I 
                   
                     E 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     2 
                   
                 
               
               = 
               
                 
                   V 
                   
                     BE 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     1 
                   
                 
                 
                   R 
                   1 
                 
               
             
           
         
       
     
     In the driver device of  FIG. 1 , the transistor T 2  is therefore used as a driver circuit of the LEDs, while the transistor T 1  has a stabilizing function. 
     The collector current I C2  of the driver transistor T 2  drives the light sources LED 1  and LED  2  and the stabilization circuit T 1 , R 1 , R 2  generates a reference voltage V BE1 —stable with reference to the input voltage VDC 1 —which, applied to the transistor T 2 , makes the current I LED  flowing through the LEDs equally stable with reference to the input voltage VDC 1 . The consequence is a stabilization of the current I LED  with reference to the input voltage VDC 1 . 
     Nevertheless, there remains a dependence of the current I LED  from the usage temperature T (because of the thermal drifts of V BE ). 
     A possible range of the operating temperatures of LED modules goes from −30° C. to +80° C., because of the different ambient conditions where they can be employed. 
     The base-emitter junction voltage of the transistor T 1  varies with the temperature T on the basis of a coefficient k (mV/° C.). 
     Considering the reference standard ambient temperature of 25° C., the current I LED  that flows in the LEDs varies on the basis of the following relation: 
     
       
         
           
             
               
                 
                   
                     
                       I 
                       LED 
                     
                     ⁡ 
                     
                       ( 
                       T 
                       ) 
                     
                   
                   = 
                     
                   ⁢ 
                   
                     
                       
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                         
                         ⁡ 
                         
                           ( 
                           T 
                           ) 
                         
                       
                       
                         R 
                         1 
                       
                     
                     = 
                     
                       
                         
                           
                             
                               V 
                               
                                 BE 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 1 
                               
                             
                             ⁡ 
                             
                               ( 
                               T0 
                               ) 
                             
                           
                           + 
                           
                             
                               k 
                               · 
                               Δ 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             T 
                           
                         
                         
                           R 
                           1 
                         
                       
                       = 
                       
                         
                           
                             
                               V 
                               
                                 BE 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 1 
                               
                             
                             ⁡ 
                             
                               ( 
                               T0 
                               ) 
                             
                           
                           + 
                           
                             k 
                             ⁡ 
                             
                               ( 
                               
                                 T 
                                 - 
                                 
                                   T 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   0 
                                 
                               
                               ) 
                             
                           
                         
                         
                           R 
                           1 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                   ⁢ 
                   
                     
                       
                         
                           
                             V 
                             
                               BE 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                             
                           
                           ⁡ 
                           
                             ( 
                             
                               T 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               0 
                             
                             ) 
                           
                         
                         
                           R 
                           1 
                         
                       
                       + 
                       
                         
                           k 
                           ⁡ 
                           
                             ( 
                             
                               T 
                               - 
                               
                                 T 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 0 
                               
                             
                             ) 
                           
                         
                         
                           R 
                           1 
                         
                       
                     
                     = 
                     
                       
                         
                           I 
                           LED 
                         
                         ⁡ 
                         
                           ( 
                           
                             T 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             0 
                           
                           ) 
                         
                       
                       + 
                       
                         Δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           
                             I 
                             LED 
                           
                           ⁡ 
                           
                             ( 
                             T 
                             ) 
                           
                         
                       
                     
                   
                 
               
             
           
         
       
     
     The value of the current I LED (T) at a given temperature T is therefore given by the value of the current I LED (T 0 ) flowing in the LEDs at the reference temperature of 25° C. with a variation ΔI LED (T) (positive or negative) which depends on the temperature T. 
     We can define: 
     
       
         
           
             
               H 
               ⁢ 
               
                   
               
               ⁢ 
               1 
             
             = 
             
               
                 
                   ⅆ 
                   
                     ⅆ 
                     T 
                   
                 
                 ⁢ 
                 
                   ( 
                   
                     
                       
                         I 
                         LED 
                       
                       ⁡ 
                       
                         ( 
                         T 
                         ) 
                       
                     
                     
                       
                         I 
                         LED 
                       
                       ⁡ 
                       
                         ( 
                         
                           T 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           0 
                         
                         ) 
                       
                     
                   
                   ) 
                 
               
               = 
               
                 k 
                 
                   V 
                   
                     BE 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     1 
                   
                 
               
             
           
         
       
     
     A typical value for the coefficient k for a p-n junction is −2 mV/° C. 
     Let us assume a numerical example where: 
     I LED (25° C.)=30 mA, 
     V BE1 (25° C.)=0.6V, 
     R 1 =20Ω. 
     In this case, the current at a low temperature of −30° C. increases up to the value of: 
                       I   LED     ⁡     (       -   30     ⁢   °   ⁢           ⁢   C     )       =           V     BE   ⁢           ⁢   1       ⁡     (     25   ⁢   °   ⁢           ⁢   C     )         R   1       +       k   ⁡     (       -   30     -   25     )         R   1                     =       30   ⁢           ⁢   mA     +           -   0.002     ⁢     (       -   30     -   25     )       20     ⁢   mA                   =     35.5   ⁢           ⁢   mA                 
while at a high temperature of +80° C. the current decreases to a value of:
 
     
       
         
           
             
               
                 
                   
                     
                       I 
                       LED 
                     
                     ⁡ 
                     
                       ( 
                       
                         80 
                         ⁢ 
                         ° 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         C 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                         
                         ⁡ 
                         
                           ( 
                           
                             25 
                             ⁢ 
                             ° 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             C 
                           
                           ) 
                         
                       
                       
                         R 
                         1 
                       
                     
                     + 
                     
                       
                         k 
                         ⁡ 
                         
                           ( 
                           
                             80 
                             - 
                             25 
                           
                           ) 
                         
                       
                       
                         R 
                         1 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     
                       30 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       mA 
                     
                     + 
                     
                       
                         
                           
                             - 
                             0.002 
                           
                           ⁢ 
                           
                             ( 
                             
                               80 
                               - 
                               25 
                             
                             ) 
                           
                         
                         20 
                       
                       ⁢ 
                       mA 
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     24.5 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     mA 
                   
                 
               
             
           
         
       
     
     In conclusion, it can be stated that the current at a temperature of −30° C. amounts approximately to 118% of the current at ambient temperature (I LED (−30° C.)=118% I LED (25° C.)), while at a temperature of +80° C. it amounts approximately to 82% of the current at ambient temperature (I LED (80° C.)=82% I LED (25° C.)). 
     Consequently, the LED module input power could have a variation of ±20% from the rated input power at a reference ambient temperature of 25° C. 
     SUMMARY 
     In various embodiments, a driver device to produce a regulated current from an input voltage is provided. The driver device may include a driver transistor for providing said regulated current; and a stabilization circuit to produce, from said input voltage, a stabilized reference voltage to be applied to said driver transistor; wherein said stabilization circuit may include: a first bipolar stabilization transistor and a second bipolar stabilization transistor, the base-emitter voltage of said second stabilization transistor determining said stabilized reference voltage for said driver transistor; a voltage divider including a first resistance and a second resistance, said voltage divider being interposed between the bases of said first and said second stabilization transistors, with the first resistor connected between the base of said second stabilization transistor and said partition point of said voltage divider and the partition point acting on the base of the first stabilization transistor; and a polarization network to determine the base-emitter voltages of said first and said second stabilization transistors, wherein: said first resistance of said voltage divider has a value lower than the value of said second resistance of said voltage divider, and the base-emitter voltage of said first stabilization transistor is higher than the base-emitter voltage of said second stabilization transistor. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
       In the drawings, like reference characters generally refer to the same parts throughout the different views. The drawings are not necessarily to scale, emphasis instead generally being placed upon illustrating the principles of the invention. In the following description, various embodiments of the invention are described with reference to the following drawings, in which: 
         FIG. 1  has been described in the foregoing; and 
         FIG. 2  shows a circuit diagram showing an embodiment of the solution described herein. 
     
    
    
     DETAILED DESCRIPTION 
     The following detailed description refers to the accompanying drawings that show, by way of illustration, specific details and embodiments in which the invention may be practiced. 
     In the following description, numerous specific details are given to provide a thorough understanding of embodiments. The embodiments can be practiced without one or more of the specific details, or with other methods, components, materials, etc. In other instances, well-known structures, materials, or operations are not shown or described in detail to avoid obscuring aspects of the embodiments. 
     Reference throughout this specification to “one embodiment” or “an embodiment” means that a particular feature, structure, or characteristic described in connection with the embodiment is included in at least one embodiment. Thus, the appearances of the phrases “in one embodiment” or “in an embodiment” in various places throughout this specification are not necessarily all referring to the same embodiment. Furthermore, the particular features, structures, or characteristics may be combined in any suitable manner in one or more embodiments. 
     The word “exemplary” is used herein to mean “serving as an example, instance, or illustration”. Any embodiment or design described herein as “exemplary” is not necessarily to be construed as preferred or advantageous over other embodiments or designs. 
     In view of the foregoing the need is felt for drivers for light sources, for example LEDs (Light Emitting Diodes), capable of performing a linear regulation of the current, while allowing to obtain a thermal stabilization of the current itself. 
     Various embodiments provide such a device. 
     The claims are an integral part of the teachings of the invention provided herein. 
     The headings provided herein are for convenience only and do not interpret the scope or meaning of the embodiments. 
     In  FIG. 2  the parts, elements or components identical or equivalent to parts, elements or components already described with reference to  FIG. 1  are indicated with the same reference numbers, which makes it unnecessary to repeat the related description. 
     Referring to  FIG. 2 , a linear current regulation driver is achieved using three bipolar transistors (BJT), respectively indicated with the references T 1   a , T 1   b  and T 2 , here of the npn type, and five resistances (resistors) R 1 , R 2 , R 3 , R 4  and RS. 
     In an embodiment, at least the transistor T 2  can be a field effect transistor, such as a MOSFET. As a consequence, the terms “emitter”, “base” and “collector”, as used herein and referring to bipolar technology, are to be construed as throughout applicable (including the claims) to the terms “source”, “gate” and “drain”, which denote the corresponding elements of a FET. 
     As in the example shown in  FIG. 1 , in this case too the circuit is supposed to drive a load made up of a light source, comprising two diodes LED connected in series, indicated with LED 1  and LED 2 . 
     Always referring to  FIG. 1 , to the positive terminal of the generator VDC 1  a first input line  30  is connected, including the resistances R 1  and R 2 , that serve primarily as polarization resistances of the transistors T 1   a  and T 1   b , particularly (as will be detailed in the following) as regards the setting of the values of the respective base-emitter voltages. 
     Specifically, the resistance R 1  is interposed between the supply line  30  (node L) and the collector C 1   b  of the transistor T 1   b , while the resistance R 2  is interposed between the supply line  30  (node M) and the collector C 1   a  of the transistor T 1   a.    
     The transistor T 1   b  has the base B 1   b  and the collector C 1   b  mutually connected (node O, that coincides with node P), so as to function essentially as a diode. 
     The resistances R 3  and R 4  form a voltage divider between the base B 1   b  of the transistor T 1   b  and the base B 1   a  of the transistor T 1   a.    
     Specifically, the resistance R 3  is interposed (as a “high” branch of the divider) between the bases B 1   a  and B 1   b  of the two transistors T 1   a  and T 1   b , while the resistance R 4  is interposed (as a “low” branch of the divider) between a node Q, arranged between the base B 1   a  of the transistor T 1   a  and the resistance R 3 , and a second supply line  40  leading to the negative terminal of the voltage generator VDC 1  (node S). 
     The emitter E 1   a  of the transistor T 1   a  is connected to the second line  40  as well, via the node V. 
     As already stated, the resistance R 2  is connected between the first supply line  30  (node M) and the node N, to which in turn the collector C 1   a  of the transistor T 1   a  and the base B 2  of the transistor T 2  are connected. The emitter E 1   b  of the transistor T 1   b  and the emitter E 2  are mutually connected in a node R, to which also the resistance RS is connected, obtaining a function of amperometric resistor sensing the current flowing through the load, i.e. through the diodes LED 1  and LED 2  (for example to perform regulation functions that are not specifically considered herein). The resistance RS is connected between said node S and node R. 
     The two diodes LED 1  and LED 2  are connected in a series configuration and the anode of LED 1  is connected to node M, while the cathode of LED 2  is connected to collector C 2  of the transistor T 2 ; therefore, through the two LEDs a current flows that equals the collector current of the transistor T 2 . 
     In the regulator presently considered, the function of thermal stabilization makes use of the fact that the base-emitter voltage drop V BE  of a bipolar transistor varies with the temperature T during use. 
     In the embodiment of  FIG. 2 , the resistance R 2  sets the basic current for the transistor T 2  (I B2  is set on the basis of β F2  and I LED ). Assuming, as it is correct to suppose, that the currents I T1  and I R4 , that flow in the emitter of T 1   a  and in the resistance R 4 , are negligible, the current in the load, i.e. in the LEDs, indicated as I LED , is given by: 
     
       
         
           
             
               I 
               LED 
             
             = 
             
               
                 
                   V 
                   
                     DC 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     1 
                   
                 
                 - 
                 
                   V 
                   LED 
                 
                 - 
                 
                   V 
                   
                     CE 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     2 
                   
                 
               
               
                 R 
                 S 
               
             
           
         
       
     
     The value of the resistance R 1  is chosen as to supply the transistor T 1   b  with such a current as to establish, between base and emitter, a voltage V BE1bON , which is the stable reference voltage of the circuit applied to the driver transistor T 2 . 
     When the transistor T 1   b  is on, through the voltage divider made up by the resistances R 3  and R 4 , the voltage V BE1aON  between the base and the emitter of transistor T 1   a  reaches the threshold voltage V BE1a  and therefore T 1   a  is switched on as well. 
     Once the transistor T 1   a  is on, the current flowing through Rs, and which is equal to the current I LED , is determined by the mesh stabilization given by R 3 , R 4 , T 1bon  and Rs. The voltage T 1bon  added to the voltage VRs, through the voltage divider made up by R 3  and R 4 , must correspond to the voltage T 1aon  (which is the same as VR 4 ). As both T ion s are set through R 1  and R 2  and are stable, the voltage VRs stabilizes and, on the basis of the value of Rs, it is possible to set the needed current I LED . 
     The stability condition is given by: 
     
       
         
           
             
               
                 
                   
                     V 
                     
                       BE 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       1 
                       ⁢ 
                       a 
                     
                   
                   = 
                   
                     
                       ( 
                       
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             bON 
                           
                         
                         + 
                         
                           V 
                           RS 
                         
                       
                       ) 
                     
                     · 
                     
                       ( 
                       
                         
                           R 
                           4 
                         
                         
                           
                             R 
                             3 
                           
                           + 
                           
                             R 
                             4 
                           
                         
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     
                       ( 
                       
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             bON 
                           
                         
                         + 
                         
                           
                             I 
                             LED 
                           
                           · 
                           
                             R 
                             S 
                           
                         
                       
                       ) 
                     
                     · 
                     
                       ( 
                       
                         
                           R 
                           4 
                         
                         
                           
                             R 
                             3 
                           
                           + 
                           
                             R 
                             4 
                           
                         
                       
                       ) 
                     
                   
                 
               
             
           
         
       
     
     The electrical and thermal behaviour of the circuit can be improved making use of the fact that bipolar transistors such as the transistors T 1   a  and T 1   b  have the same thermal behaviour, e.g. as regards their base-emitter junction, so that their thermal drifts are compensated. 
     In an embodiment, this result is achieved by choosing the transistors T 1   a  and T 1   b  mutually identical. In an embodiment, T 1   a  and T 1   b  are on the same integrated component (and so both transistors are subjected to the same temperature). In an embodiment T 1   a  and T 1   b  are a pair of integrated monolithic components. 
     When the circuit is set as to obtain V BE1a =V BE1aON , the current of the LEDs is given by (assuming that I RS ≈I LED ): 
     
       
         
           
             
               
                 I 
                 LED 
               
               ⁡ 
               
                 ( 
                 
                   T 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   0 
                 
                 ) 
               
             
             = 
             
               
                 
                   
                     V 
                     
                       BE 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       1 
                       ⁢ 
                       aON 
                     
                   
                   
                     R 
                     S 
                   
                 
                 · 
                 
                   ( 
                   
                     
                       
                         R 
                         3 
                       
                       + 
                       
                         R 
                         4 
                       
                     
                     
                       R 
                       4 
                     
                   
                   ) 
                 
               
               - 
               
                 
                   V 
                   
                     BE 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     1 
                     ⁢ 
                     bON 
                   
                 
                 
                   R 
                   S 
                 
               
             
           
         
       
     
     Consequently, Rs is given by: 
     
       
         
           
             
               R 
               S 
             
             = 
             
               
                 
                   
                     V 
                     
                       BE 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       1 
                       ⁢ 
                       aON 
                     
                   
                   
                     
                       I 
                       LED 
                     
                     ⁡ 
                     
                       ( 
                       
                         T 
                         0 
                       
                       ) 
                     
                   
                 
                 · 
                 
                   ( 
                   
                     
                       
                         R 
                         3 
                       
                       + 
                       
                         R 
                         4 
                       
                     
                     
                       R 
                       4 
                     
                   
                   ) 
                 
               
               - 
               
                 
                   V 
                   
                     BE 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     1 
                     ⁢ 
                     bON 
                   
                 
                 
                   
                     I 
                     LED 
                   
                   ⁡ 
                   
                     ( 
                     
                       T 
                       0 
                     
                     ) 
                   
                 
               
             
           
         
       
     
     where I LED (T 0 ) is the target current for the LEDs, i.e. the desired current. 
     Once V BE1bON  is set and I LED  is fixed, the values of R 3 , R 4  (and RS) may be chosen so as to minimize the current variation due to the thermal behaviour of the p-n junctions of T 1   a  and T 1   b . 
     
       
         
           
             
               Δ 
               ⁢ 
               
                   
               
               ⁢ 
               
                 I 
                 LED 
               
             
             = 
             
               
                 
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       V 
                       
                         BE 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         aON 
                       
                     
                   
                   
                     R 
                     S 
                   
                 
                 · 
                 
                   ( 
                   
                     
                       
                         R 
                         3 
                       
                       + 
                       
                         R 
                         4 
                       
                     
                     
                       R 
                       4 
                     
                   
                   ) 
                 
               
               - 
               
                 
                   Δ 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     V 
                     
                       BE 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       1 
                       ⁢ 
                       bON 
                     
                   
                 
                 
                   R 
                   S 
                 
               
             
           
         
       
     
     In this case they have the same thermal coefficient, therefore:
 
Δ V   BE1a   =ΔV   BE1b   =k ·( T−T   0 )
 
     where typically k is −0.002 mV/K. 
     Thus: 
     
       
         
           
             
               
                 
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       I 
                       LED 
                     
                   
                   = 
                   
                     
                       
                         Δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             aON 
                           
                         
                       
                       
                         R 
                         S 
                       
                     
                     · 
                     
                       ( 
                       
                         
                           
                             
                               R 
                               3 
                             
                             + 
                             
                               R 
                               4 
                             
                           
                           
                             R 
                             4 
                           
                         
                         - 
                         1 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         k 
                         · 
                         
                           ( 
                           
                             T 
                             - 
                             
                               T 
                               0 
                             
                           
                           ) 
                         
                       
                       
                         R 
                         S 
                       
                     
                     · 
                     
                       ( 
                       
                         
                           
                             
                               R 
                               3 
                             
                             + 
                             
                               R 
                               4 
                             
                           
                           
                             R 
                             4 
                           
                         
                         - 
                         1 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         k 
                         · 
                         
                           ( 
                           
                             T 
                             - 
                             
                               T 
                               0 
                             
                           
                           ) 
                         
                       
                       
                         R 
                         S 
                       
                     
                     · 
                     
                       
                         R 
                         3 
                       
                       
                         R 
                         4 
                       
                     
                   
                 
               
             
           
         
       
     
     and therefore: 
     
       
         
           
             
               
                 
                   
                     
                       I 
                       LED 
                     
                     ⁡ 
                     
                       ( 
                       T 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         I 
                         LED 
                       
                       ⁡ 
                       
                         ( 
                         
                           T 
                           0 
                         
                         ) 
                       
                     
                     + 
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         
                           I 
                           LED 
                         
                         ⁡ 
                         
                           ( 
                           T 
                           ) 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             aON 
                           
                         
                         
                           R 
                           S 
                         
                       
                       · 
                       
                         ( 
                         
                           
                             
                               R 
                               3 
                             
                             + 
                             
                               R 
                               4 
                             
                           
                           
                             R 
                             4 
                           
                         
                         ) 
                       
                     
                     - 
                     
                       
                         V 
                         
                           BE 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           1 
                           ⁢ 
                           bON 
                         
                       
                       
                         R 
                         S 
                       
                     
                     + 
                     
                       
                         
                           k 
                           · 
                           
                             ( 
                             
                               T 
                               - 
                               
                                 T 
                                 0 
                               
                             
                             ) 
                           
                         
                         
                           R 
                           S 
                         
                       
                       · 
                       
                         
                           R 
                           3 
                         
                         
                           R 
                           4 
                         
                       
                     
                   
                 
               
             
           
         
       
     
     It is therefore possible to reduce the current variation generated by a temperature variation by choosing the suitable values for the resistances R 3 , R 4  (and RS), and 
     
       
         
           
             
               
                 
                   I 
                   LED 
                 
                 ⁡ 
                 
                   ( 
                   T 
                   ) 
                 
               
               
                 
                   I 
                   LED 
                 
                 ⁡ 
                 
                   ( 
                   
                     T 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     0 
                   
                   ) 
                 
               
             
             = 
             
               1 
               + 
               
                 
                   k 
                   · 
                   
                     R 
                     3 
                   
                   · 
                   
                     ( 
                     
                       T 
                       - 
                       
                         T 
                         0 
                       
                     
                     ) 
                   
                 
                 
                   
                     
                       V 
                       
                         BE 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         aON 
                       
                     
                     · 
                     
                       ( 
                       
                         
                           R 
                           3 
                         
                         + 
                         
                           R 
                           4 
                         
                       
                       ) 
                     
                   
                   - 
                   
                     
                       V 
                       
                         BE 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         bON 
                       
                     
                     · 
                     
                       R 
                       4 
                     
                   
                 
               
             
           
         
       
     
     Moreover, independently from temperature T, it is possible to define: 
     
       
         
           
             
               H 
               ⁢ 
               
                   
               
               ⁢ 
               2 
             
             = 
             
               
                 
                   ⅆ 
                   
                     ⅆ 
                     T 
                   
                 
                 ⁢ 
                 
                   ( 
                   
                     
                       
                         I 
                         LED 
                       
                       ⁡ 
                       
                         ( 
                         T 
                         ) 
                       
                     
                     
                       
                         I 
                         LED 
                       
                       ⁡ 
                       
                         ( 
                         
                           T 
                           0 
                         
                         ) 
                       
                     
                   
                   ) 
                 
               
               = 
               
                 
                   k 
                   · 
                   
                     R 
                     3 
                   
                 
                 
                   
                     
                       V 
                       
                         BE 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         aON 
                       
                     
                     · 
                     
                       ( 
                       
                         
                           R 
                           3 
                         
                         + 
                         
                           R 
                           4 
                         
                       
                       ) 
                     
                   
                   - 
                   
                     
                       V 
                       
                         BE 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         bON 
                       
                     
                     · 
                     
                       R 
                       4 
                     
                   
                 
               
             
           
         
       
     
     Depending on the values of V BE1aON , V BE1bON , R 3  and R 4 , the denominator V BE1aON ·(R 3 +R 4 )−V BE1bON ·R 4  modifies the value of H 2  in different ways. 
     In a first case, if V BE1aON ≈N BE1bON , then 
     
       
         
           
             
               H 
               ⁢ 
               
                   
               
               ⁢ 
               2 
             
             = 
             
               
                 
                   
                     k 
                     · 
                     
                       R 
                       3 
                     
                   
                   
                     
                       
                         V 
                         
                           BE 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           1 
                           ⁢ 
                           aON 
                         
                       
                       · 
                       
                         ( 
                         
                           
                             R 
                             3 
                           
                           + 
                           
                             R 
                             4 
                           
                         
                         ) 
                       
                     
                     - 
                     
                       
                         V 
                         
                           BE 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           1 
                           ⁢ 
                           aON 
                         
                       
                       · 
                       
                         R 
                         4 
                       
                     
                   
                 
                 ≈ 
                 
                   k 
                   
                     V 
                     
                       BE 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       1 
                       ⁢ 
                       aON 
                     
                   
                 
               
               = 
               
                 H 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 1 
               
             
           
         
       
     
     and there would be no particular advantages in comparison with the standard solution. 
     In a second case, if V BE1aON &lt;V BE1bON , then particular values of R 3  and R 4  can reduce almost to zero the denominator, and H 2  can be extremely high. 
     For example, referring to an exemplary embodiment tested by the Applicant, 
     the current variation 
               I   ⁡     (   T   )         I   ⁡     (     T   ⁢           ⁢   0     )             
as a function of the voltage values V BE1aON  was obtained with the following values: R 3 =0.5 kΩ; R 4 =10 kΩ and V BE1aON −V BE1bON =−0.03V.
 
     By choosing R 3 =R 4 : 
     
       
         
           
             
               
                 
                   
                     H 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     2 
                   
                   = 
                   
                     
                       k 
                       · 
                       
                         R 
                         3 
                       
                     
                     
                       
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             aON 
                           
                         
                         · 
                         
                           ( 
                           
                             
                               R 
                               3 
                             
                             + 
                             
                               R 
                               4 
                             
                           
                           ) 
                         
                       
                       - 
                       
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             bON 
                           
                         
                         · 
                         
                           R 
                           3 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         k 
                         
                           
                             2 
                             ⁢ 
                             
                               V 
                               
                                 BE 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 1 
                                 ⁢ 
                                 aON 
                               
                             
                           
                           - 
                           
                             V 
                             
                               BE 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                               ⁢ 
                               aON 
                             
                           
                         
                       
                       ≈ 
                       
                         k 
                         
                           V 
                           
                             BE 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             aON 
                           
                         
                       
                     
                     = 
                     
                       H 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       1 
                     
                   
                 
               
             
           
         
       
     
     because V BE1aON −V BE1bON &lt;&lt;V BE1aON . 
     In a third case, if V BEiaON &gt;V BE1bON , the denominator is always positive, and therefore H 2  can be reduced by choosing suitable values for the resistances R 3  and R 4 . 
     The higher the voltage V BE1aON  is in comparison to V BE1bON , the lower H 2  turns out to be (with the suitable resistances R 3  and R 4 ). 
     The condition V BE1aON &gt;V BE1bON  represents the best solution in order to control the current variation I LED  with reference to the temperature variation T, and it is obtained only modifying the values of the resistances R 3  and R 4 . 
     In an embodiment tested by the Applicant, a current variation I LED  (−30° C.) has been detected at a temperature of −30° C., as compared to the current I LED  (25° C.) at ambient temperature in the case V BE1aON  is 0.65 V, R 4  is 10 kΩ 1  and V BE1bON  is 0.62 V. 
     If a value of R 3 &lt;1 kΩ was chosen, it was found that it is possible to keep the current value in a range of ±5% from the current value I LED  (25° C.) at ambient temperature, in the complete range of operating temperatures. 
     Table 1 that follows summarizes, as a function of different values of the quantities considered in the foregoing, the data of current variation I LED  as a function of temperature T. 
     
       
         
               
               
             
               
               
               
               
               
               
             
               
               
               
               
               
               
               
             
           
               
                   
                 TABLE 1 
               
               
                   
                   
               
             
             
               
                   
                 Values of the components 
               
               
                   
                   
               
             
          
           
               
                 R1 
                 10k 
                 30k 
                  30k 
                   30k 
                 47k 
               
               
                 R2 
                 10k 
                 10k 
                  10k 
                   10k 
                 10k 
               
               
                 R3 
                  1k 
                  1k 
                 4.7k 
                 0.56k 
                  1k 
               
               
                 R4 
                 10k 
                 10k 
                 4.7k 
                   10k 
                 10k 
               
               
                 Ratio 
                 0.907063197 
                 0.907063197 
                  0.5 
                 0.946969697 
                 0.907063197 
               
               
                 V BE1aON   
                 0.597 
                 0.629 
                  0.622 
                 0.628 
                 0.622 
               
               
                 V BE1bON   
                 0.597 
                 0.599 
                  0.591 
                 0.588 
                 0.577 
               
               
                 ΔV BEON   
                 0 
                 0.03 
                  0.031 
                 0.04 
                 0.045 
               
               
                 RS 
                 2.2 
                 3.9 
                 18 
                 1.8 
                 3.2 
               
               
                   
               
             
          
           
               
                   
                   
                   
                   
                   
                   
                 Standard 
               
               
                 T(° C.) 
                 I LED  A 
                 I LED  B 
                 I LED  C 
                 I LED  D 
                 I LED  E 
                 sol. 
               
               
                   
               
               
                 −30 
                 32.85 
                 26.15 
                 41.68 
                 37.84 
                 35.5 
                 37 
               
               
                 −15 
                 31.6 
                 25.72 
                 40.01 
                 37.45 
                 35.12 
                 35.5 
               
               
                 0 
                 30.29 
                 25.26 
                 38.2 
                 37.12 
                 34.68 
                 33.9 
               
               
                 15 
                 28.9 
                 24.77 
                 36.4 
                 36.7 
                 34.25 
                 32 
               
               
                 25 
                 27.97 
                 24.41 
                 35.24 
                 36.5 
                 34.02 
                 31 
               
               
                 30 
                 27.5 
                 24.25 
                 34.56 
                 36.3 
                 33.85 
                 30.5 
               
               
                 35 
                 27.03 
                 24.07 
                 33.96 
                 36.13 
                 33.69 
                 30 
               
               
                 40 
                 26.55 
                 23.89 
                 33.47 
                 35.84 
                 33.53 
                 29.37 
               
               
                 45 
                 26.07 
                 23.72 
                 32.88 
                 35.7 
                 33.38 
                 28.74 
               
               
                 60 
                 24.62 
                 23.19 
                 31.18 
                 35.43 
                 32.94 
                 27 
               
               
                 80 
                 22.66 
                 22.46 
                 28.75 
                 35.23 
                 32.66 
                 25 
               
               
                 −30 
                 117.4% 
                 107.1% 
                 118.3% 
                 103.7% 
                 104.4% 
                 119.35% 
               
               
                 −15 
                 113.0% 
                 105.4% 
                 113.5% 
                 102.6% 
                 103.2% 
                 114.52% 
               
               
                 0 
                 108.3% 
                 103.5% 
                 108.4% 
                 101.7% 
                 101.9% 
                 109.35% 
               
               
                 15 
                 103.3% 
                 101.5% 
                 103.3% 
                 100.5% 
                 100.7% 
                 103.23% 
               
               
                 25 
                 100.0% 
                 100.0% 
                 100.0% 
                 100.0% 
                 100.0% 
                 100.00% 
               
               
                 30 
                 98.3% 
                 99.3% 
                 98.1% 
                 99.5% 
                 99.5% 
                 98.39% 
               
               
                 35 
                 96.6% 
                 98.6% 
                 96.4% 
                 99.0% 
                 99.0% 
                 96.77% 
               
               
                 40 
                 94.9% 
                 97.9% 
                 95.0% 
                 98.2% 
                 98.6% 
                 94.74% 
               
               
                 45 
                 93.2% 
                 97.2% 
                 93.3% 
                 97.8% 
                 98.1% 
                 92.71% 
               
               
                 60 
                 88.0% 
                 95.0% 
                 88.5% 
                 97.1% 
                 96.8% 
                 87.10% 
               
               
                 80 
                 81.0% 
                 92.0% 
                 81.6% 
                 96.5% 
                 96.0% 
                 80.65% 
               
               
                   
               
             
          
         
       
     
     In various embodiments, the upper part of the Table 1 shows the values of the components and the corresponding value of the ratio corresponding to (R 4 /(R 3 +R 4 )). The central and the lower parts of the Table show on the contrary the different current values through the LEDs for different temperatures, such values being shown both in their absolute value and as a percentage of the value at 25° C. In both cases, “standard sol.” indicated the standard solution discussed in the introductory part of the description, with reference to  FIG. 1 . 
     The electrical and thermal measurements shown in Table 1 confirm the effectiveness of the described solution. 
     Specifically, it is possible to observe the following. 
     The first and the third column (I LED  A and I LED  C) show results without noteworthy advantages with reference to the standard solution, obtained in cases where:
         i) V BE1aON  and V BE1bON  are the same (ΔV BEON =0, first column), or   ii) V BE1aON  e V BE1bON  are different from each other (ΔV BEON =0.0031) but R 3  e R 4  are the same (third column).       

     The second, the fourth and the fifth column (I LED  B, I LED  D and I LED  E) show on the contrary results with remarkable advantages in comparison with the standard solution, which are obtained in cases where:
         i) V BE1aON &gt;V BE1bON  (ΔV BEON  respectively=0.003; 0.004 and 0.0045); and   ii) R 3 &lt;R 4 .       

     In various embodiments, the comparison between the results in the second, fourth and fifth column (I LED  B, I LED  D e I LED  E) shows that:
         with the same R 3 /R 4  ratio (for example 1/10 for I LED  B e I LED  E) the performance improves as ΔV BEON  increases; and   with substantially equal ΔV BEON  (see for example I LED  D and I LED  E) it is possible to obtain a further noteworthy improvement by further decreasing, to a value lower than 1/10, the R 3 /R 4  ratio, i.e. by further reducing the partition ratio of the R 3 /R 4  divider, and therefore increasing the coefficient named “ratio”. The nearer to 1 the ratio tends to, the more effective the solution is, as it decreases the current variation in the allowable temperature range.       

     It is to be noted moreover that the value of the sensing resistance RS does not in itself take on a decisive role in achieving the results, even though such a resistance determines a voltage drop between the points R and S (and therefore V) that makes it possible to fulfil the condition V BE1aON &gt;V BE1bON . 
     Without prejudice to the underlying principles of the invention, the details and the embodiments may vary, even appreciably, with respect to what has been described by way of example only, without departing from the scope of the invention as defined by the annexed claims. For example, it will be noted that, as mentioned in the foregoing, the presently described solution can be carried out in an embodiment in which at least the driver transistor T 2  is a MOSFET and/or it can be used in general to obtain a current regulator with low drop-out for applications other than light source driving. 
     While the invention has been particularly shown and described with reference to specific embodiments, it should be understood by those skilled in the art that various changes in form and detail may be made therein without departing from the spirit and scope of the invention as defined by the appended claims. The scope of the invention is thus indicated by the appended claims and all changes which come within the meaning and range of equivalency of the claims are therefore intended to be embraced.