Abstract:
Excitation power to fluid flow measurement sensor is applied in such a way so as to maintain some part of the sensor elements at a constant temperature relative to the ambient fluid temperature and some part of the sensor elements to change its temperature with fluid flow which results in a sensor output that remains constant and linear with per unit flow. This semi-constant temperature sensor excitation scheme results in higher sensor output, added sensor range and temperature insensitive flow measurement. Therefore, this sensor excitation method negates the drawbacks of smaller and non-linear output and/or thermal runaway that are associated with other excitation methods.

Description:
RELATED APPLICATION  
       [0001]     The present application is based on and claims priority to U.S. Provisional application Ser. No. 60/711,728, filed Aug. 26, 2005 and bearing Attorney Docket No. M-16145-V1US. 
     
    
     BACKGROUND  
       [0002]     1. Field of the Invention  
         [0003]     The present invention relates generally to fluid flow measurement sensors, and more particularly to methods and circuits that enable linear and constant sensor outputs.  
         [0004]     2. Related Art  
         [0005]     In a fluid flow measurement device such as a mass flow controller, a sensor  100  typically has multiple coils  102  and  104  wrapped around a sensor tube  106  as shown in  FIG. 1 . The sensor is usually excited either by a constant power, current, or voltage source. When fluid flows inside sensor tube  106  from a heated upstream coil  102  to a heated downstream coil  104  that are electrically balanced, thermal energy is transferred from the coils to the flowing fluid. For a given flow rate, the amount of thermal energy transferred from the coils to the fluid is inversely proportional to the fluid temperature. Thermal energy transfer from the upstream coil  102  and the downstream coil  104  is disproportionate because the fluid temperature is different at the upstream coil than at the downstream coil. This different rate of heat transfer from the coils to the fluid causes temperature differential between the coils which manifests itself as a change in relative resistance of the two coils. This change in resistance is directly proportional to the amount of fluid flowing through the sensor tube.  
         [0006]     The upstream and downstream sensor coils are part of a Wheatstone bridge. The circuit is configured to form a balanced bridge network with little or zero output when there is no fluid flow. The bridge network measures the flow through the sensor as a change of resistance of the coils and generates a signal corresponding to the flow rate of the fluid through the sensor tube. The problems associated with the above sensor excitation schemes are non-linear output, thermal runaway and low sensor output.  
         [0007]     Accordingly, there is a need in the art for a fluid flow measurement sensor that gives a high linear sensor output without the danger of thermal runaway.  
       SUMMARY  
       [0008]     According to one aspect of the present invention, a semi-constant temperature excitation method for fluid flow measurement sensor involves maintaining some part of the sensor elements (e.g., an upstream coil, or a downstream coil, or a portion of each coil) at a constant temperature relative to the ambient fluid temperature and some part of the sensor elements to change their temperature with fluid flow so that the sensor output remains constant and linear per flow unit. This scheme can drive the upstream coil at a constant temperature T Ru  relative to ambient or drive the downstream coil at a constant temperature T Rd  relative to ambient.  
         [0009]     Additionally, the sensor could be driven so that upstream and downstream coils are allowed to change their temperatures T Ru  and T Rd , respectively, at a certain proportional rate relative to a temperature between T Ru  and T Rd  which is maintained constant above the ambient.  
         [0010]     This sensor excitation scheme of maintaining a part of the sensor elements at a constant temperature relative to ambient has certain definite advantages as compared to conventional excitation methods.  
         [0011]     For example, in a conventional constant current drive method, the sensor output per unit flow drops as the flow increases due to uncompensated cooling of the sensor elements. In a conventional constant voltage drive method, the sensor output per unit flow increases as the flow increases due to uncompensated heating of the sensor elements. The resultant condition can lead to a thermal run-away, and subsequent sensor damage. Both drive methods result in sensor non-linearity, which degrades progressively at higher and higher flows. This requires the sensor full scale flow range to be selected at a low flow value in order to maximize output linearity.  
         [0012]     The semi-constant temperature excitation method allows a higher sensor output, higher sensor full scale flow range (since the sensor can be forced to flow more fluid without losing linearity), and inherent temperature insensitive flow measurement without the problems of low, non-linear output or thermal runaway.  
         [0013]     These and other features and advantages of the present invention will be more readily apparent from the detailed description of the preferred embodiments set forth below taken in conjunction with the accompanying drawings. 
     
    
     BRIEF DESCRIPTION OF THE FIGURES  
       [0014]      FIG. 1  shows a conventional mass flow sensor;  
         [0015]      FIG. 2  shows a half bridge circuit, according to one embodiment, with constant average temperature ((T Ru +T Rd )/2) of upstream and downstream sensor coils with respect to ambient temperature;  
         [0016]      FIG. 3  shows a half bridge circuit, according to another embodiment, with constant temperature (T Ru ) of upstream sensor coil with respect to ambient temperature;  
         [0017]      FIG. 4  shows a full bridge circuit, according to one embodiment, with constant temperature (T (f(Ru)+f(Rd)) ) with respect to ambient temperature that is adjustable between T Ru  and T Rd  of sensor coils;  
         [0018]      FIG. 5  shows a full bridge circuit, according to another embodiment, with constant temperature (T Ru ) of upstream sensor coil with respect to ambient and with slave mirror current in downstream sensor coil;  
         [0019]      FIG. 6  shows a full bridge circuit, according to yet another embodiment, with constant temperature (T Ru ) of upstream sensor coil with respect to ambient and with slave mirror voltage across downstream sensor coil; and  
         [0020]      FIG. 7  shows the fluid flow versus output of a sensor using different sensor excitation methods. 
     
    
       [0021]     Like element numbers in different figures represent the same or similar elements.  
       DETAILED DESCRIPTION  
       [0022]     According to one aspect of the present invention, a part of the sensor elements, which could be the upstream element with resistance R u , the downstream element with resistance R d , or a combination of a certain proportion of R u  and R d  (i.e., X % of R u +Y % of R d ), is kept at a constant temperature differential with respect to the ambient temperature of the fluid. This condition is maintained under all fluid flow conditions. Ambient temperature of the fluid, before it enters the sensor, is measured by R ref  or it could be measured separately from the sensor electrical circuit and the correction applied by microprocessor firmware. The rest of the sensor is allowed to change its temperature with fluid flow.  
         [0023]     A variable voltage source electrical circuit with feedback is used to excite all the elements of the sensor as well as to measure the fluid ambient temperature and keep a part of the sensor elements at a constant temperature differential with respect to the ambient temperature of the fluid.  
         [0024]     Upstream and downstream sensor elements with resistance R u  and R d , respectively, and the ambient temperature measurement element with resistance R ref  comprise of the same material with a large temperature coefficient of resistance. The ambient resistances R u  and R d  of the elements are made approximately equal.  
         [0025]     The variable voltage source electrical circuit with feedback is configured in such a way so that with no fluid flow, the voltage drop across the upstream element and downstream element is approximately equal. Therefore, an approximately equal amount of electrical power is dissipated from the upstream and the downstream elements with no fluid flow, resulting in temperature T Ru  and T Rd  being approximately equal.  
         [0026]     When a fluid flows through the sensor past the upstream element, heat energy is lost from the upstream element in proportion to the temperature difference of the element to the fluid temperature (T Ru −T Rref ) and the rate of mass flow of the fluid. The fluid temperature increases by a small amount (ΔT) and becomes T Rref +ΔT.  
         [0027]     As the fluid passes the downstream element, heat energy is lost from the downstream element in proportion to the temperature difference of the element to the fluid temperature (T Rd −T Rref −ΔT) and the rate of mass flow of the fluid.  
         [0028]     The mass flow rate of the fluid is same across the upstream and the downstream elements. However, the heat loss from the upstream element is more than the heat loss from the downstream element, because the temperature differential between the element and the fluid is greater at the upstream element than at the downstream element.  
         [0029]     This different rate of heat loss from the upstream and downstream elements manifests itself by changing the resistance of the elements in proportion to the amount of heat loss. However, the variable voltage source electrical circuit with feedback tries to maintain a part of the sensor elements at a constant temperature differential with respect to the ambient temperature of the fluid. Depending upon the electrical circuit configuration, the part of the sensor elements kept at a constant temperature differential with respect to the ambient temperature of the fluid could be the upstream element with resistance R u , the downstream element with resistance R d , or a combination of a certain proportion of R u  and R d  (X % of R u +Y % of R d ). This causes the voltage across R u  and R d  to change in proportion to the mass flow rate of the fluid. This voltage differential across R u  and R d  is directly proportional to the mass flow rate of the fluid.  
         [0030]     The variable voltage source electrical circuit also provides soft start of the sensor. At the instance when excitation power is applied, the sensor elements are at ambient temperature and the circuit is essentially operating in an open loop control mode. Without a soft start, a high current surge through the elements could destroy the sensor elements. The circuit also monitors the voltage difference across the sensor elements and limits the current through the elements if an uncontrolled fluid flow surge through the sensor occurs. Without current limiting, the sensor could get into thermal run-away mode resulting in sensor destruction.  
         [0031]      FIGS. 2-6  show various circuit configurations for implementing the above-described invention.  FIG. 2  shows a half-bridge circuit  200  according to one embodiment of the present invention. Circuit  200  includes an amplifier  202  and a current source  204 . The upstream and downstream sensor coils  206  and  208  are made in such a way so that their respective resistances R u  (for the upstream element) and R d  (for the downstream element) are approximately equal at ambient temperature. Values of R u  and R d  are selected based on the desired temperature difference of the sensor coils relative to ambient temperature T Rref  as measured by R ref  and the maximum available voltage from the variable voltage source electrical circuit. The upstream and downstream sensor elements may also be planar heating elements, such as made by thin/thick film deposition.  
         [0032]     The value for R ref  is selected so that during normal operation, its power dissipation does not affect its resistance due to self-heating. The actual component  210  associated with R ref  is thermally connected to a large thermal mass, which indicates the pre-sensor fluid temperature. R ref , R u , and R d  all comprise of the same type of material in one embodiment, thus ensuring that their temperature coefficients of resistance are equal.  
         [0033]     R 1  has a low temperature coefficient of resistance and its ohmic value is selected to set the temperature ratio between (R u +R d ) and R ref . R 2  is selected with a low temperature coefficient of resistance and an ohmic value near that of R u  or R d  resistance.  
         [0034]     Resistor R 3  and resistor R 4  form a passive bridge circuit to enable differential sensor output. R 3  and R 4  resistances have a low temperature coefficient of resistance and their ohmic values are sufficiently high to minimize thermal loading and subsequent temperature control error of the R u  and R d  pair.  
         [0035]     The ratio R 1 /R ref  sets a reference voltage (V r ) on U 1 , which in turn forces the sum of upstream resistor R u  and downstream resistor R d  to increase resistance due to thermal heating, until the ratios of R 1 /R ref  and R 2 /(R u +R d ) are approximately equal. When this is achieved, the resistance (R u +R d ), and therefore the average temperature (T Ru +T Rd )/2, will track the resistance R ref  and its temperature T Rref . Since the average temperature (T Ru +T Rd )/2 is maintained constant above the ambient and the same amount of current flows through R d  as through R u , the resistances R u  and R d  will be nearly identical with no fluid flow. As fluid flow increases, more heat energy is lost from the upstream element than from the downstream element and therefore T Ru  and R u  decrease more than T Rd  and R d , thus causing the voltage across R u  to decrease and the voltage across R d  to increase. The excitation circuit will automatically compensate for the required heat energy increase in order to maintain a constant temperature average (T Ru +T Rd )/2 above ambient. The attached Appendix shows details the various voltages and output ΔV of circuit  200 .  FIG. 2  also shows exemplary resistor values according to one embodiment.  
         [0036]     The current source I source  is normally not a part of the sensor excitation power control loop, but does limit the power to the excitation circuit when the excitation circuit is open loop (e.g. at startup and during massive fluid overflow conditions).  
         [0037]      FIG. 3  shows a half bridge circuit  300  another embodiment of the present invention. Circuit  300  includes an amplifier  202  and a current source  204 . The upstream and downstream sensor coils  302  and  304  are made in such a way so that their respective resistances R u  and R d  are approximately equal at ambient temperature. Values of R u  and R d  are selected based on the desired temperature difference of the sensor coils relative to ambient temperature T R1ref  as measured by R 1ref  and the maximum available voltage from the variable voltage source electrical circuit.  
         [0038]     R 1ref  and R 2ref  values are nearly equal in magnitude and selected so that during normal operation, their power dissipation does not affect their respective resistances due to self-heating. The actual components associated with R 1ref  and R 2ref  are thermally connected to a large thermal mass, which indicates the pre-sensor fluid temperature. R 1ref , R 2ref , R u , and R d  all comprise of the same type of material in one embodiment, thus ensuring that their temperature coefficients of resistance are equal.  
         [0039]     R 1  has a low temperature coefficient of resistance and its ohmic value is selected to set the temperature ratio between R u  and R 1ref . R 2  is selected with a low temperature coefficient of resistance and an ohmic value near that of R u  or R d  resistance.  
         [0040]     Resistor R 3  and resistor R 4  form a passive bridge circuit to enable differential sensor output. R 3  and R 4  resistances have a low temperature coefficient of resistance and their ohmic values are sufficiently high to minimize thermal loading and subsequent temperature control error of the R u  and R d  pair.  
         [0041]     The ratio of resistor R 1  to resistor R 1ref  sets a reference voltage (V r ) on U 1 , which in turn forces the sum of upstream resistor R u  and downstream resistor R d  to increase resistance due to thermal heating, until the ratios R 1 /R 1ref  and R 2 /R u  are approximately equal. When this is achieved, the resistance R u , and therefore its temperature T Ru , will track the resistance R 1ref  and its temperature T R1ref . Since the temperature T Ru  is maintained constant above the ambient and same amount of current flows through R d  as through R u , the resistances R u  and R d  will be nearly identical with no fluid flow. As fluid flow increases, more heat energy is lost from upstream element than from downstream element and therefore T Ru  and R u  decrease more than T Rd  and R d , thus causing the voltage across R u  to decrease more than the voltage across R d . The excitation circuit will automatically compensate for the required heat energy increase in order to maintain the constant temperature T Ru  above ambient. R 2ref  does not significantly affect the R u  temperature, but is included to provide an equivalent load across R d  to compensate the equivalent Thevenin loading between R 1ref  and R u . The attached Appendix shows details the various voltages and output ΔV of circuit  300 .  FIG. 3  also shows exemplary resistor values according to one embodiment  
         [0042]     The current source I source  is normally not a part of the sensor excitation power control loop, but does limit the power to the excitation circuit when the excitation circuit is open loop (e.g. at startup and during massive fluid overflow conditions).  
         [0043]      FIG. 4  shows a full bridge circuit  400  according to another embodiment of the present invention. Circuit  400  includes an amplifier  202  and a current source  204 . The upstream and downstream sensor coils  402  and  404  are made in such a way so that their respective resistances R u  and R d  are approximately equal at ambient temperature. Values of R u  and R d  are selected based on the desired temperature difference of the sensor coils relative to ambient temperature T Rref  as measured by R ref  and the maximum available voltage from the variable voltage source electrical circuit.  
         [0044]     R ref  value is selected so that during normal operation, its power dissipation does not affect its resistance due to self-heating. The actual component associated with R ref  is thermally connected to a large thermal mass, which indicates the pre-sensor fluid temperature. R ref , R u , and R d  are all made of the same type of material, thus ensuring that their temperature coefficients of resistance are equal.  
         [0045]     R 1  has a low temperature coefficient of resistance and its ohmic value is selected to set the temperature ratio between the resistance of the R u  and R d  network and R ref . R 2  and R 3  are selected with a low temperature coefficient of resistance and an ohmic value near that of R u  or R d  resistance. R 4  and R 5  have a low temperature coefficient of resistance and their ohmic values are selected so that during normal operation, their power dissipation does not affect their respective resistance due to self-heating.  
         [0046]     The ratio R 1 /R ref  sets a reference voltage (V r ) on U 1 , which in turn forces the sum of upstream resistor R u  and downstream resistor R d  to increase resistance due to thermal heating, until the voltage at the node of R 1  and R ref  is equal to the voltage at the node of R 4  and R 5 . When this is achieved, the resistance of network R u  and R d  and its average temperature T (f(Ru)+f(Rd))  will track the resistance R ref  and its corresponding temperature T Rref . Since the average temperature of R u  and R d  network is maintained constant above the ambient and if R 4  and R 5  are equal in magnitude, then the resistances R u  and R d  will be nearly identical with no flow. As fluid flow increases, more heat energy is lost from upstream element than from downstream element. Therefore T Ru  and R u  decrease more than T Rd  and R d  causing the voltage across R u  to decrease and the voltage across R d  to increase. The excitation circuit will automatically compensate for the required heat energy increase in order to maintain the average temperature T (f(Ru)+f(Rd))  of the R u  and R d  network above ambient constant. When R 4  and R 5  are equal, this circuit operates in a similar manner as a half bridge circuit with constant average temperature ((T Ru +T Rd )/2) above ambient of upstream and downstream sensor coils as shown in  FIG. 2 .  
         [0047]     This full bridge circuit allows the temperature control from 100% of constant temperature above ambient of the upstream element to 100% of constant temperature above ambient of the downstream element, or any ratio in between by varying R 4  and R 5 .  
         [0048]     This circuit uses slightly more power than the circuit in  FIG. 2  due to R 3  dissipation, but the full bridge output provides nearly 100% more differential voltage output, which improves the signal to noise and signal to error ratios accordingly. Another benefit of running the full bridge topology is the reduction of the total power supply voltage requirement, since R u  and R d  elements are in parallel instead of in series. The attached Appendix shows details the various voltages and output ΔV of circuit  400 .  FIG. 4  also shows exemplary resistor values according to one embodiment  
         [0049]      FIG. 5  shows a full bridge circuit  500  according another embodiment of the present invention. Circuit  500  includes an amplifier  202  and a current source  204 . The upstream and downstream sensor coils  502  and  504  are made in such a way so that their respective resistances R u  and R d  are approximately equal at ambient temperature. Values of R u  and R d  are selected based on the desired temperature difference of the sensor coils relative to ambient temperature T Rref  as measured by R ref  and the maximum available voltage from the variable voltage source electrical circuit.  
         [0050]     The value of R ref  is selected so that during normal operation, its power dissipation does not affect its resistance due to self-heating. The actual component associated with R ref  is thermally connected to a large thermal mass, which indicates the pre-sensor fluid temperature. R ref , R u , and R d  all comprise of the same type of material in one embodiment, thus ensuring that their temperature coefficients of resistance are equal.  
         [0051]     R 1  has a low temperature coefficient of resistance and its ohmic value is selected to set the temperature ratio between R u  and R ref . R 2  and R 3  are selected with a low temperature coefficient of resistance and an ohmic value near that of R u  or R d  resistance.  
         [0052]     The ratio R 1 /R ref  sets a reference voltage (V r ) on U 1 , which in turn forces the upstream resistor R u  to increase resistance due to thermal heating, until the ratios of R ref /R 1  and R u /R 2  are approximately equal. When this is achieved, the resistance R u  and therefore its temperature T Ru  will track the resistance R ref  and its temperature T Rref . Since the temperature T Ru  is maintained constant and element R d  is driven by a current mirror with a value equal to that of the current flowing through R u , the resistances R u  and R d  will be nearly identical with no flow. As flow increases, more heat energy is lost from R u  than from R d . Therefore the current through R u  must be increased in order to maintain a constant T Ru  above ambient. The excitation circuit will automatically compensate for the required heat energy increase in order to maintain the constant temperature T Ru  above ambient, thus resulting in voltage increase across R u . Since the same current is flowing through R d  and heat loss from R d  is less as compared to R u , the voltage increase across R d  will be higher than across R u .  
         [0053]     This circuit uses slightly higher power than the half bridge equivalent circuit (due to R 3  dissipation) as shown in  FIG. 3 , but the full bridge output provides nearly 100% more differential voltage output, which improves the signal to noise and signal to error ratios accordingly. Another benefit of running the full bridge topology is the reduction of the total power supply voltage requirement, since the R u  and R d  elements are in parallel instead of in series. The attached Appendix shows details the various voltages and output ΔV of circuit  500 .  FIG. 5  also shows exemplary resistor values according to one embodiment  
         [0054]      FIG. 6  shows a full bridge circuit  600  according another embodiment of the present invention. Circuit  600  includes an amplifier  202  and a current source  204 . The upstream and downstream sensor coils  602  and  604  are made in such a way so that their respective resistances R u  and R d  are approximately equal at ambient temperature Values of R u  and R d  are selected based on the desired temperature difference of the sensor coils relative to ambient temperature T Rref  as measured by R ref  and the maximum available voltage from the variable voltage source electrical circuit.  
         [0055]     The value of R ref  is selected so that during normal operation, its power dissipation does not affect its resistance due to self-heating. The actual component associated with R ref  is thermally connected to a large thermal mass, which indicates the pre-sensor fluid temperature. R ref , R u , and R d  are all made of the same type of material, thus ensuring that their temperature coefficients of resistance are equal.  
         [0056]     R 1  has a low temperature coefficient of resistance and its ohmic value is selected to set the temperature ratio between R u  and R ref . R 2  and R 3  are selected with a low temperature coefficient of resistance and an ohmic value near that of R u  or R d  resistance.  
         [0057]     The ratio R 1 /R ref  sets a reference voltage (V r ) on U 1 , which in turn forces the upstream resistor R u  to increase resistance due to thermal heating, until the ratios R 1 /R ref  and R 2 /R u  are approximately equal. When this is achieved, the resistance R u  and therefore its temperature T Ru  will track the resistance R ref  and its temperature T Rref . Since the temperature T Ru  is maintained constant and element R d  is driven by a voltage mirror with a value equal to that of the voltage across R u , the resistances R u  and R d  will be nearly identical with no flow. As flow increases, more heat energy is lost from R u  than from R d . Therefore, the current through R u  must be increased in order to maintain a constant T Ru  above ambient. The excitation circuit will automatically compensate for the required heat energy increase in order to maintain the constant temperature T Ru  above ambient, thus resulting in voltage increase across R u . Since the same voltage is maintained across R d  and heat loss from R d  is less as compared to R u , the current through R d  will be less than current through R u .  
         [0058]     This circuit uses slightly higher power than the half bridge equivalent circuit (due to R 3  dissipation) as shown in  FIG. 3 , but the full bridge output provides nearly 100% more differential voltage output, which improves the signal to noise and signal to error ratios accordingly. Another benefit of running the full bridge topology is the reduction of the total power supply voltage requirement, since R u  and R d  elements are in parallel instead of in series. The attached Appendix shows details the various voltages and output ΔV of circuit  600 .  FIG. 6  also shows exemplary resistor values according to one embodiment  
         [0059]      FIG. 7  shows the relative performance of a conventional constant current excitation method to one embodiment of the semi-constant temperature excitation method used in a thermal fluid flow sensor of a mass flow controller.  
         [0060]     The sensor excited with one embodiment of the present invention exhibits near perfect linearity as a function of mass flow as well as higher sensor output per flow unit as compared to the sensor excited by constant current method.  
         [0061]     This excitation scheme lends itself well to multi-range or wide range usage. The high sensor output means that the signal to error ratio is superior to other excitation methods.  
         [0062]     The sensor design lends itself well to thermal balance between the upstream and down stream coils. Therefore, the sensor position has very little effect on the output.  
         [0063]     Having thus described embodiments of the present invention, persons of ordinary skill in the art will recognize that changes may be made in form and detail without departing from the scope of the invention. Thus the invention is limited only by the following claims.  
       APPENDIX  
       [0000]      FIG. 2 :
   V=I   1 ×( R   1   +R   REF )  (1)   V   R   =I   1   ×R   REF   (2) 
 Dividing Eq. (1) by Eq. (2):  
               V     V   R       =         (       R   1     +     R   REF       )       R   REF       =         R   1       R   REF       +   1                                   R   1       R   REF       =       (     V   -     V   R       )       V   R               (   3   )               V   =       I   2     ×     [       R   2     +         (       R   U     +     R   D       )     ⁢     (       R   3     +     R   4       )         (       R   3     +     R   4     +     R   U     +     R   D       )         ]               (   4   )                 V   R     =       I   2     ×         (       R   U     +     R   D       )     ⁢     (       R   3     +     R   4       )         (       R   3     +     R   4     +     R   U     +     R   D       )                 (   5   )               
 From Eq. (3), (4) and (5):  
                         R   1       R   REF       =       [         I   2     ×     (       R   2     +         (       R   U     +     R   D       )     ⁢     (       R   3     +     R   4       )         (       R   3     +     R   4     +     R   U     +     R   D       )         )       -       I   2     ×         (       R   U     +     R   D       )     ⁢     (       R   3     +     R   4       )         (       R   3     +     R   4     +     R   U     +     R   D       )           ]         I   2     ×         (       R   U     +     R   D       )     ⁢     (       R   3     +     R   4       )         (       R   3     +     R   4     +     R   U     +     R   D       )                       =         R   2     ×     (       R   3     +     R   4     +     R   U     +     R   D       )           (       R   U     +     R   D       )     ⁢     (       R   3     +     R   4       )                     =         R   2       (       R   U     +     R   D       )       ×       (       R   3     +     R   4     +     R   U     +     R   D       )       (       R   3     +     R   4       )                 ⁢     
     ⁢       If   ⁢           ⁢     R   3       ,       R   4     ⪢     R   U       ,     R   D     ,   then             (   6   )                     R   1       R   REF       ≅       R   2       (       R   U     +     R   D       )         =     K   ⁡     (   Constant   )               (   7   )                 V   R     =       I   3     ×     (       R   U     +     R   D       )               (   8   )                 V   R     =       I   4     ×     (       R   3     +     R   4       )               (   9   )               
         [0064]     From Eq. (8) and (9):  
                 I   3     ×     (       R   U     +     R   D       )       =       I   4     ×     (       R   3     +     R   4       )                                   I   3       I   4       =       (       R   3     +     R   4       )       (       R   U     +     R   D       )               (   10   )                 V   1     =       I   3     ×     R   D               (   11   )                 V   2     =       I   4     ×     R   4               (   12   )             
 
 Subtracting Eq.(12) from Eq. (11):  
                 V   1     -     V   2       =           I   3     ×     R   D       -       I   4     ×     R   4         =       I   4     ⁡     (           I   3     ×     R   D         I   4       -     R   4       )                 (   13   )             
 
 From Eq. (9), (10) and (13):  
               Δ   ⁢           ⁢   V     =         V   R       (       R   3     +     R   4       )       ⁡     [         R   D     ⁢       (       R   3     +     R   4       )       (       R   U     +     R   D       )         -     R   4       ]                                   Δ   ⁢           ⁢   V     =         V   R       (       R   3     +     R   4       )       ⁡     [           R   D     ×     R   3       +       R   D     ×     R   4       -       R   4     ×     R   U       -       R   4     ×     R   D             R   U     +     R   D         ]         ⁢     
     ⁢         If   ⁢           ⁢     R   3       =     R   4       ,     then   ⁢           ⁢     Eq   .           ⁢     (   14   )       ⁢           ⁢   becomes               (   14   )                 Δ   ⁢           ⁢   V     =         V   R     2     ⁡     [         R   D     -     R   U           R   U     +     R   D         ]               (   15   )             
 
  FIG. 3 :
 
 V−V   R   =I   1   ×R   1   (1)
 
 V   R   −V   1   =I   1   ×R   1REF   (2)
 
 Dividing Eq (1) by Eq. (2):  
                 (     V   -     V   R       )         V   R     -     V   1         =       R   1       R     1   ⁢           ⁢   REF                 (   3   )                 V   -     V   R       =       I   2     ×     R   2               (   4   )                   V   R     -     V   1       =       I   3     ×     R   U               (   5   )             
 
 From Eq. (3), (4), and (5):  
                 R   1       R     1   ⁢           ⁢   REF         =         (     V   -     V   R       )       (       V   R     -     V   1       )       =       (       I   2     ×     R   2       )       (       I   3     ×     R   U       )                 (   6   )                 I   2     =       I   3     +     I   4               (   7   )             
 
 Since R 3  and R 4 &gt;&gt;R U  and R D , L 4 ≈0 and Eq (7) becomes
 
I 2 =I 3   (8)
 
 From Eq (6) and (8):  
                   R   1       R     1   ⁢           ⁢   REF         =         R   2       R   U       =     K   ⁡     (   Constant   )           ⁢     
     ⁢       V   R     =       I   4     ×     (       R   3     +     R   4       )                 (   9   )                 I   4     =       V   R       (       R   3     +     R   4       )               (   10   )                 V   2     =       I   4     ×     R   4               (   11   )             
 
 From Eq. (10) and (11):  
               V   2     =         V   R     ×     R   4         (       R   3     +     R   4       )               (   12   )                 V   1     =         (       I   1     +     I   3       )     ×     R   D     ×     R     2   ⁢           ⁢   REF           (       R   D     +     R     2   ⁢           ⁢   REF         )               (   13   )                   V   R     -     V   1       =         I   1     ×     R     1   ⁢           ⁢   REF         =       I   3     ×     R   U                 (   14   )                 I   1     =       (       V   R     -     V   1       )       R     1   ⁢           ⁢   REF                 (   15   )             
 
 From Eq. (14):  
               I   3     =       I   1     ⁢       R     1   ⁢           ⁢   REF         R   U                 (   16   )             
 
 From Eq. (13) and (16):  
                 V   1     =         I   1     ×     (     1   +       R     1   ⁢           ⁢   REF         R   U         )     ×     R   D     ⁢     R     2   ⁢           ⁢   REF           (       R   D     +     R     2   ⁢           ⁢   REF         )         ⁢     
     ⁢       V   1     =         I   1     ×     R   D     ×       R     2   ⁢           ⁢   REF       ⁡     (       R   U     +     R     1   ⁢           ⁢   REF         )             R   U     ×     (       R   D     +     R     2   ⁢           ⁢   REF         )                   (   17   )             
 
 From Eq. (15) and (17):  
                 V   1     =           (       V   R     -     V   1       )       R     1   ⁢           ⁢   REF         ×     R   D     ×       R     2   ⁢           ⁢   REF       ⁡     (       R   U     +     R     1   ⁢           ⁢   REF         )             R   U     ×     (       R   D     +     R     2   ⁢           ⁢   REF         )           ⁢     
     ⁢         V   1     ×     R   U     ⁢     R     1   ⁢           ⁢   REF       ×     (       R   D     +     R     2   ⁢           ⁢   REF         )       =       (       V   R     -     V   1       )     ×     R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )         ⁢     
     ⁢           V   1     ×     R   U     ×     R     1   ⁢           ⁢   REF       ×     (       R   D     +     R     2   ⁢           ⁢   REF         )       +       V   1     ×     R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )         =       V   R     ×     R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )         ⁢     
     ⁢         V   1     ⁡     [         R   U     ×     R     1   ⁢           ⁢   REF       ×     (       R   D     +     R     2   ⁢           ⁢   REF         )       +       R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )         ]       =       V   R     ×     R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )         ⁢     
     ⁢       V   1     =         V   R     ×     R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )             R   U     ×     R     1   ⁢           ⁢   REF       ×     (       R   D     +     R     2   ⁢           ⁢   REF         )       +       R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )                     (   18   )                 Δ   ⁢           ⁢   V     =       V   1     -     V   2               (   19   )             
 
 From Eq. (12), (18), and (19):  
                 Δ   ⁢           ⁢   V     =           V   R     ×     R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )             R   U     ×     R     1   ⁢           ⁢   REF       ×     (       R   D     +     R     2   ⁢           ⁢   REF         )       +       R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )           -         V   R     ×     R   4         (       R   3     +     R   4       )           ⁢     
     ⁢       Δ   ⁢           ⁢   V     =     [           ⁢                 R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )                   R   U     ×     R     1   ⁢           ⁢   REF       ×     (       R   D     +     R     2   ⁢           ⁢   REF         )       +                 R   D     ×     R     2   ⁢           ⁢   REF       ×     (       R   U     +     R     1   ⁢           ⁢   REF         )               -                 R   4       (       R   3     +     R   4       )             ⁢           ]               (   20   )                   If   ⁢           ⁢     R     1   ⁢           ⁢   REF         =         R     2   ⁢           ⁢   REF       ⁢           ⁢   and   ⁢           ⁢     R   3       =     R   4         ,           ⁢   Then                             Δ   ⁢           ⁢   V     =       V   R     ×     [           R   D     ×     (       R   U     +     R     1   ⁢           ⁢   REF         )             R   U     ×     (       R   D     +     R     1   ⁢           ⁢   REF         )       +       R   D     ×     (       R   U     +     R     1   ⁢           ⁢   REF         )           -   0.5     ]               (   21   )             
 
  FIG. 4 :
 
 V=I   1 ×( R   1   +R   REF )  (1)
 
 V   R   =I   1   R   REF   (2)
 
 Divide Eq. (1) by Eq. (2):  
                 V     V   R       =         (       R   1     +     R   REF       )       R   REF       =         R   1       R   REF       +   1         ⁢     
     ⁢         R   1       R   REF       =         (     V   -     V   R       )       V   R       =         V     I   3       -       V   R       I   3             V   R       I   3                     (   3   )               V   =       I   2     ×     (       R   2     +     R   U       )               (   4   )               V   =       I   3     ×     (       R   3     +     R   D       )               (   5   )             
 
 From Eq. (4) and (5):  
                   I   2     ×     (       R   2     +     R   U       )       =       I   3     ×     (       R   3     +     R   D       )         ⁢     
     ⁢         I   2       I   3       =       (       R   3     +     R   D       )         R   2     +     R   U                   (   6   )                 V   1     =       I   2     ×     R   U               (   7   )                 V   2     =       I   3     ×     R   D               (   8   )                   V   2     -     V   R       =         R   5     ×     (       V   2     -     V   1       )         (       R   4     +     R   5       )               (   9   )             
 
 From Eq. (7), (8), and (9):  
               V   R     =         I   3     ⨯     R   D       -           R   5     ×     (         I   3     ×     R   D       -       I   2     ×     R   U         )         (       R   4     +     R   5       )       ⁢     
     ⁢       V   R     =           I   3     ×     R   D     ×     (       R   4     +     R   5       )       -       R   5     ×     (         I   3     ×     R   D       -       I   2     ×     R   U         )           (       R   4     +     R   5       )         ⁢     
     ⁢       V   R     =           I   3     ×     R   D     ×     R   4       +       I   3     ×     R   5       -       R   5     ×     I   3     ×     R   D       +       R   5     ×     I   2     ×     R   U           (       R   4     +     R   5       )         ⁢     
     ⁢         V   R       I   3       =       (         R   4     ×     R   D       +           I   2     ×     R   5     ×     R   U       ⁢               I   3         )       (       R   4     +     R   5       )                     (   10   )             
 
 From Eq. (6) and (10):  
                   V   R       I   3       =       [         R   4     ×     R   D       +         R   5     ×       R   U     ⁡     (       R   3     +     R   D       )           (       R   2     +     R   U       )         ]       (       R   4     +     R   5       )         ⁢     
     ⁢         V   R       I   3       =       [         R   4     ×     R   D     ×     (       R   2     +     R   U       )       +       R   5     ×       R   U     ⁡     (       R   3     +     R   D       )           ]         (       R   4     +     R   5       )     ⁢     (       R   2     +     R   U       )                   (   11   )             
 
 From Eq. (5):  
               V     I   3       =     (       R   3     +     R   D       )             (   12   )             
 
 From Eq. (3), (11), and (12):  
                       R   1       R   REF       =       ⁢       (       V     I   3       -       V   R       I   3         )         V   R       I   3                     =       ⁢       [       (       R   3     +     R   D       )     -       [         R   4     ×     R   D     ×     (       R   2     +     R   U       )       +       R   5     ×     R   U     ×     (       R   3     +     R   D       )         ]         (       R   4     +     R   5       )     ⁢     (       R   2     +     R   U       )           ]         [         R   4     ×     R   D     ×     (       R   2     +     R   U       )       +       R   5     ×     R   U     ×     (       R   3     +     R   D       )         ]         (       R   4     +     R   5       )     ⁢     (       R   2     +     R   U       )                       =       ⁢       [               (       R   3     +     R   D       )     ⁢     (       R   4     +     R   5       )     ⁢     (       R   2     +     R   U       )       -       R   4     ×     R   D     ×                   (       R   2     +     R   U       )     -       R   5     ×     R   U     ×     (       R   3     +     R   D       )               ]           R   4     ×     R   D     ×     (       R   2     +     R   U       )       +       R   5     ×     R   U     ×     (       R   3     +     R   D       )                       =       ⁢     K   ⁡     (   Constant   )                     (   13   )             
 
 Subtracting Eq. (7) from Eq. (8):  
                 V   2     -     V   1       =           I   3     ×     R   D       -       I   2     ×     R   U         =       I   3     ×     (       R   D     -         I   2     ×     R   U         I   3         )                 (   14   )             
 
 From Eq. (5), (6), and (14):  
                 Δ   ⁢           ⁢   V     =       V     (       R   3     -     R   D       )       ×     [       R   D     -         R   U     ×     (       R   3     +     R   D       )         (       R   2     +     R   U       )         ]         ⁢     
     ⁢       Δ   ⁢           ⁢   V     =       V     (       R   3     +     R   D       )       ×     [           R   D     ×     (       R   2     +     R   U       )       -       R   U     ×     (       R   3     +     R   D       )           (       R   2     +     R   U       )       ]         ⁢     
     ⁢       Δ   ⁢           ⁢   V     =       V     (       R   3     +     R   D       )       ×     [           R   D     ×     R   2       +       R   D     ×     R   U       -       R   U     ×     R   3       -       R   U     ×     R   D           (       R   2     +     R   U       )       ]         ⁢     
     ⁢       Δ   ⁢           ⁢   V     =       V     (       R   3     +     R   D       )       ×     [           R   D     ×     R   2       -       R   U     ×     R   3           (       R   2     +     R   U       )       ]         ⁢     
     ⁢         Since   ⁢           ⁢     R   2       =     R   3       ,     
     ⁢       Δ   ⁢           ⁢   V     =       V     (       R   2     +     R   D       )       ×     [           R   D     ×     R   2       -       R   U     ×     R   2           (       R   2     +     R   U       )       ]           ⁢     
     ⁢       Δ   ⁢           ⁢   V     =       V   ×     R   2     ×     (       R   D     -     R   U       )           (       R   2     +     R   D       )     ⁢     (       R   2     +     R   U       )                   (   15   )                 Divide   ⁢           ⁢     Eq   .           ⁢     (   1   )       ⁢           ⁢   by   ⁢           ⁢     Eq   .           ⁢     (   2   )       ⁢     :       ⁢     
     ⁢       V     V   R       =       (       R   1     +     R   REF       )       R   REF         ⁢           ⁢     V   =       V   R     ×       (       R   1     +     R   REF       )       R   REF                   (   16   )             
 
 From Eq. (15) by Eq. (16):  
               Δ   ⁢           ⁢   V     =         V   R     ×       R   2     ⁡     (       R   1     +     R   REF       )       ×     (       R   D     -     R   U       )           R   REF     ×     (       R   2     +     R   D       )     ⁢     (       R   2     +     R   U       )                 (   17   )             
 
  FIG. 5 :
 
 V=I   1 ×( R   REF   +R   1 )  (1)
 
 V   R   =I   1   ×R   1   (2)
 
 Divide Eq. (1) by Eq. (2):  
                 V     V   R       =         (       R   REF     +     R   1       )       R   1       =         R   REF       R   1       +   1         ⁢     
     ⁢         R   REF       R   1       =       (     V   -     V   R       )       V   R                 (   3   )               V   =       I   2     ×     (       R   U     +     R   2       )               (   4   )                 V   R     =       I   2     ×     R   2               (   5   )                 V   R     =       I   3     ×     R   3               (   6   )             
 
 From Eq. (5) and (6):  
                 I   3       I   2       =       R   2       R   3               (   7   )             
 
 From Eq. (3), (4), and (5):  
                   R   REF       R   1       =           I   2     ×     (       R   U     +     R   2       )       -       I   2     ×     R   2             I   2     ×     R   2           ⁢     
     ⁢         R   REF       R   1       =             I   2     ×     R   U       +       I   2     ×     R   2       -       I   2     ×     R   2             I   2     ×     R   2         =       R   U       R   2           ⁢     
     ⁢         R   1       R   REF       =         R   2       R   U       =     K   ⁡     (   Constant   )                   (   8   )                 Δ   ⁢           ⁢   V     =       V   1     -   V             (   9   )                 V   1     =       I   3     ×     (       R   D     +     R   3       )               (   10   )               V   =       I   2     ×     (       R   U     +     R   2       )               (   11   )                 V   R     =         I   1     ×     R   1       =         I   2     ×     R   2       =       I   3     ×     R   3                   (   12   )             
 
 From Eq. (9), (10), and (11):
 
Δ V=I   3 ×( R   D   +R   3 )− I   2 ×( R   U   +R   2 )  (13)
 
 From Eq. (12) and (13):  
         Δ   ⁢           ⁢   v     =           V   R       R   3       ×     (       R   D     +     R   3       )       -         V   R       R   2       ×     (       R   U     +     R   2       )             
 
 Since R 2 =R 3 ,  
                 Δ   ⁢           ⁢   V     =         V   R       R   2       ×     (       R   D     +     R   2     -     R   U     -     R   2       )         ⁢     
     ⁢       Δ   ⁢           ⁢   V     =         V   R       R   2       ×     (       R   D     -     R   U       )                 (   14   )             
 
  FIG. 6 :
 
 V=I   1 ×( R   REF   +R   1 )  (1)
 
 V   R   =I   1   ×R   REF   (2)
 
 Divide Eq. (1) by Eq. (2):  
                 V     V   R       =         (       R   REF     +     R   1       )       R   REF       =         R   1       R   REF       +   1         ⁢     
     ⁢         R   1       R   REF       =       (     V   -     V   R       )       V   R                 (   3   )               V   =       I   2     ×     (       R   U     +     R   2       )               (   4   )                 V   R     =       I   2     ×     R   U               (   5   )             
 
 From Eq. (3), (4), and (5):  
                   R   1       R   REF       =       [         I   2     ×     (       R   U     +     R   2       )       -       I   2     ×     R   U         ]       (       I   2     ×     R   U       )         ⁢     
     ⁢         R   1       R   REF       =         [       I   2     ×     R   2       ]       (       I   2     ×     R   U       )       =         R   2       R   U       =     K   ⁡     (   Constant   )                     (   6   )                 Δ   ⁢           ⁢   V     =     V   -     V   1               (   7   )                 V   1     =       I   3     ×     (       R   D     +     R   3       )               (   8   )               V   =       I   2     ×     (       R   U     +     R   2       )               (   9   )                 V   R     =         I   1     ×     R   REF       =         I   2     ×     R   U       =       I   3     ×     R   D                   (   10   )             
 
 From Eq. (7), (8), and (9):  
                 Δ   ⁢           ⁢   V     =         I   2     ⨯     (       R   U     +     R   2       )       -       I   3     ⨯     (       R   D     +     R   3       )           ⁢     
     ⁢       Δ   ⁢           ⁢   V     =       I   3     ⁡     [           I   2       I   3       ⨯     (       R   U     +     R   2       )       -     (       R   D     +     R   3       )       ]                 (   11   )             
 
 From Eq. (10) and (11):  
               Δ   ⁢           ⁢   V     =       I   3     ⁡     [           R   D       R   U       ×     (       R   U     +     R   2       )       -     (       R   D     +     R   3       )       ]                                   Δ   ⁢           ⁢   V     =         I   3       R   U       ×     [         R   D     ×     R   U       +       R   D     ×     R   2       -       R   D     ×     R   U       -       R   U     ×     R   3         ]         ⁢     
     ⁢           S   ⁢   ince     ⁢           ⁢     R   2       =     R   3       ,                               Δ   ⁢           ⁢   V     =           I   3     ×     R   2         R   U       ×     [       R   D     -     R   U       ]               (   12   )             
 
 From Eq. (10) and (12):  
               Δ   ⁢           ⁢   V     =           V   R     ×     R   2           R   U     ×     R   D         ×     [       R   D     -     R   U       ]               (   13   )