A heat-loss vacuum gauge may be used to determine gas pressure in an environment by measuring heat transfer rates from a heated sensing element to the gas under certain conditions. In a Pirani-type heat-loss vacuum gauge (“Pirani gauge”), heat transfer rates are measured using a Wheatstone bridge network, which serves both to heat the sensing element and to measure its resistance to obtain a pressure indication.
FIG. 1A shows a simplified schematic diagram of a Wheatstone bridge network of a Pirani gauge. The Pirani gauge includes a temperature sensitive element RS connected between nodes B and C forming a temperature sensitive leg of the Wheatstone bridge. R2 is typically a temperature sensitive resistance designed to have a negligible temperature rise due to a current i2. R2 may include resistances R and RC connected in series between nodes C and D forming a temperature compensating leg of the Wheatstone bridge. R3 and R4 are typically fixed resistances. RS and typically RC are exposed to the vacuum environment (indicated by the dashed box) whose pressure is to be measured. FIG. 1B shows an alternative Wheatstone bridge configuration.
Voltage VB is automatically controlled to maintain the voltage difference between nodes A and C at zero volts. When the potential drop from A to C is zero, the Wheatstone bridge is said to be balanced. Referring again to FIG. 1A, at bridge balance the following conditions exist:iS=i2,  (1)i4=i3,  (2)iSRS=i4R4,  (3)andi2R2=i3R3.  (4)Dividing Eq. (3) by Eq. (4) and using Eqs. (1) and (2) gives
                              RS          =                      β            ⁢                                                  ⁢            R            ⁢                                                  ⁢            2                          ,                                  ⁢        where                            (        5        )                                β        =                                            R              ⁢                                                          ⁢              4                                      R              ⁢                                                          ⁢              3                                .                                    (        6        )            Thus, at bridge balance RS is a constant fraction β of R2.
To achieve a steady state condition in RS at any given pressure, Eq. (7) below must be satisfied:
                              Electrical          ⁢                                          ⁢          power          ⁢                                          ⁢          input          ⁢                                          ⁢          to          ⁢                                          ⁢          RS                =                              Power            ⁢                                                  ⁢            radiated            ⁢                                                  ⁢            by            ⁢                                                  ⁢            RS                    +                      Power            ⁢                                                  ⁢            conducted            ⁢                                                  ⁢            away            ⁢                                                  ⁢            from            ⁢                                                  ⁢            RS            ⁢                                                  ⁢            through            ⁢                                                  ⁢            its            ⁢                                                  ⁢            mechanical            ⁢                                                  ⁢            supports                    +                      Power            ⁢                                                  ⁢            lost            ⁢                                                  ⁢            to            ⁢                                                  ⁢            gas            ⁢                                                  ⁢            by            ⁢                                                  ⁢            RS                                              (        7        )            
A conventional Pirani gauge is calibrated against several known pressures to determine a relationship between unknown pressure, Px, and the power lost to the gas. Then, assuming RS mechanical conduction and radiation losses remain constant, the unknown pressure of the gas, Px, may be directly determined by the power lost to the gas or related to the bridge voltage at bridge balance.
The Pirani gauges shown in FIGS. 1A and 1B, however, suffer from ambient temperature-caused errors. All of the heat loss terms in Eq. (7) are dependent on ambient temperature and on sensing element RS temperature at any given pressure.
Any attempt at pressure measurement with a Pirani gauge without temperature correction will be confused for non-pressure dependent power losses caused by changes in ambient temperature. Thus, all modern Pirani gauges attempt to correct for errors caused by ambient temperature changes.
It has long been known to use for R2 a compensating element RC, with a similar Temperature Coefficient of Resistance (TCR) as the sensing element, in series with a temperature stable element R. This provides temperature compensation for the changes in cold resistance of the sensor, and changes in all three power losses (Eq. 7) which vary with the temperature difference between the sensing element and its surrounding environment.
This method of temperature compensation assumes that if (1) the temperature coefficients of resistance of the sensing and compensating elements are equal; and (2) the change in sensing element resistance can be made to rise in tandem with change in compensating element resistance, then (3) the temperature of the sensing element will rise in tandem with ambient temperature changes. Satisfying these assumptions is highly desirable, of course, because doing so would assure that the temperature difference between the heated sensing element and the surrounding wall at ambient temperature would remain constant as ambient temperature changes.
However, Pirani gauges which utilize a temperature stable element R in series with a temperature sensitive element RC for R2 provide only partial temperature compensation as will now be explained.
Assume that in FIG. 1A, R2 is composed of a temperature sensitive compensating element RC and a temperature stable element R so thatR2=RC+R,  (8)Thus, Eq. (5) derived above for bridge balance may be written asRS=β(RC+R)  (9)where β is defined by Eq. (6) above.
Further, assume that the sensing element RS operates at temperature Ts1, and the compensating element RC operates at temperature TC1 when the ambient temperature of the gauge environment is equal to T1. Thus, whenTAMBIENT=T1,  (10)Eq. (9) may be written asRS(T1)(1+αS(TS1−T1))=β[RC(T1)(1+αC(TC1\−T1))+R].  (11)
Here, RS(T1) is the resistance of the sensing element RS at temperature T1, αS is the TCR of RS at T1, RC(T1) is the resistance of the compensating element RC at temperature T1, and αC is the TCR of RC at T1. Thus, whenTAMBIENT=T2  (12)Eq. (9) may be written asRS(T1)(1+αS(TS2−T))=β[RC(T1)(1+αC(TC2−T1))+R].  (13)Solving Eq. (11) for TS1 gives
                              T                      S            ⁢                                                  ⁢            1                          =                                            [                                                                    β                                          RS                      ⁡                                              (                                                  T                          1                                                )                                                                              ⁡                                      [                                                                                            RC                          ⁡                                                      (                                                          T                              1                                                        )                                                                          ⁢                                                  (                                                      1                            +                                                                                          α                                C                                                            ⁡                                                              (                                                                                                      T                                                                          C                                      ⁢                                                                                                                                                          ⁢                                      1                                                                                                        -                                                                      T                                    1                                                                                                  )                                                                                                              )                                                                    +                      R                                        ]                                                  -                1                            ]                        /                          α              S                                +                                    T              1                        .                                              (        14        )            Solving Eq. (13) for TS2 gives
                              T                      S            ⁢                                                  ⁢            2                          =                                            [                                                                    β                                          RS                      ⁡                                              (                                                  T                          1                                                )                                                                              ⁡                                      [                                                                                            RC                          ⁡                                                      (                                                          T                              1                                                        )                                                                          ⁢                                                  (                                                      1                            +                                                                                          α                                C                                                            ⁡                                                              (                                                                                                      T                                                                          C                                      ⁢                                                                                                                                                          ⁢                                      2                                                                                                        -                                                                      T                                    1                                                                                                  )                                                                                                              )                                                                    +                      R                                        ]                                                  -                1                            ]                        /                          α              S                                +                                    T              1                        .                                              (        15        )            Subtracting Eq. (14) from Eq. (15) gives the temperature change ΔT in the sensing element RS when ambient temperature changes from T1 to T2. Thus,
                              Δ          ⁢                                          ⁢          T                =                                            T                              S                ⁢                                                                  ⁢                2                                      -                          T                              S                ⁢                                                                  ⁢                1                                              =                                    β              ⁡                              (                                                      RC                    ⁡                                          (                                              T                        1                                            )                                                                            RS                    ⁡                                          (                                              T                        1                                            )                                                                      )                                      ⁢                          (                                                α                  C                                                  α                  S                                            )                        ⁢                                          (                                                      T                                          C                      ⁢                                                                                          ⁢                      2                                                        -                                      T                                          C                      ⁢                                                                                          ⁢                      1                                                                      )                            .                                                          (        16        )            Note that an effective compensating element is designed so that its temperature closely follows ambient temperature. Thus, to a very good approximation,TC2−T2=TC1−T1  (17)orTC2−TC1=T2−T1.  (18)Thus, Eq. (16) may be written as
                              Δ          ⁢                                          ⁢          T                =                              β            ⁡                          (                                                RC                  ⁡                                      (                                          T                      1                                        )                                                                    RS                  ⁡                                      (                                          T                      1                                        )                                                              )                                ⁢                      (                                          α                C                                            α                S                                      )                    ⁢                                    (                                                T                  2                                -                                  T                  1                                            )                        .                                              (        19        )            
It is evident from Eq. (19) that the temperature change ΔT in the sensing element RS will be equal to the change in ambient temperature T2−T1 only if
                                          β            ⁡                          (                                                RC                  ⁡                                      (                                          T                      1                                        )                                                                    RS                  ⁡                                      (                                          T                      1                                        )                                                              )                                ⁢                      (                                          α                C                                            α                S                                      )                          =        1.                            (        20        )            
Thus, Pirani gauges using a temperature sensitive compensating element RC in series with a fixed resistance element R for R2 in FIG. 1A provide only partial temperature compensation depending on the choice of β. To further improve temperature compensation, the physical components of the Pirani gauge may be designed to satisfy Eq. 20.
For example, the Temperature Coefficient of Resistance (TCR) and absolute value of the temperature sensitive leg of the Wheatstone bridge may be adjusted (i.e., adjust αC and R) to satisfy Eq. 20. FIG. 1C shows an alternative Wheatstone bridge configuration that allows for precise adjustments to the TCR and absolute value of the temperature sensitive leg R2. The temperature sensitive leg R2 includes a temperature sensitive section 124 in series with a temperature stable section 122. The temperature sensitive section 124 may include a temperature sensitive resistive element 126 in parallel with one or more fixed resistances and the temperature stable section 222 may include one or more fixed resistances connected in parallel. The voltage between nodes A and C is fed back through amplifier 150 to node B to maintain the voltage across nodes A and C at zero volts.
The TCR and absolute value (ac and R) of the temperature sensitive leg R2 may be precisely adjusted by choosing an appropriate temperature sensitive resistive element 126 and appropriate fixed resistances in sections 122 and 124. By carefully choosing the optimum pressure and two temperatures at which the Pirani gauges are individually compensated, perfect temperature compensation is assured only at one pressure and two temperatures. For example, FIG. 1D graphically illustrates the typical performance of the conventional Pirani heat-loss gauge in which nearly perfect temperature compensation is achieved at about 400 Torr at the temperatures 25° Celsius and 44° Celsius. Even though good temperature compensation is achieved over much of the gauge's pressure range (about ±10% indication error between 5 mTorr and 50 Torr and between 200 Torr and 1000 Torr), there is much room for improvement over most of the gauge's full measurement range, especially below 1 mTorr and between 50 and 200 Torr.
Attempts have been made to improve the overall temperature compensation such as measuring the bridge current and summing the bridge current with the bridge voltage. Again, although this will improve temperature compensation at certain pressures, it does not provide temperature compensation over all pressures.
USPTO Backgrounds
USPTO Backgrounds