Automatic Transaxle Features & Operation (FS5A-EL): Other Mazda 3 BL

POWER FLOW OUTLINE

1. In the powertrain mechanism, hydraulic pressure is transmitted from the control valves or shift solenoid A, B, C (duty-cycle type), pressure control solenoid B (duty-cycle type) or shift solenoid F (ON/OFF type) operate the clutches and brakes, and the planetary gear changes the gear ratio according to the vehicle driving condition.

POWER FLOW STRUCTURE

1. The powertrain mechanism of the FS5A-EL type consists of four pairs of clutches, two pairs of brakes, band brake, two pairs of one-way clutches, and three pairs of single type planetary gears.

Scheme 536

Scheme 537

FORWARD CLUTCH, 3-4 CLUTCH, REVERSE CLUTCH, DIRECT CLUTCH, LOW AND REVERSE BRAKE, REDUCTION BRAKE OUTLINE

1. Each multi-disc type clutch and brake has the following function and operates in the gear position(s) as shown in the figure.

POWER FLOW OPERATION

Scheme 538

Scheme 539

1. The basic structure is as shown in the figure below. In figure A, the fluid is in the clutch plates (drive plates, driven plates) and the power is not transmitted because of the fluid slippage on each plate. Figure B shows the clutch condition with the hydraulic pressure acted on the piston; the drive plates and the driven plates are pressed tightly together to transmit the clutch drum rotation speed to the hub. When the hydraulic pressure in the piston is drained, the clutches are separated because of the return spring and return to the condition in figure A.
2. The dished plates used for the reverse clutch and the low and reverse brake reduce the shock caused by the sudden clutch engagement. The piston check ball built in the 2-4 brake drum (reverse clutch) drains the ATF only during freewheel to prevent the hydraulic pressure from increasing to half-engage the clutches because of the residual ATF. In the forward clutch, the 3-4 clutch and the direct clutch, the centrifugal balance chamber is installed opposite the general clutch chamber. The centrifugal balance chamber of forward clutch, 3-4 clutch is always filled with the ATF from the exclusive lubrication passage of the turbine shaft. The centrifugal balance chamber of Direct clutch is always filled with the ATF from the exclusive lubrication passage of the counter shaft.

Scheme 540

CENTRIFUGAL BALANCE CLUTCH OUTLINE

1. A centrifugal balance clutch mechanism, which cancels the centrifugal oil pressure, has been adopted to improve clutch control.
2. A bonded seal piston (press-worked component of a piston and a seal) has been adopted for each clutch and brake to reduce the piston size and weight.

CENTRIFUGAL BALANCE CLUTCH STRUCTURE

1. The centrifugal balance clutch chambers are installed opposite the clutch chamber. The centrifugal balance clutch chambers are constantly filled with ATF from an exclusive hydraulic passage of the turbine shaft.

When clutch pressure is not applied

1. When the clutch drum rotates, centrifugal force acts on the residual ATF in the clutch chamber to push against the piston. However, centrifugal force also acts on the ATF filling the centrifugal balance clutch chamber to push back the piston. As a result, the two forces are cancelled out and the piston remains stationary, thus preventing clutch engagement.

When clutch pressure is applied

1. When clutch pressure is applied to the clutch chamber, the clutch pressure overcomes the oil pressure and spring force in the opposite centrifugal balance clutch chamber, and pushes the piston to engage the clutches. Because the centrifugal force acting on the clutch pressure in the clutch chamber is canceled by another centrifugal force acting on the ATF filling the centrifugal balance clutch chamber, the influence of the centrifugal force created by the clutch drum revolution speed is eliminated. As a result, stable piston pushing force is obtained in all rotation ranges, and smoother shifts can be made.

Scheme 541

2-4 BRAKE BAND OUTLINE

1. The 2-4 brake band locks the 2-4 brake drum and fixes the rear sun gear. The 2-4 brake band operates in 2GR, 4GR or 5GR.

2-4 BRAKE BAND STRUCTURE

1. The 2-4 brake band is set to wind the 2-4 brake drum and one end of the 2-4 brake band is fixed with a band strut. The servo piston is in the transaxle case.

PLANETARY GEAR OUTLINE

1. The planetary gear is a transaxle which converts the driving force of the turbine shaft to the optimal driving force and transmits it to the output gear through the operation of each clutch and brake.
2. A double arranged gear with a single planetary gear unit is adopted as the main shifting mechanism for the planetary gear; they are the front planetary gear and the rear planetary gear (from converter side).
3. A single planetary gear unit is adopted as the sub-shifting mechanism.
4. The planetary gear consists of the internal gear, planetary carrier (pinion gears), and the sun gear.

PLANETARY GEAR STRUCTURE

1. The front planetary gear is integrated with the one-way clutch outer race and engaged with the drive plate of the low and reverse brake. Because of this, when the front planetary gear rotates, the one-way clutch outer race and the drive plate of the low and reverse brake also rotate together.
2. The front sun gear is installed inside of the front pinion gears, and the front internal gear is installed outside of the front pinion gears. The front sun gear is engaged with the forward clutch hub, and the front internal gear is engaged with the rear planetary carrier.
3. The rear planetary gear and the rear pinion gear have the rear sun gear installed inside and the rear internal gear outside. The rear sun gear is engaged with the turbine shaft via the 2-4 brake drum, and the rear internal gear is engaged with the primary gear via the front planetary carrier.
4. For the secondary planetary gear, the secondary sun gear is built inside the secondary pinion gear, and the secondary internal gear is built externally. The secondary sun gear is connected to the direct clutch drum, and the secondary gear is connected to the secondary internal gear. The secondary planetary carrier is combined with the counter shaft, and also connected with the drive plate of the clutch.

Scheme 542

Scheme 543

Scheme 544

1. The planetary gear works as a transaxle when the sun gear and the internal gear are engaged.
2. The sun gear, installed inside of the pinion gears, and the internal gear, installed outside of the pinion gears, are engaged with their respective gears. The sun gear and the internal gear rotate on the center of the planetary gear.
3. The pinion gears turn in the following two ways: On their own centers (rotation) On the center of the planetary gear (revolution)

Gear ratio of each range

1. The relation between each element of the planetary gear set and the rotation speed is generally indicated in the formula below. (Z R +Z S ) N C =Z R N R +Z S N S : formula (1) In this formula Z stands for the number of teeth, N stands for the rotation speed, and R, S, C stand internal gear for each gear element (refer to the table below).

Scheme 545

Number of teeth and symbol of each gear

GEAR RATIO RANGE

Scheme 546

Gear rotation speed

GEAR ROTATION SPEED SPECIFICATIONS

1. Suppose the reduction ratio on the main shifting side is i 1 , i 1 =N SF /N CF .
2. From the result N RF =0 in formula (1), the rotation speed of the front planetary gear unit can be calculated using the following formula: (Z RF +Z SF )N CF =Z SF N SF Therefore, i 1 =N SF /N CF =(Z RF +Z SF )/Z SF =(89+49)/49=2.8163.
3. Because the reduction ratio on the main shifting side is transmitted from the primary gear to the secondary gear, it can be calculated using the following formula: The reduction ratio of the primary/secondary gear A = the number of primary gear teeth/the number of secondary gear teeth Therefore, A=82/86=0.9535
4. Suppose the reduction ratio on the sub-shifting side is ii 1 , ii 1 =N RS /N CS .
5. From the result Nsg=0 in formula (1), the rotation speed of the secondary planetary gear unit can be calculated using the following formula. (Z RS +Z SS )N CS =Z SS N RS Therefore, ii 1 =N RS /N CS =(Z RS +Z SS )/Z RS =(89+31)/89=1.3483 And the reduction ratio of 1st gear= i 1 x A x ii 1 =2.8163 x 0.9535 x 1.3483=3.620 As a result, the reduction ratio of 1st gear is 3.620.

Scheme 547

Gear rotation speed

GEAR ROTATION SPEED SPECIFICATIONS

Note. The front internal gear and the rear planetary carrier are integrated. The front planetary carrier and the rear internal gear rotate at the same speed.

1. Suppose the reduction ratio on the main shifting side is i 2 , i 2 =N SF /N R .
2. From formula (1), the relation between the gear ratio in second gear and the rotation speeds of the front and the rear planetary gar sets is indicated in formulas (2) and (3). (Z RF +Z SF )N R =Z RF N C +Z SF N SF : (2) (Front planetary gear set) (Z RR +Z SR ) N C =Z RR N R +Z SR N SF : (3) (Rear planetary gear set)
3. From the result N SR =0 in formula (3). N C =(Z RR /(Z RR +Z SR ))N R :(4)
4. Here we substitute formula (4) in formula (2). Z SR N SF = (((Z RR +Z SR ) (Z RF +Z SF ) -Z RF Z RR ) / (Z RR +Z SR )) NR Therefore, i 2 =N SF /N R = (((Z RR +Z SR ) (Z RF +Z SF ) -Z RF Z RR ) / (Z SF (Z RR +Z SR ))) NR = ((98+37)(89+49) -89 x 98) / (49 (98+37)) =1.4978
5. Because the reduction ratio on the main shifting side is transmitted from the primary gear to the secondary gear, it can be calculated using the following formula: The reduction ratio of the primary/secondary gear A = the number of primary gear teeth/the number of secondary gear teeth Therefore, A=82/86=0.9535
6. Suppose the reduction ratio on the sub-shifting side is ii 2 , ii 2 =N RS /N CS .
7. From the result N SS =O in formula (1), the rotation speed of the secondary planetary gear unit can be calculated using the following formula. (Z RS +Z SS )N CS =Z SS N RS Therefore, ii 2 =N RS /N CS =(Z RS +Z SS )/Z RS =(89+31 )/89=1.3483 And the reduction ratio of 2nd gear= i 2 x A x ii 2 =1.4978 x 0.9535 x 1.3483=1.925 As a result, the reduction ratio of 2nd gear is 1.925.

Scheme 548

Gear rotation speed

GEAR ROTATION SPEED SPECIFICATIONS

1. Here we have the result on N RF =N SF .
2. Suppose the reduction ratio on the main shifting side is i 3 , i 3 =N SF /N CF .
3. From the result of N rF =N sf in formula (1), the relation between the gear ratio in 3rd gear and the rotation speed of the front planetary gar set is indicated in the following formula: (N RF +Z SF N CF = (Z RF +Z SF ) N RF Therefore, i 3 =N RF /N CF = (Z RF +Z SF ) / (Z RF +Z SF ) = (89+49) / (89+49) =1.000
4. Because the reduction ratio on the main shifting side is transmitted from the primary gear to the secondary gear, it can be calculated using the following formula: The reduction ratio of the primary/secondary gear A = the number of primary gear teeth/the number of secondary gear teeth Therefore, A=82/86=0.9535
5. Suppose the reduction ratio on the sub-shifting side is ii 3 , ii 3 =N RS /N CS .
6. From the result N gs =0 in formula (1), the rotation speed of the secondary planetary gear unit can be calculated using the following formula. (Z RS +Z SS )N CS =Z SS N RS Therefore, ii 3 =N RS /N CS =(Z RS +Z SS )/Z RS =(89+31 )/89=1.3483 And the reduction ratio of 3rd gear= i 3 x A x ii 3 =1.000 x 0.9535 x 1.3483=1.285 As a result, the reduction ratio of 3rd gear is 1.285.

Scheme 549

Gear rotation speed

GEAR ROTATION SPEED SPECIFICATIONS

1. Suppose gear ratio in fourth gear is i 4 , i 4 =N CR /N RR
2. From the result of N gR =O in formula (2), the relation between the gear ratio in fourth gear and the rotation speed of the rear planetary gear set is indicated in the following formula: (Z RR +Z SR ) N CR =Z RR N RR Therefore, i 4 =N CR /N RR =Z RR / (Z RR +Z SR ) =98/ (98+37) =0.7259
3. Because the reduction ratio on the main shifting side is transmitted from the primary gear to the secondary gear, it can be calculated using the following formula: The reduction ratio of the primary/secondary gear A = the number of primary gear teeth/the number of secondary gear teeth Therefore, A=82/86=0.9535
4. Suppose the reduction ratio on the sub-shifting side is ii 4 , ii 4 =N RS /N CS .
5. From the result N SS =0 in formula (1), the rotation speed of the secondary planetary gear unit can be calculated using the following formula. (Z RS +Z SS )N CS =Z SS N RS Therefore, ii 4 =N RS /N CS =(Z RS +Z SS )/Z RS =(89+31 )/89=1.3483 And the reduction ratio of 4th gear= i 4 x A x ii 4 =0.7259 x 0.9535 x 1.3483=0.933 As a result, the reduction ratio of 4th gear is 0.933.

Scheme 550

Gear rotation speed

GEAR ROTATION SPEED SPECIFICATIONS

1. Suppose gear ratio in fifth gear is i5, i 5 =N CR /N RR
2. From the result of N SR =0 in formula (2), the relation between the gear ratio in fourth gear and the rotation speed of the rear planetary gear set is indicated in the following formula: (Z RR +Z SR ) N CR =Z RR N RR Therefore, i 5 =N CR /N RR =Z RR /(Z RR +Z SR ) =98/ (98+37) =0.7259
3. Because the reduction ratio on the main shifting side is transmitted from the primary gear to the secondary gear, it can be calculated using the following formula: The reduction ratio of the primary/secondary gear A = the number of primary gear teeth/the number of secondary gear teeth Therefore, A=82/86=0.9535
4. Suppose the reduction ratio on the sub-shifting side is ii 5 , ii 5 =N RS /N CS .
5. From the result N RS = N SS in formula (1), the rotation speed of the secondary planetary gear unit can be calculated using the following formula. (Z RS +Z SS )N CS =(Z RS Z SS )N RS Therefore, ii 5 =N RS /N CS =(Z RS +Z SS )/(Z RS +Z SS )=(89+31 )/(89+31)=1.000 And the reduction ratio of 5th gear= i 5 x A x ii 5 =0.7259 x 0.9535 x 1.000=0.692 As a result, the reduction ratio of 5th gear is 0.692.

Scheme 551

Gear rotation speed

GEAR ROTATION SPEED SPECIFICATIONS

1. Suppose gear ratio in reverse gear is i REV , i REV =N SR /N RR
2. From the result of N CR =0 in formula (2), the relation between the gear ratio during reverse movement and the rotation speed of the planetary gar set is indicated in the formula below. (Z RR +Z SR ) 0=Z RR N RR +Z SR N SR Therefore, i REV =N SR /N RR =Z RR /Z SR =-98/37=-2.6486
3. Because the reduction ratio on the main shifting side is transmitted from the primary gear to the secondary gear, it can be calculated using the following formula: The reduction ratio of the primary/secondary gear A = the number of primary gear teeth/the number of secondary gear teeth Therefore, A=82/86=0.9535
4. Suppose the reduction ratio on the sub-shifting side is ii REV , ii REV =N RS /N CS .
5. From the result N ss =0 in formula (1), the rotation speed of the secondary planetary gear unit can be calculated using the following formula. (Z RS +Z SS )N CS =Z SS N RS Therefore, ii REV =N RS /N CS =(Z RS +Z SS )/Z RS =(89+31 )/89=1.3483 And the reduction ratio of reverse gear= i REV x A x ii REV =-2.6486 x 0.9535 x 1.3483=-3.405 As a result, the reduction ratio of reverse gear is -3.405.

PARKING MECHANISM OUTLINE

1. When the selector lever is shifted to P position, the parking pawl engages the parking gear and locks the output gear (i.e., rotation of the driving wheels).

PARKING MECHANISM STRUCTURE

1. The parking pawl is installed in the transaxle case via the parking pawl shaft and pushed to the support actuator by the return spring except in P position. The parking rod component is designed to slide on the support actuator and connected to the manual plate.

OUTPUT GEAR OUTLINE

1. The two-step final drive mechanism has been adopted by arranging the secondary gear and the output gear on the output gear shaft to miniaturize the transaxle.

Scheme 552

OIL PUMP OUTLINE

1. The light-weight, compact, and quiet trochoid gear type oil pump has been adopted to reduce the pump driving torque.
2. The direct drive type oil pump has been adopted and placed behind the torque converter.

Scheme 553

OIL PUMP STRUCTURE

1. The outer rotor and the inner rotor are installed in the oil pump housing.
2. The inner rotor in the oil pump housing is driven by the torque converter.

Scheme 554

FORWARD CLUTCH, 3-4 CLUTCH HYDRAULIC CIRCUIT OUTLINE

1. By designing exclusive passages for the forward clutch and the 3-4 clutch in the transaxle case, via the oil pump and end cover the hydraulic pressure passages are shortened and control during clutch engagement is improved.

Scheme 555

CONTROL VALVE BODY OUTLINE

1. The primary control valve body has been adopted as the main shifting mechanism.
2. The secondary control valve body has been adopted as the sub-shifting mechanism.
3. Because the clutch engagement pressure is controlled electronically, the hydraulic circuits are simplified, the valve types are reduced, and the control valve body is miniaturized.
4. The nonwoven fabric oil strainer is installed in the primary control valve body to prevent contamination.

Primary Control Valve Body

1. The primary control valve body is composed of three bodies: the upper control valve body, main control valve body, and the solenoid control valve body.

Scheme 556

Secondary Control Valve Body

1. The secondary control valve body is composed of two bodies: the secondary lower control valve body, and secondary main control valve body.

Scheme 557

SHIFT SOLENOID A, B AND C (DUTY-CYCLE TYPE) OUTLINE

1. A clutch pressure direct control, which supplies the clutch pressure directly to each clutch and/or brake, has been adopted. A three-way duty-cycle type solenoids with excellent controllability have been adopted, to improve response.

SHIFT SOLENOID A, B AND C (DUTY-CYCLE TYPE) FUNCTION

1. The duty-cycle type shift solenoid adjusts the amount of output pressure according to the signal from the TCM, and controls the pressure of each clutch.
2. The duty-cycle type shift solenoid, which switches on/off at 50 Hz (20 ms cycle) and controls the output pressure, is adopted. By changing the on time ratio a cycle (0-100%), the solenoid adjusts the time ratio of the open (supply) and close (drain), and maintains the clutch pressure at the designated hydraulic pressure. As a result, the clutch pressure rises when the duty ratio (50 Hz on time ratio) is reduced, and falls when the duty ratio is raised.

Scheme 558

SHIFT SOLENOID D, E AND F (ON/OFF TYPE) OUTLINE

1. A compact, light-weight three-way solenoid has been adopted for shift solenoids D, E and F to reduce consumption discharge amount.

SHIFT SOLENOID FUNCTION

SHIFT SOLENOID D, E AND F (ON/OFF TYPE) FUNCTION

1. An on/off type solenoid valve switches the supply drain of output port according to the electrical current flow switching.

Scheme 559

PRESSURE CONTROL SOLENOID A (LINEAR TYPE) OUTLINE

1. A pressure control solenoid A with high stability in hydraulic pressure has been adopted for the line pressure control.
2. Because the pressure control solenoid controls the hydraulic pressure according to the current value, the degree of freedom in control increases. The controllability is maintained even under aeration, and pressure variation can be reduced.

PRESSURE CONTROL SOLENOID B (DUTY-CYCLE TYPE) OUTLINE

1. A clutch pressure direct control, which supplies the clutch pressure directly to each clutch and/or brake, has been adopted. A three-way duty-cycle type solenoids with excellent controllability have been adopted, to improve response.

PRESSURE CONTROL SOLENOID B (DUTY-CYCLE TYPE) FUNCTION

1. The duty-cycle type shift solenoid adjusts the amount of output pressure according to the signal from the TCM, and controls the pressure of each clutch.
2. The duty-cycle type shift solenoid, which switches on/off at 50 Hz (20 ms cycle) and controls the output pressure, is adopted. By changing the on time ratio a cycle (0-100%), the solenoid adjusts the time ratio of the open (supply) and close (drain), and maintains the 4-5 duty solenoid pressure at the designated hydraulic pressure. As a result, the clutch pressure rises when the duty ratio (50 Hz on time ratio) is reduced, and falls when the duty ratio is raised.

Scheme 560