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Triggering an ultrasonic sensor
Homemade soil moisture analog value reading jumping from 0 to maxDetect movement on moving robot? PIR SensorAlgorithm to detect motion using ultrasonic sensorIs a HC-SR04 fast enough to make decisions on moving robot (4 Johnson motors 300 rpm)?Problem with my Arduino projectAmplify Piezo SensorWeird probled with TFT LCD screen and DHT22 sensorHow do I use digital pin on sensor to analog on arduino unoPowering Atmega328P-Pu externallyProblem with Pulse Sensor signal
I made a smart blind stick for my class project. It uses an ultrasonic sensor to detect obstacles and I was asked this question in VIVA...
Why is it necessary for giving a digital high for 10 μS on the trigger (Trig) pin of an ultrasonic sensor ?
Why 10 μS only?
I changed it to 5 μS then to 20 μS and to 50 μS and it still works fine! Why so?
arduino-uno sensors
edited Apr 19 at 6:40
Greenonline
2,26251839
asked Apr 19 at 6:01
manishmanish
62
New contributor
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
I made a smart blind stick for my class project. It uses an ultrasonic sensor to detect obstacles and I was asked this question in VIVA...
Why is it necessary for giving a digital high for 10 μS on the trigger (Trig) pin of an ultrasonic sensor ?
Why 10 μS only?
I changed it to 5 μS then to 20 μS and to 50 μS and it still works fine! Why so?
arduino-uno sensors
edited Apr 19 at 6:40
Greenonline
2,26251839
asked Apr 19 at 6:01
manishmanish
62
New contributor
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
obviously i googled for an explanation but couldn't find my answer for "Why 10u" and neither could i find it in your answer 😒
– manish
Apr 19 at 6:22
|
I made a smart blind stick for my class project. It uses an ultrasonic sensor to detect obstacles and I was asked this question in VIVA...
Why is it necessary for giving a digital high for 10 μS on the trigger (Trig) pin of an ultrasonic sensor ?
Why 10 μS only?
I changed it to 5 μS then to 20 μS and to 50 μS and it still works fine! Why so?
arduino-uno sensors
edited Apr 19 at 6:40
Greenonline
2,26251839
asked Apr 19 at 6:01
manishmanish
62
New contributor
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I made a smart blind stick for my class project. It uses an ultrasonic sensor to detect obstacles and I was asked this question in VIVA...
Why is it necessary for giving a digital high for 10 μS on the trigger (Trig) pin of an ultrasonic sensor ?
Why 10 μS only?
I changed it to 5 μS then to 20 μS and to 50 μS and it still works fine! Why so?
arduino-uno sensors
arduino-uno sensors
edited Apr 19 at 6:40
Greenonline
2,26251839
asked Apr 19 at 6:01
manishmanish
62
New contributor
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 19 at 6:40
Greenonline
2,26251839
asked Apr 19 at 6:01
manishmanish
62
New contributor
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 19 at 6:40
Greenonline
2,26251839
edited Apr 19 at 6:40
Greenonline
2,26251839
edited Apr 19 at 6:40
Greenonline
2,26251839
2,26251839
asked Apr 19 at 6:01
manishmanish
62
New contributor
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 19 at 6:01
manishmanish
62
asked Apr 19 at 6:01
manishmanish
62
62
New contributor
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
manish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
obviously i googled for an explanation but couldn't find my answer for "Why 10u" and neither could i find it in your answer 😒
– manish
Apr 19 at 6:22
|
obviously i googled for an explanation but couldn't find my answer for "Why 10u" and neither could i find it in your answer 😒
– manish
Apr 19 at 6:22
obviously i googled for an explanation but couldn't find my answer for "Why 10u" and neither could i find it in your answer 😒
– manish
Apr 19 at 6:22
obviously i googled for an explanation but couldn't find my answer for "Why 10u" and neither could i find it in your answer 😒
– manish
Apr 19 at 6:22
|
1 Answer
1
active
oldest
votes
The specification1 says (emphasis is mine):
using IO trigger distance, the high signal to at least 10us.
So, longer than 10 μS is not a problem.
With respect to less than 10 μS, well 5 μS may also work, but it is out-of-spec and therefore not guaranteed to work and so may not always work correctly, or as expected.
From Source 1:
From Source 2:
As to why must it be 10 μS, well that is probably (at a guess) down to the design of the logic upon the silicon and the resultant propagation delay through the logic gates. Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
To confirm this, you would need to speak to either the designers, or take a look at the mask used to fabricate the IC. Test simulations were probably run during the design phase, and they suggested a minimum trigger pulse of 10 μS. The figure could also be arrived at by simply looking at the design and counting the number of stages and multiplying that by the propagation delay per stage.
However, as Valbhav points out, the HC-SR04 is controlled by a STM8 (actually it is a STC11F), and after having looked at the STM8 programming manual, it is more related to instruction cycles. The Trig pin is, I assume, an interrupt, and from section 3.2 CPU registers
When an interrupt occurs, the CPU registers (CC, X, Y, A, PC) are pushed onto the stack.
This operation takes 9 CPU cycles and uses 9 bytes in RAM
So, 9 CPU cycles just begin to to service the interrupt, and then however many needed to actually do the rest. I can't find any timing diagrams in the Programming Manual which would show the minimum interrupt signal length. This information, is however in the Reference Manual
The length of a CPU cycle will obviously depend upon the clock frequency., which according to this schematic, from An Ultrasonic Shortcut – Getting the HC-SR04 Transducer Up and Running Fast!, is 4 MHz:
So, 10 μS is 40 CPU cycles.
1 Sources for downloading HC-SR04 pdf datasheet:
- Source 1
- Source 2
HC-SR04 Specifications
- Working Voltage: DC 5V
- Working Current: 15mA
- Working Frequency: 40Hz
- Max Range: 4m
- Min Range: 2cm
- Measuring Angle: 15 degree
- Trigger Input Signal: 10µS TTL pulse
- Echo Output Signal Input TTL lever signal and the range in proportion
- Dimension 45 * 20 * 15mm
answered Apr 19 at 6:29
GreenonlineGreenonline
2,26251839
1
But it is still a question why > 10us. Its like the logic that internally lies in ULTRASONIC SENSOR has something like it checks the TRIG pin status for 10us or more if he gets it high it will generate the 8 pulse of 40Khz.
– Vaibhav
Apr 19 at 6:45
Propagation delay through the logic gates? Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
– Greenonline
Apr 19 at 6:51
It contains an STM8 Controller
– Vaibhav
Apr 19 at 7:03
@Vaibhav - Ok, so then, in that case. it is more to do with instruction cycles rather than propagation delays. STM8 Programming manual
– Greenonline
Apr 19 at 7:11
1
It might vary, in some datasheet I had seen STM8.
– Vaibhav
Apr 19 at 7:40
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The specification1 says (emphasis is mine):
using IO trigger distance, the high signal to at least 10us.
So, longer than 10 μS is not a problem.
With respect to less than 10 μS, well 5 μS may also work, but it is out-of-spec and therefore not guaranteed to work and so may not always work correctly, or as expected.
From Source 1:
From Source 2:
As to why must it be 10 μS, well that is probably (at a guess) down to the design of the logic upon the silicon and the resultant propagation delay through the logic gates. Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
To confirm this, you would need to speak to either the designers, or take a look at the mask used to fabricate the IC. Test simulations were probably run during the design phase, and they suggested a minimum trigger pulse of 10 μS. The figure could also be arrived at by simply looking at the design and counting the number of stages and multiplying that by the propagation delay per stage.
However, as Valbhav points out, the HC-SR04 is controlled by a STM8 (actually it is a STC11F), and after having looked at the STM8 programming manual, it is more related to instruction cycles. The Trig pin is, I assume, an interrupt, and from section 3.2 CPU registers
When an interrupt occurs, the CPU registers (CC, X, Y, A, PC) are pushed onto the stack.
This operation takes 9 CPU cycles and uses 9 bytes in RAM
So, 9 CPU cycles just begin to to service the interrupt, and then however many needed to actually do the rest. I can't find any timing diagrams in the Programming Manual which would show the minimum interrupt signal length. This information, is however in the Reference Manual
The length of a CPU cycle will obviously depend upon the clock frequency., which according to this schematic, from An Ultrasonic Shortcut – Getting the HC-SR04 Transducer Up and Running Fast!, is 4 MHz:
So, 10 μS is 40 CPU cycles.
1 Sources for downloading HC-SR04 pdf datasheet:
- Source 1
- Source 2
HC-SR04 Specifications
- Working Voltage: DC 5V
- Working Current: 15mA
- Working Frequency: 40Hz
- Max Range: 4m
- Min Range: 2cm
- Measuring Angle: 15 degree
- Trigger Input Signal: 10µS TTL pulse
- Echo Output Signal Input TTL lever signal and the range in proportion
- Dimension 45 * 20 * 15mm
answered Apr 19 at 6:29
GreenonlineGreenonline
2,26251839
1
But it is still a question why > 10us. Its like the logic that internally lies in ULTRASONIC SENSOR has something like it checks the TRIG pin status for 10us or more if he gets it high it will generate the 8 pulse of 40Khz.
– Vaibhav
Apr 19 at 6:45
Propagation delay through the logic gates? Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
– Greenonline
Apr 19 at 6:51
It contains an STM8 Controller
– Vaibhav
Apr 19 at 7:03
@Vaibhav - Ok, so then, in that case. it is more to do with instruction cycles rather than propagation delays. STM8 Programming manual
– Greenonline
Apr 19 at 7:11
1
It might vary, in some datasheet I had seen STM8.
– Vaibhav
Apr 19 at 7:40
|
show 1 more comment
The specification1 says (emphasis is mine):
using IO trigger distance, the high signal to at least 10us.
So, longer than 10 μS is not a problem.
With respect to less than 10 μS, well 5 μS may also work, but it is out-of-spec and therefore not guaranteed to work and so may not always work correctly, or as expected.
From Source 1:
From Source 2:
As to why must it be 10 μS, well that is probably (at a guess) down to the design of the logic upon the silicon and the resultant propagation delay through the logic gates. Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
To confirm this, you would need to speak to either the designers, or take a look at the mask used to fabricate the IC. Test simulations were probably run during the design phase, and they suggested a minimum trigger pulse of 10 μS. The figure could also be arrived at by simply looking at the design and counting the number of stages and multiplying that by the propagation delay per stage.
However, as Valbhav points out, the HC-SR04 is controlled by a STM8 (actually it is a STC11F), and after having looked at the STM8 programming manual, it is more related to instruction cycles. The Trig pin is, I assume, an interrupt, and from section 3.2 CPU registers
When an interrupt occurs, the CPU registers (CC, X, Y, A, PC) are pushed onto the stack.
This operation takes 9 CPU cycles and uses 9 bytes in RAM
So, 9 CPU cycles just begin to to service the interrupt, and then however many needed to actually do the rest. I can't find any timing diagrams in the Programming Manual which would show the minimum interrupt signal length. This information, is however in the Reference Manual
The length of a CPU cycle will obviously depend upon the clock frequency., which according to this schematic, from An Ultrasonic Shortcut – Getting the HC-SR04 Transducer Up and Running Fast!, is 4 MHz:
So, 10 μS is 40 CPU cycles.
1 Sources for downloading HC-SR04 pdf datasheet:
- Source 1
- Source 2
HC-SR04 Specifications
- Working Voltage: DC 5V
- Working Current: 15mA
- Working Frequency: 40Hz
- Max Range: 4m
- Min Range: 2cm
- Measuring Angle: 15 degree
- Trigger Input Signal: 10µS TTL pulse
- Echo Output Signal Input TTL lever signal and the range in proportion
- Dimension 45 * 20 * 15mm
answered Apr 19 at 6:29
GreenonlineGreenonline
2,26251839
1
But it is still a question why > 10us. Its like the logic that internally lies in ULTRASONIC SENSOR has something like it checks the TRIG pin status for 10us or more if he gets it high it will generate the 8 pulse of 40Khz.
– Vaibhav
Apr 19 at 6:45
Propagation delay through the logic gates? Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
– Greenonline
Apr 19 at 6:51
It contains an STM8 Controller
– Vaibhav
Apr 19 at 7:03
@Vaibhav - Ok, so then, in that case. it is more to do with instruction cycles rather than propagation delays. STM8 Programming manual
– Greenonline
Apr 19 at 7:11
1
It might vary, in some datasheet I had seen STM8.
– Vaibhav
Apr 19 at 7:40
|
show 1 more comment
The specification1 says (emphasis is mine):
using IO trigger distance, the high signal to at least 10us.
So, longer than 10 μS is not a problem.
With respect to less than 10 μS, well 5 μS may also work, but it is out-of-spec and therefore not guaranteed to work and so may not always work correctly, or as expected.
From Source 1:
From Source 2:
As to why must it be 10 μS, well that is probably (at a guess) down to the design of the logic upon the silicon and the resultant propagation delay through the logic gates. Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
To confirm this, you would need to speak to either the designers, or take a look at the mask used to fabricate the IC. Test simulations were probably run during the design phase, and they suggested a minimum trigger pulse of 10 μS. The figure could also be arrived at by simply looking at the design and counting the number of stages and multiplying that by the propagation delay per stage.
However, as Valbhav points out, the HC-SR04 is controlled by a STM8 (actually it is a STC11F), and after having looked at the STM8 programming manual, it is more related to instruction cycles. The Trig pin is, I assume, an interrupt, and from section 3.2 CPU registers
When an interrupt occurs, the CPU registers (CC, X, Y, A, PC) are pushed onto the stack.
This operation takes 9 CPU cycles and uses 9 bytes in RAM
So, 9 CPU cycles just begin to to service the interrupt, and then however many needed to actually do the rest. I can't find any timing diagrams in the Programming Manual which would show the minimum interrupt signal length. This information, is however in the Reference Manual
The length of a CPU cycle will obviously depend upon the clock frequency., which according to this schematic, from An Ultrasonic Shortcut – Getting the HC-SR04 Transducer Up and Running Fast!, is 4 MHz:
So, 10 μS is 40 CPU cycles.
1 Sources for downloading HC-SR04 pdf datasheet:
- Source 1
- Source 2
HC-SR04 Specifications
- Working Voltage: DC 5V
- Working Current: 15mA
- Working Frequency: 40Hz
- Max Range: 4m
- Min Range: 2cm
- Measuring Angle: 15 degree
- Trigger Input Signal: 10µS TTL pulse
- Echo Output Signal Input TTL lever signal and the range in proportion
- Dimension 45 * 20 * 15mm
answered Apr 19 at 6:29
GreenonlineGreenonline
2,26251839
The specification1 says (emphasis is mine):
using IO trigger distance, the high signal to at least 10us.
So, longer than 10 μS is not a problem.
With respect to less than 10 μS, well 5 μS may also work, but it is out-of-spec and therefore not guaranteed to work and so may not always work correctly, or as expected.
From Source 1:
From Source 2:
As to why must it be 10 μS, well that is probably (at a guess) down to the design of the logic upon the silicon and the resultant propagation delay through the logic gates. Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
To confirm this, you would need to speak to either the designers, or take a look at the mask used to fabricate the IC. Test simulations were probably run during the design phase, and they suggested a minimum trigger pulse of 10 μS. The figure could also be arrived at by simply looking at the design and counting the number of stages and multiplying that by the propagation delay per stage.
However, as Valbhav points out, the HC-SR04 is controlled by a STM8 (actually it is a STC11F), and after having looked at the STM8 programming manual, it is more related to instruction cycles. The Trig pin is, I assume, an interrupt, and from section 3.2 CPU registers
When an interrupt occurs, the CPU registers (CC, X, Y, A, PC) are pushed onto the stack.
This operation takes 9 CPU cycles and uses 9 bytes in RAM
So, 9 CPU cycles just begin to to service the interrupt, and then however many needed to actually do the rest. I can't find any timing diagrams in the Programming Manual which would show the minimum interrupt signal length. This information, is however in the Reference Manual
The length of a CPU cycle will obviously depend upon the clock frequency., which according to this schematic, from An Ultrasonic Shortcut – Getting the HC-SR04 Transducer Up and Running Fast!, is 4 MHz:
So, 10 μS is 40 CPU cycles.
1 Sources for downloading HC-SR04 pdf datasheet:
- Source 1
- Source 2
HC-SR04 Specifications
- Working Voltage: DC 5V
- Working Current: 15mA
- Working Frequency: 40Hz
- Max Range: 4m
- Min Range: 2cm
- Measuring Angle: 15 degree
- Trigger Input Signal: 10µS TTL pulse
- Echo Output Signal Input TTL lever signal and the range in proportion
- Dimension 45 * 20 * 15mm
answered Apr 19 at 6:29
GreenonlineGreenonline
2,26251839
edited Apr 19 at 7:44
answered Apr 19 at 6:29
GreenonlineGreenonline
2,26251839
answered Apr 19 at 6:29
GreenonlineGreenonline
2,26251839
answered Apr 19 at 6:29
GreenonlineGreenonline
2,26251839
2,26251839
1
But it is still a question why > 10us. Its like the logic that internally lies in ULTRASONIC SENSOR has something like it checks the TRIG pin status for 10us or more if he gets it high it will generate the 8 pulse of 40Khz.
– Vaibhav
Apr 19 at 6:45
Propagation delay through the logic gates? Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
– Greenonline
Apr 19 at 6:51
It contains an STM8 Controller
– Vaibhav
Apr 19 at 7:03
@Vaibhav - Ok, so then, in that case. it is more to do with instruction cycles rather than propagation delays. STM8 Programming manual
– Greenonline
Apr 19 at 7:11
1
It might vary, in some datasheet I had seen STM8.
– Vaibhav
Apr 19 at 7:40
|
show 1 more comment
1
But it is still a question why > 10us. Its like the logic that internally lies in ULTRASONIC SENSOR has something like it checks the TRIG pin status for 10us or more if he gets it high it will generate the 8 pulse of 40Khz.
– Vaibhav
Apr 19 at 6:45
Propagation delay through the logic gates? Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
– Greenonline
Apr 19 at 6:51
It contains an STM8 Controller
– Vaibhav
Apr 19 at 7:03
@Vaibhav - Ok, so then, in that case. it is more to do with instruction cycles rather than propagation delays. STM8 Programming manual
– Greenonline
Apr 19 at 7:11
1
It might vary, in some datasheet I had seen STM8.
– Vaibhav
Apr 19 at 7:40
1
1
But it is still a question why > 10us. Its like the logic that internally lies in ULTRASONIC SENSOR has something like it checks the TRIG pin status for 10us or more if he gets it high it will generate the 8 pulse of 40Khz.
– Vaibhav
Apr 19 at 6:45
But it is still a question why > 10us. Its like the logic that internally lies in ULTRASONIC SENSOR has something like it checks the TRIG pin status for 10us or more if he gets it high it will generate the 8 pulse of 40Khz.
– Vaibhav
Apr 19 at 6:45
Propagation delay through the logic gates? Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
– Greenonline
Apr 19 at 6:51
Propagation delay through the logic gates? Most, if not all, logic ICs specify a minimum amount of time required for a signal level, in order to allow the signal to propagate through the transistors that make up the logic, so that they switch correctly. PN junctions require a finite time for the electrons and holes to move, and when there are a number of them in sequential stages, that time is multiplied by the number of stages that the signal needs to propagate through. That is a very simplified way of looking at it.
– Greenonline
Apr 19 at 6:51
It contains an STM8 Controller
– Vaibhav
Apr 19 at 7:03
It contains an STM8 Controller
– Vaibhav
Apr 19 at 7:03
@Vaibhav - Ok, so then, in that case. it is more to do with instruction cycles rather than propagation delays. STM8 Programming manual
– Greenonline
Apr 19 at 7:11
@Vaibhav - Ok, so then, in that case. it is more to do with instruction cycles rather than propagation delays. STM8 Programming manual
– Greenonline
Apr 19 at 7:11
1
1
It might vary, in some datasheet I had seen STM8.
– Vaibhav
Apr 19 at 7:40
It might vary, in some datasheet I had seen STM8.
– Vaibhav
Apr 19 at 7:40
|
show 1 more comment
obviously i googled for an explanation but couldn't find my answer for "Why 10u" and neither could i find it in your answer 😒
– manish
Apr 19 at 6:22