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What is meaning of active low input in combinational logic circuits?
Gate Output as a Switch?Cross-coupled logic gates and timingSignal-driven 3 output logic gate decoder or switch?What are the rules for combining transistors to form digital circuits?Multiplexing a CircuitWhy do I experience a voltage drop between logic gates when combining multiple gates (7408 and 7402)Explain the internals of digital multiplexersImplementing a function using decoder, encoder and some gatesHow can output from a single logic gate/DIP switch supply input for multiple gates?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I am currently doing self study on combinational logic circuits. I encountered few terms like active low output, active low input. I understood what active low output means (putting not gates at output side). I guess active low means putting not gate at input side.
It will be very helpful if some one can explain this using an example (note that I have knowledge of encoders, decoders, Multiplexers so you can use these in your example).
digital-logic integrated-circuit input
New contributor
$endgroup$
add a comment
|
$begingroup$
I am currently doing self study on combinational logic circuits. I encountered few terms like active low output, active low input. I understood what active low output means (putting not gates at output side). I guess active low means putting not gate at input side.
It will be very helpful if some one can explain this using an example (note that I have knowledge of encoders, decoders, Multiplexers so you can use these in your example).
digital-logic integrated-circuit input
New contributor
$endgroup$
$begingroup$
"Active" means ENABLED. Active_high needs a high, +3.3v, a ONE, a TRUE. Active_low needs a low, 0.0v, a ZERO, a FALSE.
$endgroup$
– analogsystemsrf
8 hours ago
$begingroup$
And, there's arguments about terminology. Because some people (me) would say that "active low" means that a low voltage is interpreted as true or one.
$endgroup$
– TimWescott
8 hours ago
$begingroup$
indeed. You can define True as any level you wish.
$endgroup$
– analogsystemsrf
7 hours ago
add a comment
|
$begingroup$
I am currently doing self study on combinational logic circuits. I encountered few terms like active low output, active low input. I understood what active low output means (putting not gates at output side). I guess active low means putting not gate at input side.
It will be very helpful if some one can explain this using an example (note that I have knowledge of encoders, decoders, Multiplexers so you can use these in your example).
digital-logic integrated-circuit input
New contributor
$endgroup$
I am currently doing self study on combinational logic circuits. I encountered few terms like active low output, active low input. I understood what active low output means (putting not gates at output side). I guess active low means putting not gate at input side.
It will be very helpful if some one can explain this using an example (note that I have knowledge of encoders, decoders, Multiplexers so you can use these in your example).
digital-logic integrated-circuit input
digital-logic integrated-circuit input
New contributor
New contributor
edited 22 mins ago
Voltage Spike
38.6k12 gold badges43 silver badges112 bronze badges
38.6k12 gold badges43 silver badges112 bronze badges
New contributor
asked 8 hours ago
Harry WilliamsonHarry Williamson
132 bronze badges
132 bronze badges
New contributor
New contributor
$begingroup$
"Active" means ENABLED. Active_high needs a high, +3.3v, a ONE, a TRUE. Active_low needs a low, 0.0v, a ZERO, a FALSE.
$endgroup$
– analogsystemsrf
8 hours ago
$begingroup$
And, there's arguments about terminology. Because some people (me) would say that "active low" means that a low voltage is interpreted as true or one.
$endgroup$
– TimWescott
8 hours ago
$begingroup$
indeed. You can define True as any level you wish.
$endgroup$
– analogsystemsrf
7 hours ago
add a comment
|
$begingroup$
"Active" means ENABLED. Active_high needs a high, +3.3v, a ONE, a TRUE. Active_low needs a low, 0.0v, a ZERO, a FALSE.
$endgroup$
– analogsystemsrf
8 hours ago
$begingroup$
And, there's arguments about terminology. Because some people (me) would say that "active low" means that a low voltage is interpreted as true or one.
$endgroup$
– TimWescott
8 hours ago
$begingroup$
indeed. You can define True as any level you wish.
$endgroup$
– analogsystemsrf
7 hours ago
$begingroup$
"Active" means ENABLED. Active_high needs a high, +3.3v, a ONE, a TRUE. Active_low needs a low, 0.0v, a ZERO, a FALSE.
$endgroup$
– analogsystemsrf
8 hours ago
$begingroup$
"Active" means ENABLED. Active_high needs a high, +3.3v, a ONE, a TRUE. Active_low needs a low, 0.0v, a ZERO, a FALSE.
$endgroup$
– analogsystemsrf
8 hours ago
$begingroup$
And, there's arguments about terminology. Because some people (me) would say that "active low" means that a low voltage is interpreted as true or one.
$endgroup$
– TimWescott
8 hours ago
$begingroup$
And, there's arguments about terminology. Because some people (me) would say that "active low" means that a low voltage is interpreted as true or one.
$endgroup$
– TimWescott
8 hours ago
$begingroup$
indeed. You can define True as any level you wish.
$endgroup$
– analogsystemsrf
7 hours ago
$begingroup$
indeed. You can define True as any level you wish.
$endgroup$
– analogsystemsrf
7 hours ago
add a comment
|
3 Answers
3
active
oldest
votes
$begingroup$
It means the signal is inverted (like a NOT gate). Let's take this 555 timer below as an example
Picture can be found here... Not my picture (and excuse the massive compression for this picture, hence the ugly pixels)
Say that a signal that goes to this pin is a 1
or HIGH
. Since Pin 4 is active low, it will end up being a 0
or LOW
for this pin. The opposite is true: If the signal leading up to the pin is 0
or LOW
, then Pin 4 will be 1
or HIGH
.
The purpose for a signal to be active low is to have some type of external logic device to turn off the signal. CPLDs are a good example of external logic that would shut off a device by sending a signal to an active low pin. You might thinking, "Why don't we just simply make it active high instead?" That's a valid question and I'm not really sure to be honest but if I had to guess, it could be to just simply save power.
$endgroup$
$begingroup$
Suppose i have i decoder , say 3 input decoder .If i connect not gates at each input respectively then it is active low input otherwise it is active high input by default ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
The active low inversion is typically built into the CMOS of the same chip. There's no external NOT gates that cause a signal to be active low. But yes, if you had that scenario, it would behave the same way (with exception of extremely small time differences).
$endgroup$
– KingDuken
8 hours ago
add a comment
|
$begingroup$
There are two things:
- The signal level
- What the signal means, ie assertion
The signal level is either digital Low or High
The signal meaning is attached to either Low or High, so we say the signal is asserted low or the signal is asserted high. Usually a bar or a slash indicates a low signal assertion level.
In the case above the reset is asserted low, so "reseting" happens when the signal is brought low. Since we could also reset while the signal is brought high, it is important to track the assertion.
It is especially important in HDL's to track the signal assertion level. Which is why you should label all of your signals. I've typically seen adding a _L or _H suffix to signal names to indicate the assertion level. In the case above it would be RESET_L. Even adding assertion suffixes in schematics can be helpful
$endgroup$
add a comment
|
$begingroup$
Active LOW means that a 0 V level is considered to be a logic 1
.
For instance, consider a logic input tied high using a pullup resistor and pulled to ground through a pushbutton switch.
Whenever the switch is not pressed, the input is at the pullup voltage, 5 V for example.
When the switch is pressed, the input is pulled to ground.
That input can be considered active low, because the low level means that the button has been pressed (logic 1)
$endgroup$
$begingroup$
This means it is like i have decoder and at input side i connect not gate ?
$endgroup$
– Harry Williamson
8 hours ago
add a comment
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It means the signal is inverted (like a NOT gate). Let's take this 555 timer below as an example
Picture can be found here... Not my picture (and excuse the massive compression for this picture, hence the ugly pixels)
Say that a signal that goes to this pin is a 1
or HIGH
. Since Pin 4 is active low, it will end up being a 0
or LOW
for this pin. The opposite is true: If the signal leading up to the pin is 0
or LOW
, then Pin 4 will be 1
or HIGH
.
The purpose for a signal to be active low is to have some type of external logic device to turn off the signal. CPLDs are a good example of external logic that would shut off a device by sending a signal to an active low pin. You might thinking, "Why don't we just simply make it active high instead?" That's a valid question and I'm not really sure to be honest but if I had to guess, it could be to just simply save power.
$endgroup$
$begingroup$
Suppose i have i decoder , say 3 input decoder .If i connect not gates at each input respectively then it is active low input otherwise it is active high input by default ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
The active low inversion is typically built into the CMOS of the same chip. There's no external NOT gates that cause a signal to be active low. But yes, if you had that scenario, it would behave the same way (with exception of extremely small time differences).
$endgroup$
– KingDuken
8 hours ago
add a comment
|
$begingroup$
It means the signal is inverted (like a NOT gate). Let's take this 555 timer below as an example
Picture can be found here... Not my picture (and excuse the massive compression for this picture, hence the ugly pixels)
Say that a signal that goes to this pin is a 1
or HIGH
. Since Pin 4 is active low, it will end up being a 0
or LOW
for this pin. The opposite is true: If the signal leading up to the pin is 0
or LOW
, then Pin 4 will be 1
or HIGH
.
The purpose for a signal to be active low is to have some type of external logic device to turn off the signal. CPLDs are a good example of external logic that would shut off a device by sending a signal to an active low pin. You might thinking, "Why don't we just simply make it active high instead?" That's a valid question and I'm not really sure to be honest but if I had to guess, it could be to just simply save power.
$endgroup$
$begingroup$
Suppose i have i decoder , say 3 input decoder .If i connect not gates at each input respectively then it is active low input otherwise it is active high input by default ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
The active low inversion is typically built into the CMOS of the same chip. There's no external NOT gates that cause a signal to be active low. But yes, if you had that scenario, it would behave the same way (with exception of extremely small time differences).
$endgroup$
– KingDuken
8 hours ago
add a comment
|
$begingroup$
It means the signal is inverted (like a NOT gate). Let's take this 555 timer below as an example
Picture can be found here... Not my picture (and excuse the massive compression for this picture, hence the ugly pixels)
Say that a signal that goes to this pin is a 1
or HIGH
. Since Pin 4 is active low, it will end up being a 0
or LOW
for this pin. The opposite is true: If the signal leading up to the pin is 0
or LOW
, then Pin 4 will be 1
or HIGH
.
The purpose for a signal to be active low is to have some type of external logic device to turn off the signal. CPLDs are a good example of external logic that would shut off a device by sending a signal to an active low pin. You might thinking, "Why don't we just simply make it active high instead?" That's a valid question and I'm not really sure to be honest but if I had to guess, it could be to just simply save power.
$endgroup$
It means the signal is inverted (like a NOT gate). Let's take this 555 timer below as an example
Picture can be found here... Not my picture (and excuse the massive compression for this picture, hence the ugly pixels)
Say that a signal that goes to this pin is a 1
or HIGH
. Since Pin 4 is active low, it will end up being a 0
or LOW
for this pin. The opposite is true: If the signal leading up to the pin is 0
or LOW
, then Pin 4 will be 1
or HIGH
.
The purpose for a signal to be active low is to have some type of external logic device to turn off the signal. CPLDs are a good example of external logic that would shut off a device by sending a signal to an active low pin. You might thinking, "Why don't we just simply make it active high instead?" That's a valid question and I'm not really sure to be honest but if I had to guess, it could be to just simply save power.
edited 8 hours ago
answered 8 hours ago
KingDukenKingDuken
1,5682 gold badges6 silver badges17 bronze badges
1,5682 gold badges6 silver badges17 bronze badges
$begingroup$
Suppose i have i decoder , say 3 input decoder .If i connect not gates at each input respectively then it is active low input otherwise it is active high input by default ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
The active low inversion is typically built into the CMOS of the same chip. There's no external NOT gates that cause a signal to be active low. But yes, if you had that scenario, it would behave the same way (with exception of extremely small time differences).
$endgroup$
– KingDuken
8 hours ago
add a comment
|
$begingroup$
Suppose i have i decoder , say 3 input decoder .If i connect not gates at each input respectively then it is active low input otherwise it is active high input by default ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
The active low inversion is typically built into the CMOS of the same chip. There's no external NOT gates that cause a signal to be active low. But yes, if you had that scenario, it would behave the same way (with exception of extremely small time differences).
$endgroup$
– KingDuken
8 hours ago
$begingroup$
Suppose i have i decoder , say 3 input decoder .If i connect not gates at each input respectively then it is active low input otherwise it is active high input by default ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
Suppose i have i decoder , say 3 input decoder .If i connect not gates at each input respectively then it is active low input otherwise it is active high input by default ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
The active low inversion is typically built into the CMOS of the same chip. There's no external NOT gates that cause a signal to be active low. But yes, if you had that scenario, it would behave the same way (with exception of extremely small time differences).
$endgroup$
– KingDuken
8 hours ago
$begingroup$
The active low inversion is typically built into the CMOS of the same chip. There's no external NOT gates that cause a signal to be active low. But yes, if you had that scenario, it would behave the same way (with exception of extremely small time differences).
$endgroup$
– KingDuken
8 hours ago
add a comment
|
$begingroup$
There are two things:
- The signal level
- What the signal means, ie assertion
The signal level is either digital Low or High
The signal meaning is attached to either Low or High, so we say the signal is asserted low or the signal is asserted high. Usually a bar or a slash indicates a low signal assertion level.
In the case above the reset is asserted low, so "reseting" happens when the signal is brought low. Since we could also reset while the signal is brought high, it is important to track the assertion.
It is especially important in HDL's to track the signal assertion level. Which is why you should label all of your signals. I've typically seen adding a _L or _H suffix to signal names to indicate the assertion level. In the case above it would be RESET_L. Even adding assertion suffixes in schematics can be helpful
$endgroup$
add a comment
|
$begingroup$
There are two things:
- The signal level
- What the signal means, ie assertion
The signal level is either digital Low or High
The signal meaning is attached to either Low or High, so we say the signal is asserted low or the signal is asserted high. Usually a bar or a slash indicates a low signal assertion level.
In the case above the reset is asserted low, so "reseting" happens when the signal is brought low. Since we could also reset while the signal is brought high, it is important to track the assertion.
It is especially important in HDL's to track the signal assertion level. Which is why you should label all of your signals. I've typically seen adding a _L or _H suffix to signal names to indicate the assertion level. In the case above it would be RESET_L. Even adding assertion suffixes in schematics can be helpful
$endgroup$
add a comment
|
$begingroup$
There are two things:
- The signal level
- What the signal means, ie assertion
The signal level is either digital Low or High
The signal meaning is attached to either Low or High, so we say the signal is asserted low or the signal is asserted high. Usually a bar or a slash indicates a low signal assertion level.
In the case above the reset is asserted low, so "reseting" happens when the signal is brought low. Since we could also reset while the signal is brought high, it is important to track the assertion.
It is especially important in HDL's to track the signal assertion level. Which is why you should label all of your signals. I've typically seen adding a _L or _H suffix to signal names to indicate the assertion level. In the case above it would be RESET_L. Even adding assertion suffixes in schematics can be helpful
$endgroup$
There are two things:
- The signal level
- What the signal means, ie assertion
The signal level is either digital Low or High
The signal meaning is attached to either Low or High, so we say the signal is asserted low or the signal is asserted high. Usually a bar or a slash indicates a low signal assertion level.
In the case above the reset is asserted low, so "reseting" happens when the signal is brought low. Since we could also reset while the signal is brought high, it is important to track the assertion.
It is especially important in HDL's to track the signal assertion level. Which is why you should label all of your signals. I've typically seen adding a _L or _H suffix to signal names to indicate the assertion level. In the case above it would be RESET_L. Even adding assertion suffixes in schematics can be helpful
edited 7 hours ago
answered 8 hours ago
Voltage SpikeVoltage Spike
38.6k12 gold badges43 silver badges112 bronze badges
38.6k12 gold badges43 silver badges112 bronze badges
add a comment
|
add a comment
|
$begingroup$
Active LOW means that a 0 V level is considered to be a logic 1
.
For instance, consider a logic input tied high using a pullup resistor and pulled to ground through a pushbutton switch.
Whenever the switch is not pressed, the input is at the pullup voltage, 5 V for example.
When the switch is pressed, the input is pulled to ground.
That input can be considered active low, because the low level means that the button has been pressed (logic 1)
$endgroup$
$begingroup$
This means it is like i have decoder and at input side i connect not gate ?
$endgroup$
– Harry Williamson
8 hours ago
add a comment
|
$begingroup$
Active LOW means that a 0 V level is considered to be a logic 1
.
For instance, consider a logic input tied high using a pullup resistor and pulled to ground through a pushbutton switch.
Whenever the switch is not pressed, the input is at the pullup voltage, 5 V for example.
When the switch is pressed, the input is pulled to ground.
That input can be considered active low, because the low level means that the button has been pressed (logic 1)
$endgroup$
$begingroup$
This means it is like i have decoder and at input side i connect not gate ?
$endgroup$
– Harry Williamson
8 hours ago
add a comment
|
$begingroup$
Active LOW means that a 0 V level is considered to be a logic 1
.
For instance, consider a logic input tied high using a pullup resistor and pulled to ground through a pushbutton switch.
Whenever the switch is not pressed, the input is at the pullup voltage, 5 V for example.
When the switch is pressed, the input is pulled to ground.
That input can be considered active low, because the low level means that the button has been pressed (logic 1)
$endgroup$
Active LOW means that a 0 V level is considered to be a logic 1
.
For instance, consider a logic input tied high using a pullup resistor and pulled to ground through a pushbutton switch.
Whenever the switch is not pressed, the input is at the pullup voltage, 5 V for example.
When the switch is pressed, the input is pulled to ground.
That input can be considered active low, because the low level means that the button has been pressed (logic 1)
answered 8 hours ago
jsotolajsotola
1,5291 gold badge8 silver badges11 bronze badges
1,5291 gold badge8 silver badges11 bronze badges
$begingroup$
This means it is like i have decoder and at input side i connect not gate ?
$endgroup$
– Harry Williamson
8 hours ago
add a comment
|
$begingroup$
This means it is like i have decoder and at input side i connect not gate ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
This means it is like i have decoder and at input side i connect not gate ?
$endgroup$
– Harry Williamson
8 hours ago
$begingroup$
This means it is like i have decoder and at input side i connect not gate ?
$endgroup$
– Harry Williamson
8 hours ago
add a comment
|
Harry Williamson is a new contributor. Be nice, and check out our Code of Conduct.
Harry Williamson is a new contributor. Be nice, and check out our Code of Conduct.
Harry Williamson is a new contributor. Be nice, and check out our Code of Conduct.
Harry Williamson is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
"Active" means ENABLED. Active_high needs a high, +3.3v, a ONE, a TRUE. Active_low needs a low, 0.0v, a ZERO, a FALSE.
$endgroup$
– analogsystemsrf
8 hours ago
$begingroup$
And, there's arguments about terminology. Because some people (me) would say that "active low" means that a low voltage is interpreted as true or one.
$endgroup$
– TimWescott
8 hours ago
$begingroup$
indeed. You can define True as any level you wish.
$endgroup$
– analogsystemsrf
7 hours ago