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What is meaning of active low input in combinational logic circuits?

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2

$begingroup$

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 explain this using an example (note that I have knowledge about encoder, decoder, Multiplexers so you can use these in your example).

digital-logic integrated-circuit input

share|improve this question

edited 6 hours ago

Voltage Spike

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Harry WilliamsonHarry Williamson

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Harry Williamson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "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
  7 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  indeed. You can define True as any level you wish.
  $endgroup$
  – analogsystemsrf
  7 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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 explain this using an example (note that I have knowledge about encoder, decoder, Multiplexers so you can use these in your example).

digital-logic integrated-circuit input

share|improve this question

edited 6 hours ago

Voltage Spike

38.6k1212 gold badges4343 silver badges112112 bronze badges

asked 8 hours ago

Harry WilliamsonHarry Williamson

1322 bronze badges

New contributor

Harry Williamson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "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
  7 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
  7 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 explain this using an example (note that I have knowledge about encoder, decoder, Multiplexers so you can use these in your example).

digital-logic integrated-circuit input

share|improve this question

edited 6 hours ago

Voltage Spike

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asked 8 hours ago

Harry WilliamsonHarry Williamson

1322 bronze badges

New contributor

Harry Williamson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 6 hours ago

Voltage Spike

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asked 8 hours ago

Harry WilliamsonHarry Williamson

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Harry Williamson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 6 hours ago

Voltage Spike

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asked 8 hours ago

Harry WilliamsonHarry Williamson

1322 bronze badges

New contributor

Harry Williamson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 6 hours ago

Voltage Spike

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edited 6 hours ago

Voltage Spike

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Harry WilliamsonHarry Williamson

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asked 8 hours ago

Harry WilliamsonHarry Williamson

1322 bronze badges

3 Answers
3

active

oldest

votes

1

$begingroup$

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.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

KingDukenKingDuken

1,56822 gold badges66 silver badges1717 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 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
  7 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Voltage SpikeVoltage Spike

38.6k1212 gold badges4343 silver badges112112 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

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)

share|improve this answer

answered 7 hours ago

jsotolajsotola

1,52911 gold badge88 silver badges1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This means it is like i have decoder and at input side i connect not gate ?
  $endgroup$
  – Harry Williamson
  7 hours ago
  

  
  
  
  

add a comment
 | 

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1

$begingroup$

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.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

KingDukenKingDuken

1,56822 gold badges66 silver badges1717 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 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
  7 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

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.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

KingDukenKingDuken

1,56822 gold badges66 silver badges1717 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 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
  7 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.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

KingDukenKingDuken

1,56822 gold badges66 silver badges1717 bronze badges

$endgroup$

share|improve this answer

edited 7 hours ago

answered 7 hours ago

KingDukenKingDuken

1,56822 gold badges66 silver badges1717 bronze badges

answered 7 hours ago

KingDukenKingDuken

1,56822 gold badges66 silver badges1717 bronze badges

answered 7 hours ago

KingDukenKingDuken

1,56822 gold badges66 silver badges1717 bronze badges

3

$begingroup$

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Voltage SpikeVoltage Spike

38.6k1212 gold badges4343 silver badges112112 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Voltage SpikeVoltage Spike

38.6k1212 gold badges4343 silver badges112112 bronze badges

$endgroup$

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Voltage SpikeVoltage Spike

38.6k1212 gold badges4343 silver badges112112 bronze badges

$endgroup$

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Voltage SpikeVoltage Spike

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answered 7 hours ago

Voltage SpikeVoltage Spike

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answered 7 hours ago

Voltage SpikeVoltage Spike

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0

$begingroup$

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)

share|improve this answer

answered 7 hours ago

jsotolajsotola

1,52911 gold badge88 silver badges1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This means it is like i have decoder and at input side i connect not gate ?
  $endgroup$
  – Harry Williamson
  7 hours ago
  

  
  
  
  

add a comment
 | 

0

$begingroup$

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)

share|improve this answer

answered 7 hours ago

jsotolajsotola

1,52911 gold badge88 silver badges1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This means it is like i have decoder and at input side i connect not gate ?
  $endgroup$
  – Harry Williamson
  7 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)

share|improve this answer

answered 7 hours ago

jsotolajsotola

1,52911 gold badge88 silver badges1111 bronze badges

$endgroup$

share|improve this answer

answered 7 hours ago

jsotolajsotola

1,52911 gold badge88 silver badges1111 bronze badges

answered 7 hours ago

jsotolajsotola

1,52911 gold badge88 silver badges1111 bronze badges

answered 7 hours ago

jsotolajsotola

1,52911 gold badge88 silver badges1111 bronze badges