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Is there a more efficient alternative to pull down resistors?

Protecting Microcontroller Input Pins from Soft Power SwitchDesigning a current limiting circuit for my projectExperiment with pull up/down resistors gives unexpected resultsTurn pull-up into pull-downSimple SPST Switch Circuit - Pull-down Resistor VS. Current Limiting ResistorIs this a reasonable way to switch a circuit on and off?Using a high resistance pull down resistorPull-up vs Pull-down on enable pinNormally closed semiconductor switchPull-down resistor confusion

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3

$begingroup$

I am building a LED spinner circuit and I am at the point of optimizing it. The whole circuit itself only draws about 10-20mA max. I was today looking at this part of the circuit:

Now as you can see, when my switch is at position 5, it turns the circuit off. But, now when my circuit is off, there is still current flowing through the pull down resistor, draining the battery. I know this is a very small current, but I was wondering if there was a way to make this switch so that it does not draw any current when switched off.

Edit:
I should have maybe put the whole circuit in.

transistors switches pulldown

share|improve this question

edited 7 hours ago

Francois landry

asked 8 hours ago

Francois landryFrancois landry

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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There will always be some kind of leakage in "off" switches. You could use bigger resistors, or a FET in place of a resistor with an extremely high open/off resistance, but you will always have some leakage.
  $endgroup$
  – schadjo
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand that with most solutions there will be leakage as well as during my on time there is wasted current going through that transistor to the pulldown resistor. I was just curious if there was a way to completely stop the current when the circuit is off and I must thank Dave for the answer to my question.
  $endgroup$
  – Francois landry
  5 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

I am building a LED spinner circuit and I am at the point of optimizing it. The whole circuit itself only draws about 10-20mA max. I was today looking at this part of the circuit:

Now as you can see, when my switch is at position 5, it turns the circuit off. But, now when my circuit is off, there is still current flowing through the pull down resistor, draining the battery. I know this is a very small current, but I was wondering if there was a way to make this switch so that it does not draw any current when switched off.

Edit:
I should have maybe put the whole circuit in.

transistors switches pulldown

share|improve this question

edited 7 hours ago

Francois landry

asked 8 hours ago

Francois landryFrancois landry

1844 bronze badges

New contributor

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

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There will always be some kind of leakage in "off" switches. You could use bigger resistors, or a FET in place of a resistor with an extremely high open/off resistance, but you will always have some leakage.
  $endgroup$
  – schadjo
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand that with most solutions there will be leakage as well as during my on time there is wasted current going through that transistor to the pulldown resistor. I was just curious if there was a way to completely stop the current when the circuit is off and I must thank Dave for the answer to my question.
  $endgroup$
  – Francois landry
  5 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

I am building a LED spinner circuit and I am at the point of optimizing it. The whole circuit itself only draws about 10-20mA max. I was today looking at this part of the circuit:

Now as you can see, when my switch is at position 5, it turns the circuit off. But, now when my circuit is off, there is still current flowing through the pull down resistor, draining the battery. I know this is a very small current, but I was wondering if there was a way to make this switch so that it does not draw any current when switched off.

Edit:
I should have maybe put the whole circuit in.

transistors switches pulldown

share|improve this question

edited 7 hours ago

Francois landry

asked 8 hours ago

Francois landryFrancois landry

1844 bronze badges

New contributor

Francois landry 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 7 hours ago

Francois landry

asked 8 hours ago

Francois landryFrancois landry

1844 bronze badges

New contributor

Francois landry 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 7 hours ago

Francois landry

asked 8 hours ago

Francois landryFrancois landry

1844 bronze badges

New contributor

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

asked 8 hours ago

Francois landryFrancois landry

1844 bronze badges

New contributor

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

asked 8 hours ago

Francois landryFrancois landry

1844 bronze badges

3 Answers
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1

$begingroup$

Note that the current is wasted regardless of whether the circuit is "on" or "off" — when it is "on" the voltage drop across R11 is only slightly less than when it is "off".

Using a PMOS transistor instead of the PNP would mean that the pulldown resistor could be on the order of megohms, reducing the "leakage" current to microamps.

Or you could use a different strategy altogether, eliminating the off-state current entirely:

^{simulate this circuit – Schematic created using CircuitLab}

share|improve this answer

answered 7 hours ago

Dave Tweed♦Dave Tweed

137k1111 gold badges175175 silver badges300300 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

• You could use a PMOS FET in place of Q1. Then R11 could be 50k or 100k instead of 10k, reducing leakage in the off position.




• You could use a separate "off" switch, or a special rotary switch with a special "off" position that disconnects VCC from the transistor altogether.

share|improve this answer

answered 8 hours ago

The PhotonThe Photon

93.9k33 gold badges113113 silver badges222222 bronze badges

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add a comment
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1

$begingroup$

You could use three Schottky rectifiers in place of the transistor and pull-down. Place anodes to switch pins 1, 2, 4, cathodes tied together to "feed main circuit." Disconnect pin 5 so it becomes "true off." The "feed main circuit" will be about 0.25v lower than Vcc.

share|improve this answer

answered 7 hours ago

rdtscrdtsc

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1

$begingroup$

Note that the current is wasted regardless of whether the circuit is "on" or "off" — when it is "on" the voltage drop across R11 is only slightly less than when it is "off".

Using a PMOS transistor instead of the PNP would mean that the pulldown resistor could be on the order of megohms, reducing the "leakage" current to microamps.

Or you could use a different strategy altogether, eliminating the off-state current entirely:

^{simulate this circuit – Schematic created using CircuitLab}

share|improve this answer

answered 7 hours ago

Dave Tweed♦Dave Tweed

137k1111 gold badges175175 silver badges300300 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

Note that the current is wasted regardless of whether the circuit is "on" or "off" — when it is "on" the voltage drop across R11 is only slightly less than when it is "off".

Using a PMOS transistor instead of the PNP would mean that the pulldown resistor could be on the order of megohms, reducing the "leakage" current to microamps.

Or you could use a different strategy altogether, eliminating the off-state current entirely:

^{simulate this circuit – Schematic created using CircuitLab}

share|improve this answer

answered 7 hours ago

Dave Tweed♦Dave Tweed

137k1111 gold badges175175 silver badges300300 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Note that the current is wasted regardless of whether the circuit is "on" or "off" — when it is "on" the voltage drop across R11 is only slightly less than when it is "off".

Using a PMOS transistor instead of the PNP would mean that the pulldown resistor could be on the order of megohms, reducing the "leakage" current to microamps.

Or you could use a different strategy altogether, eliminating the off-state current entirely:

^{simulate this circuit – Schematic created using CircuitLab}

share|improve this answer

answered 7 hours ago

Dave Tweed♦Dave Tweed

137k1111 gold badges175175 silver badges300300 bronze badges

$endgroup$

share|improve this answer

answered 7 hours ago

Dave Tweed♦Dave Tweed

137k1111 gold badges175175 silver badges300300 bronze badges

answered 7 hours ago

Dave Tweed♦Dave Tweed

137k1111 gold badges175175 silver badges300300 bronze badges

answered 7 hours ago

Dave Tweed♦Dave Tweed

137k1111 gold badges175175 silver badges300300 bronze badges

1

$begingroup$

• You could use a PMOS FET in place of Q1. Then R11 could be 50k or 100k instead of 10k, reducing leakage in the off position.




• You could use a separate "off" switch, or a special rotary switch with a special "off" position that disconnects VCC from the transistor altogether.

share|improve this answer

answered 8 hours ago

The PhotonThe Photon

93.9k33 gold badges113113 silver badges222222 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

• You could use a PMOS FET in place of Q1. Then R11 could be 50k or 100k instead of 10k, reducing leakage in the off position.




• You could use a separate "off" switch, or a special rotary switch with a special "off" position that disconnects VCC from the transistor altogether.

share|improve this answer

answered 8 hours ago

The PhotonThe Photon

93.9k33 gold badges113113 silver badges222222 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

• You could use a PMOS FET in place of Q1. Then R11 could be 50k or 100k instead of 10k, reducing leakage in the off position.




• You could use a separate "off" switch, or a special rotary switch with a special "off" position that disconnects VCC from the transistor altogether.

share|improve this answer

answered 8 hours ago

The PhotonThe Photon

93.9k33 gold badges113113 silver badges222222 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

The PhotonThe Photon

93.9k33 gold badges113113 silver badges222222 bronze badges

answered 8 hours ago

The PhotonThe Photon

93.9k33 gold badges113113 silver badges222222 bronze badges

answered 8 hours ago

The PhotonThe Photon

93.9k33 gold badges113113 silver badges222222 bronze badges

1

$begingroup$

You could use three Schottky rectifiers in place of the transistor and pull-down. Place anodes to switch pins 1, 2, 4, cathodes tied together to "feed main circuit." Disconnect pin 5 so it becomes "true off." The "feed main circuit" will be about 0.25v lower than Vcc.

share|improve this answer

answered 7 hours ago

rdtscrdtsc

5,59333 gold badges1313 silver badges3939 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

You could use three Schottky rectifiers in place of the transistor and pull-down. Place anodes to switch pins 1, 2, 4, cathodes tied together to "feed main circuit." Disconnect pin 5 so it becomes "true off." The "feed main circuit" will be about 0.25v lower than Vcc.

share|improve this answer

answered 7 hours ago

rdtscrdtsc

5,59333 gold badges1313 silver badges3939 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You could use three Schottky rectifiers in place of the transistor and pull-down. Place anodes to switch pins 1, 2, 4, cathodes tied together to "feed main circuit." Disconnect pin 5 so it becomes "true off." The "feed main circuit" will be about 0.25v lower than Vcc.

share|improve this answer

answered 7 hours ago

rdtscrdtsc

5,59333 gold badges1313 silver badges3939 bronze badges

$endgroup$

share|improve this answer

answered 7 hours ago

rdtscrdtsc

5,59333 gold badges1313 silver badges3939 bronze badges

answered 7 hours ago

rdtscrdtsc

5,59333 gold badges1313 silver badges3939 bronze badges

answered 7 hours ago

rdtscrdtsc

5,59333 gold badges1313 silver badges3939 bronze badges