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Would a 7805 5v regulator drain a 9v battery?


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2












$begingroup$


Doing some DIY as a hobby, I'm making a small humidity-temperature radio sensor.



An ATmega328 is reading from a DHT11 sensor and then transmitting data to a raspberry-pi by a STX882 radio transmitter. It is powered by a 9v battery using a 7805 5v regulator with 10µF and 100µF capacitances.



The C code on the ATmega is reading humidity and temperature and then sending it every 30 minutes:



const unsigned long DELAY = 30*60*1000UL;   //30 minutes
void loop() {
delay(DELAY);
send_data(); //maybe a little overcomplicated but I think it is not the point
}


This was working like a charm, but the battery life has been unexpectedly short. It was brand new, I made some sporadic tests with a short delay, with no abnormal heat coming from anywhere.



When I was satisfied, I put the 30 minutes delay and leave it alone (which was maybe a bit hazardous?), but after less than 24 hours the battery was 5.4v dead. The 30 minutes delay was approximately respected for its lifespan though.



What could explain such a short battery life? Could it be the 5v regulator? How could I build a long-lasting circuit?



PS: I'm still trying to Fritzing some diagram but this takes age for noobs like me...



EDIT:
I used a 6lp3146 generic brand 9v battery which apparently provided 300-500mAh at 100mA current, which is far more than what my circuit use.



Here is all information I could gather from the datasheet:



+-----------------+------------+---------+-----------+---------+
| | DHT11 | STX882 | ATmega328 | 7805reg |
+-----------------+------------+---------+-----------+---------+
| voltage | 3-5.5V | 1.2-6V | 2.7-5.5V | |
+-----------------+------------+---------+-----------+---------+
| active current | 0.5-2.5mA | 34mA | 1.5mA | |
+-----------------+------------+---------+-----------+---------+
| standby current | 0.1-0.15mA | <0.01µA | 1µA | 4-8mA* |
+-----------------+------------+---------+-----------+---------+
*"bias current"


If I understand correctly, my system is active for a few seconds every 30 minutes so the standby current is all that should matter, and it is indeed driven by the 7805 regulator.



So yes, in the worst case, with 300mAh I should be able to keep the system alive for only 40 hours.



Is there any way I could feed my system 5V for a much longer time without a much bigger size?










share|improve this question











$endgroup$












  • $begingroup$
    I suggest using the LowPower library or similar to put the ATmega to sleep between measurements.
    $endgroup$
    – calcium3000
    8 hours ago










  • $begingroup$
    How short is short?
    $endgroup$
    – Scott Seidman
    7 hours ago










  • $begingroup$
    note: a standard 9V battery contains 6 AAAA cells .... those have way less capacity than AA cells
    $endgroup$
    – jsotola
    7 hours ago










  • $begingroup$
    This Tx and uC is best paired with an 18650 Li Ion cell or 3 good quality AA Alkalines.
    $endgroup$
    – Sunnyskyguy EE75
    7 hours ago










  • $begingroup$
    I'm not familiar if the ATmega328 has a sleep mode and what current it draws. But for this sleep mode (drawing uA or less of current) you'd better feed this (or a better uP) directly from the 9V battery using a resistor and use a kind of wake up signal to turn on (the regulator or better:) the converter when measurements need to be done
    $endgroup$
    – Huisman
    6 hours ago




















2












$begingroup$


Doing some DIY as a hobby, I'm making a small humidity-temperature radio sensor.



An ATmega328 is reading from a DHT11 sensor and then transmitting data to a raspberry-pi by a STX882 radio transmitter. It is powered by a 9v battery using a 7805 5v regulator with 10µF and 100µF capacitances.



The C code on the ATmega is reading humidity and temperature and then sending it every 30 minutes:



const unsigned long DELAY = 30*60*1000UL;   //30 minutes
void loop() {
delay(DELAY);
send_data(); //maybe a little overcomplicated but I think it is not the point
}


This was working like a charm, but the battery life has been unexpectedly short. It was brand new, I made some sporadic tests with a short delay, with no abnormal heat coming from anywhere.



When I was satisfied, I put the 30 minutes delay and leave it alone (which was maybe a bit hazardous?), but after less than 24 hours the battery was 5.4v dead. The 30 minutes delay was approximately respected for its lifespan though.



What could explain such a short battery life? Could it be the 5v regulator? How could I build a long-lasting circuit?



PS: I'm still trying to Fritzing some diagram but this takes age for noobs like me...



EDIT:
I used a 6lp3146 generic brand 9v battery which apparently provided 300-500mAh at 100mA current, which is far more than what my circuit use.



Here is all information I could gather from the datasheet:



+-----------------+------------+---------+-----------+---------+
| | DHT11 | STX882 | ATmega328 | 7805reg |
+-----------------+------------+---------+-----------+---------+
| voltage | 3-5.5V | 1.2-6V | 2.7-5.5V | |
+-----------------+------------+---------+-----------+---------+
| active current | 0.5-2.5mA | 34mA | 1.5mA | |
+-----------------+------------+---------+-----------+---------+
| standby current | 0.1-0.15mA | <0.01µA | 1µA | 4-8mA* |
+-----------------+------------+---------+-----------+---------+
*"bias current"


If I understand correctly, my system is active for a few seconds every 30 minutes so the standby current is all that should matter, and it is indeed driven by the 7805 regulator.



So yes, in the worst case, with 300mAh I should be able to keep the system alive for only 40 hours.



Is there any way I could feed my system 5V for a much longer time without a much bigger size?










share|improve this question











$endgroup$












  • $begingroup$
    I suggest using the LowPower library or similar to put the ATmega to sleep between measurements.
    $endgroup$
    – calcium3000
    8 hours ago










  • $begingroup$
    How short is short?
    $endgroup$
    – Scott Seidman
    7 hours ago










  • $begingroup$
    note: a standard 9V battery contains 6 AAAA cells .... those have way less capacity than AA cells
    $endgroup$
    – jsotola
    7 hours ago










  • $begingroup$
    This Tx and uC is best paired with an 18650 Li Ion cell or 3 good quality AA Alkalines.
    $endgroup$
    – Sunnyskyguy EE75
    7 hours ago










  • $begingroup$
    I'm not familiar if the ATmega328 has a sleep mode and what current it draws. But for this sleep mode (drawing uA or less of current) you'd better feed this (or a better uP) directly from the 9V battery using a resistor and use a kind of wake up signal to turn on (the regulator or better:) the converter when measurements need to be done
    $endgroup$
    – Huisman
    6 hours ago
















2












2








2





$begingroup$


Doing some DIY as a hobby, I'm making a small humidity-temperature radio sensor.



An ATmega328 is reading from a DHT11 sensor and then transmitting data to a raspberry-pi by a STX882 radio transmitter. It is powered by a 9v battery using a 7805 5v regulator with 10µF and 100µF capacitances.



The C code on the ATmega is reading humidity and temperature and then sending it every 30 minutes:



const unsigned long DELAY = 30*60*1000UL;   //30 minutes
void loop() {
delay(DELAY);
send_data(); //maybe a little overcomplicated but I think it is not the point
}


This was working like a charm, but the battery life has been unexpectedly short. It was brand new, I made some sporadic tests with a short delay, with no abnormal heat coming from anywhere.



When I was satisfied, I put the 30 minutes delay and leave it alone (which was maybe a bit hazardous?), but after less than 24 hours the battery was 5.4v dead. The 30 minutes delay was approximately respected for its lifespan though.



What could explain such a short battery life? Could it be the 5v regulator? How could I build a long-lasting circuit?



PS: I'm still trying to Fritzing some diagram but this takes age for noobs like me...



EDIT:
I used a 6lp3146 generic brand 9v battery which apparently provided 300-500mAh at 100mA current, which is far more than what my circuit use.



Here is all information I could gather from the datasheet:



+-----------------+------------+---------+-----------+---------+
| | DHT11 | STX882 | ATmega328 | 7805reg |
+-----------------+------------+---------+-----------+---------+
| voltage | 3-5.5V | 1.2-6V | 2.7-5.5V | |
+-----------------+------------+---------+-----------+---------+
| active current | 0.5-2.5mA | 34mA | 1.5mA | |
+-----------------+------------+---------+-----------+---------+
| standby current | 0.1-0.15mA | <0.01µA | 1µA | 4-8mA* |
+-----------------+------------+---------+-----------+---------+
*"bias current"


If I understand correctly, my system is active for a few seconds every 30 minutes so the standby current is all that should matter, and it is indeed driven by the 7805 regulator.



So yes, in the worst case, with 300mAh I should be able to keep the system alive for only 40 hours.



Is there any way I could feed my system 5V for a much longer time without a much bigger size?










share|improve this question











$endgroup$




Doing some DIY as a hobby, I'm making a small humidity-temperature radio sensor.



An ATmega328 is reading from a DHT11 sensor and then transmitting data to a raspberry-pi by a STX882 radio transmitter. It is powered by a 9v battery using a 7805 5v regulator with 10µF and 100µF capacitances.



The C code on the ATmega is reading humidity and temperature and then sending it every 30 minutes:



const unsigned long DELAY = 30*60*1000UL;   //30 minutes
void loop() {
delay(DELAY);
send_data(); //maybe a little overcomplicated but I think it is not the point
}


This was working like a charm, but the battery life has been unexpectedly short. It was brand new, I made some sporadic tests with a short delay, with no abnormal heat coming from anywhere.



When I was satisfied, I put the 30 minutes delay and leave it alone (which was maybe a bit hazardous?), but after less than 24 hours the battery was 5.4v dead. The 30 minutes delay was approximately respected for its lifespan though.



What could explain such a short battery life? Could it be the 5v regulator? How could I build a long-lasting circuit?



PS: I'm still trying to Fritzing some diagram but this takes age for noobs like me...



EDIT:
I used a 6lp3146 generic brand 9v battery which apparently provided 300-500mAh at 100mA current, which is far more than what my circuit use.



Here is all information I could gather from the datasheet:



+-----------------+------------+---------+-----------+---------+
| | DHT11 | STX882 | ATmega328 | 7805reg |
+-----------------+------------+---------+-----------+---------+
| voltage | 3-5.5V | 1.2-6V | 2.7-5.5V | |
+-----------------+------------+---------+-----------+---------+
| active current | 0.5-2.5mA | 34mA | 1.5mA | |
+-----------------+------------+---------+-----------+---------+
| standby current | 0.1-0.15mA | <0.01µA | 1µA | 4-8mA* |
+-----------------+------------+---------+-----------+---------+
*"bias current"


If I understand correctly, my system is active for a few seconds every 30 minutes so the standby current is all that should matter, and it is indeed driven by the 7805 regulator.



So yes, in the worst case, with 300mAh I should be able to keep the system alive for only 40 hours.



Is there any way I could feed my system 5V for a much longer time without a much bigger size?







voltage-regulator battery-operated low-battery






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 5 hours ago







Dan Chaltiel

















asked 8 hours ago









Dan ChaltielDan Chaltiel

1186




1186












  • $begingroup$
    I suggest using the LowPower library or similar to put the ATmega to sleep between measurements.
    $endgroup$
    – calcium3000
    8 hours ago










  • $begingroup$
    How short is short?
    $endgroup$
    – Scott Seidman
    7 hours ago










  • $begingroup$
    note: a standard 9V battery contains 6 AAAA cells .... those have way less capacity than AA cells
    $endgroup$
    – jsotola
    7 hours ago










  • $begingroup$
    This Tx and uC is best paired with an 18650 Li Ion cell or 3 good quality AA Alkalines.
    $endgroup$
    – Sunnyskyguy EE75
    7 hours ago










  • $begingroup$
    I'm not familiar if the ATmega328 has a sleep mode and what current it draws. But for this sleep mode (drawing uA or less of current) you'd better feed this (or a better uP) directly from the 9V battery using a resistor and use a kind of wake up signal to turn on (the regulator or better:) the converter when measurements need to be done
    $endgroup$
    – Huisman
    6 hours ago




















  • $begingroup$
    I suggest using the LowPower library or similar to put the ATmega to sleep between measurements.
    $endgroup$
    – calcium3000
    8 hours ago










  • $begingroup$
    How short is short?
    $endgroup$
    – Scott Seidman
    7 hours ago










  • $begingroup$
    note: a standard 9V battery contains 6 AAAA cells .... those have way less capacity than AA cells
    $endgroup$
    – jsotola
    7 hours ago










  • $begingroup$
    This Tx and uC is best paired with an 18650 Li Ion cell or 3 good quality AA Alkalines.
    $endgroup$
    – Sunnyskyguy EE75
    7 hours ago










  • $begingroup$
    I'm not familiar if the ATmega328 has a sleep mode and what current it draws. But for this sleep mode (drawing uA or less of current) you'd better feed this (or a better uP) directly from the 9V battery using a resistor and use a kind of wake up signal to turn on (the regulator or better:) the converter when measurements need to be done
    $endgroup$
    – Huisman
    6 hours ago


















$begingroup$
I suggest using the LowPower library or similar to put the ATmega to sleep between measurements.
$endgroup$
– calcium3000
8 hours ago




$begingroup$
I suggest using the LowPower library or similar to put the ATmega to sleep between measurements.
$endgroup$
– calcium3000
8 hours ago












$begingroup$
How short is short?
$endgroup$
– Scott Seidman
7 hours ago




$begingroup$
How short is short?
$endgroup$
– Scott Seidman
7 hours ago












$begingroup$
note: a standard 9V battery contains 6 AAAA cells .... those have way less capacity than AA cells
$endgroup$
– jsotola
7 hours ago




$begingroup$
note: a standard 9V battery contains 6 AAAA cells .... those have way less capacity than AA cells
$endgroup$
– jsotola
7 hours ago












$begingroup$
This Tx and uC is best paired with an 18650 Li Ion cell or 3 good quality AA Alkalines.
$endgroup$
– Sunnyskyguy EE75
7 hours ago




$begingroup$
This Tx and uC is best paired with an 18650 Li Ion cell or 3 good quality AA Alkalines.
$endgroup$
– Sunnyskyguy EE75
7 hours ago












$begingroup$
I'm not familiar if the ATmega328 has a sleep mode and what current it draws. But for this sleep mode (drawing uA or less of current) you'd better feed this (or a better uP) directly from the 9V battery using a resistor and use a kind of wake up signal to turn on (the regulator or better:) the converter when measurements need to be done
$endgroup$
– Huisman
6 hours ago






$begingroup$
I'm not familiar if the ATmega328 has a sleep mode and what current it draws. But for this sleep mode (drawing uA or less of current) you'd better feed this (or a better uP) directly from the 9V battery using a resistor and use a kind of wake up signal to turn on (the regulator or better:) the converter when measurements need to be done
$endgroup$
– Huisman
6 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$


What could explain such a short battery life? Could it be the 5v regulator?




As mentioned, the 7805 has about 4mA of quiescent current. You need to find a data sheet for the battery (Eveready has nice battery datasheets, if you're using an alkaline cell). It's probably no more than 100mAh -- 100mAh / 4mA = 25 hours, so that should say something to you.




How could I build a long-lasting circuit?




The 7805 is old technology. There are better newer linear regulators out there. You should be able to easily find something that uses 10 times less quiescent current, and with digging even less than that.



To use even less power you'd use a buck converter that's specifically designed for low quiescent current -- but I gather that you're not ready to design one into a board at the component level. There may be a module out there that'll do the job, but you'll need to shop around for it. TI does have some buck converter modules, but you'll want to pay a lot of attention to their capabilities, both for maximum current delivery and quiescent current.



To use less power yet, do everything that you can to minimize the current consumption of your circuit when it is quiescent. This will require careful use of the microprocessor's sleep function, as well as managing how the board is powered (for example, if it only comes on once every 30 minutes, you may well want to turn off power to the radio and the humidity-reading portions of the circuit).



Measure the current consumption in all modes of operation, and use this to determine which modes are the worst offenders overall, then concentrate on minimizing the currents in those modes if you can.






share|improve this answer











$endgroup$













  • $begingroup$
    My first thought too: skip to regulator, use a converter as you probably draw more than only the quiescent current.
    $endgroup$
    – Huisman
    7 hours ago






  • 1




    $begingroup$
    The thing sitting quiet for 30 minutes then taking a measurement and sending it, then going back to sleep. So the quiescent current can't be ignored.
    $endgroup$
    – TimWescott
    7 hours ago










  • $begingroup$
    The 173010578 or the 173950536 from Wuerth or TSR 1-2450 from Traco power are pin-compatible replacements and have lower quiescent current, but it's not that huge difference.
    $endgroup$
    – Huisman
    6 hours ago












  • $begingroup$
    I meant: in idle mode, the 7805 will still be loaded, so it is not only drawing its own quiescent current but also some load current in a inefficient way.
    $endgroup$
    – Huisman
    6 hours ago










  • $begingroup$
    @Huisman OK -- I see what you mean. Yes, the larger question of how to reduce overall current consumption should include reducing the sleep current of the balance of the circuit. In fact, I'm going to add that...
    $endgroup$
    – TimWescott
    6 hours ago



















2












$begingroup$

The idle current of a 7805 regulator is around 4 mA so, armed with the ampere hour capacity of your battery, work how long it will last with a continuous drain of 4 mA.



If you establish that is the problem, you will find that there are plenty of regulators having a significantly lower quiescent current.






share|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$


    What could explain such a short battery life? Could it be the 5v regulator?




    As mentioned, the 7805 has about 4mA of quiescent current. You need to find a data sheet for the battery (Eveready has nice battery datasheets, if you're using an alkaline cell). It's probably no more than 100mAh -- 100mAh / 4mA = 25 hours, so that should say something to you.




    How could I build a long-lasting circuit?




    The 7805 is old technology. There are better newer linear regulators out there. You should be able to easily find something that uses 10 times less quiescent current, and with digging even less than that.



    To use even less power you'd use a buck converter that's specifically designed for low quiescent current -- but I gather that you're not ready to design one into a board at the component level. There may be a module out there that'll do the job, but you'll need to shop around for it. TI does have some buck converter modules, but you'll want to pay a lot of attention to their capabilities, both for maximum current delivery and quiescent current.



    To use less power yet, do everything that you can to minimize the current consumption of your circuit when it is quiescent. This will require careful use of the microprocessor's sleep function, as well as managing how the board is powered (for example, if it only comes on once every 30 minutes, you may well want to turn off power to the radio and the humidity-reading portions of the circuit).



    Measure the current consumption in all modes of operation, and use this to determine which modes are the worst offenders overall, then concentrate on minimizing the currents in those modes if you can.






    share|improve this answer











    $endgroup$













    • $begingroup$
      My first thought too: skip to regulator, use a converter as you probably draw more than only the quiescent current.
      $endgroup$
      – Huisman
      7 hours ago






    • 1




      $begingroup$
      The thing sitting quiet for 30 minutes then taking a measurement and sending it, then going back to sleep. So the quiescent current can't be ignored.
      $endgroup$
      – TimWescott
      7 hours ago










    • $begingroup$
      The 173010578 or the 173950536 from Wuerth or TSR 1-2450 from Traco power are pin-compatible replacements and have lower quiescent current, but it's not that huge difference.
      $endgroup$
      – Huisman
      6 hours ago












    • $begingroup$
      I meant: in idle mode, the 7805 will still be loaded, so it is not only drawing its own quiescent current but also some load current in a inefficient way.
      $endgroup$
      – Huisman
      6 hours ago










    • $begingroup$
      @Huisman OK -- I see what you mean. Yes, the larger question of how to reduce overall current consumption should include reducing the sleep current of the balance of the circuit. In fact, I'm going to add that...
      $endgroup$
      – TimWescott
      6 hours ago
















    4












    $begingroup$


    What could explain such a short battery life? Could it be the 5v regulator?




    As mentioned, the 7805 has about 4mA of quiescent current. You need to find a data sheet for the battery (Eveready has nice battery datasheets, if you're using an alkaline cell). It's probably no more than 100mAh -- 100mAh / 4mA = 25 hours, so that should say something to you.




    How could I build a long-lasting circuit?




    The 7805 is old technology. There are better newer linear regulators out there. You should be able to easily find something that uses 10 times less quiescent current, and with digging even less than that.



    To use even less power you'd use a buck converter that's specifically designed for low quiescent current -- but I gather that you're not ready to design one into a board at the component level. There may be a module out there that'll do the job, but you'll need to shop around for it. TI does have some buck converter modules, but you'll want to pay a lot of attention to their capabilities, both for maximum current delivery and quiescent current.



    To use less power yet, do everything that you can to minimize the current consumption of your circuit when it is quiescent. This will require careful use of the microprocessor's sleep function, as well as managing how the board is powered (for example, if it only comes on once every 30 minutes, you may well want to turn off power to the radio and the humidity-reading portions of the circuit).



    Measure the current consumption in all modes of operation, and use this to determine which modes are the worst offenders overall, then concentrate on minimizing the currents in those modes if you can.






    share|improve this answer











    $endgroup$













    • $begingroup$
      My first thought too: skip to regulator, use a converter as you probably draw more than only the quiescent current.
      $endgroup$
      – Huisman
      7 hours ago






    • 1




      $begingroup$
      The thing sitting quiet for 30 minutes then taking a measurement and sending it, then going back to sleep. So the quiescent current can't be ignored.
      $endgroup$
      – TimWescott
      7 hours ago










    • $begingroup$
      The 173010578 or the 173950536 from Wuerth or TSR 1-2450 from Traco power are pin-compatible replacements and have lower quiescent current, but it's not that huge difference.
      $endgroup$
      – Huisman
      6 hours ago












    • $begingroup$
      I meant: in idle mode, the 7805 will still be loaded, so it is not only drawing its own quiescent current but also some load current in a inefficient way.
      $endgroup$
      – Huisman
      6 hours ago










    • $begingroup$
      @Huisman OK -- I see what you mean. Yes, the larger question of how to reduce overall current consumption should include reducing the sleep current of the balance of the circuit. In fact, I'm going to add that...
      $endgroup$
      – TimWescott
      6 hours ago














    4












    4








    4





    $begingroup$


    What could explain such a short battery life? Could it be the 5v regulator?




    As mentioned, the 7805 has about 4mA of quiescent current. You need to find a data sheet for the battery (Eveready has nice battery datasheets, if you're using an alkaline cell). It's probably no more than 100mAh -- 100mAh / 4mA = 25 hours, so that should say something to you.




    How could I build a long-lasting circuit?




    The 7805 is old technology. There are better newer linear regulators out there. You should be able to easily find something that uses 10 times less quiescent current, and with digging even less than that.



    To use even less power you'd use a buck converter that's specifically designed for low quiescent current -- but I gather that you're not ready to design one into a board at the component level. There may be a module out there that'll do the job, but you'll need to shop around for it. TI does have some buck converter modules, but you'll want to pay a lot of attention to their capabilities, both for maximum current delivery and quiescent current.



    To use less power yet, do everything that you can to minimize the current consumption of your circuit when it is quiescent. This will require careful use of the microprocessor's sleep function, as well as managing how the board is powered (for example, if it only comes on once every 30 minutes, you may well want to turn off power to the radio and the humidity-reading portions of the circuit).



    Measure the current consumption in all modes of operation, and use this to determine which modes are the worst offenders overall, then concentrate on minimizing the currents in those modes if you can.






    share|improve this answer











    $endgroup$




    What could explain such a short battery life? Could it be the 5v regulator?




    As mentioned, the 7805 has about 4mA of quiescent current. You need to find a data sheet for the battery (Eveready has nice battery datasheets, if you're using an alkaline cell). It's probably no more than 100mAh -- 100mAh / 4mA = 25 hours, so that should say something to you.




    How could I build a long-lasting circuit?




    The 7805 is old technology. There are better newer linear regulators out there. You should be able to easily find something that uses 10 times less quiescent current, and with digging even less than that.



    To use even less power you'd use a buck converter that's specifically designed for low quiescent current -- but I gather that you're not ready to design one into a board at the component level. There may be a module out there that'll do the job, but you'll need to shop around for it. TI does have some buck converter modules, but you'll want to pay a lot of attention to their capabilities, both for maximum current delivery and quiescent current.



    To use less power yet, do everything that you can to minimize the current consumption of your circuit when it is quiescent. This will require careful use of the microprocessor's sleep function, as well as managing how the board is powered (for example, if it only comes on once every 30 minutes, you may well want to turn off power to the radio and the humidity-reading portions of the circuit).



    Measure the current consumption in all modes of operation, and use this to determine which modes are the worst offenders overall, then concentrate on minimizing the currents in those modes if you can.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 5 hours ago

























    answered 7 hours ago









    TimWescottTimWescott

    9,2941720




    9,2941720












    • $begingroup$
      My first thought too: skip to regulator, use a converter as you probably draw more than only the quiescent current.
      $endgroup$
      – Huisman
      7 hours ago






    • 1




      $begingroup$
      The thing sitting quiet for 30 minutes then taking a measurement and sending it, then going back to sleep. So the quiescent current can't be ignored.
      $endgroup$
      – TimWescott
      7 hours ago










    • $begingroup$
      The 173010578 or the 173950536 from Wuerth or TSR 1-2450 from Traco power are pin-compatible replacements and have lower quiescent current, but it's not that huge difference.
      $endgroup$
      – Huisman
      6 hours ago












    • $begingroup$
      I meant: in idle mode, the 7805 will still be loaded, so it is not only drawing its own quiescent current but also some load current in a inefficient way.
      $endgroup$
      – Huisman
      6 hours ago










    • $begingroup$
      @Huisman OK -- I see what you mean. Yes, the larger question of how to reduce overall current consumption should include reducing the sleep current of the balance of the circuit. In fact, I'm going to add that...
      $endgroup$
      – TimWescott
      6 hours ago


















    • $begingroup$
      My first thought too: skip to regulator, use a converter as you probably draw more than only the quiescent current.
      $endgroup$
      – Huisman
      7 hours ago






    • 1




      $begingroup$
      The thing sitting quiet for 30 minutes then taking a measurement and sending it, then going back to sleep. So the quiescent current can't be ignored.
      $endgroup$
      – TimWescott
      7 hours ago










    • $begingroup$
      The 173010578 or the 173950536 from Wuerth or TSR 1-2450 from Traco power are pin-compatible replacements and have lower quiescent current, but it's not that huge difference.
      $endgroup$
      – Huisman
      6 hours ago












    • $begingroup$
      I meant: in idle mode, the 7805 will still be loaded, so it is not only drawing its own quiescent current but also some load current in a inefficient way.
      $endgroup$
      – Huisman
      6 hours ago










    • $begingroup$
      @Huisman OK -- I see what you mean. Yes, the larger question of how to reduce overall current consumption should include reducing the sleep current of the balance of the circuit. In fact, I'm going to add that...
      $endgroup$
      – TimWescott
      6 hours ago
















    $begingroup$
    My first thought too: skip to regulator, use a converter as you probably draw more than only the quiescent current.
    $endgroup$
    – Huisman
    7 hours ago




    $begingroup$
    My first thought too: skip to regulator, use a converter as you probably draw more than only the quiescent current.
    $endgroup$
    – Huisman
    7 hours ago




    1




    1




    $begingroup$
    The thing sitting quiet for 30 minutes then taking a measurement and sending it, then going back to sleep. So the quiescent current can't be ignored.
    $endgroup$
    – TimWescott
    7 hours ago




    $begingroup$
    The thing sitting quiet for 30 minutes then taking a measurement and sending it, then going back to sleep. So the quiescent current can't be ignored.
    $endgroup$
    – TimWescott
    7 hours ago












    $begingroup$
    The 173010578 or the 173950536 from Wuerth or TSR 1-2450 from Traco power are pin-compatible replacements and have lower quiescent current, but it's not that huge difference.
    $endgroup$
    – Huisman
    6 hours ago






    $begingroup$
    The 173010578 or the 173950536 from Wuerth or TSR 1-2450 from Traco power are pin-compatible replacements and have lower quiescent current, but it's not that huge difference.
    $endgroup$
    – Huisman
    6 hours ago














    $begingroup$
    I meant: in idle mode, the 7805 will still be loaded, so it is not only drawing its own quiescent current but also some load current in a inefficient way.
    $endgroup$
    – Huisman
    6 hours ago




    $begingroup$
    I meant: in idle mode, the 7805 will still be loaded, so it is not only drawing its own quiescent current but also some load current in a inefficient way.
    $endgroup$
    – Huisman
    6 hours ago












    $begingroup$
    @Huisman OK -- I see what you mean. Yes, the larger question of how to reduce overall current consumption should include reducing the sleep current of the balance of the circuit. In fact, I'm going to add that...
    $endgroup$
    – TimWescott
    6 hours ago




    $begingroup$
    @Huisman OK -- I see what you mean. Yes, the larger question of how to reduce overall current consumption should include reducing the sleep current of the balance of the circuit. In fact, I'm going to add that...
    $endgroup$
    – TimWescott
    6 hours ago













    2












    $begingroup$

    The idle current of a 7805 regulator is around 4 mA so, armed with the ampere hour capacity of your battery, work how long it will last with a continuous drain of 4 mA.



    If you establish that is the problem, you will find that there are plenty of regulators having a significantly lower quiescent current.






    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      The idle current of a 7805 regulator is around 4 mA so, armed with the ampere hour capacity of your battery, work how long it will last with a continuous drain of 4 mA.



      If you establish that is the problem, you will find that there are plenty of regulators having a significantly lower quiescent current.






      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The idle current of a 7805 regulator is around 4 mA so, armed with the ampere hour capacity of your battery, work how long it will last with a continuous drain of 4 mA.



        If you establish that is the problem, you will find that there are plenty of regulators having a significantly lower quiescent current.






        share|improve this answer









        $endgroup$



        The idle current of a 7805 regulator is around 4 mA so, armed with the ampere hour capacity of your battery, work how long it will last with a continuous drain of 4 mA.



        If you establish that is the problem, you will find that there are plenty of regulators having a significantly lower quiescent current.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 8 hours ago









        Andy akaAndy aka

        248k11190437




        248k11190437






























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