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What informations can we obtain with these voltage and current measurements of a little electronic device?

Why can't a multimeter measure its own current and voltage?Correctly measure L7805ACT regulator outputFiguring out battery life from amp draw?Mixing same brand batteries with similar voltageConnecting batteries in parallel doesn't produce greater amperageWhy are these readings violating ohm's law? (Are they?)Multimeter precision on low batteryWhy does the measured potential-difference between two poles of an AA fizzle out after 5 seconds?How can a used alkaline cell rate at 1.9 volts?Replace Lithium Battery with Permanent Power Source

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

$begingroup$

I powered a little electronic device with 2 AA batteries and here are the results measured with a multimeter:

• 
  

  When using a first set of 2 AA batteries:

  
  
  
  • Voltage when device unplugged: 2.45 V
  
  
  • Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A
  
  


• 
  

  When using a second set of 2 brand new AA batteries:

  
  
  
  • Voltage when device unplugged: 3.10 V
  
  
  • Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A
  
  

Just being curious: what informations can I deduce about the device or about the batteries with these measurements?

voltage batteries multimeter

share|improve this question

edited 11 hours ago

Basj

asked 12 hours ago

BasjBasj

74222 gold badges1414 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try posting a picture of the device and get a better response instead of guesses or plainly basic information like the device's input port resistance. Do you know ohms law BTW?
  $endgroup$
  – Andy aka
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is this electronics puzzle question you got as homework :P
  $endgroup$
  – Mitu Raj
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Andyaka It's a Zoom pedal like this: sweetwater.com/store/detail/… It can be powered by 2 AA batteries or 9V DC adapter or 5V USB. Pretty convenient, so I was curious about how it works! And yes I know Ohm's law
  $endgroup$
  – Basj
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MituRaj No no ;) It's a device I have, and since it can be powered by 2 AA batteries or 5V USB or 9V DC adapter, I was curious about its powering system :)
  $endgroup$
  – Basj
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was curious about its powering system - This may be obvious enough to not warrant explanation, but the device has a voltage regulator internally that converts any of those three power sources to the correct voltage for the actual circuit to function.
  $endgroup$
  – dwizum
  2 hours ago
  

  
  
  
  

 | 
show 3 more comments

2

$begingroup$

I powered a little electronic device with 2 AA batteries and here are the results measured with a multimeter:

• 
  

  When using a first set of 2 AA batteries:

  
  
  
  • Voltage when device unplugged: 2.45 V
  
  
  • Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A
  
  


• 
  

  When using a second set of 2 brand new AA batteries:

  
  
  
  • Voltage when device unplugged: 3.10 V
  
  
  • Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A
  
  

Just being curious: what informations can I deduce about the device or about the batteries with these measurements?

voltage batteries multimeter

share|improve this question

edited 11 hours ago

Basj

asked 12 hours ago

BasjBasj

74222 gold badges1414 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try posting a picture of the device and get a better response instead of guesses or plainly basic information like the device's input port resistance. Do you know ohms law BTW?
  $endgroup$
  – Andy aka
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is this electronics puzzle question you got as homework :P
  $endgroup$
  – Mitu Raj
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Andyaka It's a Zoom pedal like this: sweetwater.com/store/detail/… It can be powered by 2 AA batteries or 9V DC adapter or 5V USB. Pretty convenient, so I was curious about how it works! And yes I know Ohm's law
  $endgroup$
  – Basj
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MituRaj No no ;) It's a device I have, and since it can be powered by 2 AA batteries or 5V USB or 9V DC adapter, I was curious about its powering system :)
  $endgroup$
  – Basj
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was curious about its powering system - This may be obvious enough to not warrant explanation, but the device has a voltage regulator internally that converts any of those three power sources to the correct voltage for the actual circuit to function.
  $endgroup$
  – dwizum
  2 hours ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

I powered a little electronic device with 2 AA batteries and here are the results measured with a multimeter:

• 
  

  When using a first set of 2 AA batteries:

  
  
  
  • Voltage when device unplugged: 2.45 V
  
  
  • Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A
  
  


• 
  

  When using a second set of 2 brand new AA batteries:

  
  
  
  • Voltage when device unplugged: 3.10 V
  
  
  • Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A
  
  

Just being curious: what informations can I deduce about the device or about the batteries with these measurements?

voltage batteries multimeter

share|improve this question

edited 11 hours ago

Basj

asked 12 hours ago

BasjBasj

74222 gold badges1414 silver badges2525 bronze badges

$endgroup$

share|improve this question

edited 11 hours ago

Basj

asked 12 hours ago

BasjBasj

74222 gold badges1414 silver badges2525 bronze badges

share|improve this question

edited 11 hours ago

Basj

asked 12 hours ago

BasjBasj

74222 gold badges1414 silver badges2525 bronze badges

asked 12 hours ago

BasjBasj

74222 gold badges1414 silver badges2525 bronze badges

asked 12 hours ago

BasjBasj

74222 gold badges1414 silver badges2525 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

2.1V * 0.3A = 0.63W

3.1V * 0.2A = 0.62W

The device seems to use constant power no matter what the input voltage is, which most likely indicates it is internally powered by a switching DC-DC converter. It could also be a flashlight with a switching LED driver.

share|improve this answer

answered 11 hours ago

peufeupeufeu

27.6k22 gold badges4040 silver badges8181 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thank you! Is there something we can say about the batteries? What does the voltage drop (0.35V for the first set of batteries, 0.10V for the second set) tell about them?
  $endgroup$
  – Basj
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Fresh batteries have lower internal resistance
  $endgroup$
  – peufeu
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Because the first batteries are getting a bit older ;) , internal resistance increases with usage over time
  $endgroup$
  – Mitu Raj
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MituRaj can we estimate the internal resistances with these measurements?
  $endgroup$
  – Basj
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ofcourse, its just the dropped value 0.35V divided by the current drawn ....
  $endgroup$
  – Mitu Raj
  10 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

*• When using a first set of 2 AA batteries:
• Voltage when device unplugged: 2.45 V
• Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A

Battery ESR = 0.35V/0.30A = 1.17 Ohms, load = 0.63W.

Battery loss= 0.3^2* 1.17= 0.1W

• When using a second set of 2 brand new AA batteries:
• Voltage when device unplugged: 3.10 V
• Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A*

Battery ESR = 0.1V/0.2A = 0.5 Ohms, load = 0.60W
Battery loss = 0.2^2* 0.5= 0.02W
Cells may or may not be balanced when reading 2.1/3/1 = 67% of no load fresh cell voltage and ESR always rises which DoD.

Load appears to be constant power yet 5% more efficient with fresh batteries while battery losses reduce from 0.1W to 0.02W with low DoD fresh cells.

So supply inefficiency rises from 0.02/0.60W= 3.3% to 0.1W/0.63W = 16% additional losses when drained. This explains more rapid decay in storage energy remaining when batteries get low with a constant power load.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

80.8k22 gold badges3030 silver badges116116 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting: Measuring Internal Resistance of Batteries.
  $endgroup$
  – Basj
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Batteries are just like batteries with a low ESR but exceptionally high C value. My answer assumes you know Ohm’s Law and ESR of dielectric ~ electrode interface exists for every battery and every capacitor. There is also a double-electric layer effect, giving 2 time constants, the longer one appears like “memory effect”
  $endgroup$
  – Sunnyskyguy EE75
  2 hours ago
  
  

  
  
  
  

add a comment
 | 

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6

$begingroup$

2.1V * 0.3A = 0.63W

3.1V * 0.2A = 0.62W

The device seems to use constant power no matter what the input voltage is, which most likely indicates it is internally powered by a switching DC-DC converter. It could also be a flashlight with a switching LED driver.

share|improve this answer

answered 11 hours ago

peufeupeufeu

27.6k22 gold badges4040 silver badges8181 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thank you! Is there something we can say about the batteries? What does the voltage drop (0.35V for the first set of batteries, 0.10V for the second set) tell about them?
  $endgroup$
  – Basj
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Fresh batteries have lower internal resistance
  $endgroup$
  – peufeu
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Because the first batteries are getting a bit older ;) , internal resistance increases with usage over time
  $endgroup$
  – Mitu Raj
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MituRaj can we estimate the internal resistances with these measurements?
  $endgroup$
  – Basj
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ofcourse, its just the dropped value 0.35V divided by the current drawn ....
  $endgroup$
  – Mitu Raj
  10 hours ago
  

  
  
  
  

add a comment
 | 

6

$begingroup$

2.1V * 0.3A = 0.63W

3.1V * 0.2A = 0.62W

The device seems to use constant power no matter what the input voltage is, which most likely indicates it is internally powered by a switching DC-DC converter. It could also be a flashlight with a switching LED driver.

share|improve this answer

answered 11 hours ago

peufeupeufeu

27.6k22 gold badges4040 silver badges8181 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thank you! Is there something we can say about the batteries? What does the voltage drop (0.35V for the first set of batteries, 0.10V for the second set) tell about them?
  $endgroup$
  – Basj
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Fresh batteries have lower internal resistance
  $endgroup$
  – peufeu
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Because the first batteries are getting a bit older ;) , internal resistance increases with usage over time
  $endgroup$
  – Mitu Raj
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MituRaj can we estimate the internal resistances with these measurements?
  $endgroup$
  – Basj
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ofcourse, its just the dropped value 0.35V divided by the current drawn ....
  $endgroup$
  – Mitu Raj
  10 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

2.1V * 0.3A = 0.63W

3.1V * 0.2A = 0.62W

The device seems to use constant power no matter what the input voltage is, which most likely indicates it is internally powered by a switching DC-DC converter. It could also be a flashlight with a switching LED driver.

share|improve this answer

answered 11 hours ago

peufeupeufeu

27.6k22 gold badges4040 silver badges8181 bronze badges

$endgroup$

share|improve this answer

answered 11 hours ago

peufeupeufeu

27.6k22 gold badges4040 silver badges8181 bronze badges

answered 11 hours ago

peufeupeufeu

27.6k22 gold badges4040 silver badges8181 bronze badges

answered 11 hours ago

peufeupeufeu

27.6k22 gold badges4040 silver badges8181 bronze badges

2

$begingroup$

*• When using a first set of 2 AA batteries:
• Voltage when device unplugged: 2.45 V
• Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A

Battery ESR = 0.35V/0.30A = 1.17 Ohms, load = 0.63W.

Battery loss= 0.3^2* 1.17= 0.1W

• When using a second set of 2 brand new AA batteries:
• Voltage when device unplugged: 3.10 V
• Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A*

Battery ESR = 0.1V/0.2A = 0.5 Ohms, load = 0.60W
Battery loss = 0.2^2* 0.5= 0.02W
Cells may or may not be balanced when reading 2.1/3/1 = 67% of no load fresh cell voltage and ESR always rises which DoD.

Load appears to be constant power yet 5% more efficient with fresh batteries while battery losses reduce from 0.1W to 0.02W with low DoD fresh cells.

So supply inefficiency rises from 0.02/0.60W= 3.3% to 0.1W/0.63W = 16% additional losses when drained. This explains more rapid decay in storage energy remaining when batteries get low with a constant power load.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

80.8k22 gold badges3030 silver badges116116 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting: Measuring Internal Resistance of Batteries.
  $endgroup$
  – Basj
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Batteries are just like batteries with a low ESR but exceptionally high C value. My answer assumes you know Ohm’s Law and ESR of dielectric ~ electrode interface exists for every battery and every capacitor. There is also a double-electric layer effect, giving 2 time constants, the longer one appears like “memory effect”
  $endgroup$
  – Sunnyskyguy EE75
  2 hours ago
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

*• When using a first set of 2 AA batteries:
• Voltage when device unplugged: 2.45 V
• Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A

Battery ESR = 0.35V/0.30A = 1.17 Ohms, load = 0.63W.

Battery loss= 0.3^2* 1.17= 0.1W

• When using a second set of 2 brand new AA batteries:
• Voltage when device unplugged: 3.10 V
• Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A*

Battery ESR = 0.1V/0.2A = 0.5 Ohms, load = 0.60W
Battery loss = 0.2^2* 0.5= 0.02W
Cells may or may not be balanced when reading 2.1/3/1 = 67% of no load fresh cell voltage and ESR always rises which DoD.

Load appears to be constant power yet 5% more efficient with fresh batteries while battery losses reduce from 0.1W to 0.02W with low DoD fresh cells.

So supply inefficiency rises from 0.02/0.60W= 3.3% to 0.1W/0.63W = 16% additional losses when drained. This explains more rapid decay in storage energy remaining when batteries get low with a constant power load.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

80.8k22 gold badges3030 silver badges116116 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting: Measuring Internal Resistance of Batteries.
  $endgroup$
  – Basj
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Batteries are just like batteries with a low ESR but exceptionally high C value. My answer assumes you know Ohm’s Law and ESR of dielectric ~ electrode interface exists for every battery and every capacitor. There is also a double-electric layer effect, giving 2 time constants, the longer one appears like “memory effect”
  $endgroup$
  – Sunnyskyguy EE75
  2 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

*• When using a first set of 2 AA batteries:
• Voltage when device unplugged: 2.45 V
• Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A

Battery ESR = 0.35V/0.30A = 1.17 Ohms, load = 0.63W.

Battery loss= 0.3^2* 1.17= 0.1W

• When using a second set of 2 brand new AA batteries:
• Voltage when device unplugged: 3.10 V
• Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A*

Battery ESR = 0.1V/0.2A = 0.5 Ohms, load = 0.60W
Battery loss = 0.2^2* 0.5= 0.02W
Cells may or may not be balanced when reading 2.1/3/1 = 67% of no load fresh cell voltage and ESR always rises which DoD.

Load appears to be constant power yet 5% more efficient with fresh batteries while battery losses reduce from 0.1W to 0.02W with low DoD fresh cells.

So supply inefficiency rises from 0.02/0.60W= 3.3% to 0.1W/0.63W = 16% additional losses when drained. This explains more rapid decay in storage energy remaining when batteries get low with a constant power load.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

80.8k22 gold badges3030 silver badges116116 bronze badges

$endgroup$

*• When using a first set of 2 AA batteries:
• Voltage when device unplugged: 2.45 V
• Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A

• When using a second set of 2 brand new AA batteries:
• Voltage when device unplugged: 3.10 V
• Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A*

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

80.8k22 gold badges3030 silver badges116116 bronze badges

answered 8 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

80.8k22 gold badges3030 silver badges116116 bronze badges

answered 8 hours ago

Sunnyskyguy EE75Sunnyskyguy EE75

80.8k22 gold badges3030 silver badges116116 bronze badges