Does battery voltage actually get lower when connected to a load, or does it just appear to do so?
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Sorry if Im wording this question strangely. I am using a 3.7V battery and my microcontroller monitors the voltage and goes to sleep if my battery voltage is too low. The issue is that it reads a lower voltage than the battery shows if I disconnect it and check it with my multimeter. For example, my microcontroller would read 3.65V when my multimeter would read my disconnected battery at 3.8V. Is my microcontroller reading the voltage incorrectly or should I treat the with-load voltage that my microcontroller is readings as the actual voltage?
voltage batteries
asked 2 hours ago
Tapatio Sombrero
262
add a comment |
up vote
4
down vote
favorite
Sorry if Im wording this question strangely. I am using a 3.7V battery and my microcontroller monitors the voltage and goes to sleep if my battery voltage is too low. The issue is that it reads a lower voltage than the battery shows if I disconnect it and check it with my multimeter. For example, my microcontroller would read 3.65V when my multimeter would read my disconnected battery at 3.8V. Is my microcontroller reading the voltage incorrectly or should I treat the with-load voltage that my microcontroller is readings as the actual voltage?
voltage batteries
asked 2 hours ago
Tapatio Sombrero
262
Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
– mbrig
1 hour ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Sorry if Im wording this question strangely. I am using a 3.7V battery and my microcontroller monitors the voltage and goes to sleep if my battery voltage is too low. The issue is that it reads a lower voltage than the battery shows if I disconnect it and check it with my multimeter. For example, my microcontroller would read 3.65V when my multimeter would read my disconnected battery at 3.8V. Is my microcontroller reading the voltage incorrectly or should I treat the with-load voltage that my microcontroller is readings as the actual voltage?
voltage batteries
asked 2 hours ago
Tapatio Sombrero
262
Sorry if Im wording this question strangely. I am using a 3.7V battery and my microcontroller monitors the voltage and goes to sleep if my battery voltage is too low. The issue is that it reads a lower voltage than the battery shows if I disconnect it and check it with my multimeter. For example, my microcontroller would read 3.65V when my multimeter would read my disconnected battery at 3.8V. Is my microcontroller reading the voltage incorrectly or should I treat the with-load voltage that my microcontroller is readings as the actual voltage?
voltage batteries
voltage batteries
asked 2 hours ago
Tapatio Sombrero
262
asked 2 hours ago
Tapatio Sombrero
262
asked 2 hours ago
Tapatio Sombrero
262
asked 2 hours ago
Tapatio Sombrero
262
asked 2 hours ago
Tapatio Sombrero
262
262
Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
– mbrig
1 hour ago
add a comment |
Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
– mbrig
1 hour ago
Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
– mbrig
1 hour ago
Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
– mbrig
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered 2 hours ago
Gonzik007
2,5971021
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
– Tapatio Sombrero
35 mins ago
add a comment |
up vote
4
down vote
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered 2 hours ago
Stefan Wyss
1,978313
add a comment |
up vote
2
down vote
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered 2 hours ago
Tony EE rocketscientist
60.8k22192
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
up vote
6
down vote
When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered 2 hours ago
Gonzik007
2,5971021
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
– Tapatio Sombrero
35 mins ago
add a comment |
up vote
6
down vote
When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered 2 hours ago
Gonzik007
2,5971021
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
– Tapatio Sombrero
35 mins ago
add a comment |
up vote
6
down vote
up vote
6
down vote
When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered 2 hours ago
Gonzik007
2,5971021
When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered 2 hours ago
Gonzik007
2,5971021
answered 2 hours ago
Gonzik007
2,5971021
answered 2 hours ago
Gonzik007
2,5971021
answered 2 hours ago
Gonzik007
2,5971021
2,5971021
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
– Tapatio Sombrero
35 mins ago
add a comment |
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
– Tapatio Sombrero
35 mins ago
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
– Tapatio Sombrero
35 mins ago
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
– Tapatio Sombrero
35 mins ago
add a comment |
up vote
4
down vote
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered 2 hours ago
Stefan Wyss
1,978313
add a comment |
up vote
4
down vote
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered 2 hours ago
Stefan Wyss
1,978313
add a comment |
up vote
4
down vote
up vote
4
down vote
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered 2 hours ago
Stefan Wyss
1,978313
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered 2 hours ago
Stefan Wyss
1,978313
answered 2 hours ago
Stefan Wyss
1,978313
answered 2 hours ago
Stefan Wyss
1,978313
answered 2 hours ago
Stefan Wyss
1,978313
1,978313
add a comment |
add a comment |
up vote
2
down vote
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered 2 hours ago
Tony EE rocketscientist
60.8k22192
add a comment |
up vote
2
down vote
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered 2 hours ago
Tony EE rocketscientist
60.8k22192
add a comment |
up vote
2
down vote
up vote
2
down vote
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered 2 hours ago
Tony EE rocketscientist
60.8k22192
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered 2 hours ago
Tony EE rocketscientist
60.8k22192
edited 1 hour ago
answered 2 hours ago
Tony EE rocketscientist
60.8k22192
answered 2 hours ago
Tony EE rocketscientist
60.8k22192
answered 2 hours ago
Tony EE rocketscientist
60.8k22192
60.8k22192
add a comment |
add a comment |
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Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
– mbrig
1 hour ago