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How do devices/appliances draw more current when needed?
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Suppose I have a desktop computer and I decide to do something that requires more processing power. In this case my computer will draw more current to increase power. How is this increase in current performed? Does my computer open up more parallel circuits so the total resistance decreases? or do they have electronic potentiometer or something completely else. Is the technique used in a desktop computer the same as if I were changing the oven's temperature?
Any help is greatly appreciated.
current drawing
asked 3 hours ago
rr1303
344
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up vote
2
down vote
favorite
Suppose I have a desktop computer and I decide to do something that requires more processing power. In this case my computer will draw more current to increase power. How is this increase in current performed? Does my computer open up more parallel circuits so the total resistance decreases? or do they have electronic potentiometer or something completely else. Is the technique used in a desktop computer the same as if I were changing the oven's temperature?
Any help is greatly appreciated.
current drawing
asked 3 hours ago
rr1303
344
Your oven works basically the same way, just on much longer timescales: it shuts down heating when the temperature is over, and enables when it is under a limit.
– PlasmaHH
3 hours ago
1
The current drawn is a consequence of what happens in the CPU. Computers don't "open up the upstream gates of current" on purpose to enable more computing power, like you seem to be assuming. It doesn't work like the throttle of a car engine.
– dim
3 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose I have a desktop computer and I decide to do something that requires more processing power. In this case my computer will draw more current to increase power. How is this increase in current performed? Does my computer open up more parallel circuits so the total resistance decreases? or do they have electronic potentiometer or something completely else. Is the technique used in a desktop computer the same as if I were changing the oven's temperature?
Any help is greatly appreciated.
current drawing
asked 3 hours ago
rr1303
344
Suppose I have a desktop computer and I decide to do something that requires more processing power. In this case my computer will draw more current to increase power. How is this increase in current performed? Does my computer open up more parallel circuits so the total resistance decreases? or do they have electronic potentiometer or something completely else. Is the technique used in a desktop computer the same as if I were changing the oven's temperature?
Any help is greatly appreciated.
current drawing
current drawing
asked 3 hours ago
rr1303
344
asked 3 hours ago
rr1303
344
asked 3 hours ago
rr1303
344
asked 3 hours ago
rr1303
344
asked 3 hours ago
rr1303
344
344
Your oven works basically the same way, just on much longer timescales: it shuts down heating when the temperature is over, and enables when it is under a limit.
– PlasmaHH
3 hours ago
1
The current drawn is a consequence of what happens in the CPU. Computers don't "open up the upstream gates of current" on purpose to enable more computing power, like you seem to be assuming. It doesn't work like the throttle of a car engine.
– dim
3 hours ago
add a comment |Â
Your oven works basically the same way, just on much longer timescales: it shuts down heating when the temperature is over, and enables when it is under a limit.
– PlasmaHH
3 hours ago
1
The current drawn is a consequence of what happens in the CPU. Computers don't "open up the upstream gates of current" on purpose to enable more computing power, like you seem to be assuming. It doesn't work like the throttle of a car engine.
– dim
3 hours ago
Your oven works basically the same way, just on much longer timescales: it shuts down heating when the temperature is over, and enables when it is under a limit.
– PlasmaHH
3 hours ago
Your oven works basically the same way, just on much longer timescales: it shuts down heating when the temperature is over, and enables when it is under a limit.
– PlasmaHH
3 hours ago
1
1
The current drawn is a consequence of what happens in the CPU. Computers don't "open up the upstream gates of current" on purpose to enable more computing power, like you seem to be assuming. It doesn't work like the throttle of a car engine.
– dim
3 hours ago
The current drawn is a consequence of what happens in the CPU. Computers don't "open up the upstream gates of current" on purpose to enable more computing power, like you seem to be assuming. It doesn't work like the throttle of a car engine.
– dim
3 hours ago
add a comment |Â
5 Answers
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oldest
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up vote
5
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accepted
I decide to do something that requires more processing power. In this case my computer will draw more current to increase power.
Other way round: the computer will do more things, and as a result it will consume more power.
Does my computer open up more parallel circuits so the total resistance decreases?
This is roughly true. Except that computers don't really operate on continuous current flow, they operate in bursts driven by their internal clock; each action involves either drawing some current to turn on a transistor, or sinking some current to turn it off again. Times a billion transistors, a billion times a second. More computation involves more transistors.
answered 3 hours ago
pjc50
32.5k33881
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up vote
3
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At a high level, yes, you're right that the computer opens up more transistors or at least switches more transistors when it consumes more current. For instance, if you have a hardware multiplier and generally you don't use it, the transistors in the multiplier won't switch on and therefore won't draw much current. If the code now requests a multiplication, the transistors in it start switching and that will lower the resistance between VDD and ground. This will draw more current. The current draw will lower the VDD voltage. Now the switching voltage regulator will detect this voltage droop and kick on at a higher duty cycle to allow high current capability and approximately a constant voltage.
At a broad high-level, circuits request more current by lowering their resistance because most circuits operate with a constant voltage source.
answered 1 hour ago
horta
10.9k1536
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up vote
2
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There are several mechanisms for power consumption at the chip level.
When circuits switch, there are internal parasitic capacitors in all transistors and interconnects (internally on the chips and externally). These capacitors have to be charged and discharged when the circuit nodes are switched from off to on (or on to off). The capacitors are tiny, but when you have billions of them switching billions of times per second it adds up. (this power is actually dissipated by circuit element resistance, including parasitic resistance in the parasitic capacitors)
All circuit elements also have resistance so current flow anywhere in the circuits creates heat and consumes power. As the circuit nodes switch, the parasitic capacitors on the load side devices have to be changed or discharged and this requires current flow which, in turn, creates heat and consumes power.
The power consumption associated with these two effects varies by the number of internal node switching operations which means the power consumption varies by the activity (and clock speed) of the processor and other elements.
Transistors and other components inside the integrated circuits also have leakage current. This creates a baseline (static) power consumption that still occurs when the processor is inactive. Many modern low power systems switch off the power to entire subsystems on the processor and other chips during sleep or inactive states to minimize this static power consumption.
There are other mechanisms of power consumption in computers (power supply quiescent power, etc) but these should help you understand why the power consumption varies and why there is still some power consumption when no work is being done.
answered 3 hours ago
Dean Franks
2,0851814
This answer is correct, but, you are on a different wavelength than OP. It's an impedance mismatch.
– Harry Svensson
3 hours ago
add a comment |Â
up vote
2
down vote
Modern computers use logic gates that are designed to use very little power when they are in a steady state, but which take a burst of power to switch them from one state to another.
If the computer is idle, the processor will be in a sleep state for much of the time. Most of the circuits will be doing nothing, and so consume little power. The same goes for other components, such as the graphics card's GPU.
If you then give it something to do, then suddenly it's performing more work. The gates are switching on and off more often, and so they take more power.
In addition, many computers, especially laptops, are designed to power down whole sections of the computer if they are not being used. For instance the webcam in a laptop will be powered off until you open an application that uses it.
answered 1 hour ago
Simon B
4,483818
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up vote
1
down vote
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
Example: Calculate the expected current consumption in idle mode with TIMER1, ADC, and SPI enabled at VCC =
2.0V and F = 1MHz. From Table Additional Current Consumption (percentage) in Active and Idle mode in
the previous section, third column, we see that we need to add 14.5% for the TIMER1, 22.1% for the
ADC, and 15.7% for the SPI module. Reading from Figure Idle Supply Current vs. Low Frequency
(0.1-1.0MHz), we find that the idle current consumption is ~0.045mA at VCC = 2.0V and F = 1MHz. The
total current consumption in idle mode with TIMER1, ADC, and SPI enabled, gives:
ICCtotal ≃ 0.045 mA⋅(1 + 0.145 + 0.221 + 0.157) ≃ 0.069 mA
(Helps to have the datasheet open to look the various tables).
For a computer, running at 3.2 GHz (200 times faster) and perhaps 1.8V core logic voltage (and 4 or 8 cores for multithreading), 3.3V IO voltage, talking to memory and video chip(s) and hard drive controller and USB controllers and ethernet or wireless controller, the calculations would be similar, with each chip adding its own amount to the total. You can see why the computer processor has a big heatsink on top with a cooling fan blowing air over it.
answered 2 hours ago
CrossRoads
4664
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
I decide to do something that requires more processing power. In this case my computer will draw more current to increase power.
Other way round: the computer will do more things, and as a result it will consume more power.
Does my computer open up more parallel circuits so the total resistance decreases?
This is roughly true. Except that computers don't really operate on continuous current flow, they operate in bursts driven by their internal clock; each action involves either drawing some current to turn on a transistor, or sinking some current to turn it off again. Times a billion transistors, a billion times a second. More computation involves more transistors.
answered 3 hours ago
pjc50
32.5k33881
add a comment |Â
up vote
5
down vote
accepted
I decide to do something that requires more processing power. In this case my computer will draw more current to increase power.
Other way round: the computer will do more things, and as a result it will consume more power.
Does my computer open up more parallel circuits so the total resistance decreases?
This is roughly true. Except that computers don't really operate on continuous current flow, they operate in bursts driven by their internal clock; each action involves either drawing some current to turn on a transistor, or sinking some current to turn it off again. Times a billion transistors, a billion times a second. More computation involves more transistors.
answered 3 hours ago
pjc50
32.5k33881
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
I decide to do something that requires more processing power. In this case my computer will draw more current to increase power.
Other way round: the computer will do more things, and as a result it will consume more power.
Does my computer open up more parallel circuits so the total resistance decreases?
This is roughly true. Except that computers don't really operate on continuous current flow, they operate in bursts driven by their internal clock; each action involves either drawing some current to turn on a transistor, or sinking some current to turn it off again. Times a billion transistors, a billion times a second. More computation involves more transistors.
answered 3 hours ago
pjc50
32.5k33881
I decide to do something that requires more processing power. In this case my computer will draw more current to increase power.
Other way round: the computer will do more things, and as a result it will consume more power.
Does my computer open up more parallel circuits so the total resistance decreases?
This is roughly true. Except that computers don't really operate on continuous current flow, they operate in bursts driven by their internal clock; each action involves either drawing some current to turn on a transistor, or sinking some current to turn it off again. Times a billion transistors, a billion times a second. More computation involves more transistors.
answered 3 hours ago
pjc50
32.5k33881
answered 3 hours ago
pjc50
32.5k33881
answered 3 hours ago
pjc50
32.5k33881
answered 3 hours ago
pjc50
32.5k33881
32.5k33881
add a comment |Â
add a comment |Â
up vote
3
down vote
At a high level, yes, you're right that the computer opens up more transistors or at least switches more transistors when it consumes more current. For instance, if you have a hardware multiplier and generally you don't use it, the transistors in the multiplier won't switch on and therefore won't draw much current. If the code now requests a multiplication, the transistors in it start switching and that will lower the resistance between VDD and ground. This will draw more current. The current draw will lower the VDD voltage. Now the switching voltage regulator will detect this voltage droop and kick on at a higher duty cycle to allow high current capability and approximately a constant voltage.
At a broad high-level, circuits request more current by lowering their resistance because most circuits operate with a constant voltage source.
answered 1 hour ago
horta
10.9k1536
add a comment |Â
up vote
3
down vote
At a high level, yes, you're right that the computer opens up more transistors or at least switches more transistors when it consumes more current. For instance, if you have a hardware multiplier and generally you don't use it, the transistors in the multiplier won't switch on and therefore won't draw much current. If the code now requests a multiplication, the transistors in it start switching and that will lower the resistance between VDD and ground. This will draw more current. The current draw will lower the VDD voltage. Now the switching voltage regulator will detect this voltage droop and kick on at a higher duty cycle to allow high current capability and approximately a constant voltage.
At a broad high-level, circuits request more current by lowering their resistance because most circuits operate with a constant voltage source.
answered 1 hour ago
horta
10.9k1536
add a comment |Â
up vote
3
down vote
up vote
3
down vote
At a high level, yes, you're right that the computer opens up more transistors or at least switches more transistors when it consumes more current. For instance, if you have a hardware multiplier and generally you don't use it, the transistors in the multiplier won't switch on and therefore won't draw much current. If the code now requests a multiplication, the transistors in it start switching and that will lower the resistance between VDD and ground. This will draw more current. The current draw will lower the VDD voltage. Now the switching voltage regulator will detect this voltage droop and kick on at a higher duty cycle to allow high current capability and approximately a constant voltage.
At a broad high-level, circuits request more current by lowering their resistance because most circuits operate with a constant voltage source.
answered 1 hour ago
horta
10.9k1536
At a high level, yes, you're right that the computer opens up more transistors or at least switches more transistors when it consumes more current. For instance, if you have a hardware multiplier and generally you don't use it, the transistors in the multiplier won't switch on and therefore won't draw much current. If the code now requests a multiplication, the transistors in it start switching and that will lower the resistance between VDD and ground. This will draw more current. The current draw will lower the VDD voltage. Now the switching voltage regulator will detect this voltage droop and kick on at a higher duty cycle to allow high current capability and approximately a constant voltage.
At a broad high-level, circuits request more current by lowering their resistance because most circuits operate with a constant voltage source.
answered 1 hour ago
horta
10.9k1536
answered 1 hour ago
horta
10.9k1536
answered 1 hour ago
horta
10.9k1536
answered 1 hour ago
horta
10.9k1536
10.9k1536
add a comment |Â
add a comment |Â
up vote
2
down vote
There are several mechanisms for power consumption at the chip level.
When circuits switch, there are internal parasitic capacitors in all transistors and interconnects (internally on the chips and externally). These capacitors have to be charged and discharged when the circuit nodes are switched from off to on (or on to off). The capacitors are tiny, but when you have billions of them switching billions of times per second it adds up. (this power is actually dissipated by circuit element resistance, including parasitic resistance in the parasitic capacitors)
All circuit elements also have resistance so current flow anywhere in the circuits creates heat and consumes power. As the circuit nodes switch, the parasitic capacitors on the load side devices have to be changed or discharged and this requires current flow which, in turn, creates heat and consumes power.
The power consumption associated with these two effects varies by the number of internal node switching operations which means the power consumption varies by the activity (and clock speed) of the processor and other elements.
Transistors and other components inside the integrated circuits also have leakage current. This creates a baseline (static) power consumption that still occurs when the processor is inactive. Many modern low power systems switch off the power to entire subsystems on the processor and other chips during sleep or inactive states to minimize this static power consumption.
There are other mechanisms of power consumption in computers (power supply quiescent power, etc) but these should help you understand why the power consumption varies and why there is still some power consumption when no work is being done.
answered 3 hours ago
Dean Franks
2,0851814
This answer is correct, but, you are on a different wavelength than OP. It's an impedance mismatch.
– Harry Svensson
3 hours ago
add a comment |Â
up vote
2
down vote
There are several mechanisms for power consumption at the chip level.
When circuits switch, there are internal parasitic capacitors in all transistors and interconnects (internally on the chips and externally). These capacitors have to be charged and discharged when the circuit nodes are switched from off to on (or on to off). The capacitors are tiny, but when you have billions of them switching billions of times per second it adds up. (this power is actually dissipated by circuit element resistance, including parasitic resistance in the parasitic capacitors)
All circuit elements also have resistance so current flow anywhere in the circuits creates heat and consumes power. As the circuit nodes switch, the parasitic capacitors on the load side devices have to be changed or discharged and this requires current flow which, in turn, creates heat and consumes power.
The power consumption associated with these two effects varies by the number of internal node switching operations which means the power consumption varies by the activity (and clock speed) of the processor and other elements.
Transistors and other components inside the integrated circuits also have leakage current. This creates a baseline (static) power consumption that still occurs when the processor is inactive. Many modern low power systems switch off the power to entire subsystems on the processor and other chips during sleep or inactive states to minimize this static power consumption.
There are other mechanisms of power consumption in computers (power supply quiescent power, etc) but these should help you understand why the power consumption varies and why there is still some power consumption when no work is being done.
answered 3 hours ago
Dean Franks
2,0851814
This answer is correct, but, you are on a different wavelength than OP. It's an impedance mismatch.
– Harry Svensson
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There are several mechanisms for power consumption at the chip level.
When circuits switch, there are internal parasitic capacitors in all transistors and interconnects (internally on the chips and externally). These capacitors have to be charged and discharged when the circuit nodes are switched from off to on (or on to off). The capacitors are tiny, but when you have billions of them switching billions of times per second it adds up. (this power is actually dissipated by circuit element resistance, including parasitic resistance in the parasitic capacitors)
All circuit elements also have resistance so current flow anywhere in the circuits creates heat and consumes power. As the circuit nodes switch, the parasitic capacitors on the load side devices have to be changed or discharged and this requires current flow which, in turn, creates heat and consumes power.
The power consumption associated with these two effects varies by the number of internal node switching operations which means the power consumption varies by the activity (and clock speed) of the processor and other elements.
Transistors and other components inside the integrated circuits also have leakage current. This creates a baseline (static) power consumption that still occurs when the processor is inactive. Many modern low power systems switch off the power to entire subsystems on the processor and other chips during sleep or inactive states to minimize this static power consumption.
There are other mechanisms of power consumption in computers (power supply quiescent power, etc) but these should help you understand why the power consumption varies and why there is still some power consumption when no work is being done.
answered 3 hours ago
Dean Franks
2,0851814
There are several mechanisms for power consumption at the chip level.
When circuits switch, there are internal parasitic capacitors in all transistors and interconnects (internally on the chips and externally). These capacitors have to be charged and discharged when the circuit nodes are switched from off to on (or on to off). The capacitors are tiny, but when you have billions of them switching billions of times per second it adds up. (this power is actually dissipated by circuit element resistance, including parasitic resistance in the parasitic capacitors)
All circuit elements also have resistance so current flow anywhere in the circuits creates heat and consumes power. As the circuit nodes switch, the parasitic capacitors on the load side devices have to be changed or discharged and this requires current flow which, in turn, creates heat and consumes power.
The power consumption associated with these two effects varies by the number of internal node switching operations which means the power consumption varies by the activity (and clock speed) of the processor and other elements.
Transistors and other components inside the integrated circuits also have leakage current. This creates a baseline (static) power consumption that still occurs when the processor is inactive. Many modern low power systems switch off the power to entire subsystems on the processor and other chips during sleep or inactive states to minimize this static power consumption.
There are other mechanisms of power consumption in computers (power supply quiescent power, etc) but these should help you understand why the power consumption varies and why there is still some power consumption when no work is being done.
answered 3 hours ago
Dean Franks
2,0851814
answered 3 hours ago
Dean Franks
2,0851814
answered 3 hours ago
Dean Franks
2,0851814
answered 3 hours ago
Dean Franks
2,0851814
2,0851814
This answer is correct, but, you are on a different wavelength than OP. It's an impedance mismatch.
– Harry Svensson
3 hours ago
add a comment |Â
This answer is correct, but, you are on a different wavelength than OP. It's an impedance mismatch.
– Harry Svensson
3 hours ago
This answer is correct, but, you are on a different wavelength than OP. It's an impedance mismatch.
– Harry Svensson
3 hours ago
This answer is correct, but, you are on a different wavelength than OP. It's an impedance mismatch.
– Harry Svensson
3 hours ago
add a comment |Â
up vote
2
down vote
Modern computers use logic gates that are designed to use very little power when they are in a steady state, but which take a burst of power to switch them from one state to another.
If the computer is idle, the processor will be in a sleep state for much of the time. Most of the circuits will be doing nothing, and so consume little power. The same goes for other components, such as the graphics card's GPU.
If you then give it something to do, then suddenly it's performing more work. The gates are switching on and off more often, and so they take more power.
In addition, many computers, especially laptops, are designed to power down whole sections of the computer if they are not being used. For instance the webcam in a laptop will be powered off until you open an application that uses it.
answered 1 hour ago
Simon B
4,483818
add a comment |Â
up vote
2
down vote
Modern computers use logic gates that are designed to use very little power when they are in a steady state, but which take a burst of power to switch them from one state to another.
If the computer is idle, the processor will be in a sleep state for much of the time. Most of the circuits will be doing nothing, and so consume little power. The same goes for other components, such as the graphics card's GPU.
If you then give it something to do, then suddenly it's performing more work. The gates are switching on and off more often, and so they take more power.
In addition, many computers, especially laptops, are designed to power down whole sections of the computer if they are not being used. For instance the webcam in a laptop will be powered off until you open an application that uses it.
answered 1 hour ago
Simon B
4,483818
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Modern computers use logic gates that are designed to use very little power when they are in a steady state, but which take a burst of power to switch them from one state to another.
If the computer is idle, the processor will be in a sleep state for much of the time. Most of the circuits will be doing nothing, and so consume little power. The same goes for other components, such as the graphics card's GPU.
If you then give it something to do, then suddenly it's performing more work. The gates are switching on and off more often, and so they take more power.
In addition, many computers, especially laptops, are designed to power down whole sections of the computer if they are not being used. For instance the webcam in a laptop will be powered off until you open an application that uses it.
answered 1 hour ago
Simon B
4,483818
Modern computers use logic gates that are designed to use very little power when they are in a steady state, but which take a burst of power to switch them from one state to another.
If the computer is idle, the processor will be in a sleep state for much of the time. Most of the circuits will be doing nothing, and so consume little power. The same goes for other components, such as the graphics card's GPU.
If you then give it something to do, then suddenly it's performing more work. The gates are switching on and off more often, and so they take more power.
In addition, many computers, especially laptops, are designed to power down whole sections of the computer if they are not being used. For instance the webcam in a laptop will be powered off until you open an application that uses it.
answered 1 hour ago
Simon B
4,483818
answered 1 hour ago
Simon B
4,483818
answered 1 hour ago
Simon B
4,483818
answered 1 hour ago
Simon B
4,483818
4,483818
add a comment |Â
add a comment |Â
up vote
1
down vote
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
Example: Calculate the expected current consumption in idle mode with TIMER1, ADC, and SPI enabled at VCC =
2.0V and F = 1MHz. From Table Additional Current Consumption (percentage) in Active and Idle mode in
the previous section, third column, we see that we need to add 14.5% for the TIMER1, 22.1% for the
ADC, and 15.7% for the SPI module. Reading from Figure Idle Supply Current vs. Low Frequency
(0.1-1.0MHz), we find that the idle current consumption is ~0.045mA at VCC = 2.0V and F = 1MHz. The
total current consumption in idle mode with TIMER1, ADC, and SPI enabled, gives:
ICCtotal ≃ 0.045 mA⋅(1 + 0.145 + 0.221 + 0.157) ≃ 0.069 mA
(Helps to have the datasheet open to look the various tables).
For a computer, running at 3.2 GHz (200 times faster) and perhaps 1.8V core logic voltage (and 4 or 8 cores for multithreading), 3.3V IO voltage, talking to memory and video chip(s) and hard drive controller and USB controllers and ethernet or wireless controller, the calculations would be similar, with each chip adding its own amount to the total. You can see why the computer processor has a big heatsink on top with a cooling fan blowing air over it.
answered 2 hours ago
CrossRoads
4664
add a comment |Â
up vote
1
down vote
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
Example: Calculate the expected current consumption in idle mode with TIMER1, ADC, and SPI enabled at VCC =
2.0V and F = 1MHz. From Table Additional Current Consumption (percentage) in Active and Idle mode in
the previous section, third column, we see that we need to add 14.5% for the TIMER1, 22.1% for the
ADC, and 15.7% for the SPI module. Reading from Figure Idle Supply Current vs. Low Frequency
(0.1-1.0MHz), we find that the idle current consumption is ~0.045mA at VCC = 2.0V and F = 1MHz. The
total current consumption in idle mode with TIMER1, ADC, and SPI enabled, gives:
ICCtotal ≃ 0.045 mA⋅(1 + 0.145 + 0.221 + 0.157) ≃ 0.069 mA
(Helps to have the datasheet open to look the various tables).
For a computer, running at 3.2 GHz (200 times faster) and perhaps 1.8V core logic voltage (and 4 or 8 cores for multithreading), 3.3V IO voltage, talking to memory and video chip(s) and hard drive controller and USB controllers and ethernet or wireless controller, the calculations would be similar, with each chip adding its own amount to the total. You can see why the computer processor has a big heatsink on top with a cooling fan blowing air over it.
answered 2 hours ago
CrossRoads
4664
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
Example: Calculate the expected current consumption in idle mode with TIMER1, ADC, and SPI enabled at VCC =
2.0V and F = 1MHz. From Table Additional Current Consumption (percentage) in Active and Idle mode in
the previous section, third column, we see that we need to add 14.5% for the TIMER1, 22.1% for the
ADC, and 15.7% for the SPI module. Reading from Figure Idle Supply Current vs. Low Frequency
(0.1-1.0MHz), we find that the idle current consumption is ~0.045mA at VCC = 2.0V and F = 1MHz. The
total current consumption in idle mode with TIMER1, ADC, and SPI enabled, gives:
ICCtotal ≃ 0.045 mA⋅(1 + 0.145 + 0.221 + 0.157) ≃ 0.069 mA
(Helps to have the datasheet open to look the various tables).
For a computer, running at 3.2 GHz (200 times faster) and perhaps 1.8V core logic voltage (and 4 or 8 cores for multithreading), 3.3V IO voltage, talking to memory and video chip(s) and hard drive controller and USB controllers and ethernet or wireless controller, the calculations would be similar, with each chip adding its own amount to the total. You can see why the computer processor has a big heatsink on top with a cooling fan blowing air over it.
answered 2 hours ago
CrossRoads
4664
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
The different ICs in the computer will each have different current draw. Here is some data from the Atmega328P, a simple 8-bit 16 MHz microcontroller used in the Arduino Uno and other similar boards.
Example: Calculate the expected current consumption in idle mode with TIMER1, ADC, and SPI enabled at VCC =
2.0V and F = 1MHz. From Table Additional Current Consumption (percentage) in Active and Idle mode in
the previous section, third column, we see that we need to add 14.5% for the TIMER1, 22.1% for the
ADC, and 15.7% for the SPI module. Reading from Figure Idle Supply Current vs. Low Frequency
(0.1-1.0MHz), we find that the idle current consumption is ~0.045mA at VCC = 2.0V and F = 1MHz. The
total current consumption in idle mode with TIMER1, ADC, and SPI enabled, gives:
ICCtotal ≃ 0.045 mA⋅(1 + 0.145 + 0.221 + 0.157) ≃ 0.069 mA
(Helps to have the datasheet open to look the various tables).
For a computer, running at 3.2 GHz (200 times faster) and perhaps 1.8V core logic voltage (and 4 or 8 cores for multithreading), 3.3V IO voltage, talking to memory and video chip(s) and hard drive controller and USB controllers and ethernet or wireless controller, the calculations would be similar, with each chip adding its own amount to the total. You can see why the computer processor has a big heatsink on top with a cooling fan blowing air over it.
answered 2 hours ago
CrossRoads
4664
answered 2 hours ago
CrossRoads
4664
answered 2 hours ago
CrossRoads
4664
answered 2 hours ago
CrossRoads
4664
4664
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Your oven works basically the same way, just on much longer timescales: it shuts down heating when the temperature is over, and enables when it is under a limit.
– PlasmaHH
3 hours ago
1
The current drawn is a consequence of what happens in the CPU. Computers don't "open up the upstream gates of current" on purpose to enable more computing power, like you seem to be assuming. It doesn't work like the throttle of a car engine.
– dim
3 hours ago