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How does an ARM MCU run faster than the external crystal?
ARM Cortex (M3-M4): manufacturer and development IDEDoes the CPU of an ARM Cortex M halt during flash self-programming?Can a wrong crystal oscilator capacitor burn the MCU?ARM Processor board and ProgrammingLockup when executing relocated ISR in ARM Cortex-M0+Using “oscillator” rather than crystal for MCUWhen should one use an external crystal for this MCU given that the internal oscillator is much faster?How should I connect the external crystal in MPU6050LPC1768 Baud rateRectifying failed MCU crystal oscillator design
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
So before this I only worked with simple 8 bit Atmel MCUs and I realized on my development board schematics it has only a 12Mhz crystal, yet the MCU operates at up to 100MHz. (I think the default is 80MHz. I only made it higher once for fun. It's just a simple line in the code.)
How does it do that? Why does an Atmega328, for example, run at the used crystal speed?
arm crystal cortex-m
edited May 26 at 12:11
JRE
25.8k64786
asked May 26 at 11:55
Amy GambleAmy Gamble
694
New contributor
Amy Gamble is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
So before this I only worked with simple 8 bit Atmel MCUs and I realized on my development board schematics it has only a 12Mhz crystal, yet the MCU operates at up to 100MHz. (I think the default is 80MHz. I only made it higher once for fun. It's just a simple line in the code.)
How does it do that? Why does an Atmega328, for example, run at the used crystal speed?
arm crystal cortex-m
edited May 26 at 12:11
JRE
25.8k64786
asked May 26 at 11:55
Amy GambleAmy Gamble
694
New contributor
Amy Gamble is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
1) mHz means milli Hertz so 1/1000th of a Hertz, use MHz (capital M) when you mean Mega Hertz. 2) what 8 Bit MCU? 3) What development board? 4) The ATMega 328 doesn't have to run at the crystal's speed, read the section of clocking in the datasheet to see what is possible. 5) Some ICs have a PLL which can be used to multiply the external clock frequency.
$endgroup$
– Bimpelrekkie
May 26 at 12:03
1
$begingroup$
@Bimpelrekkie: Most of that comment is the answer. Would you like to make an answer of it?
$endgroup$
– JRE
May 26 at 12:12
$begingroup$
VCO divider to compare with 12 MHz ends up multiplying f. Thats what a PLL freq synth does.
$endgroup$
– Sunnyskyguy EE75
May 26 at 12:43
$begingroup$
@Bimpelrekkie looks like someone already fixed it for me. I already mentioned Atmel and atmega328. but it doesn't matter as that's an example. afaik atmega328p doesn't have a PPL? I was mostly curious about how you can just simply multiply it. thankfully Marcus Muller already explained it very well.
$endgroup$
– Amy Gamble
May 26 at 14:14
$begingroup$
the processor in the computer you are reading this on is likely using a 100Mhz reference clock or slower yet the core runs in the ghz.
$endgroup$
– old_timer
May 26 at 18:43
|
show 1 more comment
$begingroup$
So before this I only worked with simple 8 bit Atmel MCUs and I realized on my development board schematics it has only a 12Mhz crystal, yet the MCU operates at up to 100MHz. (I think the default is 80MHz. I only made it higher once for fun. It's just a simple line in the code.)
How does it do that? Why does an Atmega328, for example, run at the used crystal speed?
arm crystal cortex-m
edited May 26 at 12:11
JRE
25.8k64786
asked May 26 at 11:55
Amy GambleAmy Gamble
694
New contributor
Amy Gamble is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
So before this I only worked with simple 8 bit Atmel MCUs and I realized on my development board schematics it has only a 12Mhz crystal, yet the MCU operates at up to 100MHz. (I think the default is 80MHz. I only made it higher once for fun. It's just a simple line in the code.)
How does it do that? Why does an Atmega328, for example, run at the used crystal speed?
arm crystal cortex-m
arm crystal cortex-m
edited May 26 at 12:11
JRE
25.8k64786
asked May 26 at 11:55
Amy GambleAmy Gamble
694
New contributor
Amy Gamble is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 26 at 12:11
JRE
25.8k64786
asked May 26 at 11:55
Amy GambleAmy Gamble
694
New contributor
Amy Gamble is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 26 at 12:11
JRE
25.8k64786
edited May 26 at 12:11
JRE
25.8k64786
edited May 26 at 12:11
JRE
25.8k64786
25.8k64786
asked May 26 at 11:55
Amy GambleAmy Gamble
694
New contributor
Amy Gamble is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 26 at 11:55
Amy GambleAmy Gamble
694
asked May 26 at 11:55
Amy GambleAmy Gamble
694
694
New contributor
Amy Gamble is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Amy Gamble is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
$begingroup$
1) mHz means milli Hertz so 1/1000th of a Hertz, use MHz (capital M) when you mean Mega Hertz. 2) what 8 Bit MCU? 3) What development board? 4) The ATMega 328 doesn't have to run at the crystal's speed, read the section of clocking in the datasheet to see what is possible. 5) Some ICs have a PLL which can be used to multiply the external clock frequency.
$endgroup$
– Bimpelrekkie
May 26 at 12:03
1
$begingroup$
@Bimpelrekkie: Most of that comment is the answer. Would you like to make an answer of it?
$endgroup$
– JRE
May 26 at 12:12
$begingroup$
VCO divider to compare with 12 MHz ends up multiplying f. Thats what a PLL freq synth does.
$endgroup$
– Sunnyskyguy EE75
May 26 at 12:43
$begingroup$
@Bimpelrekkie looks like someone already fixed it for me. I already mentioned Atmel and atmega328. but it doesn't matter as that's an example. afaik atmega328p doesn't have a PPL? I was mostly curious about how you can just simply multiply it. thankfully Marcus Muller already explained it very well.
$endgroup$
– Amy Gamble
May 26 at 14:14
$begingroup$
the processor in the computer you are reading this on is likely using a 100Mhz reference clock or slower yet the core runs in the ghz.
$endgroup$
– old_timer
May 26 at 18:43
|
show 1 more comment
5
$begingroup$
1) mHz means milli Hertz so 1/1000th of a Hertz, use MHz (capital M) when you mean Mega Hertz. 2) what 8 Bit MCU? 3) What development board? 4) The ATMega 328 doesn't have to run at the crystal's speed, read the section of clocking in the datasheet to see what is possible. 5) Some ICs have a PLL which can be used to multiply the external clock frequency.
$endgroup$
– Bimpelrekkie
May 26 at 12:03
1
$begingroup$
@Bimpelrekkie: Most of that comment is the answer. Would you like to make an answer of it?
$endgroup$
– JRE
May 26 at 12:12
$begingroup$
VCO divider to compare with 12 MHz ends up multiplying f. Thats what a PLL freq synth does.
$endgroup$
– Sunnyskyguy EE75
May 26 at 12:43
$begingroup$
@Bimpelrekkie looks like someone already fixed it for me. I already mentioned Atmel and atmega328. but it doesn't matter as that's an example. afaik atmega328p doesn't have a PPL? I was mostly curious about how you can just simply multiply it. thankfully Marcus Muller already explained it very well.
$endgroup$
– Amy Gamble
May 26 at 14:14
$begingroup$
the processor in the computer you are reading this on is likely using a 100Mhz reference clock or slower yet the core runs in the ghz.
$endgroup$
– old_timer
May 26 at 18:43
5
5
$begingroup$
1) mHz means milli Hertz so 1/1000th of a Hertz, use MHz (capital M) when you mean Mega Hertz. 2) what 8 Bit MCU? 3) What development board? 4) The ATMega 328 doesn't have to run at the crystal's speed, read the section of clocking in the datasheet to see what is possible. 5) Some ICs have a PLL which can be used to multiply the external clock frequency.
$endgroup$
– Bimpelrekkie
May 26 at 12:03
$begingroup$
1) mHz means milli Hertz so 1/1000th of a Hertz, use MHz (capital M) when you mean Mega Hertz. 2) what 8 Bit MCU? 3) What development board? 4) The ATMega 328 doesn't have to run at the crystal's speed, read the section of clocking in the datasheet to see what is possible. 5) Some ICs have a PLL which can be used to multiply the external clock frequency.
$endgroup$
– Bimpelrekkie
May 26 at 12:03
1
1
$begingroup$
@Bimpelrekkie: Most of that comment is the answer. Would you like to make an answer of it?
$endgroup$
– JRE
May 26 at 12:12
$begingroup$
@Bimpelrekkie: Most of that comment is the answer. Would you like to make an answer of it?
$endgroup$
– JRE
May 26 at 12:12
$begingroup$
VCO divider to compare with 12 MHz ends up multiplying f. Thats what a PLL freq synth does.
$endgroup$
– Sunnyskyguy EE75
May 26 at 12:43
$begingroup$
VCO divider to compare with 12 MHz ends up multiplying f. Thats what a PLL freq synth does.
$endgroup$
– Sunnyskyguy EE75
May 26 at 12:43
$begingroup$
@Bimpelrekkie looks like someone already fixed it for me. I already mentioned Atmel and atmega328. but it doesn't matter as that's an example. afaik atmega328p doesn't have a PPL? I was mostly curious about how you can just simply multiply it. thankfully Marcus Muller already explained it very well.
$endgroup$
– Amy Gamble
May 26 at 14:14
$begingroup$
@Bimpelrekkie looks like someone already fixed it for me. I already mentioned Atmel and atmega328. but it doesn't matter as that's an example. afaik atmega328p doesn't have a PPL? I was mostly curious about how you can just simply multiply it. thankfully Marcus Muller already explained it very well.
$endgroup$
– Amy Gamble
May 26 at 14:14
$begingroup$
the processor in the computer you are reading this on is likely using a 100Mhz reference clock or slower yet the core runs in the ghz.
$endgroup$
– old_timer
May 26 at 18:43
$begingroup$
the processor in the computer you are reading this on is likely using a 100Mhz reference clock or slower yet the core runs in the ghz.
$endgroup$
– old_timer
May 26 at 18:43
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
This doesn't have anything to do with the core being an ARM processor; it's about how the clocking circuitry works:
In many systems like microcontrollers, RF chips, audio chips, … you need to generate a faster clock that is an exact multiple of some reference clock (for example, an external crystal).
You do that by having a voltage-controlled oscillator (VCO) that you can adjust in frequency by in- or decreasing a control voltage.
Now, by just setting any control voltage, you can bring that to oscillate at a frequency roughly in the right "ballpark", but not at an exact multiple of the input frequency. Especially, VCOs can be a bit drifty, so that frequency will also continously "wander" all over the place. You need to control that oscillator by comparing it to the reference oscillator.
The way to do that is by employing a Phase-Locked Loop. The idea is simple:
- Divide the frequency that comes out of the VCO by a factor $N$; that's the factor that we want the VCO to be faster than the reference. Doing that is easy: You can, for example, simply use a digital counter that counts to N and only then changes the output.
- Compare that $f_textVCO/N$ clock with the reference clock at $f_textref$.
If one is faster than the other, adjust the frequency accordingly. You can do that in a digital way by just XOR'ing both clocks – ideally, if they are identical, the result is a constant 0, but if one is faster than the other, then there will be a growing amount of times when the XOR of both clocks is 1; slow down or speed up the VCO accordingly.
The above is a control loop, locked to the phase of – hence the name.
For "rich" microcontrollers, which have a lot of peripherals and hence benefit from having multiple clocks internally, it's usual to have at least 1 PLL. The ATMega328 is a bit strange in that respect: It's a relatively power-hungry, relatively peripheral-rich microcontroller that still doesn't have a PLL.
answered May 26 at 12:37
Marcus MüllerMarcus Müller
37k364104
$endgroup$
1
$begingroup$
Thanks you :) this explains it very well! Instead of poking fun or focusing on my silly mistakes you just told me what I'm obviously asking. Some things are VERY hard for me to pay attention to with my Dyslexia. Even if I read trough my post 2-3 times I miss a lot of makes or accidentally put capitalization where I don't need to. Not to mention English isn't my first language.
$endgroup$
– Amy Gamble
May 26 at 14:07
6
$begingroup$
@AmyGamble your English is very good, however! I think the first comment you got under your post was because very many young engineers simply forget about capitalization of units – which can become very problematic later on :) I hope you never feel discouraged!
$endgroup$
– Marcus Müller
May 26 at 20:03
add a comment |
$begingroup$
Some devices have a PLL in them that can multiply the crystal frequency to higher frequencies. The ATMega328 does not have a PLL, it uses the crystal directly.
answered May 26 at 12:13
JustmeJustme
4,0732615
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
This doesn't have anything to do with the core being an ARM processor; it's about how the clocking circuitry works:
In many systems like microcontrollers, RF chips, audio chips, … you need to generate a faster clock that is an exact multiple of some reference clock (for example, an external crystal).
You do that by having a voltage-controlled oscillator (VCO) that you can adjust in frequency by in- or decreasing a control voltage.
Now, by just setting any control voltage, you can bring that to oscillate at a frequency roughly in the right "ballpark", but not at an exact multiple of the input frequency. Especially, VCOs can be a bit drifty, so that frequency will also continously "wander" all over the place. You need to control that oscillator by comparing it to the reference oscillator.
The way to do that is by employing a Phase-Locked Loop. The idea is simple:
- Divide the frequency that comes out of the VCO by a factor $N$; that's the factor that we want the VCO to be faster than the reference. Doing that is easy: You can, for example, simply use a digital counter that counts to N and only then changes the output.
- Compare that $f_textVCO/N$ clock with the reference clock at $f_textref$.
If one is faster than the other, adjust the frequency accordingly. You can do that in a digital way by just XOR'ing both clocks – ideally, if they are identical, the result is a constant 0, but if one is faster than the other, then there will be a growing amount of times when the XOR of both clocks is 1; slow down or speed up the VCO accordingly.
The above is a control loop, locked to the phase of – hence the name.
For "rich" microcontrollers, which have a lot of peripherals and hence benefit from having multiple clocks internally, it's usual to have at least 1 PLL. The ATMega328 is a bit strange in that respect: It's a relatively power-hungry, relatively peripheral-rich microcontroller that still doesn't have a PLL.
answered May 26 at 12:37
Marcus MüllerMarcus Müller
37k364104
$endgroup$
1
$begingroup$
Thanks you :) this explains it very well! Instead of poking fun or focusing on my silly mistakes you just told me what I'm obviously asking. Some things are VERY hard for me to pay attention to with my Dyslexia. Even if I read trough my post 2-3 times I miss a lot of makes or accidentally put capitalization where I don't need to. Not to mention English isn't my first language.
$endgroup$
– Amy Gamble
May 26 at 14:07
6
$begingroup$
@AmyGamble your English is very good, however! I think the first comment you got under your post was because very many young engineers simply forget about capitalization of units – which can become very problematic later on :) I hope you never feel discouraged!
$endgroup$
– Marcus Müller
May 26 at 20:03
add a comment |
$begingroup$
This doesn't have anything to do with the core being an ARM processor; it's about how the clocking circuitry works:
In many systems like microcontrollers, RF chips, audio chips, … you need to generate a faster clock that is an exact multiple of some reference clock (for example, an external crystal).
You do that by having a voltage-controlled oscillator (VCO) that you can adjust in frequency by in- or decreasing a control voltage.
Now, by just setting any control voltage, you can bring that to oscillate at a frequency roughly in the right "ballpark", but not at an exact multiple of the input frequency. Especially, VCOs can be a bit drifty, so that frequency will also continously "wander" all over the place. You need to control that oscillator by comparing it to the reference oscillator.
The way to do that is by employing a Phase-Locked Loop. The idea is simple:
- Divide the frequency that comes out of the VCO by a factor $N$; that's the factor that we want the VCO to be faster than the reference. Doing that is easy: You can, for example, simply use a digital counter that counts to N and only then changes the output.
- Compare that $f_textVCO/N$ clock with the reference clock at $f_textref$.
If one is faster than the other, adjust the frequency accordingly. You can do that in a digital way by just XOR'ing both clocks – ideally, if they are identical, the result is a constant 0, but if one is faster than the other, then there will be a growing amount of times when the XOR of both clocks is 1; slow down or speed up the VCO accordingly.
The above is a control loop, locked to the phase of – hence the name.
For "rich" microcontrollers, which have a lot of peripherals and hence benefit from having multiple clocks internally, it's usual to have at least 1 PLL. The ATMega328 is a bit strange in that respect: It's a relatively power-hungry, relatively peripheral-rich microcontroller that still doesn't have a PLL.
answered May 26 at 12:37
Marcus MüllerMarcus Müller
37k364104
$endgroup$
1
$begingroup$
Thanks you :) this explains it very well! Instead of poking fun or focusing on my silly mistakes you just told me what I'm obviously asking. Some things are VERY hard for me to pay attention to with my Dyslexia. Even if I read trough my post 2-3 times I miss a lot of makes or accidentally put capitalization where I don't need to. Not to mention English isn't my first language.
$endgroup$
– Amy Gamble
May 26 at 14:07
6
$begingroup$
@AmyGamble your English is very good, however! I think the first comment you got under your post was because very many young engineers simply forget about capitalization of units – which can become very problematic later on :) I hope you never feel discouraged!
$endgroup$
– Marcus Müller
May 26 at 20:03
add a comment |
$begingroup$
This doesn't have anything to do with the core being an ARM processor; it's about how the clocking circuitry works:
In many systems like microcontrollers, RF chips, audio chips, … you need to generate a faster clock that is an exact multiple of some reference clock (for example, an external crystal).
You do that by having a voltage-controlled oscillator (VCO) that you can adjust in frequency by in- or decreasing a control voltage.
Now, by just setting any control voltage, you can bring that to oscillate at a frequency roughly in the right "ballpark", but not at an exact multiple of the input frequency. Especially, VCOs can be a bit drifty, so that frequency will also continously "wander" all over the place. You need to control that oscillator by comparing it to the reference oscillator.
The way to do that is by employing a Phase-Locked Loop. The idea is simple:
- Divide the frequency that comes out of the VCO by a factor $N$; that's the factor that we want the VCO to be faster than the reference. Doing that is easy: You can, for example, simply use a digital counter that counts to N and only then changes the output.
- Compare that $f_textVCO/N$ clock with the reference clock at $f_textref$.
If one is faster than the other, adjust the frequency accordingly. You can do that in a digital way by just XOR'ing both clocks – ideally, if they are identical, the result is a constant 0, but if one is faster than the other, then there will be a growing amount of times when the XOR of both clocks is 1; slow down or speed up the VCO accordingly.
The above is a control loop, locked to the phase of – hence the name.
For "rich" microcontrollers, which have a lot of peripherals and hence benefit from having multiple clocks internally, it's usual to have at least 1 PLL. The ATMega328 is a bit strange in that respect: It's a relatively power-hungry, relatively peripheral-rich microcontroller that still doesn't have a PLL.
answered May 26 at 12:37
Marcus MüllerMarcus Müller
37k364104
$endgroup$
This doesn't have anything to do with the core being an ARM processor; it's about how the clocking circuitry works:
In many systems like microcontrollers, RF chips, audio chips, … you need to generate a faster clock that is an exact multiple of some reference clock (for example, an external crystal).
You do that by having a voltage-controlled oscillator (VCO) that you can adjust in frequency by in- or decreasing a control voltage.
Now, by just setting any control voltage, you can bring that to oscillate at a frequency roughly in the right "ballpark", but not at an exact multiple of the input frequency. Especially, VCOs can be a bit drifty, so that frequency will also continously "wander" all over the place. You need to control that oscillator by comparing it to the reference oscillator.
The way to do that is by employing a Phase-Locked Loop. The idea is simple:
- Divide the frequency that comes out of the VCO by a factor $N$; that's the factor that we want the VCO to be faster than the reference. Doing that is easy: You can, for example, simply use a digital counter that counts to N and only then changes the output.
- Compare that $f_textVCO/N$ clock with the reference clock at $f_textref$.
If one is faster than the other, adjust the frequency accordingly. You can do that in a digital way by just XOR'ing both clocks – ideally, if they are identical, the result is a constant 0, but if one is faster than the other, then there will be a growing amount of times when the XOR of both clocks is 1; slow down or speed up the VCO accordingly.
The above is a control loop, locked to the phase of – hence the name.
For "rich" microcontrollers, which have a lot of peripherals and hence benefit from having multiple clocks internally, it's usual to have at least 1 PLL. The ATMega328 is a bit strange in that respect: It's a relatively power-hungry, relatively peripheral-rich microcontroller that still doesn't have a PLL.
answered May 26 at 12:37
Marcus MüllerMarcus Müller
37k364104
answered May 26 at 12:37
Marcus MüllerMarcus Müller
37k364104
answered May 26 at 12:37
Marcus MüllerMarcus Müller
37k364104
answered May 26 at 12:37
Marcus MüllerMarcus Müller
37k364104
37k364104
1
$begingroup$
Thanks you :) this explains it very well! Instead of poking fun or focusing on my silly mistakes you just told me what I'm obviously asking. Some things are VERY hard for me to pay attention to with my Dyslexia. Even if I read trough my post 2-3 times I miss a lot of makes or accidentally put capitalization where I don't need to. Not to mention English isn't my first language.
$endgroup$
– Amy Gamble
May 26 at 14:07
6
$begingroup$
@AmyGamble your English is very good, however! I think the first comment you got under your post was because very many young engineers simply forget about capitalization of units – which can become very problematic later on :) I hope you never feel discouraged!
$endgroup$
– Marcus Müller
May 26 at 20:03
add a comment |
1
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Thanks you :) this explains it very well! Instead of poking fun or focusing on my silly mistakes you just told me what I'm obviously asking. Some things are VERY hard for me to pay attention to with my Dyslexia. Even if I read trough my post 2-3 times I miss a lot of makes or accidentally put capitalization where I don't need to. Not to mention English isn't my first language.
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– Amy Gamble
May 26 at 14:07
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@AmyGamble your English is very good, however! I think the first comment you got under your post was because very many young engineers simply forget about capitalization of units – which can become very problematic later on :) I hope you never feel discouraged!
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– Marcus Müller
May 26 at 20:03
1
1
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Thanks you :) this explains it very well! Instead of poking fun or focusing on my silly mistakes you just told me what I'm obviously asking. Some things are VERY hard for me to pay attention to with my Dyslexia. Even if I read trough my post 2-3 times I miss a lot of makes or accidentally put capitalization where I don't need to. Not to mention English isn't my first language.
$endgroup$
– Amy Gamble
May 26 at 14:07
$begingroup$
Thanks you :) this explains it very well! Instead of poking fun or focusing on my silly mistakes you just told me what I'm obviously asking. Some things are VERY hard for me to pay attention to with my Dyslexia. Even if I read trough my post 2-3 times I miss a lot of makes or accidentally put capitalization where I don't need to. Not to mention English isn't my first language.
$endgroup$
– Amy Gamble
May 26 at 14:07
6
6
$begingroup$
@AmyGamble your English is very good, however! I think the first comment you got under your post was because very many young engineers simply forget about capitalization of units – which can become very problematic later on :) I hope you never feel discouraged!
$endgroup$
– Marcus Müller
May 26 at 20:03
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@AmyGamble your English is very good, however! I think the first comment you got under your post was because very many young engineers simply forget about capitalization of units – which can become very problematic later on :) I hope you never feel discouraged!
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– Marcus Müller
May 26 at 20:03
add a comment |
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Some devices have a PLL in them that can multiply the crystal frequency to higher frequencies. The ATMega328 does not have a PLL, it uses the crystal directly.
answered May 26 at 12:13
JustmeJustme
4,0732615
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add a comment |
$begingroup$
Some devices have a PLL in them that can multiply the crystal frequency to higher frequencies. The ATMega328 does not have a PLL, it uses the crystal directly.
answered May 26 at 12:13
JustmeJustme
4,0732615
$endgroup$
add a comment |
$begingroup$
Some devices have a PLL in them that can multiply the crystal frequency to higher frequencies. The ATMega328 does not have a PLL, it uses the crystal directly.
answered May 26 at 12:13
JustmeJustme
4,0732615
$endgroup$
Some devices have a PLL in them that can multiply the crystal frequency to higher frequencies. The ATMega328 does not have a PLL, it uses the crystal directly.
answered May 26 at 12:13
JustmeJustme
4,0732615
answered May 26 at 12:13
JustmeJustme
4,0732615
answered May 26 at 12:13
JustmeJustme
4,0732615
answered May 26 at 12:13
JustmeJustme
4,0732615
4,0732615
add a comment |
add a comment |
Amy Gamble is a new contributor. Be nice, and check out our Code of Conduct.
Amy Gamble is a new contributor. Be nice, and check out our Code of Conduct.
Amy Gamble is a new contributor. Be nice, and check out our Code of Conduct.
Amy Gamble is a new contributor. Be nice, and check out our Code of Conduct.
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1) mHz means milli Hertz so 1/1000th of a Hertz, use MHz (capital M) when you mean Mega Hertz. 2) what 8 Bit MCU? 3) What development board? 4) The ATMega 328 doesn't have to run at the crystal's speed, read the section of clocking in the datasheet to see what is possible. 5) Some ICs have a PLL which can be used to multiply the external clock frequency.
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– Bimpelrekkie
May 26 at 12:03
1
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@Bimpelrekkie: Most of that comment is the answer. Would you like to make an answer of it?
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– JRE
May 26 at 12:12
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VCO divider to compare with 12 MHz ends up multiplying f. Thats what a PLL freq synth does.
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– Sunnyskyguy EE75
May 26 at 12:43
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@Bimpelrekkie looks like someone already fixed it for me. I already mentioned Atmel and atmega328. but it doesn't matter as that's an example. afaik atmega328p doesn't have a PPL? I was mostly curious about how you can just simply multiply it. thankfully Marcus Muller already explained it very well.
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– Amy Gamble
May 26 at 14:14
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the processor in the computer you are reading this on is likely using a 100Mhz reference clock or slower yet the core runs in the ghz.
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– old_timer
May 26 at 18:43