Cheap oscilloscope showing 16 MHz square waveCheap 1MHz oscilloscopeOscilloscope trace not quite squareWhat can be the reason that oscilloscope is showing voltages 10 times more than expected?Why would a 150 MHz probe work fine for a 100 MHz oscilloscope?Bouncing at some points in square waveOscilloscope displaying two voltage levels (on the same channel) for square wave
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Cheap oscilloscope showing 16 MHz square wave
Cheap 1MHz oscilloscopeOscilloscope trace not quite squareWhat can be the reason that oscilloscope is showing voltages 10 times more than expected?Why would a 150 MHz probe work fine for a 100 MHz oscilloscope?Bouncing at some points in square waveOscilloscope displaying two voltage levels (on the same channel) for square wave
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$begingroup$
I own a cheap oscilloscope Hantek DSO4102C. It's rated bandwidth is 100 MHz, and sample rate is 1 GSa/s. Some info about the tool can be found here: http://hantek.com/en/ProductDetail_3_4163.html
Now I have an Atmega328P MCU running from an external quartz at 16 MHz, without any code it it (chip erased by usbasp), only CKOUT fuse bit is set. So I supposed to see a square wave at PB0 pin, but my scope shows it quite distorted:
MCU's datasheet doesn't mention a pin rise time, which was a big surprise to me, so I cannot check if measured 9.5 ns is a valid value. But judging by Pk-Pk voltage exceeding 6 volts (and even going below zero for a good 560 mV), I believe there's a problem with the scope. Am I right?
oscilloscope
asked 8 hours ago
ZhenekZhenek
596 bronze badges
$endgroup$
add a comment |
$begingroup$
I own a cheap oscilloscope Hantek DSO4102C. It's rated bandwidth is 100 MHz, and sample rate is 1 GSa/s. Some info about the tool can be found here: http://hantek.com/en/ProductDetail_3_4163.html
Now I have an Atmega328P MCU running from an external quartz at 16 MHz, without any code it it (chip erased by usbasp), only CKOUT fuse bit is set. So I supposed to see a square wave at PB0 pin, but my scope shows it quite distorted:
MCU's datasheet doesn't mention a pin rise time, which was a big surprise to me, so I cannot check if measured 9.5 ns is a valid value. But judging by Pk-Pk voltage exceeding 6 volts (and even going below zero for a good 560 mV), I believe there's a problem with the scope. Am I right?
oscilloscope
asked 8 hours ago
ZhenekZhenek
596 bronze badges
$endgroup$
$begingroup$
Are your probes compensated correctly? Also, can you try with a different probe?
$endgroup$
– Steve G
7 hours ago
$begingroup$
Could you add a photo of how you're probing the signal? That is, how exactly your probe is connected to the circuit.
$endgroup$
– marcelm
5 hours ago
$begingroup$
Make sure your probe is in the x10 position, compensation adjustment is done and the ground lead is connected to a plane very close to the MCU ground. You can also run probe wizard and self cal routines.
$endgroup$
– Spehro Pefhany
5 hours ago
add a comment |
$begingroup$
I own a cheap oscilloscope Hantek DSO4102C. It's rated bandwidth is 100 MHz, and sample rate is 1 GSa/s. Some info about the tool can be found here: http://hantek.com/en/ProductDetail_3_4163.html
Now I have an Atmega328P MCU running from an external quartz at 16 MHz, without any code it it (chip erased by usbasp), only CKOUT fuse bit is set. So I supposed to see a square wave at PB0 pin, but my scope shows it quite distorted:
MCU's datasheet doesn't mention a pin rise time, which was a big surprise to me, so I cannot check if measured 9.5 ns is a valid value. But judging by Pk-Pk voltage exceeding 6 volts (and even going below zero for a good 560 mV), I believe there's a problem with the scope. Am I right?
oscilloscope
asked 8 hours ago
ZhenekZhenek
596 bronze badges
$endgroup$
I own a cheap oscilloscope Hantek DSO4102C. It's rated bandwidth is 100 MHz, and sample rate is 1 GSa/s. Some info about the tool can be found here: http://hantek.com/en/ProductDetail_3_4163.html
Now I have an Atmega328P MCU running from an external quartz at 16 MHz, without any code it it (chip erased by usbasp), only CKOUT fuse bit is set. So I supposed to see a square wave at PB0 pin, but my scope shows it quite distorted:
MCU's datasheet doesn't mention a pin rise time, which was a big surprise to me, so I cannot check if measured 9.5 ns is a valid value. But judging by Pk-Pk voltage exceeding 6 volts (and even going below zero for a good 560 mV), I believe there's a problem with the scope. Am I right?
oscilloscope
oscilloscope
asked 8 hours ago
ZhenekZhenek
596 bronze badges
asked 8 hours ago
ZhenekZhenek
596 bronze badges
asked 8 hours ago
ZhenekZhenek
596 bronze badges
asked 8 hours ago
ZhenekZhenek
596 bronze badges
asked 8 hours ago
ZhenekZhenek
596 bronze badges
596 bronze badges
$begingroup$
Are your probes compensated correctly? Also, can you try with a different probe?
$endgroup$
– Steve G
7 hours ago
$begingroup$
Could you add a photo of how you're probing the signal? That is, how exactly your probe is connected to the circuit.
$endgroup$
– marcelm
5 hours ago
$begingroup$
Make sure your probe is in the x10 position, compensation adjustment is done and the ground lead is connected to a plane very close to the MCU ground. You can also run probe wizard and self cal routines.
$endgroup$
– Spehro Pefhany
5 hours ago
add a comment |
$begingroup$
Are your probes compensated correctly? Also, can you try with a different probe?
$endgroup$
– Steve G
7 hours ago
$begingroup$
Could you add a photo of how you're probing the signal? That is, how exactly your probe is connected to the circuit.
$endgroup$
– marcelm
5 hours ago
$begingroup$
Make sure your probe is in the x10 position, compensation adjustment is done and the ground lead is connected to a plane very close to the MCU ground. You can also run probe wizard and self cal routines.
$endgroup$
– Spehro Pefhany
5 hours ago
$begingroup$
Are your probes compensated correctly? Also, can you try with a different probe?
$endgroup$
– Steve G
7 hours ago
$begingroup$
Are your probes compensated correctly? Also, can you try with a different probe?
$endgroup$
– Steve G
7 hours ago
$begingroup$
Could you add a photo of how you're probing the signal? That is, how exactly your probe is connected to the circuit.
$endgroup$
– marcelm
5 hours ago
$begingroup$
Could you add a photo of how you're probing the signal? That is, how exactly your probe is connected to the circuit.
$endgroup$
– marcelm
5 hours ago
$begingroup$
Make sure your probe is in the x10 position, compensation adjustment is done and the ground lead is connected to a plane very close to the MCU ground. You can also run probe wizard and self cal routines.
$endgroup$
– Spehro Pefhany
5 hours ago
$begingroup$
Make sure your probe is in the x10 position, compensation adjustment is done and the ground lead is connected to a plane very close to the MCU ground. You can also run probe wizard and self cal routines.
$endgroup$
– Spehro Pefhany
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I believe there's a problem with the scope. Am I right?
Don't think so. Overshoot is a perfectly normal phenomenon when measuring a fast-edge signal with a high-impedance probe. (Also, these signals look about as sharp as I'd expect them to be.)
There's many tutorials on sensing high-speed signals: this is the perfect time to read one!
Oh, and there's Gibb's phenomenon, which says that any band-limited observation of a theoretical perfect (or far less band-limited) edge will have some 9% of overshoot; to understand that, I'd recommend looking at the cosine series representation of the square wave and consider what you'll cut off when you get rid of anything above 5× 16 MHz (=the fundamental frequency of your square wave).
answered 8 hours ago
Marcus MüllerMarcus Müller
40.4k3 gold badges66 silver badges110 bronze badges
$endgroup$
1
$begingroup$
On OP's measure function: I'd believe 16.00MHz frequency (these 'scopes use a crystal time-base). But 9.500 ns risetime? That's suspect, especially with 1ps resolution? And 6.16V Pk-Pk often goes through the entire sample record to find the maximum extent...(I make out about 5.2V, after settling). So Marcus' verdict is reasonable - more careful probing likely gives different results - learn to trust some measure functions, dis-trust others.
$endgroup$
– glen_geek
7 hours ago
add a comment |
$begingroup$
Keep in mind that if you have a 100MHz brick-wall filter (ideal case) with a perfect 16MHz square wave in, the only harmonics you'll see are 1 (16MHz), 3 (48MHz) and 5 (80MHz). That's an ideal case, but if you do the calculations, you'll see the result isn't too far from what you're seeing.
In the nonideal case, of course, probe loading and compensation will have further distorting effects, and the waveform isn't going to be perfectly square to start with.
answered 6 hours ago
Cristobol PolychronopolisCristobol Polychronopolis
2,2403 silver badges10 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe there's a problem with the scope. Am I right?
Don't think so. Overshoot is a perfectly normal phenomenon when measuring a fast-edge signal with a high-impedance probe. (Also, these signals look about as sharp as I'd expect them to be.)
There's many tutorials on sensing high-speed signals: this is the perfect time to read one!
Oh, and there's Gibb's phenomenon, which says that any band-limited observation of a theoretical perfect (or far less band-limited) edge will have some 9% of overshoot; to understand that, I'd recommend looking at the cosine series representation of the square wave and consider what you'll cut off when you get rid of anything above 5× 16 MHz (=the fundamental frequency of your square wave).
answered 8 hours ago
Marcus MüllerMarcus Müller
40.4k3 gold badges66 silver badges110 bronze badges
$endgroup$
1
$begingroup$
On OP's measure function: I'd believe 16.00MHz frequency (these 'scopes use a crystal time-base). But 9.500 ns risetime? That's suspect, especially with 1ps resolution? And 6.16V Pk-Pk often goes through the entire sample record to find the maximum extent...(I make out about 5.2V, after settling). So Marcus' verdict is reasonable - more careful probing likely gives different results - learn to trust some measure functions, dis-trust others.
$endgroup$
– glen_geek
7 hours ago
add a comment |
$begingroup$
I believe there's a problem with the scope. Am I right?
Don't think so. Overshoot is a perfectly normal phenomenon when measuring a fast-edge signal with a high-impedance probe. (Also, these signals look about as sharp as I'd expect them to be.)
There's many tutorials on sensing high-speed signals: this is the perfect time to read one!
Oh, and there's Gibb's phenomenon, which says that any band-limited observation of a theoretical perfect (or far less band-limited) edge will have some 9% of overshoot; to understand that, I'd recommend looking at the cosine series representation of the square wave and consider what you'll cut off when you get rid of anything above 5× 16 MHz (=the fundamental frequency of your square wave).
answered 8 hours ago
Marcus MüllerMarcus Müller
40.4k3 gold badges66 silver badges110 bronze badges
$endgroup$
1
$begingroup$
On OP's measure function: I'd believe 16.00MHz frequency (these 'scopes use a crystal time-base). But 9.500 ns risetime? That's suspect, especially with 1ps resolution? And 6.16V Pk-Pk often goes through the entire sample record to find the maximum extent...(I make out about 5.2V, after settling). So Marcus' verdict is reasonable - more careful probing likely gives different results - learn to trust some measure functions, dis-trust others.
$endgroup$
– glen_geek
7 hours ago
add a comment |
$begingroup$
I believe there's a problem with the scope. Am I right?
Don't think so. Overshoot is a perfectly normal phenomenon when measuring a fast-edge signal with a high-impedance probe. (Also, these signals look about as sharp as I'd expect them to be.)
There's many tutorials on sensing high-speed signals: this is the perfect time to read one!
Oh, and there's Gibb's phenomenon, which says that any band-limited observation of a theoretical perfect (or far less band-limited) edge will have some 9% of overshoot; to understand that, I'd recommend looking at the cosine series representation of the square wave and consider what you'll cut off when you get rid of anything above 5× 16 MHz (=the fundamental frequency of your square wave).
answered 8 hours ago
Marcus MüllerMarcus Müller
40.4k3 gold badges66 silver badges110 bronze badges
$endgroup$
I believe there's a problem with the scope. Am I right?
Don't think so. Overshoot is a perfectly normal phenomenon when measuring a fast-edge signal with a high-impedance probe. (Also, these signals look about as sharp as I'd expect them to be.)
There's many tutorials on sensing high-speed signals: this is the perfect time to read one!
Oh, and there's Gibb's phenomenon, which says that any band-limited observation of a theoretical perfect (or far less band-limited) edge will have some 9% of overshoot; to understand that, I'd recommend looking at the cosine series representation of the square wave and consider what you'll cut off when you get rid of anything above 5× 16 MHz (=the fundamental frequency of your square wave).
answered 8 hours ago
Marcus MüllerMarcus Müller
40.4k3 gold badges66 silver badges110 bronze badges
edited 3 hours ago
answered 8 hours ago
Marcus MüllerMarcus Müller
40.4k3 gold badges66 silver badges110 bronze badges
answered 8 hours ago
Marcus MüllerMarcus Müller
40.4k3 gold badges66 silver badges110 bronze badges
answered 8 hours ago
Marcus MüllerMarcus Müller
40.4k3 gold badges66 silver badges110 bronze badges
40.4k3 gold badges66 silver badges110 bronze badges
1
$begingroup$
On OP's measure function: I'd believe 16.00MHz frequency (these 'scopes use a crystal time-base). But 9.500 ns risetime? That's suspect, especially with 1ps resolution? And 6.16V Pk-Pk often goes through the entire sample record to find the maximum extent...(I make out about 5.2V, after settling). So Marcus' verdict is reasonable - more careful probing likely gives different results - learn to trust some measure functions, dis-trust others.
$endgroup$
– glen_geek
7 hours ago
add a comment |
1
$begingroup$
On OP's measure function: I'd believe 16.00MHz frequency (these 'scopes use a crystal time-base). But 9.500 ns risetime? That's suspect, especially with 1ps resolution? And 6.16V Pk-Pk often goes through the entire sample record to find the maximum extent...(I make out about 5.2V, after settling). So Marcus' verdict is reasonable - more careful probing likely gives different results - learn to trust some measure functions, dis-trust others.
$endgroup$
– glen_geek
7 hours ago
1
1
$begingroup$
On OP's measure function: I'd believe 16.00MHz frequency (these 'scopes use a crystal time-base). But 9.500 ns risetime? That's suspect, especially with 1ps resolution? And 6.16V Pk-Pk often goes through the entire sample record to find the maximum extent...(I make out about 5.2V, after settling). So Marcus' verdict is reasonable - more careful probing likely gives different results - learn to trust some measure functions, dis-trust others.
$endgroup$
– glen_geek
7 hours ago
$begingroup$
On OP's measure function: I'd believe 16.00MHz frequency (these 'scopes use a crystal time-base). But 9.500 ns risetime? That's suspect, especially with 1ps resolution? And 6.16V Pk-Pk often goes through the entire sample record to find the maximum extent...(I make out about 5.2V, after settling). So Marcus' verdict is reasonable - more careful probing likely gives different results - learn to trust some measure functions, dis-trust others.
$endgroup$
– glen_geek
7 hours ago
add a comment |
$begingroup$
Keep in mind that if you have a 100MHz brick-wall filter (ideal case) with a perfect 16MHz square wave in, the only harmonics you'll see are 1 (16MHz), 3 (48MHz) and 5 (80MHz). That's an ideal case, but if you do the calculations, you'll see the result isn't too far from what you're seeing.
In the nonideal case, of course, probe loading and compensation will have further distorting effects, and the waveform isn't going to be perfectly square to start with.
answered 6 hours ago
Cristobol PolychronopolisCristobol Polychronopolis
2,2403 silver badges10 bronze badges
$endgroup$
add a comment |
$begingroup$
Keep in mind that if you have a 100MHz brick-wall filter (ideal case) with a perfect 16MHz square wave in, the only harmonics you'll see are 1 (16MHz), 3 (48MHz) and 5 (80MHz). That's an ideal case, but if you do the calculations, you'll see the result isn't too far from what you're seeing.
In the nonideal case, of course, probe loading and compensation will have further distorting effects, and the waveform isn't going to be perfectly square to start with.
answered 6 hours ago
Cristobol PolychronopolisCristobol Polychronopolis
2,2403 silver badges10 bronze badges
$endgroup$
add a comment |
$begingroup$
Keep in mind that if you have a 100MHz brick-wall filter (ideal case) with a perfect 16MHz square wave in, the only harmonics you'll see are 1 (16MHz), 3 (48MHz) and 5 (80MHz). That's an ideal case, but if you do the calculations, you'll see the result isn't too far from what you're seeing.
In the nonideal case, of course, probe loading and compensation will have further distorting effects, and the waveform isn't going to be perfectly square to start with.
answered 6 hours ago
Cristobol PolychronopolisCristobol Polychronopolis
2,2403 silver badges10 bronze badges
$endgroup$
Keep in mind that if you have a 100MHz brick-wall filter (ideal case) with a perfect 16MHz square wave in, the only harmonics you'll see are 1 (16MHz), 3 (48MHz) and 5 (80MHz). That's an ideal case, but if you do the calculations, you'll see the result isn't too far from what you're seeing.
In the nonideal case, of course, probe loading and compensation will have further distorting effects, and the waveform isn't going to be perfectly square to start with.
answered 6 hours ago
Cristobol PolychronopolisCristobol Polychronopolis
2,2403 silver badges10 bronze badges
answered 6 hours ago
Cristobol PolychronopolisCristobol Polychronopolis
2,2403 silver badges10 bronze badges
answered 6 hours ago
Cristobol PolychronopolisCristobol Polychronopolis
2,2403 silver badges10 bronze badges
answered 6 hours ago
Cristobol PolychronopolisCristobol Polychronopolis
2,2403 silver badges10 bronze badges
2,2403 silver badges10 bronze badges
add a comment |
add a comment |
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$begingroup$
Are your probes compensated correctly? Also, can you try with a different probe?
$endgroup$
– Steve G
7 hours ago
$begingroup$
Could you add a photo of how you're probing the signal? That is, how exactly your probe is connected to the circuit.
$endgroup$
– marcelm
5 hours ago
$begingroup$
Make sure your probe is in the x10 position, compensation adjustment is done and the ground lead is connected to a plane very close to the MCU ground. You can also run probe wizard and self cal routines.
$endgroup$
– Spehro Pefhany
5 hours ago