Author Topic: Difference between waveforms when using x10 attenuator  (Read 11068 times)

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Offline slashguitarTopic starter

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Difference between waveforms when using x10 attenuator
« on: September 30, 2012, 08:29:02 pm »
Hi,
I was playing around with some LC filters on a SMPS to see how much the ripple improves and i noticed that it is considerably bigger when using the x10 attenuator.
Ch1 is before the filter and Ch2 is after the filter. Which is closer to reality X1 or X10?
 

Offline Psi

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Re: Difference between waveforms when using x10 attenuator
« Reply #1 on: September 30, 2012, 08:32:46 pm »
When you used x1/x10 did you set the oscilloscope to x1/x10 as well? or only change the switch on the probe?
If both channels were set to X10 on the scope and one probe was set to x1 then the x1 probe will read 10x larger than reality.
« Last Edit: September 30, 2012, 08:49:41 pm by Psi »
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Offline slashguitarTopic starter

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Re: Difference between waveforms when using x10 attenuator
« Reply #2 on: September 30, 2012, 08:43:17 pm »
Yes I also set the scope for X10. What's strange is that with the probes shorted out there is only a slight incerase in noise when I switch to X10, so I don't think it's garbage picked up from the enviroment (there are some rather big antennas across the street )
 

alm

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Re: Difference between waveforms when using x10 attenuator
« Reply #3 on: September 30, 2012, 08:43:56 pm »
It's not clear what x10 attenuator you mean. Switching the probe to 10x also increases the bandwidth and decreases the capacitive loading, although that's unlikely to play a role at these frequencies (compare your probe's specs in x1 and x10 mode). Make sure the probe is well compensated in the x10 mode with the probe compensation signal. If the attenuation setting on probe and scope match, then the scope is effectively amplifying 10x more in x10 mode. This might show up as increased noise in the signal.
 

Offline Psi

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Re: Difference between waveforms when using x10 attenuator
« Reply #4 on: September 30, 2012, 08:48:36 pm »
have you tried swapping the probes around? Maybe there is a broken ground on one of them
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Offline slashguitarTopic starter

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Re: Difference between waveforms when using x10 attenuator
« Reply #5 on: September 30, 2012, 09:02:23 pm »
I've swapped the probes around and compensated them but nothing changed. I'm doing a selfcal now.
Quote
(compare your probe's specs in x1 and x10 mode)
It's  a DS1052E, i'll search some specs for the probes online
 

Offline Psi

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Re: Difference between waveforms when using x10 attenuator
« Reply #6 on: September 30, 2012, 09:15:09 pm »
What happens if you load the input signal with a 1M resistor to ground. Does it change the x10 waveform at all?

I'm just wondering if your signal is such high impedance that the probe on 1x is pulling it down.
« Last Edit: September 30, 2012, 09:18:52 pm by Psi »
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Offline Mechatrommer

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Re: Difference between waveforms when using x10 attenuator
« Reply #7 on: October 01, 2012, 05:20:00 am »
rigol is known to give you some black magic stuff, for example plugging anything to the probe connector (even a fully shielded terminator) will give you rise in noise for no reason and different level between channels. so this one maybe it. btw you didnt mention which picture for 1X and which for 10X?
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Offline robrenz

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Re: Difference between waveforms when using x10 attenuator
« Reply #8 on: October 01, 2012, 11:08:18 am »
Consider the probe resistance at 1x as part of the circuit and consider the probe resistance and capacitance as part of the circuit at 10x. They are part of the circuit and that may be the difference you are seeing.  To prove it you could add a resistor and capacitor that are close to what the 10x probe is to your circuit and measure with the 1x probe.

Offline w2aew

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Re: Difference between waveforms when using x10 attenuator
« Reply #9 on: October 01, 2012, 11:45:58 pm »
Consider the probe resistance at 1x as part of the circuit and consider the probe resistance and capacitance as part of the circuit at 10x. They are part of the circuit and that may be the difference you are seeing.  To prove it you could add a resistor and capacitor that are close to what the 10x probe is to your circuit and measure with the 1x probe.

The 1x probe loading will be substantially worse from a capacitance standpoint - the input capacitance plus the the capacitance of the probe and cable.  Likely over 100pF.  This is why the BW of 1x probes is always substantially less than the 10x probes.
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Offline Mechatrommer

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Re: Difference between waveforms when using x10 attenuator
« Reply #10 on: October 02, 2012, 05:07:41 am »
Consider the probe resistance at 1x as part of the circuit and consider the probe resistance and capacitance as part of the circuit at 10x. They are part of the circuit and that may be the difference you are seeing.  To prove it you could add a resistor and capacitor that are close to what the 10x probe is to your circuit and measure with the 1x probe.
The 1x probe loading will be substantially worse from a capacitance standpoint - the input capacitance plus the the capacitance of the probe and cable.  Likely over 100pF.  This is why the BW of 1x probes is always substantially less than the 10x probes.
1X or 10X setup, there will be always 1X probe/cable there, as long as we use a normal coaxial cable (not resistive coaxial), normally 15pF/ft IIRC. what makes it 10X setting and higher bandwidth is the attenuator and compensation network (resistor and parallel capacitor) at the probe tip. i believe this is what robrenz tried to suggest for the OP to simulate, ie compensation/attenuator network/circuit in the middle in between 1X cable/probe and the DUT. he tried to highlight the idea that, using the same 1X probe, you'll get different reading (noise profile) for the DUT alone compared to "the DUT + compensation network". but getting "correctly compensated" network built-into the DUT circuit (just as exactly or similarly as the one in the 10X probe tip) is not easy i believe for this trivial case and especially if the OP hasnt done or understand any compensation circuit, as far as the probe bandwidth is concerned... i'm still learning, so i'll keep saying this so it stays firm in my brain
« Last Edit: October 02, 2012, 05:26:15 am by Mechatrommer »
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Offline robrenz

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Re: Difference between waveforms when using x10 attenuator
« Reply #11 on: October 02, 2012, 11:46:38 am »
Thank you Mechatrommer,  that is exactly what I was trying to say but i was to tired to try to explain it. I think you did a better job than I would anyway.  :)
« Last Edit: October 02, 2012, 05:07:24 pm by robrenz »
 

alm

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Re: Difference between waveforms when using x10 attenuator
« Reply #12 on: October 02, 2012, 05:04:26 pm »
Just modeling the probe as the given load model (parallel RC combination) would probably be sufficient to highlight the problem. Try adding a series resistor between the 'probe' and signal source to see the effect of circuit loading at high source impedances (those datasheet specs are for 25 ohm source impedances). If you prefer solder to simulation, try adding a 100 pF capacitor in parallel while leaving the probe set to 10x.
 


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