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Spectrum Analyzer, very basic question.

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palmeiras

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Hello Guys,

When connecting my circuit (oscillator, for instance) to the spectrum analyzer:

1 Do I need to remove the DC signal? Allowing only the AC signal? Why? Using a decoupling cap.

2- What is the usually peak to peak voltage permitted by it?

Thank you very much,
 

1- Most of the Spectrum Analyzers have this option implemented on their hardware. But when using DC coupling you should be very careful with bias conditions, as it may disturb the analyzer. Actually the thing you lost is the low frequency response in most general case.

2- Refer to the manual. I frankly do not think this is specified by any standard so different analyzers have different range. But this is not a general limitation in most of the time. If required just use an attenuator.
 
Hello kemiyum,
Thanks again for your support.
Regarding 1): what do you mean with “this option is often implemented on their hardware”? Are you saying that the most spectrum analyzers have already a decoupling cap there? Or I should provide it?
I was checking the manual of one analyzer now, and on its input, it is written: 50 ohms, 0 VDC, DC coupled. So, is it mandatory to remove the DC signal of my output? Am I right?
What does it mean with “DC coupled”?

Thanks you,
 

Well there should be a switch in most general case. I do not think all of them have it but I'm pretty sure most of them have this kind of selection. If there is not and you want AC coupling you should provide the capacitor.

50 ohms means it is perfectly matched for 50 ohm. 0Vdc means the biasing condition of the input is 0Vdc which means the ground (If you are trying to measure something with one supply you might have problems with this.) DC coupled means it is DC coupled :smile: so there is no AC coupling implemented on your analyzer. DC coupled means connected directly. I mean the actual connection may not be just a wire but the working condition of the hardware resembles a DC coupling. I'm not sure what do you mean by removing the DC signal but yes you need to take precautions to make a robust measurement.
 
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Heya palmeiras!

Aye, to echo kemiyun - there's absolutely no standardisation amongst the test and measurement gear out there. Generally though, I've found the manufacturers spell it out for you in no uncertain terms on the front panel. Take for instance, the two different R&S instruments on my bench:



The VNA can tolerate a DC component present on the input port (up to 30V), whereas the spectrum analyser can't. (VNA's generally allow a small DC bias in order that active devices under test can be biassed to their appropriate operating points). I've found a series DC blocking capacitor to always be sufficient for my spectrum analysis measurements, bearing in mind that the capacitor forms a single pole RC high-pass filter with the (nominally 50 ohm) input impedance of the analyser [which can catch you out with LF measurements!]

As to the maximum allowable peak-to-peak input voltage - assuming a single frequency sinusoidal input, this is given by:

Vpp(max) = sqrt(0.4 * 10^(dBm/10))

(Which is simply the end result of rearranging P=V^2/R, and the relationships between dBm and power, RMS and peak)

Thus, for the two instruments shown, my VNA can tolerate 14.1 Vpp [superimposed on up to 30V of DC offset], and the spectrum analyser 20 Vpp [with no DC component].

Hope that helps :)
 
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it is written: 50 ohms, 0 VDC, DC coupled. So, is it mandatory to remove the DC signal of my output? Am I right?
What does it mean with “DC coupled”?
It means, that the input is directly connected to the mixer and does not tolerate larger amounts of superimposed DC.

Small (mV range) DC offsets don't harm however. The critical point is, that a DC input voltage won't be detected by the automatic attenuator control. So the safe DC level will be determined by the mixer diode rating rather than the attenuator power handling capability. Apart from a risk of damage, a DC input voltage will also affect correct mixer operation.

Wide band spectrum analyzers mostly expose a DC coupled input, because both RF performance reduction and high-pass effect of a built-in AC block are unwanted.
 
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Hello Kemiyun and thylacine!

It is perfectly clear now! Thanks very much for your replies.

Thanks FvM, for your reply too.
 

And regarding the input capacitance of the Spectrum Analyzer? is it neglectable?
Thank you, guys
 

The input shouldn't look capacitive. It should look like 50 ohms resistive. Any capacitive component should be balanced by other components so the result is a matched 50 ohms.

Keith
 
...and indeed it does (look like 50 ohms resistive).

The VNA sits right next to the spectrum analyser in the lab... and I suddenly realised I'd never connected the two together before! So now I have :)


I suspect the match is even better than the plot suggests, and the spread of values around 50R is a consequence of someone having made off with my nice VNA test probes (and leaving me a piece of coax in its place!) :)
 
Hello Guys,

If possible, one last question about it. I intend to use this spectrum analyzer to measure the output voltage of an oscillator directly on the wafer, using probe station (not package). I don’t know how the probe station is. I’ll have access to the spectrum analyzer and the probe station in the future.
Is it easy to connect the RF coaxial cable (Spectrum analyzer) on it? How can I do this? I know that is a stupid question, but, the ground is the external part of the coaxial cable, right? So, I’ll have two cables coming from the probe station that must be connected to a RF coaxial cable, and I'm wondering how I can connect them.

Thanks in advance.
 

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