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Wednesday, January 13, 2010

Measuring Power

Anyone who takes care of NMR equipment knows that visits from service engineers are very expensive. These visits can often be avoided by becoming familiar with the components of the NMR spectrometer and learning how to make simple diagnostic measurements. These measurements can be sent to service engineers who can provide advice on replacement parts. One such measurement is the determination of the output power from the amplifiers of the spectrometer. This measurement requires an oscilloscope with a band width greater than the output frequency from the amplifier. Since properly functioning amplifiers put out tens to hundreds of watts, the output must be attenuated in order to prevent damage to the oscilloscope. An attenuator of 30 or 40 dB (rated for at least 10 watts of CW power) is suitable. Also, the measurement must be made at 50 Ω impedance. The spectrometer must be set up to take pulses at regular intervals (e.g. 10 µsec pulses every second). For oscilloscopes with only a 1 MΩ input impedance setting, the measurement can be made according to the following figure using a "T" connector and a low power 50 Ω terminator to match the impedance. The "T" connector and 50 Ω terminator are not required if the oscilloscope has in input impedance setting of 50 Ω. In this case, the connections can be made according to the following figure. The output power (in Watts) is determined by the peak to peak voltage, Vpp , of the pulse as follows.

12 comments:

Kyle said...

Hi,

Sometimes books and literature state NMR power in Hz (or kHz), which I don't quite understand. How is this measured and what does it mean? Any feedback is appreciated.

Kyle

Glenn Facey said...

Kyle,

Yes, you are not the first one to wonder about this. I intend to post a BLOG entry on that subject in the near future, but essentially the power expressed in frequency units is equal to one quarter of the reciprocal of the 90 degree pulse duration.

Glenn

joshua hill said...

Hi,
Thanks for the useful post about measuring power.
Does the output power and reflected power depend on your Q factor? Is the power absorbed by the circuit just the output-reflected?thanks.



Joshua

Glenn Facey said...

Joshua,

Thank you for your comment. Both the amount of power seen by the sample and the reflected power depend on the Q factor of the probe.

Glenn

power hungry said...

With regard to power levels, why is -6dB considered full power, and not 0 dB? A related question would be "what is a 0 dB attenuator"? When setting up a new cortab entry, the prompt instructs to insert a 0 dB attenuator between the SGU RF out and the RF in on the receiver.

Glenn Facey said...

Power Hungry,

I am not sure why Bruker uses -6dB as full power. I understand that this will change in future versions of their software.

A 0 dB attenuator would mean no attenuation i.e. a direct connection between the SGU and the receiver.

Glenn

Anonymous said...

You mention that the Q factor depends on the power seen by the probe/sample and that it also depends on reflected voltage.

In order to obtain the same Q from sample to sample, would just measuring the same reflected voltage with an oscilloscope be a reliable way to ensure that Q is the same?
Measuring Q with a spectrum analyzer doesn't seem to be as accurate as measuring the voltage.

thanks.

Kyle

Glenn Facey said...

Anonymous,

Minimizing the reflected power is an acceptable way of tuning and matching an NMR probe.

Glenn

Nick F. said...

Would you please elaborate on your equations at the end of the post? How could they be derived for arbitrary attenuation?

Glenn Facey said...

Nick F.

Thank you for the question. The general formula is:

Po=((10^(X/10))/400)*((Vpp)^2)

where X is the value of the attenuator in dB.

The derivation is quite simple. It is based on the definition of a dB, The definition of power in terms of voltage (assuming 50 Ohm impedance) and the definition of peak to peak voltage in terms of RMS voltage. If you require further details, send me an email and I will send you the details of the derivation.

Glenn

Anonymous said...

Hi,

I notice that the Q factor changes depending on what type of sample I place in the coil.Sometimes, very drastically. Why is that? What types of materials/samples give high versus low Q?

Thanks,
Leo

Glenn Facey said...

Leo,

Thank you for the question. Samples with high ionic strength will drastically affect the Q of the probes circuit. Such samples are called lossy samples. See this link: http://u-of-o-nmr-facility.blogspot.ca/2008/07/tuning-problems-for-samples-of-high.html

Glenn