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www.thinkSRS.com About Lock-In Amplifiers

About Lock-In AmplifiersApplication Note # Research Systemsphone: (408) Amplifiers are used to detect and measure very smallAC signals all the way down to a few nanovolts. Accuratemeasurements may be made even when the small signal isobscured by noise sources many thousands of times Amplifiers use a technique known as phase-sensitivedetection to single out the component of the signal at aspecific reference frequency and phase. noise signals, atfrequencies other than the reference frequency, are rejectedand do not affect the Use a Lock-In ? Let's consider an example. Suppose the signal is a 10 nV sinewave at 10 kHz.

specific reference frequency and phase. Noise signals, at frequencies other than the reference frequency, are rejected and do not affect the measurement. Why Use a Lock-In? Let's consider an example. Suppose the signal is a 10 nV sine wave at 10 kHz. Clearly some amplification is required to bring the signal above the noise. A good low-noise ...

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Transcription of www.thinkSRS.com About Lock-In Amplifiers

1 About Lock-In AmplifiersApplication Note # Research Systemsphone: (408) Amplifiers are used to detect and measure very smallAC signals all the way down to a few nanovolts. Accuratemeasurements may be made even when the small signal isobscured by noise sources many thousands of times Amplifiers use a technique known as phase-sensitivedetection to single out the component of the signal at aspecific reference frequency and phase. noise signals, atfrequencies other than the reference frequency, are rejectedand do not affect the Use a Lock-In ? Let's consider an example. Suppose the signal is a 10 nV sinewave at 10 kHz.

2 Clearly some amplification is required tobring the signal above the noise . A good low- noise amplifierwill have About 5 nV/ Hz of input noise . If the amplifierbandwidth is 100 kHz and the gain is 1000, we can expect ouroutput to be 10 V of signal (10 nV 1000) and mV ofbroadband noise (5 nV/ Hz 100 kHz 1000). We won'thave much luck measuring the output signal unless we singleout the frequency of we follow the amplifier with a band pass filter with a Q=100(a VERY good filter) centered at 10 kHz, any signal in a100 Hz bandwidth will be detected (10 kHz/Q). The noise inthe filter pass band will be 50 V (5 nV/ Hz 100 Hz 1000),and the signal will still be 10 V.

3 The output noise is muchgreater than the signal, and an accurate measurement can notbe made. Further gain will not help the try following the amplifier with a phase-sensitivedetector (PSD). The PSD can detect the signal at 10 kHz witha bandwidth as narrow as Hz! In this case, the noise inthe detection bandwidth will be V (5 nV/ Hz .01 Hz 1000), while the signal is still 10 V. The signal-to-noiseratio is now 20, and an accurate measurement of the signal is Phase-Sensitive Detection? Lock-In measurements require a frequency , an experiment is excited at a fixed frequency (froman oscillator or function generator), and the Lock-In detects theresponse from the experiment at the reference frequency.

4 Inthe following diagram, the reference signal is a square wave atfrequency r. This might be the sync output from a functiongenerator. If the sine output from the function generator isused to excite the experiment, the response might be the signalwaveform shown below. The signal is Vsigsin( rt + sig) whereVsigis the signal amplitude, ris the signal frequency, and sigis the signal s Amplifiers generate their own internal referencesignal usually by a phase-locked-loop locked to the externalreference. In the diagram, the external reference, the Lock-In sreference, and the signal are all shown. The internal referenceis VLsin( Lt + ref).

5 The Lock-In amplifies the signal and then multiplies it by thelock-in reference using a phase-sensitive detector ormultiplier. The output of the PSD is simply the product of twosine = VsigVLsin( rt + sig)sin( Lt + ref)= VsigVLcos([ r L]t + sig ref) VsigVLcos([ r+ L]t + sig+ ref)The PSD output is two AC signals, one at the differencefrequency ( r L) and the other at the sum frequency ( r + L). If the PSD output is passed through a low pass filter, the ACsignals are removed. What will be left? In the general case,nothing. However, if requals L, the difference frequencycomponent will be a DC signal. In this case, the filtered PSDoutput will be:Vpsd= VsigVLcos( sig ref)This is a very nice signal it is a DC signal proportional to thesignal amplitude.

6 It s important to consider the physical nature of thismultiplication and filtering process in different types oflock-ins. In traditional analog lock-ins, the signal andreference are analog voltage signals. The signal and referenceare multiplied in an analog multiplier, and the result is filteredwith one or more stages of RC filters. In a digital Lock-In , suchas the SR830 or SR850, the signal and reference arerepresented by sequences of numbers. Multiplication andfiltering are performed mathematically by a digital signalprocessing (DSP) chip. We ll discuss this in more detail Band DetectionLet s return to our generic Lock-In example.

7 Suppose thatinstead of being a pure sine wave, the input is made up ofsignal plus noise . The PSD and low pass filter only detectReferenceSignalLock-in ref sigAbout Lock-In Amplifierssignals whose frequencies are very close to the lock-inreference frequency. noise signals, at frequencies far from thereference, are attenuated at the PSD output by the low passfilter (neither noise refnor noise + refare close to DC). noise at frequencies very close to the reference frequency willresult in very low frequency AC outputs from the PSD (| noise ref|is small). Their attenuation depends upon the low pass filterbandwidth and rolloff.

8 A narrower bandwidth will removenoise sources very close to the reference frequency; a widerbandwidth allows these signals to pass. The low pass filterbandwidth determines the bandwidth of detection. Only thesignal at the reference frequency will result in a true DCoutput and be unaffected by the low pass filter. This is thesignal we want to Does the Lock-In Reference Come From?We need to make the Lock-In reference the same as the signalfrequency, r= L. Not only do the frequencies have to bethe same, the phase between the signals can not change withtime. Otherwise, cos( sig ref) will change and Vpsdwill notbe a DC signal.

9 In other words, the Lock-In reference needs tobe phase-locked to the signal reference. Lock-In Amplifiers use a phase-locked loop (PLL) to generatethe reference signal. An external reference signal (in this case,the reference square wave) is provided to the Lock-In . The PLLin the Lock-In amplifier locks the internal reference oscillatorto this external reference, resulting in a reference sine wave at rwith a fixed phase shift of ref. Since the PLL activelytracks the external reference, changes in the external referencefrequency do not affect the Reference SourcesIn the case just discussed, the reference is provided by theexcitation source (the function generator).

10 This is called anexternal reference source. In many situations the Lock-In sinternal oscillator may be used instead. The internal oscillatoris just like a function generator (with variable sine output anda TTL sync) which is always phase-locked to the referenceoscillator. Magnitude and PhaseRemember that the PSD output is proportional to Vsigcos ,where = ( sig ref). is the phase difference between thesignal and the Lock-In reference oscillator. By adjusting refwecan make equal to zero. In which case we can measureVsig(cos = 1). Conversely, if is 90 , there will be no outputat all. A Lock-In with a single PSD is called a single-phaselock-in and its output is Vsigcos.


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