측정 부하, 증폭기 이득 및 탐지기 반응
도입됨| English | 의미 |
|---|---|
| dead time/ded taɪm/ | 死时间: 死时间 |
| input impedance/ˈɪnpʊt ɪmˈpiːdəns/ | 输入阻抗: 输入阻抗 |
A decision before an answer
- A voltmeter changes the circuit it measures; the reading is a divider output, not the open-circuit voltage.
- Your goal: Calculate measurement loading from finite meter and source impedances.
Model the measurement
- Model the measured circuit as a Thevenin source V_th with resistance R_th and the instrument as input resistance R_m. The reading is V_th·R_m/(R_m+R_th), with the same sign as V_th and magnitude no greater than |V_th| for positive resistances.
- A 10 V source with R_th=100 kΩ read by a 1 MΩ meter gives 10×(1/1.1)=9.09 V, a 9% low bias. A 10× oscilloscope probe raises the effective input resistance and divides the signal by ten; quote both effects.
A 12 V source with internal resistance 200 kΩ is measured by a voltmeter with input resistance 1 MΩ. The meter reads:
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Divider: 12×(1.0/1.2)=10.0 V; the finite meter loads the source.
Set gain and cutoff
- An ideal non-inverting op-amp stage has gain 1+R2/R1; the inverting stage has −R2/R1, with the sign carried explicitly.
- A first-order RC low-pass has cutoff f_c=1/(2πRC); its amplitude ratio is 1/sqrt(1+(f/f_c)²) and its phase is −arctan(f/f_c). At f=f_c the amplitude is 1/√2 (the −3 dB point), not one half. For R=1 kΩ and C=100 nF, f_c≈1.59 kHz.
A first-order RC low-pass driven exactly at its cutoff frequency passes an amplitude fraction of:
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1/sqrt(1+(f/f_c)²)=1/√2 at f=f_c, the −3 dB point.
Rate the detector
- Detector efficiency ε is detected events divided by incident events; energy resolution is quoted as FWHM/E, such as 13.2 keV on a 662 keV line, about 2%.
- A non-paralyzable detector with dead time τ records m=n/(1+nτ) from true rate n; inverting gives n=m/(1−mτ), which fails as mτ→1. State which dead-time model you assume; paralyzable behaviour differs.
A 12 V source with 200 kΩ internal resistance read by a 1 MΩ meter shows 12×(1/1.2)=10.0 V. An RC filter with f_c=1 kHz driven at 2 kHz passes amplitude 1/√5≈0.447. A counter with τ=10 μs at true rate 10⁴ /s records 9091 /s; a recorded 8000 /s with τ=20 μs implies n=8000/(1−0.16)≈9524 /s.
An ideal non-inverting op-amp stage with R1=1 kΩ and R2=9 kΩ has gain ____.
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1+R2/R1=1+9=10.
Calibrate with its limits
- A linear calibration against known standards can correct offset and scale error; it does not remove noise or guarantee correction of nonlinearity. A calibration curve maps indicated to true values; interpolation between points assumes local smoothness.
- Report resolution and efficiency separately: sharp peaks with poor efficiency still miss events, and high efficiency with broad resolution still mixes nearby lines.
Treating the 1/√2 cutoff as one half, ignoring that a finite meter reads low through the divider, or applying the non-paralyzable correction beyond its range.
Which answer fits this case?
Calculate measurement loading from finite meter and source impedances
Increasing a detector dead time increases the recorded count rate at fixed true rate.
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Dead time removes events: m=n/(1+nτ) falls as τ grows.
Keep the distinctions
- dead time 死时间 — The minimum interval after one recorded event during which the detector cannot record another.
- input impedance 输入阻抗 — The effective load a measuring instrument presents to the circuit under test.
- Calculate measurement loading from finite meter and source impedances.
- Determine simple amplifier gain and first-order filter response.
- Interpret detector efficiency, resolution, dead time and calibration bias.
Match each term with its precise meaning in this lesson.
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Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
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The order follows the stated process; check each stage before the next.