Measurement loading, amplifier gain and detector response
Introduced| English |
|---|
| dead time/ded taɪm/ |
| input impedance/ˈɪnpʊt ɪmˈpiːdəns/ |
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:
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:
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 ____.
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.
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.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.