Diode Peak Detector Reading Low at 10 kHz
A technician builds a peak detector to capture the amplitude of a 10 kHz sine wave. The source is 4.0 Vpp, or 2.0 V peak, with a 600 Ω source impedance. The circuit is a 1N4148 diode in series with a 0.1 µF capacitor to ground, and a 1 MΩ resistor across the capacitor. A 10 MΩ DMM reads 1.38 V DC at the capacitor. The technician expected 2.0 V and suspects the diode is leaky. The 1N4148 forward drop is about 0.55 V at 1 mA and 0.7 V at 10 mA; reverse leakage is 25 nA at 20 V. The DMM input resistance is 10 MΩ. The load that will use this signal has a 100 kΩ input impedance. The technician must decide whether to replace the diode, change the capacitor, or add a buffer.
1. Explain the 1.38 V reading. Is the 1N4148 faulty?
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The capacitor charges through the diode to the positive peak minus the diode forward drop. Near the peak the 1N4148 carries a small current and drops roughly 0.5–0.6 V. Expected DC is about 2.0 − 0.6 = 1.4 V, so 1.38 V is normal, not a leaky diode. The 1 MΩ load and 10 MΩ meter produce a small droop: Rtotal = 1M || 10M = 0.909 MΩ, τ = 0.909M × 0.1 µF = 0.0909 s. Over one 100 µs period the droop is about 1.4 × 100µ/0.0909 = 1.5 mV. Reverse leakage of 25 nA through 0.909 MΩ would be only 23 mV, and that is worst case at high temperature and reverse voltage; it cannot explain 0.6 V. Replacing the diode with another 1N4148 will not change the reading.
2. Calculate the droop per cycle with the 1 MΩ resistor and with a 100 kΩ load at 10 kHz and at 50 Hz.
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Droop per cycle is ΔV ≈ (V/R) × T/C. At 10 kHz, T = 100 µs. With the 1 MΩ resistor and 10 MΩ meter, R = 0.909 MΩ: ΔV = (1.4/0.909M) × (100µ/0.1µ) = 1.54 mV, negligible. With a 100 kΩ load, R = 90.9 kΩ: ΔV = 15.4 mV, about 1%, still acceptable. At 50 Hz, T = 20 ms. With 100 kΩ, the linear estimate would be 3.1 V, which is impossible; the exponential gives 1.4 × e^(−20/9.09) = 0.155 V, an 89% collapse. With 1 MΩ, τ = 0.0909 s and 1.4 × e^(−0.22) = 1.12 V, about a 20% droop. The circuit is adequate at 10 kHz but not at low frequency with a heavy load. Also note the 600 Ω source and 0.1 µF give a 60 µs charge time constant, so at 10 kHz the capacitor may not fully recharge each cycle.
3. Recommend a design change to measure the true 2.0 V peak accurately, including component values and trade-offs.
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Eliminate the diode drop by placing the diode inside an op-amp feedback loop, forming a precision peak detector, or use an active rectifier. Add a high-input-impedance buffer, such as a JFET or CMOS op-amp, before the load so the hold capacitor is not discharged by the 100 kΩ load. With a buffer, the hold resistor can be 1 MΩ or higher, and the capacitor can be chosen so R × C is much greater than the period: for 50 Hz and 1 MΩ, C = 1 µF gives τ = 1 s and negligible droop, while the 600 Ω source charges it with a 0.6 ms time constant, short compared with 20 ms. For 10 kHz, 0.01 µF is enough and charges faster. A biased diode can compensate 0.6 V but drifts with temperature. Verify the op-amp's input bias current and bandwidth, and use a 10× scope probe to confirm the waveform.