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Avalanche photodiode preamplifier. Start point. Old test preamplifier -> with OpAmp AD8067 and photodiode BPX65 It has transimpedance resistance 30kOm, and -> photoelectric sensitivity 5kV/W noise characteristic of that amplifier is : 700mkV rms (on 10 MHz) and ->1.4x10e-7W rms.
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Start point • Old test preamplifier -> with OpAmp AD8067 and photodiode BPX65 • It has transimpedance resistance 30kOm, and -> photoelectric sensitivity 5kV/W • noise characteristic of that amplifier is : 700mkV rms (on 10 MHz) and ->1.4x10e-7W rms • Next step -> to change photodiode BPX65 to avalanche photodiode S5343 • And also to compare OpAmp AD8067 from ADI and OPA657 from TI
More usable gain for APD is 50. It give us for =420nm photosensitivity=13 A/W For this gain reverse voltage on photodiode must be ~150V For this reverse voltage terminal capacitance Ct=15pF Avalanche photodiode
Choice of OpAmp with MathCad • Choice from AD8067 and OPA657 • Amplifier with OPA657 have bigger transimpedance resistence(120 kOm and 14kOm) • For OPA657 SNR best then for AD8067 approximately on 2 times (In=27nA and 46nA accordingly)
Amplifier with S5343 and OPA657 • Photoelectric sensitivity is 13A/W x 120kOm=1.6MV/W • Output noise is 3.3mV rms, and input noise is 2nW(better, than for amplifier with BPX65 and AD8067 – 140nW).
APD module – C5331-11. • Active area 1mm • Board dimension 80x50x23mm • Bandwidth from 4k to 100MHz • Feedback resistance 3.9kOm • Photosensitivity (=420nm) 7A/W • NEP=1pW/Hz -> input noise = 12nW • Vn=350 mkV rms • Cost 400 EUR
C5331 measurements. • Frequency response is frequency response of LED • Vnoise = 0.4mV rms (0.35 mV in theory) Frequency response
Amplifier measurements. Frequency response Main results is: • Signal frequensy response • Gain-voltage dependance • Noise value Gain vs. reverse voltage
SNR optimization. • Output noise consist from 2 part: OpAmp noise and shot noise • Noise without constant part of signal - 3 mV SNR • Noise with constant part - from 3.5 mV to 5 mV
Estimation of light power on avalanche diode in our block • We suppose, that : • Uniformly light in spatial cone with angle 45° • Without reflection from black material. • Measurement conditions: L=50mm, I=300mV/300Om=1mA; f=100kHz; Active area 1mm • LED: =350mcd; View angle = 45°; max=470nm • Light power on photodiode: P=4*10e-8W • Vout=13A/W * 200kOm * 4*10e-8W=104mV In measurements on this conditions we see: Vout=80mV