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Designing two-wire remote sensor preamp

14 Feb 2014  | Vladimir Rentyuk

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This article implements a remote sensor preamp (e.g., for a piezoelectric transducer) which transfers both signal and power over a single wire pair or coax.

Figure 1: A remote preamplifier.

The AD822ARZ is a true single-supply operational amplifier with rail-to-rail output, and with very low input current and low frequency noise, ideal for operating with high impedance signal sources. The AD822 has single-supply capability from 5V, making it a good choice here.

R6 provides a matching load to the piezoelectric sensor. R5 and D1 protect IC1-1 from high voltage spikes, which are possible from piezoelectric sensors. IC1-1 provides primary gain (approximately 1+R7/R8 in the operating frequency range) and part of the gain-frequency characteristic. R8 & C6 suppress subsonic frequencies (cut-off frequency of 1/2πR8C6) and provide linearisation of the frequency response of the sensor. The combination R7-C5 suppresses frequencies above the operating frequency range (a cut-off frequency 1/2πR7C5). Additional lowpass filters are R10-C9 and R13-C11. The main highpass filter is a second-order filter which provides suppression of subsonic frequencies, built around IC1-2. The output of the preamp is the open collector of Q2. A load for Q2 (Rg: 1.5kΩ) is placed at the receiver.

The preamp's power supply involved the following considerations: The divider R2, R9, C7 provides offsetting voltage of half the supply (R2=R9) for both parts of IC1 (through R6, R11). The quiescent current Iq of the AD822 assuming a 5V power supply is 1.6mA maximum. Investigation has shown that a collector current Ic of the output transistor (without signal) should be several times Iq. R14 is used to set Iq, calculated as R14 < (Vs/2—VEB)/Iq = (5V/2—0.68V)/1.6mA = 1.14kΩ (0.68V is a type emitter-base voltage (VEB) for these transistors in active mode). So, let R14 be 560Ω. Thus, the collector current Ic of the output transistor is (Vs/2—VEB)/R14. The base current of Q2 is small enough to be ignored, given the hFE of a BC847C is not less than 420. So, Ic = (5V/2—0.68V)/560Ω = 3.25mA. The maximum current through the collector load (without signal) is: Imax = Ic+Iq = 3.25mA+1.6mA = 4.85mA.

Investigations show that using typical voltage regulator ICs is not feasible due to the high level of noise which will be reflected in the load resistor. Simulations do not show this phenomenon. The best solution is the simple Q1 circuit. The output voltage Vs at Q1's emitter is:

Vs = (Vext—Rg*Imax)*R3/(R3+R4)—VEB

(Vext is the external DC voltage (15V), VEB = 0.68V.)

Vs = (15V—1.5kΩ*4.85mA)*68kΩ/(20kΩ+68kΩ)—0.68V = 5.3V

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