- #1

- 14

- 0

## Homework Statement

An accelerometer is connected via a length of coaxial cable to an amplifier. The arrangement is modelled by:

(a) a Norton generator in parallel with a capacitor (CP) representing the piezoelectric crystal within the accelerometer;

(b) a lumped capacitor (CC) representing the coaxial cable; and

(c) a load (RL) representing the amplifier.

See attached for circuit diagram.

The current generated by the accelerometer (

*i*P) is proportional to the rate of displacement of the piezoelectric crystal. Hence

*i*P = K d

*x*/d

*t*. In laplace form

*i*P(s) = KsΔ

*x*(s).

(1) Derive an expression for the laplace transfer function T(s) = Δ

*v*L(s) / Δ

*i*P(s).

(2) Express Δ

*v*L as a function of time (i.e. the transient response of the voltage Δ

*v*L ) when

*i*P is subject to a step change.

(3) Using the values given in TABLE A, estimate the time taken for the voltage

*v*L to reach its steady state value if the current

*i*P is subject to a step change of 2 nA.

CP=1400 pF

CC=250 pF

RL=5 MΩ

TABLE A

## Homework Equations

## The Attempt at a Solution

Q(1).

I believe (based on internet research) that after redrawing the circuit with an equivalent single parallel capacitor, C, that the answer is

Δ

*v*L(s) / Δ

*i*P(s) = R / 1+sRC

However, I’m struggling with the derivation. In time-domain

*v*L =

*i*PR(1-e^-t/CR). But when I inverse laplace transform R / 1+sRC I get 1/C(1-e^-t/CR). I don’t understand how I can replace R with 1/C in the time-domain. Can anyone help?

Q(2).

*v*L(t) =

*i*PR(1-e^-t/CR).

Q(3).

C = CP+CC = 1400*10^-12 + 250*10^-12 = 1.65*10^-9

Time for

*v*L to reach steady state = 5CR = 5*5*10^6*1.65*10^-9 = 41.25*10^-3 s

However, if this is correct the magnitude of the step current is irrelevant to the steady state time for

*v*L. Since the problem specifies a step current of 2nA I’m wary of dismissing this variable as irrelevant. Am I missing something?

Any help with the above would be greatly appreciated.