Question

2. (15 points) The piezoelectric sensor can be equivalent to a charge generator in parallel with capacitance as below: (a), at some time moment, the piezoelectric sensor generated an amount of charge of Q, what is the output voltage? (b), below is the simplified and ideal circuit of a charge amplifier for the piezoelectric sensor, at some time moment, the piezoelectric sensor generated an amount of charge of Q, what is the output voltage from this charge amplifier? Lo Vout (c), if you got a real charge amplifier based on piezoelectric sensor, what circuit element(s) do you think should be added to the circuit shown in (b)? Please draw and explain your new circuit with details.

Answer 1

(2) (a) tr> Cuita, Ideal circuit) as the above zircuit has a current source in parallel with capacitor. so the current pones to capacitor and it stores the energy till some time, but after sometime, the enengy has to be discharged but in the above circuit there u no path for duscharge adding a So, I'm resistor to above circuit cirant (2) W (Real/ practical circuit)

-> According Kirchoff's node rule (ka) i= icti -> we know that aurrent is so Vout a ip (R) Cirsi-ic де = Vout= (da - cdvc) R de tirant (2) as above is for -> for cirenit Vout a ve (Cince voltage in paraul & same) icec dve jiedt: ve - Vc = I do do You a v

LF OF Itout (- feedback capacitor) - 2 things you need to know about opamps u that - vt. v -> currents never input terminals of op-amp. they either go through feedback on will flow to ground > simply op-amp is like a high input renstor and how output renstor. . Current I = If flows through a to output - therefore Vout= Ž Jade (Q. Changes through ) Just like I have added a resistor to circuit ces which gave rise to circuit (2), (for discharge purpose) here also add a neristor to capacitor , so that it can discharge

In above, the black line over and around the op-amp indicates the direction and flow of current. Note that the Vc is the voltage at capacitor.

- also due to v and vt being ov (due to ground) the current source 'd' might frow through capacitor & and then to ground. To avoid thi, add a voltage supply of ponitive enough value. -

As I have explained above, the modified and practical circuit is

(2v to 50

Hope this helps.

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