# Circuit schematic

### Circuit Schematic

The complete circuit schematic is shown above.

To find Rdc, given Idc must not exceed 100uA, Rdc = Vdc/Idc Rdc = 0.5/100u

Rdc = 5000 ohms

And we assume the op-amp is an ideal op-amp, therefore there's no current flowing into the V+, thus i can conclude that Idc = Iac. Rdc must be less than or equals to 5k ohms.

Graph above shows the output of the schematic circuit. Where V(4) is the output of the summing amplifier and V(6) is the final output after being amplified 5 times.

Graph above shows the input of the summing amplifier

The Idc is shown below

2 conditions are to be fulfilled in this question,

Vi is amplified by 5times

The output signal after amplification and shifting should fluctuate between 0 V to 5 V

To solve i) , i used a non-inverting amplifier as the final output signal should not be inverted. And to solve ii), the sin wave input can be shifted upwards by summing it to a dc source, in this case it is 0.5v as we used 0.5 Vac. Therefore summing amplifier is used to sum up these two Voltages.

Firstly, as shown in the circuit schematic, i sum up the two voltages FIRST and the output of the summing amplifier is fed to the input of the non-inverting amplifier. Reason being to use a lower Vdc, if the Vac is amplified first then i would have to sum it with a higher Vdc value in order to shift it above 0. In this case 0.5 to 2.5 maybe very few, but imagine if voltage are not to be amplified by 5 times, if we amplify it much more, it would be a waste to supply that equal high amount of Vdc. This design will be inpractical. Therefore, by summing Vdc and Vac first, Vdc can be minimized.
calculations involved were very basic.

(Vo-V-)/Rf =V-/R1 *V- = inverting voltage of the amplifier

Vo = V-(Rf/R1 + 1)

V- = Vo / (Rf/R1 + 1 )

Assuming this is an ideal op-amp, V+ = V- and ZERO current flowing into V+

( Vac - V+ ) / R + ( Vdc - V+ ) / R = 0,

Vac + Vdc = 2V+

Vac + Vdc = 2 Vo / ( Rf/R1 + 1 )

Vo = ½ [( Vac + Vdc )/ (Rf/R1 + 1 )]

Rf =1k, R1 =1k, Vac = 0.5sinwt v ,Vdc = 0.5v

Vo = 0.5 sinwt + 0.5 V

this if the answer for the summing amplifier part.

Again using the same analysis,

Since V- = V+ = Vi ,

Vo = Vi (Rf1/ R2 + 1 ) * Vi = 0.5 sin wt + 0.5 v , Rf1 =4k , R2 =1k

Vo = (0.5sinwt + 0.5 ) 5

Vo = 2.5sinwt + 2.5 v

this is the final output, a 2.5 peak sin wave is being shifted up by 2.5 dc value which makes the 2.5sin wave voltage fluctuate between 0 V to 5 V

Question 2, i)

Given cut-off frequency is 20kHz, i calculated Wo as 125663.71 rad/s and K = 1, so i used a voltage follower instead of feedback resistors. K= 1 + Rf/R1, since K = 1, Rf = R1 = 0.To design the 1st order low pass filter, i first set R as 5k Ohms, then using the formula ,I got C = 1.6n So i used R=5k and C = 1.6n as my resistor and capacitor values.
a low pass filter passes frequencies from DC (0 Hz) to a desired frequency fo and

attenuates or stops frequencies above fo. Therefore i should be expecting a falling graph after fo.

Theoretical value of my roll-off rate is -20db using the formula taking n = 1 as i am using first order low pass and f is 200k, fo is 20k. It can be increased by increasing the value of n, or in other words increasing the order of the low pass filter.

A modification of part a circuit, given the same situation, again i set my Resistor value as 5k ohms, i now calculate my C value using the same formula . But in this case it is a 2nd order low pass, Resistors value are the same but not the two capacitors, C2 must be 2 times of C3 from the derivation of Q, proven that . K= 1 + Rf/R1, since K = 1, Rf = R1 = 0

Fo = 20k Hz, set R = 5k

C = 1.6nF

1.6n =

C3 = 1.125nF

C2 = 2.251nF

The schematic is designed using R = 5k ohms C2 = 2.251n F and C3 = 1.125n F

2nd order low pass circuit

3dB cut-off frequency matches the desired cut-off frequency fo = 20 kHz

The roll off rate is shown in the table labelled dy = -37.0687 *it is not negative because of the

x2-x1

Part C

K= 1 + Rf/R1, since K = 1, Rf = R1 = 0

KHP = KLP

for High pass, set R = 5k for Low pass, set R= 5k

given FL = 10k given FH = 20k

using the formula

C = 3.1831n C = 1.6nF

3.1831n = 1.6n =

C1 = 2.251n F C3 = 1.125nF

C2 = 4.502n F C4 = 2.251nF

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