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ESP1104 Summary 6 © Lim Fang Jeng

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Operational Amplifier

OP AMPs are design to have high gain Differential input

v_id = v_{1− v2}

Common mode input (average) v_icm =¹

2^{(v1+ v2)} Ideal OP AMP

- Infinite gain of differential mode input - Zero gain of common-mode input - Infinite input impedance (10^{13 Ω)} - Zero output impedance (1 Ω) - Infinite bandwidth (2.8 MHz)

DC OPAMP Analysis

OP AMP often designed to be connected with negative feedback to achieve different type of amplifier

Assume high input impedance, iin=0, v_node = v⁻

−^vS_R^{− v−}

S ⁻

v_{out − v}⁻ R_F ^{= 0}

v_out = A_{OL v}⁺_{− v}⁻ _{= AOL 0 − v}⁻ v⁻=₋^vout

A_OL

v_S =_−voutR_S ¹ R_SA_OL⁺

1 R_F ⁺

1 R_FA_OL But A_OL is very large, so

� ⁼⁻ �

Hence, for ideal OP AMP, we assume

i_in = 0, v⁻= v⁺ The input impedance is

Z_in =^vS i_S ⁼

A_v =₋^Reff R_in PROBLEM ANALYSIS

When an inverting OPAMP is the problem - _{Assume i}in=iout=0, vid=0

- If possible, find the R^eff of the feedback resistor

- _{If the R}f terminal is grounded, you can connect it to any grounded region to analyse the circuit

- Use node voltage analysis to relate the current with the voltages

ESP1104 Summary 6 © Lim Fang Jeng

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AC Analysis for OPAMP

Since we have the assumption, v_i = 0_{→ vin} = v₁ v_in = v₁= ^R1

R₁+ R₂^�0→ v₀ v_in ^{= 1 +}

R₂ R₁ This time the OP AMP is not inverting.

The voltage output is in phase with the voltage input

We can short the negative input of the OP AMP to the voltage output to construct a buffer circuit which gives the voltage gain of 1

For non-inverting amplifier, we can freely choose R1 and R2. However, we cannot choose them to be too large or too small.

If we choose them to be too small, the current flow through the OP AMP would damage the OPAMP as it exceeds its power rating

If the values are too large, it will be unstable and it will pick up external interference especially in humid environment

Gain-Bandwidth Limitation

Most OP AMP circuits has the open-loop again of the following (A0OL= DC open loop gain)

A_{OL f =} ^A0OL 1 + j _f ^f

BOL

A_0CL = ^A0OL 1 +_βA0OL f_BCL = f_{BOL 1 + βA0OL} Since AOL is large, we see that ACL is decreased but the fBCL increases Negative feedback reduces the voltage gain but increases the bandwidth

A_0CLf_BCL = A_0OLf_BOL This equation is known as the gain-bandwidth of the OP AMP

A_v = 1 +^R2 R₁ PROBLEM ANALYSIS

When the problem is non-inverting amplifier,

- Assume ideal diode, summing constraint properties apply

- Analyse the feedback resistor with node voltage analysis (It HELPS A LOT though…) - Find the relationships with each voltages and currents, relate it with vⁱⁿ

- Then apply the equation

ESP1104 Summary 6 © Lim Fang Jeng

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Instrumental Circuits

AC filters

A_v= 1 + ^R2 R₁+ X_C Filter low frequency or DC variations

This is a high pass filter, which filter the DC inputs of the source

Choosing of Coupling capacitor

Choose the value at which it is 0.2 times the resistor value. X_{C ≤ 0.2R}

Integrator Circuit

Given switch is closed for a long time prior t=0, the capacitor will start to charge up

i_in =^vin R i_c = i_in = C^dvout

dt v_out =¹

C ^{iin dt}

t

0

=¹ C

v_in R ^dt

t

0

= ¹

RC ^{vin dt}

t 0

Differentiator Circuit

i_in = C^dvin dt i_out =₋^vo

R ^{= C} dv_in

dt v₀=_−RC^dvin

dt

ESP1104 Summary 6 © Lim Fang Jeng

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Charge Amplifier

The relationship between the transducer and the charge is as follows:

q = Kx_i

By using this in an OP AMP, we can analyse the output voltage, i =^dq

dt ^{=−iF =−CF} dv_out

dt dq =_−CF dv_out

v_out =₋^Kxi C_F

Differential Amplifier

To amplify a signal with two different inputs, we can use the following OP AMP circuit,

v⁺= ^R4

R₃+R₄^{�1= v}

−

−^{v+− v2} R₁ ⁻

v⁺_{− vo} R₂ ^{= 0}

v⁺=^{v2R2+ v0R1} R₁+ R₂ R₄

R₃+ R₄^{�1 =}

v₂R₂+ v₀R₁ R₁+ R₂ v₁

R₃ R₄^{+ 1}

= v₂^R_R²

1^{+ v0}

1 +^R_R²

1

Apply ^R2

R₁ ⁼ R₄ R₃^,

� ⁼− −

Instrumental- Quality Differential Amplifier

v_ox1 v₂ ^{= 1 +}

R₂ R₁

v_ox2 v₁ ^{= 1 +}

R₂ R₁ For X3,

v_o

v_{ox 1− vox 2} ⁼⁻ R R v_o = v_{ox2− vox1} v_o = _{1 +}^R2

R₁ ^{v1− v2}

This circuit is independent of the resitors of OPAMP X3 but merely depends on the difference of input voltages.

However, in practice, the two resistors R1 can be looked as one resistor. The analysis is exactly the same. However, this differential applifier has a weakness that the input impedance of v2 depends on the voltage v1, this is a weakness of this OP AMP.

However, an improved circuit known as instrumental amplifier is implemented