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Operational Amplifiers Flashcards

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1
Q

General uses of Operational Amplifiers (Op-Amps)

A

Signal conditioning including, amplification, level shift, addition and subtraction of signals

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2
Q

Amplifier Design Considerations

A

The input resistance 𝑅𝑖𝑛. We usually want it to be as large as possible to reduce the impact of loading.

We usually want the output resistance π‘…π‘œπ‘’π‘‘ to be as small as possible to ensure that power is not wasted as heat in the amplifier and the output voltage is as close as possible to the no-load voltage.

A: Amplifier/No load voltage gain. We usually want A>1. A: Amplification Factor, no units

G: Circuit gain. With no load, G = A and with load β‰  A

Vin =
Rin
⎯⎯⎯⎯⎯⎯ Γ— Vsensor
Rsensor + Rin
Vout =
Rload
⎯⎯⎯⎯⎯⎯ Γ— A Γ— Vin
Rload + Rout

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3
Q

No load gain vs gain with load

A

No load: π‘‰π‘œπ‘’π‘‘ = A Γ— 𝑉𝑖𝑛
(open circuit ∴ no current through Rout)

With load: Voltage Gain(G) =
Output Voltage Vout
⎯⎯⎯⎯⎯⎯⎯ = ⎯⎯⎯
Input Voltage Vin
∴ Vout =
RL
⎯⎯⎯⎯⎯⎯ Γ— A Γ— Vin
RL + Rout

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4
Q

Design Circuits with a Large Gain

A

Select an amplifier with a large amplifier gain, or connect many amplifiers with small amplifier gain (Cascade-connect)

Cascade-connect one after the other is usually done:

G1 = 𝑉2/𝑉𝑖𝑛
G2 = 𝑉3/𝑉2
G3 = π‘‰π‘œπ‘’π‘‘/𝑉3

G1 Γ— G2 Γ— G3 = G

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5
Q

Operational Amplifier Connections

A
  1. V2 Non-inverting
    terminal
  2. V1 Inverting
    terminal
  3. 𝑉𝑐𝑐(+)
  4. 𝑉𝑐𝑐(βˆ’)
  5. ground
  6. Vout
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6
Q

Ideal Op-Amp characteristics

A

1) Rin = ∞
2) Rout = 0
3) A = ∞
4) Bandwidth (range of frequency
that an amplifier can amplify) = ∞
Note: Most unrealistic
5) Resistance to external factors
(temperature/humidity)

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7
Q

Two properties of Op-Amp to simplify the circuit analysis

A

Virtual earth principle (VEP):
𝑉(+) = 𝑉(βˆ’)

Since A = ∞:
Vout = 𝐴 Γ— (𝑉(+) βˆ’ 𝑉(βˆ’)), ∴ Vout / A = (𝑉(+) βˆ’ 𝑉(βˆ’)) = 0
The particular equation only ocurres during negative feedback

The Op-Amp does whatever is necessary to keep the input terminals at the same potential. There is a virtual short circuit from 𝑉(+) to 𝑉(βˆ’)

Infinite input resistance:

R𝑖𝑛 = ∞, i𝑖𝑛 = 0

If R𝑖𝑛 = ∞, Rin is as large as it can be, like an open circuit
Universal rule

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8
Q

Derivation steps for gain of inverting amplifier

A

We want to find G = Vout / Vin in terms of the circuit components input resistor R1 and feedback resistor R2
Node A goes from Vin to V(-) (through resistor R1) and Vout (through resistor R2). V(+) is connected to ground.
β˜… KCL at node A:
i2 + iin = i1 = i2 + 0 ∴ i1 = i2
β˜… Write current as potential difference over resistance
Vin - V(-) V(-) - Vout
⎯⎯⎯ = ⎯⎯⎯
R1 R2
iin = 0 ∴ Vin / R1 = -Vout / R2
β˜… Γ— R1:
Vin = - VoutR1 / R2
β˜… -1:
1 / Vin = - R2 / VoutR1
β˜… Γ— Vout:
Vout - R2
G = ⎯⎯⎯ = ⎯⎯⎯
Vin R1

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9
Q

Inverting amplifier loading effect derivation, implications and solutions

A

Rin(cct) = Vin / i1
Vin R1
Rin(cct) = ⎯⎯⎯ = Vin Γ— ⎯ ∴
Vin / R1 Vin
Rin(cct) = R1

Inverting amplifier has the tendency to load the previous stage when we try to control (make large)
the gain

Rf can’t be too large (because of noise, bias currents, bandwidth limits), you end up reducing R1 to get higher gain. That makes Rin smaller β†’ heavier load on the previous stage. This can:

β€’	distort or attenuate the signal from a high-impedance source,
β€’	increase power consumption,
β€’	cause unwanted interaction between stages.

How to avoid the problem
β€’ Use a non-inverting amplifier (its input impedance ≫ R₁, so it doesn’t load the source).
β€’ Or buffer the signal with a voltage follower (unity-gain op-amp) before the inverting stage.

β€œstage” = one amplifier or circuit block, whatever is driving the input

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10
Q

Derivation steps for gain of Non-inverting amplifier

A

We want to find G = Vout / Vin in terms of the circuit components input resistor R1 and feedback resistor R2
Vin goes directly to
Node A goes from ground to V(-) (through resistor R1) and Vout (through resistor R2 as feedback)
β˜… KCL at node A:
i1 + iin = i2 = i1 + 0 ∴ i1 = i2
β˜… Write current as potential difference over resistance
V(-) - VGRD Vout - V(-)
⎯⎯⎯ = ⎯⎯⎯
R1 R2
VGRD = 0 & Vin = V(+) = V(-)∴
Vin / R1 = (Vout - Vin) / R2
β˜… + Vin / R2
Vin Γ— (R1-1 + R2-1) = Vout / R2
β˜… Γ— R2, Γ· Vin
G = Vout / Vin = 1 + (R2 / R1)

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11
Q

Non-Inverting amplifier loading effect derivation, implications and solutions

A

Rin(cct) = Vin / iin = Vin / 0 = ∞

Non-Inverting amplifier is good since it doesn’t load the previous stage but can’t be used by itself if we want to invert the signal

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12
Q

Derivation steps for finding the relationship between π‘‰π‘œπ‘’π‘‘ and 𝑉𝑖𝑛 of Inverting Summing Amplifier

A

We want to find Vout in terms of input resistors R1 and R2 and the feedback resistor RF
Node A goes from V1 and V2 (through resistors R1 and R2, respectively) to V(-) and Vout (through resistor RF, as feedback).V(+) is connected to ground.
β˜… KCL at node A:
i1 + i2 = iin + iF= iF + 0 ∴ i1 + i2= iF
β˜… Write current as potential difference over resistance
V(-) - Vout V1 - V(-) V2 - V(-)
⎯⎯⎯⎯ = ⎯⎯⎯⎯ = ⎯⎯⎯
RF R1 R2
V(-) = V(+) = 0 ∴
- Vout / RF = V1 / R1 + V2 / R2
β˜… Γ— - RF
RF RF
Vout = - (⎯⎯ V1 + ⎯⎯ V2)
R1 R2
if R1 = R2 = RF ∴ Vout = - (V1 + V2)

Generally
RF RF
Vout = - (⎯⎯ V1 + ⎯⎯ Vn + …)
R1 Rn

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13
Q

Derivation for Differential Amplifier (general form)

A

Node A goes from V1 to V(-) through R1 and to Vout thought R2 (feedback)
Node B goes from V2 to V(+) thorugh R3 and to GRD through R4
β˜… Write current as potential difference over resistance:
β˜… Get Vout with i1 = i2
Don’t take V(-) as 0 because we will need it for i3 = i4
V1 - V(-) V(-) - Vout
⎯⎯⎯ = ⎯⎯⎯
R1 R2
R2V1-R2V(-) = R1V(-)-R1Vout
R1Vout = (R1+R2)V(-)-R2V1
(R1+R2)V(-) R2V1
Vout = ⎯⎯⎯⎯⎯⎯ - ⎯⎯⎯
R1 R1

β˜… Get V(-) with i3 = i4
V2 - V(+) V(+) - GRD
⎯⎯⎯ = ⎯⎯⎯
R3 R4
R4V2 - R4V(-) =R3V(-)
R4V2 = V(-)(R3 + R4)
R4V2
V(-) = ⎯⎯⎯
R3 + R4
β˜… Substitu V(-) from i1 = 12 derivation to obtain general form
(R1 + R2) R4 R2
Vout = ⎯⎯⎯⎯⎯⎯ V2 - ⎯⎯ V1
R1 (R3 + R4) R1

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14
Q

How to make V1 and V2 equally weighted for a differential amplifier

A

β˜… If 𝑅1 = 𝑅3 and 𝑅2 = 𝑅4∴
β€˜β€˜β€˜
(R1 + R2) R2 R2
Vout = ⎯⎯⎯⎯⎯⎯ V2 - ⎯⎯ V1
R1 (R1 + R2) R1
β€˜β€˜β€˜

R1 R2 R22 R2
⎯⎯⎯⎯⎯ = ⎯⎯⎯ ∴
R1 R2 R12 R1
R2
Vout = ⎯⎯ (V2 - V1)
R1

V2 and V1 are equally weighted

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15
Q

How to get G = 1 for a differential amplifier

A

β˜… If 𝑅1 = 𝑅2 = 𝑅3 = 𝑅4
Vout = V2 - V1 ∴
G = 1

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16
Q

Summary of the 4 topologies (Inverting, Non-Inverting, Inverting Summing and Differential)

Study These Flashcards
A

Topology I: Inverting Amplifier

G = -R2 / R1
Rin(cct) = R1

Inverting amplifier has the tendency
to load the previous stage when we
try to control (make large) the gain

Topology II: Non-inverting Amplifier

G = 1 + R2 / R1
Rin(cct) = ∞

Good, but can’t
use if we want to
invert the signal

Topology III: Inverting Summing Amplifier
β€˜β€˜β€˜
RF RF
Vout = - (⎯⎯ V1 + ⎯⎯ Vn + …)
R1 Rn
β€˜β€˜β€˜
Good, we can sum
and invert the signal

Topology IV: Differential Amplifier
R1 + R2 R4 R2
Vout = ⎯⎯⎯ Γ— ⎯⎯⎯ Γ— V2 Γ— - ⎯⎯ V1
R1 R3 + R4 R1

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