Unit 4
Differential Multistage Operational Amplifier
Differential Amplifier is a device which is used to amplify the difference between the voltages applied at its inputs. Such circuits can be of two types viz.,
Figure 1. BJT Differential Amplifier
Here the input signals (V1 and V2) are applied to the base of the transistors while the output is collected across their collector terminals (Vo1 and Vo2).
Figure 2. Differential Amplifier using Op-amp
Here V1 and V2 represent the voltages applied at its inverting and non-inverting input terminals and Ad refers to its differential gain.
The output of the differential amplifier is given as
Vo = Ad (V1 – V2) + Ac (V1 + V2/2)
where AC is called the common mode gain of the amplifier.
Hence its output voltage will be equal to the sum of the output voltages produced by the Op-Amp circuit operating as an inverting amplifier and the Op-Amp circuit operating as a non-inverting amplifier.
Thus, one gets,
Vo = -Rf/R1 V1 + V2. Rf / R2 + R3 (1 + Rf/R1)
Now, if R1 = R2 and R3 = Rf, then
Vo = -Rf/R1 V1 + V2 . Rf/ R1 + Rf ( R1 + Rf/R1)
Vo = - Rf/R1 . V1 + V2 . Rf /R1
Vo = - Rf/R1 (V1 – V2)
This implies that the gain of the differential amplifier circuit is given by -Rf/R1.
Power amplifiers are basically used to enhance the power level of the input signal. Power amplifier is also called large signal amplifiers, as in order to get large power at the output, the input signal voltage required must also be large.
Key terms of Power Amplifier
Stages of Power Amplifiers:
To provide the necessary power amplification, Power amplifiers consist of the following three stages as shown below:
Figure 3. Stages of Power Amplifier
Stage 1: Voltage amplification stage: As the input signal developed by the transducer is of very low value and a higher value signal is needed at the output so input signal is amplified at the very first stage of power amplifiers.
Here, we have used two stages of voltage amplifier so as to amplify the low-value input to the desired level.
Stage 2: Driver stage: The amplified voltage obtained from the voltage amplifier is fed to the driver stage to provide maximum power gain and to facilitate impedance matching.
Stage 3: Output stage: This stage essentially consists of power amplifiers and is responsible to transfer maximum power to the output device.
As no coupling devices are used, the coupling of the amplifier stages is done directly and hence called as Direct coupled amplifier.
The figure below indicates the three stage direct coupled transistor amplifier. The output of first stage transistor T1 is connected to the input of second stage transistor T2.
Figure 4. DC coupled Multistage Amplifier
The transistor in the first stage will be an NPN transistor, while the transistor in the next stage will be a PNP transistor and so on. This is because, the variations in one transistor tend to cancel the variations in the other. The rise in the collector current and the variation in β of one transistor gets cancelled by the decrease in the other.
Operation
The input signal when applied at the base of transistor T1, gets amplified due to the transistor action and the amplified output appears at the collector resistor Rc of transistor T1. This output is applied to the base of transistor T2 which further amplifies the signal. In this way, a signal is amplified in a direct coupled amplifier circuit.
Advantages
The advantages of direct coupled amplifier are as follows.
Disadvantages
The disadvantages of direct coupled amplifier are as follows.
Applications
The applications of direct coupled amplifier are as follows.
Classifications
According to technology : Bipolar
BiFET, BiMOS
MOSFET
According to application Universal
Fast
Precision
Small supply current , voltage , noise.
Figure 5. Internal structure of Op-amp
The T12 - R5 - T11 is the input leg of two current mirrors:
1. the T11 - T10 and T9 - T8 current mirrors set the operation point current of the input differential amplifier.
2. the T12 - T13 current mirror sets the operation point current of the CE main amplifier with a Darlington pair of transistors.
The I1 current of the input leg (R5 = 39 k): I1 = (2 . 15 - 2 . 0,6)/39 = 0.7 mA
A part of this is the total base current of transistors T3, T4: I4 = 5 µA, the remaining part I3 = I5 = 30 µA providing the total op. point current of the input stage, thus the current of two legs of the diff. amplifier: I0 = I5/2 = 15µA.
This combined with the bias current given by the catalogue as Ib = 100 nA gives the current gain of the input transistors T1 and T2: ß ˜ B = 15 000/100 = 150.
2. Towards the main amplifier the current transfer is 1:1, thus the op. point current of T16 (and T14): 0.7 mA (neglecting the current of T15 and that of the divider R6 – R7).
3. The operation class of the power amplifier is set by the circuit of T14 (R6 = 4.5 k;
R7 = 7.5 k): 2VB = 0.6(R6 + R7)/R7 = … = 0.96 V
This is greater than the double of the BE threshold voltage (2 [0.35…0,4] = 0.7…0.8 V), but smaller than 2 . 0.6 = 1.2 V, thus the setting is class AB.
An op amp should also have very high open loop gain. In ideal cases, the input resistance and open loop gain of an op amp should be infinity whereas the output resistance would be zero. So, an ideal op amp should have following characteristics.
Characteristics | Value |
Open loop gain (A) | ∞ |
Input Resistance | ∞ |
Output Resistance | 0 |
Bandwidth Operation | ∞ |
Offset Voltage | 0 |
An ideal op amp is defined as, a differential amplifier with infinite open loop gain, infinite input resistance and zero output resistance. The ideal op amp has zero input current. This is because of infinite input resistance. As the input resistance of ideal op amp is infinite, an open circuit exists at input, hence current at both input terminals is zero.
Figure 6. Ideal Op-amp
There is no current through the input resistance, there will be no voltage drop between the input terminals. Hence no offset voltage appears across the inputs of an ideal operational amplifier.
If v1 and v2 are the voltages of inverting and non-inverting terminals of op amp, and v1 = v2 then in ideal case,
The bandwidth of operation of an ideal op-amp is also infinite. That means the op-amp perform its function for all ranges of frequencies of operation.
Non-Ideal Op-Amp
The non-ideal characteristics include:
Finite open loop gain:
Figure 7. Non ideal amp
vo = A v id = A ( vs – v1) = A( vs – β vo)
Av = vo/ vs = A/ 1 + A β
A β is called loop gain.
For A β >> 1
A ideal = 1/ β = 1 + R2/R1
Vid = vs – v1 = vs – β vo
= vs – A β / 1 + A β vs = vs / 1 + A β
V1 = R1/ R1 + R2 . vo = β vo
Β = R1 / R1 + R2 is the feedback resistor
Gain Error is given by GE= (ideal gain)-(actual gain)
For non-inverting amplifier,
GE = 1/ β = A / 1 + A β = 1/ β( 1+ A β)
Gain error is also expressed as a fractional or percentage error.
FGE = 1/β – A / 1 + A β / 1/ β = 1/ 1 + A β = 1/ A β
Problem:
Find ideal and actual gain and gain error is percent
Given data: Closed-loop gain of 200 (46 dB), open-loop gain of op amp is 10,000 (80 dB).
Amplifier is designed to give ideal gain and deviations from ideal case are determined.
Hence, . R1 and R2 aren’t designed to compensate for finite open-loop gain of amplifier.
Analysis
Av = A / 1 + A β = 10 4 / 1 + 10 4 /200 = 200 -196/200 = 0.02
Nonzero Output Resistance
Output terminal is driven by test source vx and current ix is calculated to determine output resistance. The equivalent circuit is same For both inverting and non-inverting amplifiers:
Rout = vx / ix
ix = io + i2
io = vx – Av id / Ro . i2 = vx / R1 + R2
Also, vid = -v1 and
v 1 = R1 / R1 + R2 vx = β vx
1/ Rout = ix / vx = 1 + A β/ Ro + 1/ R1 + R2
The shunt feedback at output reduces to
Rout = Ro / 1 + A β ( R1 + R2)
Since Ro / (1 + Aβ) << (R1 + R2)
Rout = Ro / 1 + A β
If A is infinite Rout =0
Design non-inverting amplifier and find open-loop gain
• Given Data: A v=35 dB, Rout =0.2 Ω, R o = 250 Ω
| Av| = 10 35 db / 20 db = 56.2
β = 1/ Av = 1/56.2
Rout = Ro / 1+ A β < 0.2
A ≥ 1/β ( Ro / Rout -1) = 56.2
= (250/0.2 -1 ) = 7.03 x 10 4 = 96.9 dB
Finite Input Resistance:
Test voltage source vx is applied to input and current ix is calculated.
ix = Vx – V1/ Rid v1 = i1 R1 = i2 R1
Assuming i-<<i2 implies i1=i2.
v 1 = R1 / R1 + R2 vo = β vo
= β Av id = A β (vx – v1 )
v 1 = R1 / R1 + R2 . vo = β . vo
= β A v id = A β ( vx – v1)
v 1 = A β/ 1 + A β vx / Rid = vx / (1+Aβ) Rid
Rin = Rid ( 1 + A β)
Finite Common-Mode Rejection Ratio (CMRR)
A real amplifier responds to signal common to both inputs, called common-mode input voltage. In general,
Vo = A ( v1 – v2 ) + Acm ( v1 + v2/2)
= A (vid) + Acm(vic)
In the case of the ideal op-amp, the DC voltage of the VIN(+) and VIN(-) terminals match exactly when the input voltage (Vi) is 0 V. In reality, however, there are differences in input impedance and input bias current between the VIN(+) and VIN(-) terminals, causing a slight difference in their voltages.
This difference called input offset voltage
The voltage seen at the output of an op-amp even when both the inputs are grounded is called the output offset voltage.
Input bias current
All op-amps have the input bias current, which is the current drawn by the input terminals. It is a leakage current that sources or sinks at both input terminals. Depending on the type of input transistor, the bias current can flow in or out of the input terminals.
Input offset current
The input offset current (IOS) is equal to the difference between the input bias current at the non-inverting terminal (IB+) minus the input bias current at the inverting (IB- ) terminal of the amplifier. Offset current is typically an order of magnitude less than bias current.
Slew rate
Slew rate is the maximum rate of voltage change that can be generated by the op-amp’s output circuitry. It is measured as voltage relative to time, and the typical unit used in datasheets is volts per microsecond (V/µs).
Gain bandwidth product
The op amp gain bandwidth product is
GBP = Av x f
GBP = op amp gain bandwidth product
Av = voltage gain
f = cutoff frequency (Hz)
The op amp gain bandwidth product is constant for voltage-feedback amplifiers.
References: