Concept:
The gain in dB is defined as:
\(P(dB)=10~log_{10}\frac{P_{out}}{P_{in}}\)
Power in dBm is expressed as:
\(P(dBm)=10log_{10}(\frac{P}{1m})\)
P = Power to be expressed in dBm
Calculation:
Given
Pin = 1 μW
P(dB) = 30 dB
Putting these values in the above equation, we get:
\(30=10~log_{10}\frac{P_{out}}{1~\mu W}\)
\(3=log_{10}\frac{P_{out}}{1~\mu W}\)
Taking Antilog on both sides, we get:
\(10^3=\frac{P_{out}}{1~\mu W}\)
P_{out} = 1 mW
\(P(dBm)=10log_{10}(\frac{1m}{1m})\)
P(dBm) = 10log_{10}(1)
P(dBm) = 0
Cascode amplifier:
Important points:
Amplifier type |
Configuration |
---|---|
CE - CC |
Cascade Amplifier |
CC - CC |
Darlington pair |
CE – CB |
Cascode Amplifier |
Two or more amplifiers circuits can be cascaded to increase the gain of an ac signal.
The overall gain A_{v} can be calculated by simply multiplying each gain together. This is as shown:
Mathematically, this is defined as:
\({{\rm{A}}_{\rm{v}}} = {{\rm{A}}_{{{\rm{v}}_1}}}.{{\rm{A}}_{{{\rm{v}}_2}}}.{{\rm{A}}_{{{\rm{v}}_3}}} \ldots .\)
In dB, the overall gain is defined as:
\({A_v}\left( {dB} \right) = 20\;log{A_v}\)
This can be written as:
\({A_v}\left( {dB} \right) = 20\;log{A_v} = 20\;{\rm{log}}({A_{{v_1}}}.{A_{{v_2}}}.{A_{{v_3}}} \ldots .\;\)
\({A_v}\left( {dB} \right) = 20log{A_{v1}} + 20\;log{A_{v2}} + \;20\;log{A_{v3}}\; + \ldots \)
Hence,
\({A_v}\left( {dB} \right) = {A_{v1\left( {dB} \right)}} + {A_{v2\left( {dB} \right)}} + \;{A_{v3\left( {dB} \right)}}\; + \; \ldots \)
∴ If the gain of each amplifier stage is expressed in decibels (dB), the total gain will be the sum of the gains of individual stages.
Concept:
If two amplifiers of gain A_{V1} and A_{V2} are cascaded then overall gain is given as,
\({A_v} = {A_{v1}}.{A_{v2}}\)
If gain is given in dB, then
\(({A_v})dB = ({A_{v1}})dB + \left( {{A_{v2}}} \right)dB\)
Calculation:
Given Gain = 100 of each stage, i.e.
(Gain)_{dB} = 20 log_{10} 100 = 40 dB
Also, it is given that the gain of each stage drops by 3 dB at lower frequency, i.e.
Gain at 1^{st} stage = 40 dB \(\left( {{A_{{v_1}}}} \right)\)
Gain at the output of the 1^{st} stage = 40 – 3 = 34 dB.
Gain at the output of the 2^{nd} stage = 37 – 3 = 34 dB (A_{V2})
∴ The overall gain wil be:
\(\left( {{A_v}} \right) = {A_{{v_1}}} + {A_{{v_2}}}\)
A_{v} = 74 dB
Effect of cascading on the cutoff frequency and Bandwidth:
If n-Identical non-interacting amplifiers are cascaded and each amplifier has an individual cutoff frequency fL1 and fH1 then the overall cutoff freq. of the cascaded amplifier will be:
\({f_L} = \frac{{{f_{{L_1}}}}}{{\sqrt {{2^{1/N}} - 1} }}\)
\({f_H^*} = {f_{{H}}} \times \sqrt {{2^{1/N}} - 1} \)
For,
\(\sqrt {{2^{1/N}} - 1} < 1\)
\({f_L} > {f_{{L_1}}}\;\;\& \;\;{f_H} < {f_{{H_1}}}\)
Note: Cascading Increases lower cutoff and decreases higher cutoff frequency hence bandwidth decreases.
In cascading connection voltage gain increases which in turn reduces the bandwidth to maintain a constant gain-bandwidth product.
The same analysis is valid for cascading n = 2 amplifiers.
Concept:
Two or more amplifier circuits can be cascaded to increase the gain of an ac signal.
The overall gain Av can be calculated by simply multiplying each gain together. This is as shown:
Mathematically, this is defined as:
\({{\rm{A}}_{\rm{v}}} = {{\rm{A}}_{{{\rm{v}}_1}}}\times {{\rm{A}}_{{{\rm{v}}_2}}}\times {{\rm{A}}_{{{\rm{v}}_3}}} \ldots .\)
In dB, the overall gain is defined as:
\({A_v}\left( {dB} \right) = 20\;log{A_v}\)
This can be written as:
\({A_v}\left( {dB} \right) = 20\;log{A_v} = 20\;{\rm{log}}({A_{{v_1}}}\times {A_{{v_2}}}\times {A_{{v_3}}} \ldots .\;\)
\({A_v}\left( {dB} \right) = 20log{A_{v1}} + 20\;log{A_{v2}} + \;20\;log{A_{v3}}\; + \ldots \)
Hence,
\({A_v}\left( {dB} \right) = {A_{v1\left( {dB} \right)}} + {A_{v2\left( {dB} \right)}} + \;{A_{v3\left( {dB} \right)}}\; + \; \ldots \)
∴ If the gain of each amplifier stage is expressed in decibels (dB), the total gain will be the sum of the gains of individual stages.
Calculation:
The voltage gain of 1st amplifier = 30
The voltage gain of 2nd amplifier = 40
Voltage gain of the combined amplifier = 30 x 40 = 1200
Concept:
Effect of cascading on the cutoff frequency and Bandwidth:
If n-Identical non-interacting amplifiers are cascaded and each amplifier has an individual cutoff frequency fL1 and fH1 then the overall cutoff freq. of the cascaded amplifier will be:
\({f_L} = \frac{{{f_{{L_1}}}}}{{\sqrt {{2^{1/N}} - 1} }}\)
\({f_H} = {f_{{H_1}}} \times \sqrt {{2^{1/N}} - 1} \)
For, \(\sqrt {{2^{1/N}} - 1} < 1\)
\({f_L} > {f_{{L_1}}}\;\;\& \;\;{f_H} < {f_{{H_1}}}\)
Note: Cascading Increases lower cutoff and decreases higher cutoff frequency hence bandwidth decreases.
Calculation:
Given:
\({f_{{L_1}}} = 100\;Hz\)
\({f_{{H_1}}} = 500\;kHz\)
For 3-stages N = 3,
The overall cutoff freq. (f_{L} & f_{H}) is given as,
\({f_L} = \frac{{{f_{{L_1}}}}}{{\sqrt {{2^{1/N}} - 1} }} \\= \frac{{100\;Hz}}{{\sqrt {{2^{1/3}} - 1} }}\\ \approx 196\;Hz\)
\({f_H} = {f_{{H_1}}}\;\sqrt {{2^{1/N}} - 1} \\= 500\;kHz\;\sqrt {{2^{1/3}} - 1}\\ \approx 255\;kHz\;\)
For a cascade of interacting amplifiers, net higher cutoff freq. is calculated as:
\(\frac{1}{{{f_H}}} = 1.1 \times \sqrt {\frac{1}{{f_1^2}} + \frac{1}{{f_2^2}} + \ldots } \)
‘OR’
\(\frac{1}{{{f_H}}} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} + \ldots \)
Where, f_{1}, f_{2}, f_{3,} etc are pole frequencies, which are calculated from the Transfer function of a cascaded amplifier.
Net or overall lower cutoff is calculated as:
\({f_L} = 1.1 \times \sqrt {f_{{L_1}}^2 + f_{{L_2}}^2 + \ldots }\)
A cascade of two voltage amplifiers A_{1} & A_{2} is shown in the figure the open-loop gain \({A_{{v_0}}}\), input resistance R_{in}, and output resistance R_{0} for A1 and A2 are as follows:
A_{1}: \({A_{{v_0}}}=10\), R_{in} = 10 kΩ, R_{0} = 1 kΩ
A_{2} : \({A_{{v_0}}}=5\), R_{in} = 5 kΩ, R_{0} = 200 Ω
Concept:
Representation of voltage amplifier.
Circuit Diagram of Given cascaded Amplifier is as shown:
In the output circuit, by using the voltage division rule, we get:
\({V_{out}} = \frac{{{A_{{v_2}}}\;{V_{i{n_2}}}\;{R_L}}}{{{R_{{0_2}}} + {R_L}}}\)
\({V_{out}} = \left( {\frac{{{A_{{v_2}}}\;{R_L}}}{{{R_{{0_2}}} + {R_L}}}} \right){V_{i{n_2}}}\) ---(1)
And at the cascaded point, i.e. at the input of amplifier 2, applying voltage division rule, we get:
\({V_{i{n_2}}} = \frac{{{A_{{v_1}}}\;{V_{in}}{R_{{i_2}}}}}{{{R_{{0_1}}} + {R_{{i_2}}}}}\;\) ---(2)
From (1) and (2), we get:
\({V_{out}} = \left( {\frac{{{A_{{v_2}}}{R_L}}}{{{R_{{0_2}}} + {R_L}}}} \right)\left( {\frac{{{A_{{v_1}}}\;{V_{in}}\;{R_{{i_2}}}}}{{{R_{{0_1}}} + {R_{{i_2}}}}}} \right)\)
\(\frac{{{V_{out}}}}{{{V_{in}}}} = \frac{{{A_{{v_2}}}\;{A_{{v_1}}}\;{R_L}\;{R_{{i_2}}}}}{{\left( {{R_{{0_2}}} + {R_L}} \right)\left( {{R_{{0_1}}} + {R_{{i_2}}}} \right)}}\)
Calculation:
Putting all the values given in question in the above expression, we get:
\(\frac{{{V_{out}}}}{{{V_{in}}}} = \frac{{5 \times 10 \times 1000 \times 5 \times {{10}^3}}}{{\left( {200 + 1000} \right)\left( {1000 + 5000} \right)}}\)
\(\frac{{{V_{out}}}}{{{V_{in}}}} = \frac{{5 \times 5 \times {{10}^7}}}{{1200 \times 6000}} = \frac{{25 \times {{10}^7}}}{{75 \times {{10}^5}}} = \frac{{2500}}{{72}}\)
\(\frac{{{V_{out}}}}{{{V_{in}}}} = 34.72\)
Students who are at the earlier stage of learning then tend to multiply the gain of both amplifiers to find the overall gain. But this approach is wrong because we need to consider the loading effect of amplifier 2 at the output of amplifier 1.
Input Resistance of Amplifier 2 acts as a load to amplifier 1, which decreases the voltage gain.
When two identical stages with upper cutoff frequency ω_{H} are cascaded, the overall cutoff frequency is at:
Concept:
Effect of cascading on the cutoff frequency and Bandwidth:
If n-Identical non-interacting amplifiers are cascaded and each amplifier has an individual cutoff frequency f_{L1} and f_{H1} then the overall cutoff freq. of the cascaded amplifier will be:
\({f_L} = \frac{{{f_{{L_1}}}}}{{\sqrt {{2^{1/N}} - 1} }}\)
\({f_H^*} = {f_{{H}}} \times \sqrt {{2^{1/N}} - 1} \)
For, \(\sqrt {{2^{1/N}} - 1} < 1\)
\({f_L} > {f_{{L_1}}}\;\;\& \;\;{f_H} < {f_{{H_1}}}\)
Note: Cascading Increases lower cutoff and decreases higher cutoff frequency hence bandwidth decreases.
Calculation:
Given: N = 2
\({f_H^*} = {f_{{H}}} \times \sqrt {{2^{1/N}} - 1} \)
f_{H}^{*} = 0.64f_{H}
For a cascade of interacting amplifiers, net higher cutoff freq. is calculated as:
\(\frac{1}{{{f_H}}} = 1.1 \times \sqrt {\frac{1}{{f_1^2}} + \frac{1}{{f_2^2}} + \ldots } \)
‘OR’
\(\frac{1}{{{f_H}}} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} + \ldots \)
Where, f1, f2, f3, etc are pole frequencies, which are calculated from the Transfer function of a cascaded amplifier.
Net or overall lower cutoff is calculated as:
\({f_L} = 1.1 \times \sqrt {f_{{L_1}}^2 + f_{{L_2}}^2 + \ldots }\)
The quality factor is the ratio of resonant frequency to the bandwidth.
\(Q = \frac{{{f_r}}}{{B.W}},\;Q \propto \frac{1}{{B.W}}\)
The Q of a single-stage-single-tuned amplifier is doubled, i.e. Q’ = 2Q
\(\frac{{Q'}}{Q} = \frac{{B.W}}{{B.W'}}\)
\(\frac{{2Q}}{Q} = \frac{{B.W}}{{B.W'}}\)
\(B.W' = \frac{{B.W}}{2}\)
Therefore, bandwidth will become half.
Double-tuned circuit:
Double tuned circuit and its frequency response:
The resonant frequency is given as:
\({f_r} = \frac{1}{{2\pi \sqrt {{L_1}{C_1}} }} = \frac{1}{{2\pi \sqrt {{L_2}{C_2}} }}\)
Double-tuned circuit:
So based on the above points, option 1 is correct.
Double tuned circuit and its frequency response:
The resonant frequency is given as;
\({f_r} = \frac{1}{{2\pi \sqrt {{L_1}{C_1}} }} = \frac{1}{{2\pi \sqrt {{L_2}{C_2}} }}\)
In order to increase the bandwidth of tuned amplifiers, one can use:
1. Tuned circuit with inductance having high Q factor
2. Double tuned amplifier with two tuned circuits coupled by mutual inductance
3. Staggered tuned amplifiers in which different tuned circuits which are cascaded are tuned to slightly different frequencies
The tuned circuit is capable of amplifying a signal over a narrow band of frequencies centered at f_{r}.
The quality factor is the ratio of resonant frequency to the bandwidth.
\(Q = \frac{{{f_r}}}{{B.W}},\;Q \propto \frac{1}{{B.W}}\)
Higher the Q-factor lesser the bandwidth.Double-tuned circuit:
Double tuned circuit and its frequency response:
The resonant frequency is given as:
\({f_r} = \frac{1}{{2\pi \sqrt {{L_1}{C_1}} }} = \frac{1}{{2\pi \sqrt {{L_2}{C_2}} }}\)
Stagger-tuned amplifier:
In these two single tuned cascaded amplifiers having certain BW are taken and their resonant frequencies are so adjusted that they are separated by an amount equal to BW of each stage.
Since the resonant frequencies are staggered or displaced.
They are known as stagger tuned amplifier. The advantage is better flat, wideband characteristics.
If N-Identical non-interacting amplifiers are cascaded and each amplifier has an individual cutoff frequency \({f_{{L_1}}}\;\& \;{f_{{H_1}}}\) then the overall cutoff freq. of the cascaded amplifier will be:
\({f_L} = \frac{{{f_{{L_1}}}}}{{\sqrt {{2^{1/N}} - 1} }}\)
\({f_H} = {f_{{H_1}}} \times \sqrt {{2^{1/N}} - 1} \)
For, \(\sqrt {{2^{1/N}} - 1} < 1\)
\({f_L} > {f_{{L_1}}}\;\;\& \;\;{f_H} < {f_{{H_1}}}\)
Note: Cascading Increases lower cutoff and decreases higher cutoff frequency hence bandwidth decreases.
For a cascade of interacting amplifiers, net higher cutoff freq. is calculated as:
\({f_H} = {f_{{H_1}}} \times \sqrt {{2^{1/N}} - 1} \)
‘OR’
\(\sqrt {{2^{1/N}} - 1} < 1\)
Where, f1, f2, f3, etc are pole frequencies, which are calculated from the Transfer function of a cascaded amplifier.
Net or overall lower cutoff is calculated as:
\({f_L} = 1.1 \times \sqrt {f_{{L_1}}^2 + f_{{L_2}}^2 + \ldots }\)
Concept:
Loading effect:
If two amplifiers are cascaded net load for A_{1} is calculated by considering the effect of the input impedance of A_{2}.
\(R_{L{_1}}'=R_{L_{1}}||R_{L_{2}}'\) (Net Load)
(1). If R_{L}_{2}' is smaller/medium (Ω/kΩ) then,
R_{L1}' Decreases.
This decrease in a voltage gain of A_{1} due to the input impedance of A_{2} is known as the loading effect.
The loading effect occurs when amplifiers having low & medium o/p resistance are cascaded. Such amplifiers are called Interacting Amplifiers
(2). If R_{12}’ is very large,
R_{L1}' is approximately equal to R_{L1}
The voltage gain of A_{1} does not decrease or the loading effect doest not occur.
Calculation:
The voltage gain of A_{1} decreases due to the loading effect.
\(A_{V_{1}}=\frac{V_{0_{1}}}{V_{i_{1}}}\frac{A_vR_{i_{2}}}{R_{0_{1}}+R_{i_{2}}}\)
\(A_{V_{1}}=\frac{100\times1}{0.5+1}=66.66\)
Whereas,
A_{V1} = 100
Overall gain is given as:
A_{v} = A_{V1} x A_{V2}
A_{v} = 6666.66
To amplify the signal that has frequency components in 0 Hz to 20 Hz range via a multistage amplifier, the direct coupling method is best to couples the individual's amplifier.
Direct coupling is used in a multistage amplifier because it connects the output of one stage directly to the input of the next stage. (Option (2) is correct)
In multistage amplifier design, generally, three kinds of coupling are used:
1) Resistor-capacitor (RC) coupling
2) Transformer coupling
3) Direct coupling
Important Points:
Comparison between different type of coupling method
Characteristics |
RC coupling |
Transformer coupling |
Direct coupling |
Impedance matching |
Not good |
Very good |
Good |
Space and weight |
Less |
More |
Least |
Cost |
Less |
More |
Least |
Frequency response |
Good |
Very poor |
Best among all |
Use |
Voltage amplification |
Power amplification |
To amplify extremely low frequency |
Multi-stage amplifier is cascade connected individual amplifiers which provide desired gain, power and full-fill all the requirement of an ideal amplifier.
Example: Cascade amplifier, Darlington-Amplifier.
Concept:
Mid-frequency gain of a non- inverting amplifier is given as:
\({A_v} = \left( {1 + \frac{{{R_f}}}{{{R_1}}}} \right)\)
If the two amplifiers of gain A_{V1} and A_{V2} are cascaded, the overall gain is given by:
\({A_v} = {A_{v1}}.{A_{v2}}\)
If the gain is given in dB, then
\(({A_v})dB = ({A_{v1}})dB + \left( {{A_{v2}}} \right)dB\)
Where,
\(({A_{v1}})dB = 20\log {A_{v1}}\)
\(({A_{v2}})dB = 20\log {A_{v2}}\)
Calculation:
With R_{F }= 220 kΩ and R_{1 }= 10 kΩ, the gain is obtained as:
\({A_v} = \left( {1 + \frac{{{R_f}}}{{{R_1}}}} \right)\)
Since both the amplifiers are the same, we can write:
\({A_{v1}} = {A_{v2}} = \left( {1 + \frac{{{R_f}}}{{{R_1}}}} \right)\)
\( = \left( {1 + \frac{{220}}{{10}}} \right) = 23\)
Now,
\(({A_{v1}})dB = ({A_{v2}})dB = 20\log 23 = 27.23\)
Two such amplifiers are cascaded so,
\(A_v = ({A_{v1}})dB + ({A_{v2}})dB \)
\(= 27.23 + 27.23=54.46\)
Hence the overall gain (Av) is 54.46 dB