Why does the small signal analysis work? (Intuition) The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Small signal models of MOS amplifiersSmall-signal analysis of inductor/capacitorWhy does my ground current look like this (and why is my current so low)?MOSFET amplifier mid-point biasSimple circuit analysis with dependent sourceWhy does the signal MOSFET keep failing in this circuit?R in and R out mosfet calrificationInput/out impedance and voltage gainQuestion on MOSFET - small signal modelCan't figure out how book on Trannys gets hfe of CE amplifier when using T-Equivalent and Hybrid Pi Circuits
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Why does the small signal analysis work? (Intuition)
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Small signal models of MOS amplifiersSmall-signal analysis of inductor/capacitorWhy does my ground current look like this (and why is my current so low)?MOSFET amplifier mid-point biasSimple circuit analysis with dependent sourceWhy does the signal MOSFET keep failing in this circuit?R in and R out mosfet calrificationInput/out impedance and voltage gainQuestion on MOSFET - small signal modelCan't figure out how book on Trannys gets hfe of CE amplifier when using T-Equivalent and Hybrid Pi Circuits
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Consider the Large and small signal model of the MOSFET amplifier,
simulate this circuit – Schematic created using CircuitLab
How is this transformation valid, I get the linearization of the MOSFET, but then how can a source become a short and how can we reason the simplifications of other elements. I mean, the original circuit looks totally different from the small signal model. Please provide an intuitive answer, as I get the math part (i.e.) for small signal variations, the behavior of various elements are studied and then approximated, My question is how is this valid? For example, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle that we supply, similarly wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
Is there a non-mathematical explanation for shorting voltage source during small signal analysis?
mosfet circuit-analysis amplifier small-signal
asked Apr 8 at 4:20
Aravindh VasuAravindh Vasu
305
$endgroup$
add a comment |
$begingroup$
Consider the Large and small signal model of the MOSFET amplifier,
simulate this circuit – Schematic created using CircuitLab
How is this transformation valid, I get the linearization of the MOSFET, but then how can a source become a short and how can we reason the simplifications of other elements. I mean, the original circuit looks totally different from the small signal model. Please provide an intuitive answer, as I get the math part (i.e.) for small signal variations, the behavior of various elements are studied and then approximated, My question is how is this valid? For example, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle that we supply, similarly wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
Is there a non-mathematical explanation for shorting voltage source during small signal analysis?
mosfet circuit-analysis amplifier small-signal
asked Apr 8 at 4:20
Aravindh VasuAravindh Vasu
305
$endgroup$
$begingroup$
We assume ZERO impedance between VDD (or Vs) and Ground. Hence the need for bypass capacitors.
$endgroup$
– analogsystemsrf
Apr 8 at 4:23
$begingroup$
Have you learned the Taylor series in math yet?
$endgroup$
– The Photon
Apr 8 at 5:06
$begingroup$
Yeah, I did. but then, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I think I can answer that question without referring to the Taylor series, but the more general question in your title ("Why does the small signal analysis work?") basically comes down to the Taylor series.
$endgroup$
– The Photon
Apr 8 at 5:12
add a comment |
$begingroup$
Consider the Large and small signal model of the MOSFET amplifier,
simulate this circuit – Schematic created using CircuitLab
How is this transformation valid, I get the linearization of the MOSFET, but then how can a source become a short and how can we reason the simplifications of other elements. I mean, the original circuit looks totally different from the small signal model. Please provide an intuitive answer, as I get the math part (i.e.) for small signal variations, the behavior of various elements are studied and then approximated, My question is how is this valid? For example, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle that we supply, similarly wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
Is there a non-mathematical explanation for shorting voltage source during small signal analysis?
mosfet circuit-analysis amplifier small-signal
asked Apr 8 at 4:20
Aravindh VasuAravindh Vasu
305
$endgroup$
Consider the Large and small signal model of the MOSFET amplifier,
simulate this circuit – Schematic created using CircuitLab
How is this transformation valid, I get the linearization of the MOSFET, but then how can a source become a short and how can we reason the simplifications of other elements. I mean, the original circuit looks totally different from the small signal model. Please provide an intuitive answer, as I get the math part (i.e.) for small signal variations, the behavior of various elements are studied and then approximated, My question is how is this valid? For example, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle that we supply, similarly wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
Is there a non-mathematical explanation for shorting voltage source during small signal analysis?
mosfet circuit-analysis amplifier small-signal
mosfet circuit-analysis amplifier small-signal
asked Apr 8 at 4:20
Aravindh VasuAravindh Vasu
305
asked Apr 8 at 4:20
Aravindh VasuAravindh Vasu
305
edited Apr 8 at 5:10
Aravindh Vasu
asked Apr 8 at 4:20
Aravindh VasuAravindh Vasu
305
asked Apr 8 at 4:20
Aravindh VasuAravindh Vasu
305
asked Apr 8 at 4:20
Aravindh VasuAravindh Vasu
305
305
$begingroup$
We assume ZERO impedance between VDD (or Vs) and Ground. Hence the need for bypass capacitors.
$endgroup$
– analogsystemsrf
Apr 8 at 4:23
$begingroup$
Have you learned the Taylor series in math yet?
$endgroup$
– The Photon
Apr 8 at 5:06
$begingroup$
Yeah, I did. but then, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I think I can answer that question without referring to the Taylor series, but the more general question in your title ("Why does the small signal analysis work?") basically comes down to the Taylor series.
$endgroup$
– The Photon
Apr 8 at 5:12
add a comment |
$begingroup$
We assume ZERO impedance between VDD (or Vs) and Ground. Hence the need for bypass capacitors.
$endgroup$
– analogsystemsrf
Apr 8 at 4:23
$begingroup$
Have you learned the Taylor series in math yet?
$endgroup$
– The Photon
Apr 8 at 5:06
$begingroup$
Yeah, I did. but then, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I think I can answer that question without referring to the Taylor series, but the more general question in your title ("Why does the small signal analysis work?") basically comes down to the Taylor series.
$endgroup$
– The Photon
Apr 8 at 5:12
$begingroup$
We assume ZERO impedance between VDD (or Vs) and Ground. Hence the need for bypass capacitors.
$endgroup$
– analogsystemsrf
Apr 8 at 4:23
$begingroup$
We assume ZERO impedance between VDD (or Vs) and Ground. Hence the need for bypass capacitors.
$endgroup$
– analogsystemsrf
Apr 8 at 4:23
$begingroup$
Have you learned the Taylor series in math yet?
$endgroup$
– The Photon
Apr 8 at 5:06
$begingroup$
Have you learned the Taylor series in math yet?
$endgroup$
– The Photon
Apr 8 at 5:06
$begingroup$
Yeah, I did. but then, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
Yeah, I did. but then, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I think I can answer that question without referring to the Taylor series, but the more general question in your title ("Why does the small signal analysis work?") basically comes down to the Taylor series.
$endgroup$
– The Photon
Apr 8 at 5:12
$begingroup$
I think I can answer that question without referring to the Taylor series, but the more general question in your title ("Why does the small signal analysis work?") basically comes down to the Taylor series.
$endgroup$
– The Photon
Apr 8 at 5:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In small signal-analysis, the behavior of a non-linear device is approximated as linear about a DC operating point (Quiescent point). Basically we put in a small 'wiggle' signal that doesn't change things to much from the DC level. With a linear model we can apply the principle of superposition.
That is, the total response of the system can be viewed as the sum of the responses to each source individually. In your left picture there are two sources, a signal input from the left, and what is usually a DC power supply at the top. In the right hand picture, this source is simply being zeroed so the response to the input signal can be more easily determined. Therefore it is replaced by a short circuit to ground.
The actual behavior of the circuit is then this response, plus the response to the DC source alone.
answered Apr 8 at 4:30
jramsay42jramsay42
585127
$endgroup$
$begingroup$
So, do you say that the DC source is removed to apply superposition? Wait, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle, similarly wouldn't the Voltage source be unaltered during the 'wiggle'?
$endgroup$
– Aravindh Vasu
Apr 8 at 4:35
$begingroup$
@AravindhVasu, the small signal equivalent of a linear resistor is just the same resistor. But the small signal equivalent of a nonlinear resistor (for example, a PN-junction diode considered in the low-frequency limit) is a different resistor at every operating point.
$endgroup$
– The Photon
Apr 8 at 5:05
$begingroup$
Okay, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I get the fact that we are linearly approximating a device , but I just cant digest(reason) vanishing the source.
$endgroup$
– Aravindh Vasu
Apr 8 at 5:09
add a comment |
$begingroup$
wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
When we talk about the small signal circuit, we are asking how much will the voltages and currents in the circuit change due to a small change in some input excitation (usually due to a voltage or current source).
If you change the input voltage to a circuit that contains an ideal voltage source, the voltage across that source doesn't change at all (that's why we call it ideal).
Since the voltage change is 0, its equivalent component in the small-signal circuit is a 0 V source.
The voltage assigned to the source (or any other component in the circuit) in a small-signal model is not its actual voltage, it's the voltage change due to the small-signal excitation.
answered Apr 8 at 5:10
The PhotonThe Photon
87.3k398203
$endgroup$
$begingroup$
But wouldn't shorting the DC voltage change the bias point and stuff? Please Bare with me here :)
$endgroup$
– Aravindh Vasu
Apr 8 at 5:14
$begingroup$
You don't short the voltage in the DC model used to find the operating point, only in the small-signal model used to find the effects of small changes in the inputs.
$endgroup$
– The Photon
Apr 8 at 5:14
2
$begingroup$
Yeah okay, so to study small changes in the input, we form a separate linear circuit by substituting for each component its small signal component, which is the response that the specific component gives during the small wiggle, and the voltage source doesn't contribute anything to the change in output, so we short it ,am I right?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:18
$begingroup$
@AravindhVasu, yes, that's it.
$endgroup$
– The Photon
Apr 8 at 5:19
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In small signal-analysis, the behavior of a non-linear device is approximated as linear about a DC operating point (Quiescent point). Basically we put in a small 'wiggle' signal that doesn't change things to much from the DC level. With a linear model we can apply the principle of superposition.
That is, the total response of the system can be viewed as the sum of the responses to each source individually. In your left picture there are two sources, a signal input from the left, and what is usually a DC power supply at the top. In the right hand picture, this source is simply being zeroed so the response to the input signal can be more easily determined. Therefore it is replaced by a short circuit to ground.
The actual behavior of the circuit is then this response, plus the response to the DC source alone.
answered Apr 8 at 4:30
jramsay42jramsay42
585127
$endgroup$
$begingroup$
So, do you say that the DC source is removed to apply superposition? Wait, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle, similarly wouldn't the Voltage source be unaltered during the 'wiggle'?
$endgroup$
– Aravindh Vasu
Apr 8 at 4:35
$begingroup$
@AravindhVasu, the small signal equivalent of a linear resistor is just the same resistor. But the small signal equivalent of a nonlinear resistor (for example, a PN-junction diode considered in the low-frequency limit) is a different resistor at every operating point.
$endgroup$
– The Photon
Apr 8 at 5:05
$begingroup$
Okay, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I get the fact that we are linearly approximating a device , but I just cant digest(reason) vanishing the source.
$endgroup$
– Aravindh Vasu
Apr 8 at 5:09
add a comment |
$begingroup$
In small signal-analysis, the behavior of a non-linear device is approximated as linear about a DC operating point (Quiescent point). Basically we put in a small 'wiggle' signal that doesn't change things to much from the DC level. With a linear model we can apply the principle of superposition.
That is, the total response of the system can be viewed as the sum of the responses to each source individually. In your left picture there are two sources, a signal input from the left, and what is usually a DC power supply at the top. In the right hand picture, this source is simply being zeroed so the response to the input signal can be more easily determined. Therefore it is replaced by a short circuit to ground.
The actual behavior of the circuit is then this response, plus the response to the DC source alone.
answered Apr 8 at 4:30
jramsay42jramsay42
585127
$endgroup$
$begingroup$
So, do you say that the DC source is removed to apply superposition? Wait, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle, similarly wouldn't the Voltage source be unaltered during the 'wiggle'?
$endgroup$
– Aravindh Vasu
Apr 8 at 4:35
$begingroup$
@AravindhVasu, the small signal equivalent of a linear resistor is just the same resistor. But the small signal equivalent of a nonlinear resistor (for example, a PN-junction diode considered in the low-frequency limit) is a different resistor at every operating point.
$endgroup$
– The Photon
Apr 8 at 5:05
$begingroup$
Okay, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I get the fact that we are linearly approximating a device , but I just cant digest(reason) vanishing the source.
$endgroup$
– Aravindh Vasu
Apr 8 at 5:09
add a comment |
$begingroup$
In small signal-analysis, the behavior of a non-linear device is approximated as linear about a DC operating point (Quiescent point). Basically we put in a small 'wiggle' signal that doesn't change things to much from the DC level. With a linear model we can apply the principle of superposition.
That is, the total response of the system can be viewed as the sum of the responses to each source individually. In your left picture there are two sources, a signal input from the left, and what is usually a DC power supply at the top. In the right hand picture, this source is simply being zeroed so the response to the input signal can be more easily determined. Therefore it is replaced by a short circuit to ground.
The actual behavior of the circuit is then this response, plus the response to the DC source alone.
answered Apr 8 at 4:30
jramsay42jramsay42
585127
$endgroup$
In small signal-analysis, the behavior of a non-linear device is approximated as linear about a DC operating point (Quiescent point). Basically we put in a small 'wiggle' signal that doesn't change things to much from the DC level. With a linear model we can apply the principle of superposition.
That is, the total response of the system can be viewed as the sum of the responses to each source individually. In your left picture there are two sources, a signal input from the left, and what is usually a DC power supply at the top. In the right hand picture, this source is simply being zeroed so the response to the input signal can be more easily determined. Therefore it is replaced by a short circuit to ground.
The actual behavior of the circuit is then this response, plus the response to the DC source alone.
answered Apr 8 at 4:30
jramsay42jramsay42
585127
answered Apr 8 at 4:30
jramsay42jramsay42
585127
answered Apr 8 at 4:30
jramsay42jramsay42
585127
answered Apr 8 at 4:30
jramsay42jramsay42
585127
585127
$begingroup$
So, do you say that the DC source is removed to apply superposition? Wait, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle, similarly wouldn't the Voltage source be unaltered during the 'wiggle'?
$endgroup$
– Aravindh Vasu
Apr 8 at 4:35
$begingroup$
@AravindhVasu, the small signal equivalent of a linear resistor is just the same resistor. But the small signal equivalent of a nonlinear resistor (for example, a PN-junction diode considered in the low-frequency limit) is a different resistor at every operating point.
$endgroup$
– The Photon
Apr 8 at 5:05
$begingroup$
Okay, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I get the fact that we are linearly approximating a device , but I just cant digest(reason) vanishing the source.
$endgroup$
– Aravindh Vasu
Apr 8 at 5:09
add a comment |
$begingroup$
So, do you say that the DC source is removed to apply superposition? Wait, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle, similarly wouldn't the Voltage source be unaltered during the 'wiggle'?
$endgroup$
– Aravindh Vasu
Apr 8 at 4:35
$begingroup$
@AravindhVasu, the small signal equivalent of a linear resistor is just the same resistor. But the small signal equivalent of a nonlinear resistor (for example, a PN-junction diode considered in the low-frequency limit) is a different resistor at every operating point.
$endgroup$
– The Photon
Apr 8 at 5:05
$begingroup$
Okay, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I get the fact that we are linearly approximating a device , but I just cant digest(reason) vanishing the source.
$endgroup$
– Aravindh Vasu
Apr 8 at 5:09
$begingroup$
So, do you say that the DC source is removed to apply superposition? Wait, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle, similarly wouldn't the Voltage source be unaltered during the 'wiggle'?
$endgroup$
– Aravindh Vasu
Apr 8 at 4:35
$begingroup$
So, do you say that the DC source is removed to apply superposition? Wait, the small signal model of a resistor is simply the resistor itself because the resistor produces a corresponding voltage drop during the small wiggle, similarly wouldn't the Voltage source be unaltered during the 'wiggle'?
$endgroup$
– Aravindh Vasu
Apr 8 at 4:35
$begingroup$
@AravindhVasu, the small signal equivalent of a linear resistor is just the same resistor. But the small signal equivalent of a nonlinear resistor (for example, a PN-junction diode considered in the low-frequency limit) is a different resistor at every operating point.
$endgroup$
– The Photon
Apr 8 at 5:05
$begingroup$
@AravindhVasu, the small signal equivalent of a linear resistor is just the same resistor. But the small signal equivalent of a nonlinear resistor (for example, a PN-junction diode considered in the low-frequency limit) is a different resistor at every operating point.
$endgroup$
– The Photon
Apr 8 at 5:05
$begingroup$
Okay, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
Okay, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I get the fact that we are linearly approximating a device , but I just cant digest(reason) vanishing the source.
$endgroup$
– Aravindh Vasu
Apr 8 at 5:09
$begingroup$
I get the fact that we are linearly approximating a device , but I just cant digest(reason) vanishing the source.
$endgroup$
– Aravindh Vasu
Apr 8 at 5:09
add a comment |
$begingroup$
wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
When we talk about the small signal circuit, we are asking how much will the voltages and currents in the circuit change due to a small change in some input excitation (usually due to a voltage or current source).
If you change the input voltage to a circuit that contains an ideal voltage source, the voltage across that source doesn't change at all (that's why we call it ideal).
Since the voltage change is 0, its equivalent component in the small-signal circuit is a 0 V source.
The voltage assigned to the source (or any other component in the circuit) in a small-signal model is not its actual voltage, it's the voltage change due to the small-signal excitation.
answered Apr 8 at 5:10
The PhotonThe Photon
87.3k398203
$endgroup$
$begingroup$
But wouldn't shorting the DC voltage change the bias point and stuff? Please Bare with me here :)
$endgroup$
– Aravindh Vasu
Apr 8 at 5:14
$begingroup$
You don't short the voltage in the DC model used to find the operating point, only in the small-signal model used to find the effects of small changes in the inputs.
$endgroup$
– The Photon
Apr 8 at 5:14
2
$begingroup$
Yeah okay, so to study small changes in the input, we form a separate linear circuit by substituting for each component its small signal component, which is the response that the specific component gives during the small wiggle, and the voltage source doesn't contribute anything to the change in output, so we short it ,am I right?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:18
$begingroup$
@AravindhVasu, yes, that's it.
$endgroup$
– The Photon
Apr 8 at 5:19
add a comment |
$begingroup$
wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
When we talk about the small signal circuit, we are asking how much will the voltages and currents in the circuit change due to a small change in some input excitation (usually due to a voltage or current source).
If you change the input voltage to a circuit that contains an ideal voltage source, the voltage across that source doesn't change at all (that's why we call it ideal).
Since the voltage change is 0, its equivalent component in the small-signal circuit is a 0 V source.
The voltage assigned to the source (or any other component in the circuit) in a small-signal model is not its actual voltage, it's the voltage change due to the small-signal excitation.
answered Apr 8 at 5:10
The PhotonThe Photon
87.3k398203
$endgroup$
$begingroup$
But wouldn't shorting the DC voltage change the bias point and stuff? Please Bare with me here :)
$endgroup$
– Aravindh Vasu
Apr 8 at 5:14
$begingroup$
You don't short the voltage in the DC model used to find the operating point, only in the small-signal model used to find the effects of small changes in the inputs.
$endgroup$
– The Photon
Apr 8 at 5:14
2
$begingroup$
Yeah okay, so to study small changes in the input, we form a separate linear circuit by substituting for each component its small signal component, which is the response that the specific component gives during the small wiggle, and the voltage source doesn't contribute anything to the change in output, so we short it ,am I right?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:18
$begingroup$
@AravindhVasu, yes, that's it.
$endgroup$
– The Photon
Apr 8 at 5:19
add a comment |
$begingroup$
wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
When we talk about the small signal circuit, we are asking how much will the voltages and currents in the circuit change due to a small change in some input excitation (usually due to a voltage or current source).
If you change the input voltage to a circuit that contains an ideal voltage source, the voltage across that source doesn't change at all (that's why we call it ideal).
Since the voltage change is 0, its equivalent component in the small-signal circuit is a 0 V source.
The voltage assigned to the source (or any other component in the circuit) in a small-signal model is not its actual voltage, it's the voltage change due to the small-signal excitation.
answered Apr 8 at 5:10
The PhotonThe Photon
87.3k398203
$endgroup$
wouldn't the Voltage source be unaltered during the 'wiggle', so shouldn't the small signal model of the Voltage source be itself ?
When we talk about the small signal circuit, we are asking how much will the voltages and currents in the circuit change due to a small change in some input excitation (usually due to a voltage or current source).
If you change the input voltage to a circuit that contains an ideal voltage source, the voltage across that source doesn't change at all (that's why we call it ideal).
Since the voltage change is 0, its equivalent component in the small-signal circuit is a 0 V source.
The voltage assigned to the source (or any other component in the circuit) in a small-signal model is not its actual voltage, it's the voltage change due to the small-signal excitation.
answered Apr 8 at 5:10
The PhotonThe Photon
87.3k398203
edited Apr 8 at 5:14
answered Apr 8 at 5:10
The PhotonThe Photon
87.3k398203
answered Apr 8 at 5:10
The PhotonThe Photon
87.3k398203
answered Apr 8 at 5:10
The PhotonThe Photon
87.3k398203
87.3k398203
$begingroup$
But wouldn't shorting the DC voltage change the bias point and stuff? Please Bare with me here :)
$endgroup$
– Aravindh Vasu
Apr 8 at 5:14
$begingroup$
You don't short the voltage in the DC model used to find the operating point, only in the small-signal model used to find the effects of small changes in the inputs.
$endgroup$
– The Photon
Apr 8 at 5:14
2
$begingroup$
Yeah okay, so to study small changes in the input, we form a separate linear circuit by substituting for each component its small signal component, which is the response that the specific component gives during the small wiggle, and the voltage source doesn't contribute anything to the change in output, so we short it ,am I right?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:18
$begingroup$
@AravindhVasu, yes, that's it.
$endgroup$
– The Photon
Apr 8 at 5:19
add a comment |
$begingroup$
But wouldn't shorting the DC voltage change the bias point and stuff? Please Bare with me here :)
$endgroup$
– Aravindh Vasu
Apr 8 at 5:14
$begingroup$
You don't short the voltage in the DC model used to find the operating point, only in the small-signal model used to find the effects of small changes in the inputs.
$endgroup$
– The Photon
Apr 8 at 5:14
2
$begingroup$
Yeah okay, so to study small changes in the input, we form a separate linear circuit by substituting for each component its small signal component, which is the response that the specific component gives during the small wiggle, and the voltage source doesn't contribute anything to the change in output, so we short it ,am I right?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:18
$begingroup$
@AravindhVasu, yes, that's it.
$endgroup$
– The Photon
Apr 8 at 5:19
$begingroup$
But wouldn't shorting the DC voltage change the bias point and stuff? Please Bare with me here :)
$endgroup$
– Aravindh Vasu
Apr 8 at 5:14
$begingroup$
But wouldn't shorting the DC voltage change the bias point and stuff? Please Bare with me here :)
$endgroup$
– Aravindh Vasu
Apr 8 at 5:14
$begingroup$
You don't short the voltage in the DC model used to find the operating point, only in the small-signal model used to find the effects of small changes in the inputs.
$endgroup$
– The Photon
Apr 8 at 5:14
$begingroup$
You don't short the voltage in the DC model used to find the operating point, only in the small-signal model used to find the effects of small changes in the inputs.
$endgroup$
– The Photon
Apr 8 at 5:14
2
2
$begingroup$
Yeah okay, so to study small changes in the input, we form a separate linear circuit by substituting for each component its small signal component, which is the response that the specific component gives during the small wiggle, and the voltage source doesn't contribute anything to the change in output, so we short it ,am I right?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:18
$begingroup$
Yeah okay, so to study small changes in the input, we form a separate linear circuit by substituting for each component its small signal component, which is the response that the specific component gives during the small wiggle, and the voltage source doesn't contribute anything to the change in output, so we short it ,am I right?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:18
$begingroup$
@AravindhVasu, yes, that's it.
$endgroup$
– The Photon
Apr 8 at 5:19
$begingroup$
@AravindhVasu, yes, that's it.
$endgroup$
– The Photon
Apr 8 at 5:19
add a comment |
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$begingroup$
We assume ZERO impedance between VDD (or Vs) and Ground. Hence the need for bypass capacitors.
$endgroup$
– analogsystemsrf
Apr 8 at 4:23
$begingroup$
Have you learned the Taylor series in math yet?
$endgroup$
– The Photon
Apr 8 at 5:06
$begingroup$
Yeah, I did. but then, is there a non-mathematical explanation for shorting voltage source during small signal analysis?
$endgroup$
– Aravindh Vasu
Apr 8 at 5:07
$begingroup$
I think I can answer that question without referring to the Taylor series, but the more general question in your title ("Why does the small signal analysis work?") basically comes down to the Taylor series.
$endgroup$
– The Photon
Apr 8 at 5:12