Cause of triode mode in MOSFET
$begingroup$
My reference text introduces the MOFSET as follows:
The text says the induced channel has uniform width as long as $v_{DS}$ is small, because in that case $v_{GD} = v_{GS} - v_{DS} approx v_{GS}$, and if this is violated, then the width of the channel near the drain decreases. I don't really understand why, nor how this is relevant to the width of the channel. Shouldn't $v_{DS}$ be uniform across the drain? Then wouldn't all components of the electric field be horizontal?
voltage transistors mosfet semiconductors
asked Dec 29 '18 at 9:00
jeanlucjeanluc
1223
$endgroup$
add a comment |
$begingroup$
My reference text introduces the MOFSET as follows:
The text says the induced channel has uniform width as long as $v_{DS}$ is small, because in that case $v_{GD} = v_{GS} - v_{DS} approx v_{GS}$, and if this is violated, then the width of the channel near the drain decreases. I don't really understand why, nor how this is relevant to the width of the channel. Shouldn't $v_{DS}$ be uniform across the drain? Then wouldn't all components of the electric field be horizontal?
voltage transistors mosfet semiconductors
asked Dec 29 '18 at 9:00
jeanlucjeanluc
1223
$endgroup$
1
$begingroup$
Read this ittc.ku.edu/~jstiles/312/handouts/… (page 5,6 ). And this also can help ittc.ku.edu/~jstiles/312/handouts/…
$endgroup$
– G36
Dec 29 '18 at 9:23
add a comment |
$begingroup$
My reference text introduces the MOFSET as follows:
The text says the induced channel has uniform width as long as $v_{DS}$ is small, because in that case $v_{GD} = v_{GS} - v_{DS} approx v_{GS}$, and if this is violated, then the width of the channel near the drain decreases. I don't really understand why, nor how this is relevant to the width of the channel. Shouldn't $v_{DS}$ be uniform across the drain? Then wouldn't all components of the electric field be horizontal?
voltage transistors mosfet semiconductors
asked Dec 29 '18 at 9:00
jeanlucjeanluc
1223
$endgroup$
My reference text introduces the MOFSET as follows:
The text says the induced channel has uniform width as long as $v_{DS}$ is small, because in that case $v_{GD} = v_{GS} - v_{DS} approx v_{GS}$, and if this is violated, then the width of the channel near the drain decreases. I don't really understand why, nor how this is relevant to the width of the channel. Shouldn't $v_{DS}$ be uniform across the drain? Then wouldn't all components of the electric field be horizontal?
voltage transistors mosfet semiconductors
voltage transistors mosfet semiconductors
asked Dec 29 '18 at 9:00
jeanlucjeanluc
1223
asked Dec 29 '18 at 9:00
jeanlucjeanluc
1223
asked Dec 29 '18 at 9:00
jeanlucjeanluc
1223
asked Dec 29 '18 at 9:00
jeanlucjeanluc
1223
asked Dec 29 '18 at 9:00
jeanlucjeanluc
1223
1223
1
$begingroup$
Read this ittc.ku.edu/~jstiles/312/handouts/… (page 5,6 ). And this also can help ittc.ku.edu/~jstiles/312/handouts/…
$endgroup$
– G36
Dec 29 '18 at 9:23
add a comment |
1
$begingroup$
Read this ittc.ku.edu/~jstiles/312/handouts/… (page 5,6 ). And this also can help ittc.ku.edu/~jstiles/312/handouts/…
$endgroup$
– G36
Dec 29 '18 at 9:23
1
1
$begingroup$
Read this ittc.ku.edu/~jstiles/312/handouts/… (page 5,6 ). And this also can help ittc.ku.edu/~jstiles/312/handouts/…
$endgroup$
– G36
Dec 29 '18 at 9:23
$begingroup$
Read this ittc.ku.edu/~jstiles/312/handouts/… (page 5,6 ). And this also can help ittc.ku.edu/~jstiles/312/handouts/…
$endgroup$
– G36
Dec 29 '18 at 9:23
add a comment |
2 Answers
2
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oldest
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The width of the conducting channel at any point is a function of the voltage between that point in the channel and the gate.
If Vds is zero, source and drain are at the same potential. So the potential at each point across the channel will also be the same (zero). The channel has constant width.
If Vds is (say) 1V, now there is a 1V potential distribution across the channel. Now the potential between gate and each point in the channel is not equal - it decreases as you get closer to the more positive end (drain). Let's say Vgs = 5V - that means that Vgd would be only 4V.
Hence the width of the conducting channel will decrease. This is what they are showing in the figure.
answered Dec 29 '18 at 11:24
dmbdmb
1,485512
$endgroup$
add a comment |
$begingroup$
The triode, or ohmic region, is a function of both forward, $I_F$, and reverse, $I_R$, currents. The capacitance in the channel is an approximation, as you have a varying capacitance per unit area; however, we pretty much just use
$$Cequiv(C_{ox}+C_{dep})=frac{partial Q_m}{partial psi_s}$$
If the magnitudes of $I_F$ and $I_R$ are comparable, then the channel current depends strongly both on the source voltage and on the drain voltage. In this case, the transistor is in the ohmic region. Using the approximations made my Mary Ann Maher and Carver Mead, we can say that
$$I=I_F-I_R=frac{Smu}{2C}left(Q_S^2-2CU_TQ_Sright)-frac{Smu}{2C}left(Q_D^2-2CU_TQ_Dright)$$
If the forward current component is much larger than the reverse current component, then the channel current no longer depends significantly on the drain voltage
and the transistor is saturated.
As an aside, I dislike that picture because you'll never form a complete channel unless you overdrive the gate, this is why "pinch off" occurs. The most succinct discussions on the behavior of MOSFET devices are from Andy Grove, Mary Maher or Carver Mead.
answered Dec 29 '18 at 18:55
b degnanb degnan
2,1031722
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2 Answers
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2 Answers
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$begingroup$
The width of the conducting channel at any point is a function of the voltage between that point in the channel and the gate.
If Vds is zero, source and drain are at the same potential. So the potential at each point across the channel will also be the same (zero). The channel has constant width.
If Vds is (say) 1V, now there is a 1V potential distribution across the channel. Now the potential between gate and each point in the channel is not equal - it decreases as you get closer to the more positive end (drain). Let's say Vgs = 5V - that means that Vgd would be only 4V.
Hence the width of the conducting channel will decrease. This is what they are showing in the figure.
answered Dec 29 '18 at 11:24
dmbdmb
1,485512
$endgroup$
add a comment |
$begingroup$
The width of the conducting channel at any point is a function of the voltage between that point in the channel and the gate.
If Vds is zero, source and drain are at the same potential. So the potential at each point across the channel will also be the same (zero). The channel has constant width.
If Vds is (say) 1V, now there is a 1V potential distribution across the channel. Now the potential between gate and each point in the channel is not equal - it decreases as you get closer to the more positive end (drain). Let's say Vgs = 5V - that means that Vgd would be only 4V.
Hence the width of the conducting channel will decrease. This is what they are showing in the figure.
answered Dec 29 '18 at 11:24
dmbdmb
1,485512
$endgroup$
add a comment |
$begingroup$
The width of the conducting channel at any point is a function of the voltage between that point in the channel and the gate.
If Vds is zero, source and drain are at the same potential. So the potential at each point across the channel will also be the same (zero). The channel has constant width.
If Vds is (say) 1V, now there is a 1V potential distribution across the channel. Now the potential between gate and each point in the channel is not equal - it decreases as you get closer to the more positive end (drain). Let's say Vgs = 5V - that means that Vgd would be only 4V.
Hence the width of the conducting channel will decrease. This is what they are showing in the figure.
answered Dec 29 '18 at 11:24
dmbdmb
1,485512
$endgroup$
The width of the conducting channel at any point is a function of the voltage between that point in the channel and the gate.
If Vds is zero, source and drain are at the same potential. So the potential at each point across the channel will also be the same (zero). The channel has constant width.
If Vds is (say) 1V, now there is a 1V potential distribution across the channel. Now the potential between gate and each point in the channel is not equal - it decreases as you get closer to the more positive end (drain). Let's say Vgs = 5V - that means that Vgd would be only 4V.
Hence the width of the conducting channel will decrease. This is what they are showing in the figure.
answered Dec 29 '18 at 11:24
dmbdmb
1,485512
answered Dec 29 '18 at 11:24
dmbdmb
1,485512
answered Dec 29 '18 at 11:24
dmbdmb
1,485512
answered Dec 29 '18 at 11:24
dmbdmb
1,485512
1,485512
add a comment |
add a comment |
$begingroup$
The triode, or ohmic region, is a function of both forward, $I_F$, and reverse, $I_R$, currents. The capacitance in the channel is an approximation, as you have a varying capacitance per unit area; however, we pretty much just use
$$Cequiv(C_{ox}+C_{dep})=frac{partial Q_m}{partial psi_s}$$
If the magnitudes of $I_F$ and $I_R$ are comparable, then the channel current depends strongly both on the source voltage and on the drain voltage. In this case, the transistor is in the ohmic region. Using the approximations made my Mary Ann Maher and Carver Mead, we can say that
$$I=I_F-I_R=frac{Smu}{2C}left(Q_S^2-2CU_TQ_Sright)-frac{Smu}{2C}left(Q_D^2-2CU_TQ_Dright)$$
If the forward current component is much larger than the reverse current component, then the channel current no longer depends significantly on the drain voltage
and the transistor is saturated.
As an aside, I dislike that picture because you'll never form a complete channel unless you overdrive the gate, this is why "pinch off" occurs. The most succinct discussions on the behavior of MOSFET devices are from Andy Grove, Mary Maher or Carver Mead.
answered Dec 29 '18 at 18:55
b degnanb degnan
2,1031722
$endgroup$
add a comment |
$begingroup$
The triode, or ohmic region, is a function of both forward, $I_F$, and reverse, $I_R$, currents. The capacitance in the channel is an approximation, as you have a varying capacitance per unit area; however, we pretty much just use
$$Cequiv(C_{ox}+C_{dep})=frac{partial Q_m}{partial psi_s}$$
If the magnitudes of $I_F$ and $I_R$ are comparable, then the channel current depends strongly both on the source voltage and on the drain voltage. In this case, the transistor is in the ohmic region. Using the approximations made my Mary Ann Maher and Carver Mead, we can say that
$$I=I_F-I_R=frac{Smu}{2C}left(Q_S^2-2CU_TQ_Sright)-frac{Smu}{2C}left(Q_D^2-2CU_TQ_Dright)$$
If the forward current component is much larger than the reverse current component, then the channel current no longer depends significantly on the drain voltage
and the transistor is saturated.
As an aside, I dislike that picture because you'll never form a complete channel unless you overdrive the gate, this is why "pinch off" occurs. The most succinct discussions on the behavior of MOSFET devices are from Andy Grove, Mary Maher or Carver Mead.
answered Dec 29 '18 at 18:55
b degnanb degnan
2,1031722
$endgroup$
add a comment |
$begingroup$
The triode, or ohmic region, is a function of both forward, $I_F$, and reverse, $I_R$, currents. The capacitance in the channel is an approximation, as you have a varying capacitance per unit area; however, we pretty much just use
$$Cequiv(C_{ox}+C_{dep})=frac{partial Q_m}{partial psi_s}$$
If the magnitudes of $I_F$ and $I_R$ are comparable, then the channel current depends strongly both on the source voltage and on the drain voltage. In this case, the transistor is in the ohmic region. Using the approximations made my Mary Ann Maher and Carver Mead, we can say that
$$I=I_F-I_R=frac{Smu}{2C}left(Q_S^2-2CU_TQ_Sright)-frac{Smu}{2C}left(Q_D^2-2CU_TQ_Dright)$$
If the forward current component is much larger than the reverse current component, then the channel current no longer depends significantly on the drain voltage
and the transistor is saturated.
As an aside, I dislike that picture because you'll never form a complete channel unless you overdrive the gate, this is why "pinch off" occurs. The most succinct discussions on the behavior of MOSFET devices are from Andy Grove, Mary Maher or Carver Mead.
answered Dec 29 '18 at 18:55
b degnanb degnan
2,1031722
$endgroup$
The triode, or ohmic region, is a function of both forward, $I_F$, and reverse, $I_R$, currents. The capacitance in the channel is an approximation, as you have a varying capacitance per unit area; however, we pretty much just use
$$Cequiv(C_{ox}+C_{dep})=frac{partial Q_m}{partial psi_s}$$
If the magnitudes of $I_F$ and $I_R$ are comparable, then the channel current depends strongly both on the source voltage and on the drain voltage. In this case, the transistor is in the ohmic region. Using the approximations made my Mary Ann Maher and Carver Mead, we can say that
$$I=I_F-I_R=frac{Smu}{2C}left(Q_S^2-2CU_TQ_Sright)-frac{Smu}{2C}left(Q_D^2-2CU_TQ_Dright)$$
If the forward current component is much larger than the reverse current component, then the channel current no longer depends significantly on the drain voltage
and the transistor is saturated.
As an aside, I dislike that picture because you'll never form a complete channel unless you overdrive the gate, this is why "pinch off" occurs. The most succinct discussions on the behavior of MOSFET devices are from Andy Grove, Mary Maher or Carver Mead.
answered Dec 29 '18 at 18:55
b degnanb degnan
2,1031722
edited Dec 29 '18 at 20:02
answered Dec 29 '18 at 18:55
b degnanb degnan
2,1031722
answered Dec 29 '18 at 18:55
b degnanb degnan
2,1031722
answered Dec 29 '18 at 18:55
b degnanb degnan
2,1031722
2,1031722
add a comment |
add a comment |
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$begingroup$
Read this ittc.ku.edu/~jstiles/312/handouts/… (page 5,6 ). And this also can help ittc.ku.edu/~jstiles/312/handouts/…
$endgroup$
– G36
Dec 29 '18 at 9:23