Don't understand MOSFET as amplifierHow is possible that with same Ibase there is more than one Vce?My amplifier produced a negative gain when it should be positive. Why?Designing a BJT Amplifier given some constraintsAoE3: Grounded Emitter Amplifier DistortionGain and output impedance of an amplifierHow does the voltage difference develop when MOSFET is biased with Current Source?Common source amplifier using mosfet: Why is my Id dependent on my Rd?Difficulties in biasing MOSFET push-pull amplifierHow to identify which transistor is in active region and which one is in saturation?About the output resistance of a voltage amplifier
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Don't understand MOSFET as amplifier
How is possible that with same Ibase there is more than one Vce?My amplifier produced a negative gain when it should be positive. Why?Designing a BJT Amplifier given some constraintsAoE3: Grounded Emitter Amplifier DistortionGain and output impedance of an amplifierHow does the voltage difference develop when MOSFET is biased with Current Source?Common source amplifier using mosfet: Why is my Id dependent on my Rd?Difficulties in biasing MOSFET push-pull amplifierHow to identify which transistor is in active region and which one is in saturation?About the output resistance of a voltage amplifier
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When a MOSFET is working in its saturation region, it might be used as an amplifier. But, as far as I understand, the more you increase vGS (voltage between the MOSFET's gate and ground) the less output voltage you get, which is the opposite of amplifying. Let me explain:
Having this (supposed) amplifier:
when it is in saturation region, it behaves as a VCCS, which, apparently, leads to an amplifiers' behaviour:
In the above picture, vO will be VS - eL, if eL is the voltage drop in RL. Following the iD equation, the more vIN incrases, the more voltage drop you will have in RL, right? That means that the more vIN increases, the more vO decreases. That isn't amplifying, but the opposite.
What am I missing?
mosfet amplifier
asked 8 hours ago
MartelMartel
155 bronze badges
$endgroup$
add a comment |
$begingroup$
When a MOSFET is working in its saturation region, it might be used as an amplifier. But, as far as I understand, the more you increase vGS (voltage between the MOSFET's gate and ground) the less output voltage you get, which is the opposite of amplifying. Let me explain:
Having this (supposed) amplifier:
when it is in saturation region, it behaves as a VCCS, which, apparently, leads to an amplifiers' behaviour:
In the above picture, vO will be VS - eL, if eL is the voltage drop in RL. Following the iD equation, the more vIN incrases, the more voltage drop you will have in RL, right? That means that the more vIN increases, the more vO decreases. That isn't amplifying, but the opposite.
What am I missing?
mosfet amplifier
asked 8 hours ago
MartelMartel
155 bronze badges
$endgroup$
3
$begingroup$
This is an inverting amplifier, so the AC gain has a negative sign.
$endgroup$
– glen_geek
8 hours ago
$begingroup$
If |ΔVo/ΔVin| >1 then it is amplifying.
$endgroup$
– Phil G
8 hours ago
$begingroup$
electronics.stackexchange.com/questions/355899/…
$endgroup$
– G36
8 hours ago
add a comment |
$begingroup$
When a MOSFET is working in its saturation region, it might be used as an amplifier. But, as far as I understand, the more you increase vGS (voltage between the MOSFET's gate and ground) the less output voltage you get, which is the opposite of amplifying. Let me explain:
Having this (supposed) amplifier:
when it is in saturation region, it behaves as a VCCS, which, apparently, leads to an amplifiers' behaviour:
In the above picture, vO will be VS - eL, if eL is the voltage drop in RL. Following the iD equation, the more vIN incrases, the more voltage drop you will have in RL, right? That means that the more vIN increases, the more vO decreases. That isn't amplifying, but the opposite.
What am I missing?
mosfet amplifier
asked 8 hours ago
MartelMartel
155 bronze badges
$endgroup$
When a MOSFET is working in its saturation region, it might be used as an amplifier. But, as far as I understand, the more you increase vGS (voltage between the MOSFET's gate and ground) the less output voltage you get, which is the opposite of amplifying. Let me explain:
Having this (supposed) amplifier:
when it is in saturation region, it behaves as a VCCS, which, apparently, leads to an amplifiers' behaviour:
In the above picture, vO will be VS - eL, if eL is the voltage drop in RL. Following the iD equation, the more vIN incrases, the more voltage drop you will have in RL, right? That means that the more vIN increases, the more vO decreases. That isn't amplifying, but the opposite.
What am I missing?
mosfet amplifier
mosfet amplifier
asked 8 hours ago
MartelMartel
155 bronze badges
asked 8 hours ago
MartelMartel
155 bronze badges
asked 8 hours ago
MartelMartel
155 bronze badges
asked 8 hours ago
MartelMartel
155 bronze badges
asked 8 hours ago
MartelMartel
155 bronze badges
155 bronze badges
3
$begingroup$
This is an inverting amplifier, so the AC gain has a negative sign.
$endgroup$
– glen_geek
8 hours ago
$begingroup$
If |ΔVo/ΔVin| >1 then it is amplifying.
$endgroup$
– Phil G
8 hours ago
$begingroup$
electronics.stackexchange.com/questions/355899/…
$endgroup$
– G36
8 hours ago
add a comment |
3
$begingroup$
This is an inverting amplifier, so the AC gain has a negative sign.
$endgroup$
– glen_geek
8 hours ago
$begingroup$
If |ΔVo/ΔVin| >1 then it is amplifying.
$endgroup$
– Phil G
8 hours ago
$begingroup$
electronics.stackexchange.com/questions/355899/…
$endgroup$
– G36
8 hours ago
3
3
$begingroup$
This is an inverting amplifier, so the AC gain has a negative sign.
$endgroup$
– glen_geek
8 hours ago
$begingroup$
This is an inverting amplifier, so the AC gain has a negative sign.
$endgroup$
– glen_geek
8 hours ago
$begingroup$
If |ΔVo/ΔVin| >1 then it is amplifying.
$endgroup$
– Phil G
8 hours ago
$begingroup$
If |ΔVo/ΔVin| >1 then it is amplifying.
$endgroup$
– Phil G
8 hours ago
$begingroup$
electronics.stackexchange.com/questions/355899/…
$endgroup$
– G36
8 hours ago
$begingroup$
electronics.stackexchange.com/questions/355899/…
$endgroup$
– G36
8 hours ago
add a comment |
1 Answer
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It is an amplifier because you put a wiggly signal in, and you get a bigger wiggly signal out. The fact that the output wiggles down when the input wiggles up is a trivial detail, that -- if it's a problem at all -- can be solved in any number of ways (not least of which is following one stage by another, because two negative gains, when multiplied, result in a positive gain).
answered 8 hours ago
TimWescottTimWescott
12.3k1 gold badge9 silver badges24 bronze badges
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1 Answer
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1 Answer
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$begingroup$
It is an amplifier because you put a wiggly signal in, and you get a bigger wiggly signal out. The fact that the output wiggles down when the input wiggles up is a trivial detail, that -- if it's a problem at all -- can be solved in any number of ways (not least of which is following one stage by another, because two negative gains, when multiplied, result in a positive gain).
answered 8 hours ago
TimWescottTimWescott
12.3k1 gold badge9 silver badges24 bronze badges
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add a comment |
$begingroup$
It is an amplifier because you put a wiggly signal in, and you get a bigger wiggly signal out. The fact that the output wiggles down when the input wiggles up is a trivial detail, that -- if it's a problem at all -- can be solved in any number of ways (not least of which is following one stage by another, because two negative gains, when multiplied, result in a positive gain).
answered 8 hours ago
TimWescottTimWescott
12.3k1 gold badge9 silver badges24 bronze badges
$endgroup$
add a comment |
$begingroup$
It is an amplifier because you put a wiggly signal in, and you get a bigger wiggly signal out. The fact that the output wiggles down when the input wiggles up is a trivial detail, that -- if it's a problem at all -- can be solved in any number of ways (not least of which is following one stage by another, because two negative gains, when multiplied, result in a positive gain).
answered 8 hours ago
TimWescottTimWescott
12.3k1 gold badge9 silver badges24 bronze badges
$endgroup$
It is an amplifier because you put a wiggly signal in, and you get a bigger wiggly signal out. The fact that the output wiggles down when the input wiggles up is a trivial detail, that -- if it's a problem at all -- can be solved in any number of ways (not least of which is following one stage by another, because two negative gains, when multiplied, result in a positive gain).
answered 8 hours ago
TimWescottTimWescott
12.3k1 gold badge9 silver badges24 bronze badges
answered 8 hours ago
TimWescottTimWescott
12.3k1 gold badge9 silver badges24 bronze badges
answered 8 hours ago
TimWescottTimWescott
12.3k1 gold badge9 silver badges24 bronze badges
answered 8 hours ago
TimWescottTimWescott
12.3k1 gold badge9 silver badges24 bronze badges
12.3k1 gold badge9 silver badges24 bronze badges
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$begingroup$
This is an inverting amplifier, so the AC gain has a negative sign.
$endgroup$
– glen_geek
8 hours ago
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If |ΔVo/ΔVin| >1 then it is amplifying.
$endgroup$
– Phil G
8 hours ago
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electronics.stackexchange.com/questions/355899/…
$endgroup$
– G36
8 hours ago