Gain in current but not in voltage -transistor “voltage follower” set up
Suppose we have the circuit set up as shown in the image. Also suppose that $$V_s=overline{V_s}+v_S$$, where the smal v is for the variable component of the signal which varies much faster than the time constant of both capacitors. After some analisys we get that there's no gain in the variable component of the source voltage, that is, there's unitary gain, however from the equation of a bipolar transistor:
$$i_Eapprox i_c=beta i_s$$
thus there's a gain in current, how is it possible to hava a gain in current, and consequently a gain in power, but not in voltage? This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
transistors amplifier bipolar
asked yesterday
BidonBidon
183
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Suppose we have the circuit set up as shown in the image. Also suppose that $$V_s=overline{V_s}+v_S$$, where the smal v is for the variable component of the signal which varies much faster than the time constant of both capacitors. After some analisys we get that there's no gain in the variable component of the source voltage, that is, there's unitary gain, however from the equation of a bipolar transistor:
$$i_Eapprox i_c=beta i_s$$
thus there's a gain in current, how is it possible to hava a gain in current, and consequently a gain in power, but not in voltage? This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
transistors amplifier bipolar
asked yesterday
BidonBidon
183
New contributor
Bidon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Hmmm… Looks like you have a strong misunderstanding of electricity. Anyway, if you connect a load resistor between $V_{textrm{out}}$ and ground it will drain some current. The current that is drained from $V_S$ may be significantly lower, hence the current gain.
– user2233709
yesterday
The input impedance and output impedance are very different.
– Dave Tweed♦
yesterday
Moreover, you’d generally want to polarize your transistor’s base with a pair of resistors.
– user2233709
yesterday
1
@user2233709 you're right about the misunderstanding of electricity. I don't understand how we can see current and voltage as two distinct things, on the one hand they are, but they are also so strongly connected by Ohm's law that I don't see how
– Bidon
yesterday
The way Ohm's law is generally taught to beginners ($V = RI$) assumes that $R$ is constant, which is true for ideal resistors. Other components can change resistance (more general term: impedance) depending on pretty much anything: voltage, temperature, frequency, input on another pin, ...
– Geier
yesterday
add a comment |
Suppose we have the circuit set up as shown in the image. Also suppose that $$V_s=overline{V_s}+v_S$$, where the smal v is for the variable component of the signal which varies much faster than the time constant of both capacitors. After some analisys we get that there's no gain in the variable component of the source voltage, that is, there's unitary gain, however from the equation of a bipolar transistor:
$$i_Eapprox i_c=beta i_s$$
thus there's a gain in current, how is it possible to hava a gain in current, and consequently a gain in power, but not in voltage? This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
transistors amplifier bipolar
asked yesterday
BidonBidon
183
New contributor
Bidon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Suppose we have the circuit set up as shown in the image. Also suppose that $$V_s=overline{V_s}+v_S$$, where the smal v is for the variable component of the signal which varies much faster than the time constant of both capacitors. After some analisys we get that there's no gain in the variable component of the source voltage, that is, there's unitary gain, however from the equation of a bipolar transistor:
$$i_Eapprox i_c=beta i_s$$
thus there's a gain in current, how is it possible to hava a gain in current, and consequently a gain in power, but not in voltage? This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
transistors amplifier bipolar
transistors amplifier bipolar
asked yesterday
BidonBidon
183
New contributor
Bidon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
BidonBidon
183
New contributor
Bidon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
BidonBidon
183
New contributor
Bidon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
BidonBidon
183
asked yesterday
BidonBidon
183
183
New contributor
Bidon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Bidon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Bidon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Hmmm… Looks like you have a strong misunderstanding of electricity. Anyway, if you connect a load resistor between $V_{textrm{out}}$ and ground it will drain some current. The current that is drained from $V_S$ may be significantly lower, hence the current gain.
– user2233709
yesterday
The input impedance and output impedance are very different.
– Dave Tweed♦
yesterday
Moreover, you’d generally want to polarize your transistor’s base with a pair of resistors.
– user2233709
yesterday
1
@user2233709 you're right about the misunderstanding of electricity. I don't understand how we can see current and voltage as two distinct things, on the one hand they are, but they are also so strongly connected by Ohm's law that I don't see how
– Bidon
yesterday
The way Ohm's law is generally taught to beginners ($V = RI$) assumes that $R$ is constant, which is true for ideal resistors. Other components can change resistance (more general term: impedance) depending on pretty much anything: voltage, temperature, frequency, input on another pin, ...
– Geier
yesterday
add a comment |
2
Hmmm… Looks like you have a strong misunderstanding of electricity. Anyway, if you connect a load resistor between $V_{textrm{out}}$ and ground it will drain some current. The current that is drained from $V_S$ may be significantly lower, hence the current gain.
– user2233709
yesterday
The input impedance and output impedance are very different.
– Dave Tweed♦
yesterday
Moreover, you’d generally want to polarize your transistor’s base with a pair of resistors.
– user2233709
yesterday
1
@user2233709 you're right about the misunderstanding of electricity. I don't understand how we can see current and voltage as two distinct things, on the one hand they are, but they are also so strongly connected by Ohm's law that I don't see how
– Bidon
yesterday
The way Ohm's law is generally taught to beginners ($V = RI$) assumes that $R$ is constant, which is true for ideal resistors. Other components can change resistance (more general term: impedance) depending on pretty much anything: voltage, temperature, frequency, input on another pin, ...
– Geier
yesterday
2
2
Hmmm… Looks like you have a strong misunderstanding of electricity. Anyway, if you connect a load resistor between $V_{textrm{out}}$ and ground it will drain some current. The current that is drained from $V_S$ may be significantly lower, hence the current gain.
– user2233709
yesterday
Hmmm… Looks like you have a strong misunderstanding of electricity. Anyway, if you connect a load resistor between $V_{textrm{out}}$ and ground it will drain some current. The current that is drained from $V_S$ may be significantly lower, hence the current gain.
– user2233709
yesterday
The input impedance and output impedance are very different.
– Dave Tweed♦
yesterday
The input impedance and output impedance are very different.
– Dave Tweed♦
yesterday
Moreover, you’d generally want to polarize your transistor’s base with a pair of resistors.
– user2233709
yesterday
Moreover, you’d generally want to polarize your transistor’s base with a pair of resistors.
– user2233709
yesterday
1
1
@user2233709 you're right about the misunderstanding of electricity. I don't understand how we can see current and voltage as two distinct things, on the one hand they are, but they are also so strongly connected by Ohm's law that I don't see how
– Bidon
yesterday
@user2233709 you're right about the misunderstanding of electricity. I don't understand how we can see current and voltage as two distinct things, on the one hand they are, but they are also so strongly connected by Ohm's law that I don't see how
– Bidon
yesterday
The way Ohm's law is generally taught to beginners ($V = RI$) assumes that $R$ is constant, which is true for ideal resistors. Other components can change resistance (more general term: impedance) depending on pretty much anything: voltage, temperature, frequency, input on another pin, ...
– Geier
yesterday
The way Ohm's law is generally taught to beginners ($V = RI$) assumes that $R$ is constant, which is true for ideal resistors. Other components can change resistance (more general term: impedance) depending on pretty much anything: voltage, temperature, frequency, input on another pin, ...
– Geier
yesterday
add a comment |
4 Answers
4
active
oldest
votes
Just because 2 quantities are related by an equation does not make them the same. For example, F=ma relatesd force and acceleration by a fixed parameter, mass, but they are still distinct concepts. Ohm's law is not a general rule, Kirchoff's laws are more general and do not imply that current and voltage are thje same: apply them to more complicated circuits than a single resistor and you would see this. Ohm's law has little to do with amplifiers. In the case of a voltage follwer, the output voltage is approximately the same as the input voltage but the available current is higher due to the current gain of the transistor. As already noted, the input resistance of the voltage follower is much higher than the output resistance so Ohm's law does not apply since these resistances are different.
answered yesterday
BarryBarry
9,86711415
I think I see where my mistake was, I was not accounting for the difference in impedance.
– Bidon
yesterday
add a comment |
... thus there's a gain in current, how is it possible to have a gain in current, and consequently a gain in power, but not in voltage?
By having a change in impedance.
In very crude terms the base current sees a resistance of βRE whereas the output impedance is RE.
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
No. A transistor isn't a resistor.
answered yesterday
TransistorTransistor
80.9k778174
add a comment |
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
A transistor is an active device. It modulates the energy of an external power source to mimic the input signal. While Ohm's Law can be use to describe the behavior of certain aspects of active devices in some applications, keep in mind that it was derived to describe the relationships at play in a simple resistor. For anything beyond that you have to pay attention to the details.
By its definition, a bipolar transistor delivers current gain only. It is the circuit architecture and components around the transistor that translate this into voltage gain- sometimes.
answered yesterday
AnalogKidAnalogKid
1,57625
add a comment |
Ic =hFE*Ib is then shared by both I(Re) and I(cout).
However for DC control, I(Cout) can never take all of the current of Ic . As Re serves a purpose to DC pull down the emitter and so R load must always be greater than Re for full swing voltages.
The result when overloaded, is the cap current pulls up the emitter so Vbe reduces then Ib reduces and the lower level signal cannot be reached.
Ohm’s Law still works here.
One simply uses Xc for the cap being much smaller so Rload must be greater than Re for linear operation
also Vce must be Vce saturation where hFE drops significantly as a another condition.
the power gain comes from hFE * Rb/Re equivalent loads
answered yesterday
Sunnyskyguy EE75Sunnyskyguy EE75
62.8k22194
add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
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active
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active
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votes
Just because 2 quantities are related by an equation does not make them the same. For example, F=ma relatesd force and acceleration by a fixed parameter, mass, but they are still distinct concepts. Ohm's law is not a general rule, Kirchoff's laws are more general and do not imply that current and voltage are thje same: apply them to more complicated circuits than a single resistor and you would see this. Ohm's law has little to do with amplifiers. In the case of a voltage follwer, the output voltage is approximately the same as the input voltage but the available current is higher due to the current gain of the transistor. As already noted, the input resistance of the voltage follower is much higher than the output resistance so Ohm's law does not apply since these resistances are different.
answered yesterday
BarryBarry
9,86711415
I think I see where my mistake was, I was not accounting for the difference in impedance.
– Bidon
yesterday
add a comment |
Just because 2 quantities are related by an equation does not make them the same. For example, F=ma relatesd force and acceleration by a fixed parameter, mass, but they are still distinct concepts. Ohm's law is not a general rule, Kirchoff's laws are more general and do not imply that current and voltage are thje same: apply them to more complicated circuits than a single resistor and you would see this. Ohm's law has little to do with amplifiers. In the case of a voltage follwer, the output voltage is approximately the same as the input voltage but the available current is higher due to the current gain of the transistor. As already noted, the input resistance of the voltage follower is much higher than the output resistance so Ohm's law does not apply since these resistances are different.
answered yesterday
BarryBarry
9,86711415
I think I see where my mistake was, I was not accounting for the difference in impedance.
– Bidon
yesterday
add a comment |
Just because 2 quantities are related by an equation does not make them the same. For example, F=ma relatesd force and acceleration by a fixed parameter, mass, but they are still distinct concepts. Ohm's law is not a general rule, Kirchoff's laws are more general and do not imply that current and voltage are thje same: apply them to more complicated circuits than a single resistor and you would see this. Ohm's law has little to do with amplifiers. In the case of a voltage follwer, the output voltage is approximately the same as the input voltage but the available current is higher due to the current gain of the transistor. As already noted, the input resistance of the voltage follower is much higher than the output resistance so Ohm's law does not apply since these resistances are different.
answered yesterday
BarryBarry
9,86711415
Just because 2 quantities are related by an equation does not make them the same. For example, F=ma relatesd force and acceleration by a fixed parameter, mass, but they are still distinct concepts. Ohm's law is not a general rule, Kirchoff's laws are more general and do not imply that current and voltage are thje same: apply them to more complicated circuits than a single resistor and you would see this. Ohm's law has little to do with amplifiers. In the case of a voltage follwer, the output voltage is approximately the same as the input voltage but the available current is higher due to the current gain of the transistor. As already noted, the input resistance of the voltage follower is much higher than the output resistance so Ohm's law does not apply since these resistances are different.
answered yesterday
BarryBarry
9,86711415
answered yesterday
BarryBarry
9,86711415
answered yesterday
BarryBarry
9,86711415
answered yesterday
BarryBarry
9,86711415
9,86711415
I think I see where my mistake was, I was not accounting for the difference in impedance.
– Bidon
yesterday
add a comment |
I think I see where my mistake was, I was not accounting for the difference in impedance.
– Bidon
yesterday
I think I see where my mistake was, I was not accounting for the difference in impedance.
– Bidon
yesterday
I think I see where my mistake was, I was not accounting for the difference in impedance.
– Bidon
yesterday
add a comment |
... thus there's a gain in current, how is it possible to have a gain in current, and consequently a gain in power, but not in voltage?
By having a change in impedance.
In very crude terms the base current sees a resistance of βRE whereas the output impedance is RE.
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
No. A transistor isn't a resistor.
answered yesterday
TransistorTransistor
80.9k778174
add a comment |
... thus there's a gain in current, how is it possible to have a gain in current, and consequently a gain in power, but not in voltage?
By having a change in impedance.
In very crude terms the base current sees a resistance of βRE whereas the output impedance is RE.
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
No. A transistor isn't a resistor.
answered yesterday
TransistorTransistor
80.9k778174
add a comment |
... thus there's a gain in current, how is it possible to have a gain in current, and consequently a gain in power, but not in voltage?
By having a change in impedance.
In very crude terms the base current sees a resistance of βRE whereas the output impedance is RE.
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
No. A transistor isn't a resistor.
answered yesterday
TransistorTransistor
80.9k778174
... thus there's a gain in current, how is it possible to have a gain in current, and consequently a gain in power, but not in voltage?
By having a change in impedance.
In very crude terms the base current sees a resistance of βRE whereas the output impedance is RE.
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
No. A transistor isn't a resistor.
answered yesterday
TransistorTransistor
80.9k778174
answered yesterday
TransistorTransistor
80.9k778174
answered yesterday
TransistorTransistor
80.9k778174
answered yesterday
TransistorTransistor
80.9k778174
80.9k778174
add a comment |
add a comment |
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
A transistor is an active device. It modulates the energy of an external power source to mimic the input signal. While Ohm's Law can be use to describe the behavior of certain aspects of active devices in some applications, keep in mind that it was derived to describe the relationships at play in a simple resistor. For anything beyond that you have to pay attention to the details.
By its definition, a bipolar transistor delivers current gain only. It is the circuit architecture and components around the transistor that translate this into voltage gain- sometimes.
answered yesterday
AnalogKidAnalogKid
1,57625
add a comment |
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
A transistor is an active device. It modulates the energy of an external power source to mimic the input signal. While Ohm's Law can be use to describe the behavior of certain aspects of active devices in some applications, keep in mind that it was derived to describe the relationships at play in a simple resistor. For anything beyond that you have to pay attention to the details.
By its definition, a bipolar transistor delivers current gain only. It is the circuit architecture and components around the transistor that translate this into voltage gain- sometimes.
answered yesterday
AnalogKidAnalogKid
1,57625
add a comment |
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
A transistor is an active device. It modulates the energy of an external power source to mimic the input signal. While Ohm's Law can be use to describe the behavior of certain aspects of active devices in some applications, keep in mind that it was derived to describe the relationships at play in a simple resistor. For anything beyond that you have to pay attention to the details.
By its definition, a bipolar transistor delivers current gain only. It is the circuit architecture and components around the transistor that translate this into voltage gain- sometimes.
answered yesterday
AnalogKidAnalogKid
1,57625
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
A transistor is an active device. It modulates the energy of an external power source to mimic the input signal. While Ohm's Law can be use to describe the behavior of certain aspects of active devices in some applications, keep in mind that it was derived to describe the relationships at play in a simple resistor. For anything beyond that you have to pay attention to the details.
By its definition, a bipolar transistor delivers current gain only. It is the circuit architecture and components around the transistor that translate this into voltage gain- sometimes.
answered yesterday
AnalogKidAnalogKid
1,57625
answered yesterday
AnalogKidAnalogKid
1,57625
answered yesterday
AnalogKidAnalogKid
1,57625
answered yesterday
AnalogKidAnalogKid
1,57625
1,57625
add a comment |
add a comment |
Ic =hFE*Ib is then shared by both I(Re) and I(cout).
However for DC control, I(Cout) can never take all of the current of Ic . As Re serves a purpose to DC pull down the emitter and so R load must always be greater than Re for full swing voltages.
The result when overloaded, is the cap current pulls up the emitter so Vbe reduces then Ib reduces and the lower level signal cannot be reached.
Ohm’s Law still works here.
One simply uses Xc for the cap being much smaller so Rload must be greater than Re for linear operation
also Vce must be Vce saturation where hFE drops significantly as a another condition.
the power gain comes from hFE * Rb/Re equivalent loads
answered yesterday
Sunnyskyguy EE75Sunnyskyguy EE75
62.8k22194
add a comment |
Ic =hFE*Ib is then shared by both I(Re) and I(cout).
However for DC control, I(Cout) can never take all of the current of Ic . As Re serves a purpose to DC pull down the emitter and so R load must always be greater than Re for full swing voltages.
The result when overloaded, is the cap current pulls up the emitter so Vbe reduces then Ib reduces and the lower level signal cannot be reached.
Ohm’s Law still works here.
One simply uses Xc for the cap being much smaller so Rload must be greater than Re for linear operation
also Vce must be Vce saturation where hFE drops significantly as a another condition.
the power gain comes from hFE * Rb/Re equivalent loads
answered yesterday
Sunnyskyguy EE75Sunnyskyguy EE75
62.8k22194
add a comment |
Ic =hFE*Ib is then shared by both I(Re) and I(cout).
However for DC control, I(Cout) can never take all of the current of Ic . As Re serves a purpose to DC pull down the emitter and so R load must always be greater than Re for full swing voltages.
The result when overloaded, is the cap current pulls up the emitter so Vbe reduces then Ib reduces and the lower level signal cannot be reached.
Ohm’s Law still works here.
One simply uses Xc for the cap being much smaller so Rload must be greater than Re for linear operation
also Vce must be Vce saturation where hFE drops significantly as a another condition.
the power gain comes from hFE * Rb/Re equivalent loads
answered yesterday
Sunnyskyguy EE75Sunnyskyguy EE75
62.8k22194
Ic =hFE*Ib is then shared by both I(Re) and I(cout).
However for DC control, I(Cout) can never take all of the current of Ic . As Re serves a purpose to DC pull down the emitter and so R load must always be greater than Re for full swing voltages.
The result when overloaded, is the cap current pulls up the emitter so Vbe reduces then Ib reduces and the lower level signal cannot be reached.
Ohm’s Law still works here.
One simply uses Xc for the cap being much smaller so Rload must be greater than Re for linear operation
also Vce must be Vce saturation where hFE drops significantly as a another condition.
the power gain comes from hFE * Rb/Re equivalent loads
answered yesterday
Sunnyskyguy EE75Sunnyskyguy EE75
62.8k22194
edited 19 hours ago
answered yesterday
Sunnyskyguy EE75Sunnyskyguy EE75
62.8k22194
answered yesterday
Sunnyskyguy EE75Sunnyskyguy EE75
62.8k22194
answered yesterday
Sunnyskyguy EE75Sunnyskyguy EE75
62.8k22194
62.8k22194
add a comment |
add a comment |
Bidon is a new contributor. Be nice, and check out our Code of Conduct.
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Hmmm… Looks like you have a strong misunderstanding of electricity. Anyway, if you connect a load resistor between $V_{textrm{out}}$ and ground it will drain some current. The current that is drained from $V_S$ may be significantly lower, hence the current gain.
– user2233709
yesterday
The input impedance and output impedance are very different.
– Dave Tweed♦
yesterday
Moreover, you’d generally want to polarize your transistor’s base with a pair of resistors.
– user2233709
yesterday
1
@user2233709 you're right about the misunderstanding of electricity. I don't understand how we can see current and voltage as two distinct things, on the one hand they are, but they are also so strongly connected by Ohm's law that I don't see how
– Bidon
yesterday
The way Ohm's law is generally taught to beginners ($V = RI$) assumes that $R$ is constant, which is true for ideal resistors. Other components can change resistance (more general term: impedance) depending on pretty much anything: voltage, temperature, frequency, input on another pin, ...
– Geier
yesterday