Negative Resistance Announcing the arrival of Valued Associate #679: Cesar Manara ...
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Negative Resistance
Announcing the arrival of Valued Associate #679: Cesar Manara
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}
$begingroup$
I am a bit confused about the physical meaning of negative resistance.
Mathematically, a component which has negative resistance shows a decreasing voltage across its terminal when the current inside it grows, and vice versa. But how is this physically possible?
Somewhere I have read that an example of component with negative resistance is a voltage source. But I do not understand this statement, since a voltage source is a component which at most shows a (positive) internal resistance.
voltage current resistors resistance voltage-source
edited 46 mins ago
Marcus Müller
35.5k363101
asked 58 mins ago
Kinka-ByoKinka-Byo
662
$endgroup$
add a comment |
$begingroup$
I am a bit confused about the physical meaning of negative resistance.
Mathematically, a component which has negative resistance shows a decreasing voltage across its terminal when the current inside it grows, and vice versa. But how is this physically possible?
Somewhere I have read that an example of component with negative resistance is a voltage source. But I do not understand this statement, since a voltage source is a component which at most shows a (positive) internal resistance.
voltage current resistors resistance voltage-source
edited 46 mins ago
Marcus Müller
35.5k363101
asked 58 mins ago
Kinka-ByoKinka-Byo
662
$endgroup$
$begingroup$
Maybe if you see a circuit with two resistors in series (voltage divider), having in the middle 2.5V, a component with negative resistance can be said to 'add voltage' instead of removing voltage... but I leave a real answer to the experts here ;-)
$endgroup$
– Michel Keijzers
52 mins ago
$begingroup$
Minus R will provide power, not dissipate power.
$endgroup$
– analogsystemsrf
22 mins ago
add a comment |
$begingroup$
I am a bit confused about the physical meaning of negative resistance.
Mathematically, a component which has negative resistance shows a decreasing voltage across its terminal when the current inside it grows, and vice versa. But how is this physically possible?
Somewhere I have read that an example of component with negative resistance is a voltage source. But I do not understand this statement, since a voltage source is a component which at most shows a (positive) internal resistance.
voltage current resistors resistance voltage-source
edited 46 mins ago
Marcus Müller
35.5k363101
asked 58 mins ago
Kinka-ByoKinka-Byo
662
$endgroup$
I am a bit confused about the physical meaning of negative resistance.
Mathematically, a component which has negative resistance shows a decreasing voltage across its terminal when the current inside it grows, and vice versa. But how is this physically possible?
Somewhere I have read that an example of component with negative resistance is a voltage source. But I do not understand this statement, since a voltage source is a component which at most shows a (positive) internal resistance.
voltage current resistors resistance voltage-source
voltage current resistors resistance voltage-source
edited 46 mins ago
Marcus Müller
35.5k363101
asked 58 mins ago
Kinka-ByoKinka-Byo
662
edited 46 mins ago
Marcus Müller
35.5k363101
asked 58 mins ago
Kinka-ByoKinka-Byo
662
edited 46 mins ago
Marcus Müller
35.5k363101
edited 46 mins ago
Marcus Müller
35.5k363101
edited 46 mins ago
Marcus Müller
35.5k363101
35.5k363101
asked 58 mins ago
Kinka-ByoKinka-Byo
662
asked 58 mins ago
Kinka-ByoKinka-Byo
662
asked 58 mins ago
Kinka-ByoKinka-Byo
662
662
$begingroup$
Maybe if you see a circuit with two resistors in series (voltage divider), having in the middle 2.5V, a component with negative resistance can be said to 'add voltage' instead of removing voltage... but I leave a real answer to the experts here ;-)
$endgroup$
– Michel Keijzers
52 mins ago
$begingroup$
Minus R will provide power, not dissipate power.
$endgroup$
– analogsystemsrf
22 mins ago
add a comment |
$begingroup$
Maybe if you see a circuit with two resistors in series (voltage divider), having in the middle 2.5V, a component with negative resistance can be said to 'add voltage' instead of removing voltage... but I leave a real answer to the experts here ;-)
$endgroup$
– Michel Keijzers
52 mins ago
$begingroup$
Minus R will provide power, not dissipate power.
$endgroup$
– analogsystemsrf
22 mins ago
$begingroup$
Maybe if you see a circuit with two resistors in series (voltage divider), having in the middle 2.5V, a component with negative resistance can be said to 'add voltage' instead of removing voltage... but I leave a real answer to the experts here ;-)
$endgroup$
– Michel Keijzers
52 mins ago
$begingroup$
Maybe if you see a circuit with two resistors in series (voltage divider), having in the middle 2.5V, a component with negative resistance can be said to 'add voltage' instead of removing voltage... but I leave a real answer to the experts here ;-)
$endgroup$
– Michel Keijzers
52 mins ago
$begingroup$
Minus R will provide power, not dissipate power.
$endgroup$
– analogsystemsrf
22 mins ago
$begingroup$
Minus R will provide power, not dissipate power.
$endgroup$
– analogsystemsrf
22 mins ago
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
There are a number of mechanisms that result in a region where locally increasing voltage results in locally decreasing current. For example, an Esaki (tunnel) diode.
A common example would be a switching power supply with a steady load. Assuming the efficiency is more-or-less constant, increasing the input voltage results in less current being drawn. It is always consuming energy though.
A stand-alone component that exhibits negative resistance (rather than negative differential resistance) is not possible without some kind of energy source within the component, otherwise it would violate conservation of energy ($P = E^2/R$) and negative P would indicate it is acting as a power source.
If you want to play with a negative resistance effect, one way (assuming you don't mind one end being grounded) is to use a negative impedance converter:
simulate this circuit – Schematic created using CircuitLab
The above circuit acts like a -10K resistor with one end grounded (within its linear range), and works down to about zero volts. Any power it produces comes from the op-amp supplies.
answered 48 mins ago
Spehro PefhanySpehro Pefhany
215k5164438
$endgroup$
1
$begingroup$
That is really a fine choice of an example device you picked.
$endgroup$
– The Photon
40 mins ago
$begingroup$
@ThePhoton LOL, great minds and all that.
$endgroup$
– Spehro Pefhany
19 mins ago
add a comment |
$begingroup$
Anything that drops in voltage with a rise in current has a negative resistance.
Power sources have this property. The passive components with incremental negative resistance include; any gas discharge bulb or arc, Avalanche effect diodes, Tunnel Diodes, SCR's during trigger phase.
https://en.wikipedia.org/wiki/Negative_resistance
answered 52 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
72.2k227103
$endgroup$
add a comment |
$begingroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
answered 24 mins ago
LvWLvW
14.9k21330
$endgroup$
add a comment |
$begingroup$
A perfect negative resistor is impossible, but a device can have negative resistance characteristics over a limited range.
The resistance of a non-linear device varies and at a given voltage the equivalent resistance is equal to the slope of the line. If the slope is negative in a range, that range has negative resistance.
answered 44 mins ago
Mattman944Mattman944
3015
$endgroup$
add a comment |
$begingroup$
But how is this physically possible?
Some components, like Esaki diodes and glow tubes, have an I-V curve that is entirely in the I and III quadrants, but has a negative slope region over a limited range. In this region, a small-signal model of the device will have negative resistance.
(image source)
In the Esaki diode, this behavior is caused by tunneling current that is possible at low bias but not at higher bias voltage.
It's also possible to make an op-amp circuit with negative input resistance over a limited range. There the I-V curve can even pass through the II and IV quadrants since power can be supplied from the op-amp's power terminals.
Somewhere I have read that an example of component with negative resistance is a voltage source.
Looking at the input side of a regulated switching supply with a fixed load, it will often appear as a negative resistance.
This is because it is a constant power load. If the input voltage drops, the regulator circuit will increase the current drawn in order to continue supplying the load with the desired output voltage.
answered 41 mins ago
The PhotonThe Photon
87.9k399205
$endgroup$
add a comment |
$begingroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
simulate this circuit – Schematic created using CircuitLab
Comments: The shown NIC is stable as long as the source resistance of the voltage source (not shown in the figure) is smaller than R1. These NIC blocks are use for undamping filters, oscillators and other systems with unwanted positive (parasitic) resistances. Mathematically, they can be treated as "normal" resistors in series and parallel combinations - however, with a negative sign, of course.
A very popular application is the "NIC integrator" (or "Deboo integrator"), where an NIC block is connected to the common node of a simple R-C lowpass. In this case, the NIC can compensate the pos. resistor R - thus resembling a current source which loads the intergating capacitor.
answered 14 mins ago
LvWLvW
14.9k21330
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are a number of mechanisms that result in a region where locally increasing voltage results in locally decreasing current. For example, an Esaki (tunnel) diode.
A common example would be a switching power supply with a steady load. Assuming the efficiency is more-or-less constant, increasing the input voltage results in less current being drawn. It is always consuming energy though.
A stand-alone component that exhibits negative resistance (rather than negative differential resistance) is not possible without some kind of energy source within the component, otherwise it would violate conservation of energy ($P = E^2/R$) and negative P would indicate it is acting as a power source.
If you want to play with a negative resistance effect, one way (assuming you don't mind one end being grounded) is to use a negative impedance converter:
simulate this circuit – Schematic created using CircuitLab
The above circuit acts like a -10K resistor with one end grounded (within its linear range), and works down to about zero volts. Any power it produces comes from the op-amp supplies.
answered 48 mins ago
Spehro PefhanySpehro Pefhany
215k5164438
$endgroup$
1
$begingroup$
That is really a fine choice of an example device you picked.
$endgroup$
– The Photon
40 mins ago
$begingroup$
@ThePhoton LOL, great minds and all that.
$endgroup$
– Spehro Pefhany
19 mins ago
add a comment |
$begingroup$
There are a number of mechanisms that result in a region where locally increasing voltage results in locally decreasing current. For example, an Esaki (tunnel) diode.
A common example would be a switching power supply with a steady load. Assuming the efficiency is more-or-less constant, increasing the input voltage results in less current being drawn. It is always consuming energy though.
A stand-alone component that exhibits negative resistance (rather than negative differential resistance) is not possible without some kind of energy source within the component, otherwise it would violate conservation of energy ($P = E^2/R$) and negative P would indicate it is acting as a power source.
If you want to play with a negative resistance effect, one way (assuming you don't mind one end being grounded) is to use a negative impedance converter:
simulate this circuit – Schematic created using CircuitLab
The above circuit acts like a -10K resistor with one end grounded (within its linear range), and works down to about zero volts. Any power it produces comes from the op-amp supplies.
answered 48 mins ago
Spehro PefhanySpehro Pefhany
215k5164438
$endgroup$
1
$begingroup$
That is really a fine choice of an example device you picked.
$endgroup$
– The Photon
40 mins ago
$begingroup$
@ThePhoton LOL, great minds and all that.
$endgroup$
– Spehro Pefhany
19 mins ago
add a comment |
$begingroup$
There are a number of mechanisms that result in a region where locally increasing voltage results in locally decreasing current. For example, an Esaki (tunnel) diode.
A common example would be a switching power supply with a steady load. Assuming the efficiency is more-or-less constant, increasing the input voltage results in less current being drawn. It is always consuming energy though.
A stand-alone component that exhibits negative resistance (rather than negative differential resistance) is not possible without some kind of energy source within the component, otherwise it would violate conservation of energy ($P = E^2/R$) and negative P would indicate it is acting as a power source.
If you want to play with a negative resistance effect, one way (assuming you don't mind one end being grounded) is to use a negative impedance converter:
simulate this circuit – Schematic created using CircuitLab
The above circuit acts like a -10K resistor with one end grounded (within its linear range), and works down to about zero volts. Any power it produces comes from the op-amp supplies.
answered 48 mins ago
Spehro PefhanySpehro Pefhany
215k5164438
$endgroup$
There are a number of mechanisms that result in a region where locally increasing voltage results in locally decreasing current. For example, an Esaki (tunnel) diode.
A common example would be a switching power supply with a steady load. Assuming the efficiency is more-or-less constant, increasing the input voltage results in less current being drawn. It is always consuming energy though.
A stand-alone component that exhibits negative resistance (rather than negative differential resistance) is not possible without some kind of energy source within the component, otherwise it would violate conservation of energy ($P = E^2/R$) and negative P would indicate it is acting as a power source.
If you want to play with a negative resistance effect, one way (assuming you don't mind one end being grounded) is to use a negative impedance converter:
simulate this circuit – Schematic created using CircuitLab
The above circuit acts like a -10K resistor with one end grounded (within its linear range), and works down to about zero volts. Any power it produces comes from the op-amp supplies.
answered 48 mins ago
Spehro PefhanySpehro Pefhany
215k5164438
edited 14 mins ago
answered 48 mins ago
Spehro PefhanySpehro Pefhany
215k5164438
answered 48 mins ago
Spehro PefhanySpehro Pefhany
215k5164438
answered 48 mins ago
Spehro PefhanySpehro Pefhany
215k5164438
215k5164438
1
$begingroup$
That is really a fine choice of an example device you picked.
$endgroup$
– The Photon
40 mins ago
$begingroup$
@ThePhoton LOL, great minds and all that.
$endgroup$
– Spehro Pefhany
19 mins ago
add a comment |
1
$begingroup$
That is really a fine choice of an example device you picked.
$endgroup$
– The Photon
40 mins ago
$begingroup$
@ThePhoton LOL, great minds and all that.
$endgroup$
– Spehro Pefhany
19 mins ago
1
1
$begingroup$
That is really a fine choice of an example device you picked.
$endgroup$
– The Photon
40 mins ago
$begingroup$
That is really a fine choice of an example device you picked.
$endgroup$
– The Photon
40 mins ago
$begingroup$
@ThePhoton LOL, great minds and all that.
$endgroup$
– Spehro Pefhany
19 mins ago
$begingroup$
@ThePhoton LOL, great minds and all that.
$endgroup$
– Spehro Pefhany
19 mins ago
add a comment |
$begingroup$
Anything that drops in voltage with a rise in current has a negative resistance.
Power sources have this property. The passive components with incremental negative resistance include; any gas discharge bulb or arc, Avalanche effect diodes, Tunnel Diodes, SCR's during trigger phase.
https://en.wikipedia.org/wiki/Negative_resistance
answered 52 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
72.2k227103
$endgroup$
add a comment |
$begingroup$
Anything that drops in voltage with a rise in current has a negative resistance.
Power sources have this property. The passive components with incremental negative resistance include; any gas discharge bulb or arc, Avalanche effect diodes, Tunnel Diodes, SCR's during trigger phase.
https://en.wikipedia.org/wiki/Negative_resistance
answered 52 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
72.2k227103
$endgroup$
add a comment |
$begingroup$
Anything that drops in voltage with a rise in current has a negative resistance.
Power sources have this property. The passive components with incremental negative resistance include; any gas discharge bulb or arc, Avalanche effect diodes, Tunnel Diodes, SCR's during trigger phase.
https://en.wikipedia.org/wiki/Negative_resistance
answered 52 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
72.2k227103
$endgroup$
Anything that drops in voltage with a rise in current has a negative resistance.
Power sources have this property. The passive components with incremental negative resistance include; any gas discharge bulb or arc, Avalanche effect diodes, Tunnel Diodes, SCR's during trigger phase.
https://en.wikipedia.org/wiki/Negative_resistance
answered 52 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
72.2k227103
answered 52 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
72.2k227103
answered 52 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
72.2k227103
answered 52 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
72.2k227103
72.2k227103
add a comment |
add a comment |
$begingroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
answered 24 mins ago
LvWLvW
14.9k21330
$endgroup$
add a comment |
$begingroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
answered 24 mins ago
LvWLvW
14.9k21330
$endgroup$
add a comment |
$begingroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
answered 24 mins ago
LvWLvW
14.9k21330
$endgroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
answered 24 mins ago
LvWLvW
14.9k21330
answered 24 mins ago
LvWLvW
14.9k21330
answered 24 mins ago
LvWLvW
14.9k21330
answered 24 mins ago
LvWLvW
14.9k21330
14.9k21330
add a comment |
add a comment |
$begingroup$
A perfect negative resistor is impossible, but a device can have negative resistance characteristics over a limited range.
The resistance of a non-linear device varies and at a given voltage the equivalent resistance is equal to the slope of the line. If the slope is negative in a range, that range has negative resistance.
answered 44 mins ago
Mattman944Mattman944
3015
$endgroup$
add a comment |
$begingroup$
A perfect negative resistor is impossible, but a device can have negative resistance characteristics over a limited range.
The resistance of a non-linear device varies and at a given voltage the equivalent resistance is equal to the slope of the line. If the slope is negative in a range, that range has negative resistance.
answered 44 mins ago
Mattman944Mattman944
3015
$endgroup$
add a comment |
$begingroup$
A perfect negative resistor is impossible, but a device can have negative resistance characteristics over a limited range.
The resistance of a non-linear device varies and at a given voltage the equivalent resistance is equal to the slope of the line. If the slope is negative in a range, that range has negative resistance.
answered 44 mins ago
Mattman944Mattman944
3015
$endgroup$
A perfect negative resistor is impossible, but a device can have negative resistance characteristics over a limited range.
The resistance of a non-linear device varies and at a given voltage the equivalent resistance is equal to the slope of the line. If the slope is negative in a range, that range has negative resistance.
answered 44 mins ago
Mattman944Mattman944
3015
answered 44 mins ago
Mattman944Mattman944
3015
answered 44 mins ago
Mattman944Mattman944
3015
answered 44 mins ago
Mattman944Mattman944
3015
3015
add a comment |
add a comment |
$begingroup$
But how is this physically possible?
Some components, like Esaki diodes and glow tubes, have an I-V curve that is entirely in the I and III quadrants, but has a negative slope region over a limited range. In this region, a small-signal model of the device will have negative resistance.
(image source)
In the Esaki diode, this behavior is caused by tunneling current that is possible at low bias but not at higher bias voltage.
It's also possible to make an op-amp circuit with negative input resistance over a limited range. There the I-V curve can even pass through the II and IV quadrants since power can be supplied from the op-amp's power terminals.
Somewhere I have read that an example of component with negative resistance is a voltage source.
Looking at the input side of a regulated switching supply with a fixed load, it will often appear as a negative resistance.
This is because it is a constant power load. If the input voltage drops, the regulator circuit will increase the current drawn in order to continue supplying the load with the desired output voltage.
answered 41 mins ago
The PhotonThe Photon
87.9k399205
$endgroup$
add a comment |
$begingroup$
But how is this physically possible?
Some components, like Esaki diodes and glow tubes, have an I-V curve that is entirely in the I and III quadrants, but has a negative slope region over a limited range. In this region, a small-signal model of the device will have negative resistance.
(image source)
In the Esaki diode, this behavior is caused by tunneling current that is possible at low bias but not at higher bias voltage.
It's also possible to make an op-amp circuit with negative input resistance over a limited range. There the I-V curve can even pass through the II and IV quadrants since power can be supplied from the op-amp's power terminals.
Somewhere I have read that an example of component with negative resistance is a voltage source.
Looking at the input side of a regulated switching supply with a fixed load, it will often appear as a negative resistance.
This is because it is a constant power load. If the input voltage drops, the regulator circuit will increase the current drawn in order to continue supplying the load with the desired output voltage.
answered 41 mins ago
The PhotonThe Photon
87.9k399205
$endgroup$
add a comment |
$begingroup$
But how is this physically possible?
Some components, like Esaki diodes and glow tubes, have an I-V curve that is entirely in the I and III quadrants, but has a negative slope region over a limited range. In this region, a small-signal model of the device will have negative resistance.
(image source)
In the Esaki diode, this behavior is caused by tunneling current that is possible at low bias but not at higher bias voltage.
It's also possible to make an op-amp circuit with negative input resistance over a limited range. There the I-V curve can even pass through the II and IV quadrants since power can be supplied from the op-amp's power terminals.
Somewhere I have read that an example of component with negative resistance is a voltage source.
Looking at the input side of a regulated switching supply with a fixed load, it will often appear as a negative resistance.
This is because it is a constant power load. If the input voltage drops, the regulator circuit will increase the current drawn in order to continue supplying the load with the desired output voltage.
answered 41 mins ago
The PhotonThe Photon
87.9k399205
$endgroup$
But how is this physically possible?
Some components, like Esaki diodes and glow tubes, have an I-V curve that is entirely in the I and III quadrants, but has a negative slope region over a limited range. In this region, a small-signal model of the device will have negative resistance.
(image source)
In the Esaki diode, this behavior is caused by tunneling current that is possible at low bias but not at higher bias voltage.
It's also possible to make an op-amp circuit with negative input resistance over a limited range. There the I-V curve can even pass through the II and IV quadrants since power can be supplied from the op-amp's power terminals.
Somewhere I have read that an example of component with negative resistance is a voltage source.
Looking at the input side of a regulated switching supply with a fixed load, it will often appear as a negative resistance.
This is because it is a constant power load. If the input voltage drops, the regulator circuit will increase the current drawn in order to continue supplying the load with the desired output voltage.
answered 41 mins ago
The PhotonThe Photon
87.9k399205
answered 41 mins ago
The PhotonThe Photon
87.9k399205
answered 41 mins ago
The PhotonThe Photon
87.9k399205
answered 41 mins ago
The PhotonThe Photon
87.9k399205
87.9k399205
add a comment |
add a comment |
$begingroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
simulate this circuit – Schematic created using CircuitLab
Comments: The shown NIC is stable as long as the source resistance of the voltage source (not shown in the figure) is smaller than R1. These NIC blocks are use for undamping filters, oscillators and other systems with unwanted positive (parasitic) resistances. Mathematically, they can be treated as "normal" resistors in series and parallel combinations - however, with a negative sign, of course.
A very popular application is the "NIC integrator" (or "Deboo integrator"), where an NIC block is connected to the common node of a simple R-C lowpass. In this case, the NIC can compensate the pos. resistor R - thus resembling a current source which loads the intergating capacitor.
answered 14 mins ago
LvWLvW
14.9k21330
$endgroup$
add a comment |
$begingroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
simulate this circuit – Schematic created using CircuitLab
Comments: The shown NIC is stable as long as the source resistance of the voltage source (not shown in the figure) is smaller than R1. These NIC blocks are use for undamping filters, oscillators and other systems with unwanted positive (parasitic) resistances. Mathematically, they can be treated as "normal" resistors in series and parallel combinations - however, with a negative sign, of course.
A very popular application is the "NIC integrator" (or "Deboo integrator"), where an NIC block is connected to the common node of a simple R-C lowpass. In this case, the NIC can compensate the pos. resistor R - thus resembling a current source which loads the intergating capacitor.
answered 14 mins ago
LvWLvW
14.9k21330
$endgroup$
add a comment |
$begingroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
simulate this circuit – Schematic created using CircuitLab
Comments: The shown NIC is stable as long as the source resistance of the voltage source (not shown in the figure) is smaller than R1. These NIC blocks are use for undamping filters, oscillators and other systems with unwanted positive (parasitic) resistances. Mathematically, they can be treated as "normal" resistors in series and parallel combinations - however, with a negative sign, of course.
A very popular application is the "NIC integrator" (or "Deboo integrator"), where an NIC block is connected to the common node of a simple R-C lowpass. In this case, the NIC can compensate the pos. resistor R - thus resembling a current source which loads the intergating capacitor.
answered 14 mins ago
LvWLvW
14.9k21330
$endgroup$
In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance. My following answer concerns only the static negative resistor:
Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.
Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).
simulate this circuit – Schematic created using CircuitLab
Comments: The shown NIC is stable as long as the source resistance of the voltage source (not shown in the figure) is smaller than R1. These NIC blocks are use for undamping filters, oscillators and other systems with unwanted positive (parasitic) resistances. Mathematically, they can be treated as "normal" resistors in series and parallel combinations - however, with a negative sign, of course.
A very popular application is the "NIC integrator" (or "Deboo integrator"), where an NIC block is connected to the common node of a simple R-C lowpass. In this case, the NIC can compensate the pos. resistor R - thus resembling a current source which loads the intergating capacitor.
answered 14 mins ago
LvWLvW
14.9k21330
edited 1 min ago
answered 14 mins ago
LvWLvW
14.9k21330
answered 14 mins ago
LvWLvW
14.9k21330
answered 14 mins ago
LvWLvW
14.9k21330
14.9k21330
add a comment |
add a comment |
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$begingroup$
Maybe if you see a circuit with two resistors in series (voltage divider), having in the middle 2.5V, a component with negative resistance can be said to 'add voltage' instead of removing voltage... but I leave a real answer to the experts here ;-)
$endgroup$
– Michel Keijzers
52 mins ago
$begingroup$
Minus R will provide power, not dissipate power.
$endgroup$
– analogsystemsrf
22 mins ago