Will linear voltage regulator step up current?
$begingroup$
I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.
power-supply linear-regulator
asked 8 hours ago
Jayant PahujaJayant Pahuja
91
New contributor
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$endgroup$
add a comment |
$begingroup$
I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.
power-supply linear-regulator
asked 8 hours ago
Jayant PahujaJayant Pahuja
91
New contributor
Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
$endgroup$
– Sunnyskyguy EE75
8 hours ago
add a comment |
$begingroup$
I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.
power-supply linear-regulator
asked 8 hours ago
Jayant PahujaJayant Pahuja
91
New contributor
Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.
power-supply linear-regulator
power-supply linear-regulator
asked 8 hours ago
Jayant PahujaJayant Pahuja
91
New contributor
Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Jayant PahujaJayant Pahuja
91
New contributor
Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Jayant PahujaJayant Pahuja
91
New contributor
Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Jayant PahujaJayant Pahuja
91
asked 8 hours ago
Jayant PahujaJayant Pahuja
91
91
New contributor
Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
$endgroup$
– Sunnyskyguy EE75
8 hours ago
add a comment |
$begingroup$
With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
$endgroup$
– Sunnyskyguy EE75
8 hours ago
$begingroup$
With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
$endgroup$
– Sunnyskyguy EE75
8 hours ago
$begingroup$
With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
$endgroup$
– Sunnyskyguy EE75
8 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.
$P_{in} = P_{out}$
but for a linear regulator it looks like this:
$V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$
The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:
$V_{in} times I_{in} = V_{in} times I_{out}$
Therefore:
$I_{in} = I_{out}$
answered 8 hours ago
ToorToor
4218
New contributor
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I fixed your minor grammar errors. Takes time to learn details of English. +1 for solid answer.
$endgroup$
– Sparky256
7 hours ago
$begingroup$
Thanks. I changed the sentence structure halfway through but missed changing "burn".
$endgroup$
– Toor
7 hours ago
$begingroup$
@Toor * means convolution and not a multiplication.
$endgroup$
– Jan
7 hours ago
$begingroup$
Right. How lemme try and figure out how to get a multiplication sign in there.
$endgroup$
– Toor
7 hours ago
$begingroup$
Changed. It's rather inelegant to need to go "times"
$endgroup$
– Toor
7 hours ago
|
show 5 more comments
$begingroup$
No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.
However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.
A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)
A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)
They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.
answered 6 hours ago
laptop2dlaptop2d
25.3k123278
$endgroup$
add a comment |
$begingroup$
since power = voltage * current.
It is generally true, but only for devices that transform electricity (as AC transformers or more sophisticated transformers known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.
In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.
answered 3 hours ago
Ale..chenskiAle..chenski
27.5k11865
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.
$P_{in} = P_{out}$
but for a linear regulator it looks like this:
$V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$
The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:
$V_{in} times I_{in} = V_{in} times I_{out}$
Therefore:
$I_{in} = I_{out}$
answered 8 hours ago
ToorToor
4218
New contributor
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I fixed your minor grammar errors. Takes time to learn details of English. +1 for solid answer.
$endgroup$
– Sparky256
7 hours ago
$begingroup$
Thanks. I changed the sentence structure halfway through but missed changing "burn".
$endgroup$
– Toor
7 hours ago
$begingroup$
@Toor * means convolution and not a multiplication.
$endgroup$
– Jan
7 hours ago
$begingroup$
Right. How lemme try and figure out how to get a multiplication sign in there.
$endgroup$
– Toor
7 hours ago
$begingroup$
Changed. It's rather inelegant to need to go "times"
$endgroup$
– Toor
7 hours ago
|
show 5 more comments
$begingroup$
No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.
$P_{in} = P_{out}$
but for a linear regulator it looks like this:
$V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$
The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:
$V_{in} times I_{in} = V_{in} times I_{out}$
Therefore:
$I_{in} = I_{out}$
answered 8 hours ago
ToorToor
4218
New contributor
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I fixed your minor grammar errors. Takes time to learn details of English. +1 for solid answer.
$endgroup$
– Sparky256
7 hours ago
$begingroup$
Thanks. I changed the sentence structure halfway through but missed changing "burn".
$endgroup$
– Toor
7 hours ago
$begingroup$
@Toor * means convolution and not a multiplication.
$endgroup$
– Jan
7 hours ago
$begingroup$
Right. How lemme try and figure out how to get a multiplication sign in there.
$endgroup$
– Toor
7 hours ago
$begingroup$
Changed. It's rather inelegant to need to go "times"
$endgroup$
– Toor
7 hours ago
|
show 5 more comments
$begingroup$
No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.
$P_{in} = P_{out}$
but for a linear regulator it looks like this:
$V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$
The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:
$V_{in} times I_{in} = V_{in} times I_{out}$
Therefore:
$I_{in} = I_{out}$
answered 8 hours ago
ToorToor
4218
New contributor
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.
$P_{in} = P_{out}$
but for a linear regulator it looks like this:
$V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$
The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:
$V_{in} times I_{in} = V_{in} times I_{out}$
Therefore:
$I_{in} = I_{out}$
answered 8 hours ago
ToorToor
4218
New contributor
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
answered 8 hours ago
ToorToor
4218
New contributor
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
ToorToor
4218
answered 8 hours ago
ToorToor
4218
4218
New contributor
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I fixed your minor grammar errors. Takes time to learn details of English. +1 for solid answer.
$endgroup$
– Sparky256
7 hours ago
$begingroup$
Thanks. I changed the sentence structure halfway through but missed changing "burn".
$endgroup$
– Toor
7 hours ago
$begingroup$
@Toor * means convolution and not a multiplication.
$endgroup$
– Jan
7 hours ago
$begingroup$
Right. How lemme try and figure out how to get a multiplication sign in there.
$endgroup$
– Toor
7 hours ago
$begingroup$
Changed. It's rather inelegant to need to go "times"
$endgroup$
– Toor
7 hours ago
|
show 5 more comments
$begingroup$
I fixed your minor grammar errors. Takes time to learn details of English. +1 for solid answer.
$endgroup$
– Sparky256
7 hours ago
$begingroup$
Thanks. I changed the sentence structure halfway through but missed changing "burn".
$endgroup$
– Toor
7 hours ago
$begingroup$
@Toor * means convolution and not a multiplication.
$endgroup$
– Jan
7 hours ago
$begingroup$
Right. How lemme try and figure out how to get a multiplication sign in there.
$endgroup$
– Toor
7 hours ago
$begingroup$
Changed. It's rather inelegant to need to go "times"
$endgroup$
– Toor
7 hours ago
$begingroup$
I fixed your minor grammar errors. Takes time to learn details of English. +1 for solid answer.
$endgroup$
– Sparky256
7 hours ago
$begingroup$
I fixed your minor grammar errors. Takes time to learn details of English. +1 for solid answer.
$endgroup$
– Sparky256
7 hours ago
$begingroup$
Thanks. I changed the sentence structure halfway through but missed changing "burn".
$endgroup$
– Toor
7 hours ago
$begingroup$
Thanks. I changed the sentence structure halfway through but missed changing "burn".
$endgroup$
– Toor
7 hours ago
$begingroup$
@Toor * means convolution and not a multiplication.
$endgroup$
– Jan
7 hours ago
$begingroup$
@Toor * means convolution and not a multiplication.
$endgroup$
– Jan
7 hours ago
$begingroup$
Right. How lemme try and figure out how to get a multiplication sign in there.
$endgroup$
– Toor
7 hours ago
$begingroup$
Right. How lemme try and figure out how to get a multiplication sign in there.
$endgroup$
– Toor
7 hours ago
$begingroup$
Changed. It's rather inelegant to need to go "times"
$endgroup$
– Toor
7 hours ago
$begingroup$
Changed. It's rather inelegant to need to go "times"
$endgroup$
– Toor
7 hours ago
|
show 5 more comments
$begingroup$
No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.
However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.
A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)
A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)
They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.
answered 6 hours ago
laptop2dlaptop2d
25.3k123278
$endgroup$
add a comment |
$begingroup$
No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.
However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.
A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)
A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)
They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.
answered 6 hours ago
laptop2dlaptop2d
25.3k123278
$endgroup$
add a comment |
$begingroup$
No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.
However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.
A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)
A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)
They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.
answered 6 hours ago
laptop2dlaptop2d
25.3k123278
$endgroup$
No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.
However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.
A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)
A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)
They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.
answered 6 hours ago
laptop2dlaptop2d
25.3k123278
answered 6 hours ago
laptop2dlaptop2d
25.3k123278
answered 6 hours ago
laptop2dlaptop2d
25.3k123278
answered 6 hours ago
laptop2dlaptop2d
25.3k123278
25.3k123278
add a comment |
add a comment |
$begingroup$
since power = voltage * current.
It is generally true, but only for devices that transform electricity (as AC transformers or more sophisticated transformers known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.
In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.
answered 3 hours ago
Ale..chenskiAle..chenski
27.5k11865
$endgroup$
add a comment |
$begingroup$
since power = voltage * current.
It is generally true, but only for devices that transform electricity (as AC transformers or more sophisticated transformers known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.
In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.
answered 3 hours ago
Ale..chenskiAle..chenski
27.5k11865
$endgroup$
add a comment |
$begingroup$
since power = voltage * current.
It is generally true, but only for devices that transform electricity (as AC transformers or more sophisticated transformers known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.
In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.
answered 3 hours ago
Ale..chenskiAle..chenski
27.5k11865
$endgroup$
since power = voltage * current.
It is generally true, but only for devices that transform electricity (as AC transformers or more sophisticated transformers known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.
In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.
answered 3 hours ago
Ale..chenskiAle..chenski
27.5k11865
answered 3 hours ago
Ale..chenskiAle..chenski
27.5k11865
answered 3 hours ago
Ale..chenskiAle..chenski
27.5k11865
answered 3 hours ago
Ale..chenskiAle..chenski
27.5k11865
27.5k11865
add a comment |
add a comment |
Jayant Pahuja is a new contributor. Be nice, and check out our Code of Conduct.
Jayant Pahuja is a new contributor. Be nice, and check out our Code of Conduct.
Jayant Pahuja is a new contributor. Be nice, and check out our Code of Conduct.
Jayant Pahuja is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
$endgroup$
– Sunnyskyguy EE75
8 hours ago