Will linear voltage regulator step up current?
Clash Royale CLAN TAG#URR8PPP
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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
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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
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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
Feb 20 at 20:32
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
$endgroup$
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 Feb 20 at 20:19
Jayant PahujaJayant Pahuja
743
743
$begingroup$
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
Feb 20 at 20:32
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
Feb 20 at 20:32
$begingroup$
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
Feb 20 at 20:32
$begingroup$
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
Feb 20 at 20:32
add a comment |
3 Answers
3
active
oldest
votes
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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:
$I_in = I_out$
$endgroup$
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.
$endgroup$
3
$begingroup$
"drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
$endgroup$
– quetzalcoatl
Feb 21 at 10:16
add a comment |
$begingroup$
since power = voltage * current.
It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices 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.
$endgroup$
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:
$I_in = I_out$
$endgroup$
add a comment |
$begingroup$
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:
$I_in = I_out$
$endgroup$
add a comment |
$begingroup$
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:
$I_in = I_out$
$endgroup$
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:
$I_in = I_out$
edited Feb 20 at 22:30
answered Feb 20 at 20:22
ToorToor
1,14319
1,14319
add a comment |
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.
$endgroup$
3
$begingroup$
"drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
$endgroup$
– quetzalcoatl
Feb 21 at 10:16
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.
$endgroup$
3
$begingroup$
"drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
$endgroup$
– quetzalcoatl
Feb 21 at 10:16
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.
$endgroup$
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 Feb 20 at 22:07
laptop2dlaptop2d
26.5k123383
26.5k123383
3
$begingroup$
"drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
$endgroup$
– quetzalcoatl
Feb 21 at 10:16
add a comment |
3
$begingroup$
"drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
$endgroup$
– quetzalcoatl
Feb 21 at 10:16
3
3
$begingroup$
"drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
$endgroup$
– quetzalcoatl
Feb 21 at 10:16
$begingroup$
"drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
$endgroup$
– quetzalcoatl
Feb 21 at 10:16
add a comment |
$begingroup$
since power = voltage * current.
It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices 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.
$endgroup$
add a comment |
$begingroup$
since power = voltage * current.
It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices 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.
$endgroup$
add a comment |
$begingroup$
since power = voltage * current.
It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices 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.
$endgroup$
since power = voltage * current.
It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices 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.
edited Feb 22 at 1:24
Volker Siegel
5661417
5661417
answered Feb 21 at 1:02
Ale..chenskiAle..chenski
28.4k11866
28.4k11866
add a comment |
add a comment |
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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
Feb 20 at 20:32