Must I reduce USB's 1A output to 450mA?
-2
I have a bluetooth headset that charges at direct current 5 Volt and 450 Milliampere. I charge it with its original AC-DC-converter and I want to be able to charge it over USB (e.g. from a portable pre-charged battery/charger or a solar USB charger).
I could easily cut the cable and solder the one end to a charging USB cable, but with this I would almost always give 5 Volt and 1 Ampere since that is the standard, right?
- Are there modified USB "cables" with extra electrical components on the market that use 500mA from 1A?
- Is there a DIY way to take out those 650mA or 500mA?
usb charging ac-adapter
asked Oct 18 '16 at 2:09
scjorgescjorge
19013
|
show 9 more comments
-2
I have a bluetooth headset that charges at direct current 5 Volt and 450 Milliampere. I charge it with its original AC-DC-converter and I want to be able to charge it over USB (e.g. from a portable pre-charged battery/charger or a solar USB charger).
I could easily cut the cable and solder the one end to a charging USB cable, but with this I would almost always give 5 Volt and 1 Ampere since that is the standard, right?
- Are there modified USB "cables" with extra electrical components on the market that use 500mA from 1A?
- Is there a DIY way to take out those 650mA or 500mA?
usb charging ac-adapter
asked Oct 18 '16 at 2:09
scjorgescjorge
19013
6
The device will only draw 500ma
– Ramhound
Oct 18 '16 at 2:14
1
This logic must be based on Grassman Algebra or something...
– Ale..chenski
Oct 18 '16 at 3:07
1
@DanielB, The simple Ohm or Kirchhoff formulas are applicable to circuit components with constant parameters. The AD-DC adapter is not a linear voltage source, and it will either abruptly reduce the voltage output when connected load exceeds its rated DC current capability (either along some artificially defined cut-off function, or just turn off). The device is also a non-linear load, starting with constant current, which will be reduced when the internal battery enters a constant voltage stage of charging. So these simple formulas do not apply.
– Ale..chenski
Oct 18 '16 at 6:55
1
@DanielB, I believe that the details are necessary because otherwise some curious people get confused with the discrepancy between their expectations and measured results.
– Ale..chenski
Oct 18 '16 at 7:20
1
@fixer1234, certainly the load is defined by Ohm's Law, if you mean that to consume 450mA from 5V it must have about 11 Ohms of effective resistance. However, there is a bunch of active elements that make this "impression", and this "resistor" will change in time as charging mode changes. So, given an unknown state of battery in the device when you plug it in first, you never know what its resistance is. How does the Ohm Law help here? So I am unsure what the objection is.
– Ale..chenski
Oct 18 '16 at 22:07
|
show 9 more comments
-2
-2
-2
I have a bluetooth headset that charges at direct current 5 Volt and 450 Milliampere. I charge it with its original AC-DC-converter and I want to be able to charge it over USB (e.g. from a portable pre-charged battery/charger or a solar USB charger).
I could easily cut the cable and solder the one end to a charging USB cable, but with this I would almost always give 5 Volt and 1 Ampere since that is the standard, right?
- Are there modified USB "cables" with extra electrical components on the market that use 500mA from 1A?
- Is there a DIY way to take out those 650mA or 500mA?
usb charging ac-adapter
asked Oct 18 '16 at 2:09
scjorgescjorge
19013
I have a bluetooth headset that charges at direct current 5 Volt and 450 Milliampere. I charge it with its original AC-DC-converter and I want to be able to charge it over USB (e.g. from a portable pre-charged battery/charger or a solar USB charger).
I could easily cut the cable and solder the one end to a charging USB cable, but with this I would almost always give 5 Volt and 1 Ampere since that is the standard, right?
- Are there modified USB "cables" with extra electrical components on the market that use 500mA from 1A?
- Is there a DIY way to take out those 650mA or 500mA?
usb charging ac-adapter
usb charging ac-adapter
asked Oct 18 '16 at 2:09
scjorgescjorge
19013
asked Oct 18 '16 at 2:09
scjorgescjorge
19013
edited Jan 7 at 15:07
scjorge
asked Oct 18 '16 at 2:09
scjorgescjorge
19013
asked Oct 18 '16 at 2:09
scjorgescjorge
19013
asked Oct 18 '16 at 2:09
scjorgescjorge
19013
19013
6
The device will only draw 500ma
– Ramhound
Oct 18 '16 at 2:14
1
This logic must be based on Grassman Algebra or something...
– Ale..chenski
Oct 18 '16 at 3:07
1
@DanielB, The simple Ohm or Kirchhoff formulas are applicable to circuit components with constant parameters. The AD-DC adapter is not a linear voltage source, and it will either abruptly reduce the voltage output when connected load exceeds its rated DC current capability (either along some artificially defined cut-off function, or just turn off). The device is also a non-linear load, starting with constant current, which will be reduced when the internal battery enters a constant voltage stage of charging. So these simple formulas do not apply.
– Ale..chenski
Oct 18 '16 at 6:55
1
@DanielB, I believe that the details are necessary because otherwise some curious people get confused with the discrepancy between their expectations and measured results.
– Ale..chenski
Oct 18 '16 at 7:20
1
@fixer1234, certainly the load is defined by Ohm's Law, if you mean that to consume 450mA from 5V it must have about 11 Ohms of effective resistance. However, there is a bunch of active elements that make this "impression", and this "resistor" will change in time as charging mode changes. So, given an unknown state of battery in the device when you plug it in first, you never know what its resistance is. How does the Ohm Law help here? So I am unsure what the objection is.
– Ale..chenski
Oct 18 '16 at 22:07
|
show 9 more comments
6
The device will only draw 500ma
– Ramhound
Oct 18 '16 at 2:14
1
This logic must be based on Grassman Algebra or something...
– Ale..chenski
Oct 18 '16 at 3:07
1
@DanielB, The simple Ohm or Kirchhoff formulas are applicable to circuit components with constant parameters. The AD-DC adapter is not a linear voltage source, and it will either abruptly reduce the voltage output when connected load exceeds its rated DC current capability (either along some artificially defined cut-off function, or just turn off). The device is also a non-linear load, starting with constant current, which will be reduced when the internal battery enters a constant voltage stage of charging. So these simple formulas do not apply.
– Ale..chenski
Oct 18 '16 at 6:55
1
@DanielB, I believe that the details are necessary because otherwise some curious people get confused with the discrepancy between their expectations and measured results.
– Ale..chenski
Oct 18 '16 at 7:20
1
@fixer1234, certainly the load is defined by Ohm's Law, if you mean that to consume 450mA from 5V it must have about 11 Ohms of effective resistance. However, there is a bunch of active elements that make this "impression", and this "resistor" will change in time as charging mode changes. So, given an unknown state of battery in the device when you plug it in first, you never know what its resistance is. How does the Ohm Law help here? So I am unsure what the objection is.
– Ale..chenski
Oct 18 '16 at 22:07
6
6
The device will only draw 500ma
– Ramhound
Oct 18 '16 at 2:14
The device will only draw 500ma
– Ramhound
Oct 18 '16 at 2:14
1
1
This logic must be based on Grassman Algebra or something...
– Ale..chenski
Oct 18 '16 at 3:07
This logic must be based on Grassman Algebra or something...
– Ale..chenski
Oct 18 '16 at 3:07
1
1
@DanielB, The simple Ohm or Kirchhoff formulas are applicable to circuit components with constant parameters. The AD-DC adapter is not a linear voltage source, and it will either abruptly reduce the voltage output when connected load exceeds its rated DC current capability (either along some artificially defined cut-off function, or just turn off). The device is also a non-linear load, starting with constant current, which will be reduced when the internal battery enters a constant voltage stage of charging. So these simple formulas do not apply.
– Ale..chenski
Oct 18 '16 at 6:55
@DanielB, The simple Ohm or Kirchhoff formulas are applicable to circuit components with constant parameters. The AD-DC adapter is not a linear voltage source, and it will either abruptly reduce the voltage output when connected load exceeds its rated DC current capability (either along some artificially defined cut-off function, or just turn off). The device is also a non-linear load, starting with constant current, which will be reduced when the internal battery enters a constant voltage stage of charging. So these simple formulas do not apply.
– Ale..chenski
Oct 18 '16 at 6:55
1
1
@DanielB, I believe that the details are necessary because otherwise some curious people get confused with the discrepancy between their expectations and measured results.
– Ale..chenski
Oct 18 '16 at 7:20
@DanielB, I believe that the details are necessary because otherwise some curious people get confused with the discrepancy between their expectations and measured results.
– Ale..chenski
Oct 18 '16 at 7:20
1
1
@fixer1234, certainly the load is defined by Ohm's Law, if you mean that to consume 450mA from 5V it must have about 11 Ohms of effective resistance. However, there is a bunch of active elements that make this "impression", and this "resistor" will change in time as charging mode changes. So, given an unknown state of battery in the device when you plug it in first, you never know what its resistance is. How does the Ohm Law help here? So I am unsure what the objection is.
– Ale..chenski
Oct 18 '16 at 22:07
@fixer1234, certainly the load is defined by Ohm's Law, if you mean that to consume 450mA from 5V it must have about 11 Ohms of effective resistance. However, there is a bunch of active elements that make this "impression", and this "resistor" will change in time as charging mode changes. So, given an unknown state of battery in the device when you plug it in first, you never know what its resistance is. How does the Ohm Law help here? So I am unsure what the objection is.
– Ale..chenski
Oct 18 '16 at 22:07
|
show 9 more comments
1 Answer
1
active
oldest
votes
4
You don't need to reduce anything. The headset is designed with a particular internal battery that has particular charging specifications, and its circuitry will not take more than whatever it is designed for (450mA in your case), no matter how capable the 5V source is, 1A or 10A or 30A.
FYI, USB cables do not "demand" anything. At most, if they have Type-C connector, they can inform the power provider and consumer about its wire gauge (current carrying capability). But you will be had pressed to find any of them yet.
answered Oct 18 '16 at 3:38
Ale..chenskiAle..chenski
8,54841731
I updated my question so I think the 2nd part of your answer is superluos. I suggested an edit and would then mark the answer with a green tick
– scjorge
Oct 18 '16 at 6:26
I am not sure I should do this. You have expressed a fairly common misconception about USB cables, I've seen this a half dozen times on stackexchange. Even in your edited form you still did not get it - a cable is a mere wire, it can't "block" 500mA.
– Ale..chenski
Oct 18 '16 at 6:37
by modifyed "cable" I mean: wire plus electrical component
– scjorge
Oct 18 '16 at 6:47
1
@scjorge I have rejected your edit. The second bullet could still useful for others who visit with a similar question to you
– Dave
Oct 18 '16 at 7:06
@AliChen... this issue of the mA has been going arround my mind these last days. What makes it possible that my phone and tablet charge faster with an AC-DC converter from HTC with 1.5A printed on than with their original (non-HTC) AC-DC converter with 1A printed on? should I open another question regarding this matter?
– scjorge
Oct 24 '16 at 5:57
|
show 2 more comments
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4
You don't need to reduce anything. The headset is designed with a particular internal battery that has particular charging specifications, and its circuitry will not take more than whatever it is designed for (450mA in your case), no matter how capable the 5V source is, 1A or 10A or 30A.
FYI, USB cables do not "demand" anything. At most, if they have Type-C connector, they can inform the power provider and consumer about its wire gauge (current carrying capability). But you will be had pressed to find any of them yet.
answered Oct 18 '16 at 3:38
Ale..chenskiAle..chenski
8,54841731
I updated my question so I think the 2nd part of your answer is superluos. I suggested an edit and would then mark the answer with a green tick
– scjorge
Oct 18 '16 at 6:26
I am not sure I should do this. You have expressed a fairly common misconception about USB cables, I've seen this a half dozen times on stackexchange. Even in your edited form you still did not get it - a cable is a mere wire, it can't "block" 500mA.
– Ale..chenski
Oct 18 '16 at 6:37
by modifyed "cable" I mean: wire plus electrical component
– scjorge
Oct 18 '16 at 6:47
1
@scjorge I have rejected your edit. The second bullet could still useful for others who visit with a similar question to you
– Dave
Oct 18 '16 at 7:06
@AliChen... this issue of the mA has been going arround my mind these last days. What makes it possible that my phone and tablet charge faster with an AC-DC converter from HTC with 1.5A printed on than with their original (non-HTC) AC-DC converter with 1A printed on? should I open another question regarding this matter?
– scjorge
Oct 24 '16 at 5:57
|
show 2 more comments
4
You don't need to reduce anything. The headset is designed with a particular internal battery that has particular charging specifications, and its circuitry will not take more than whatever it is designed for (450mA in your case), no matter how capable the 5V source is, 1A or 10A or 30A.
FYI, USB cables do not "demand" anything. At most, if they have Type-C connector, they can inform the power provider and consumer about its wire gauge (current carrying capability). But you will be had pressed to find any of them yet.
answered Oct 18 '16 at 3:38
Ale..chenskiAle..chenski
8,54841731
I updated my question so I think the 2nd part of your answer is superluos. I suggested an edit and would then mark the answer with a green tick
– scjorge
Oct 18 '16 at 6:26
I am not sure I should do this. You have expressed a fairly common misconception about USB cables, I've seen this a half dozen times on stackexchange. Even in your edited form you still did not get it - a cable is a mere wire, it can't "block" 500mA.
– Ale..chenski
Oct 18 '16 at 6:37
by modifyed "cable" I mean: wire plus electrical component
– scjorge
Oct 18 '16 at 6:47
1
@scjorge I have rejected your edit. The second bullet could still useful for others who visit with a similar question to you
– Dave
Oct 18 '16 at 7:06
@AliChen... this issue of the mA has been going arround my mind these last days. What makes it possible that my phone and tablet charge faster with an AC-DC converter from HTC with 1.5A printed on than with their original (non-HTC) AC-DC converter with 1A printed on? should I open another question regarding this matter?
– scjorge
Oct 24 '16 at 5:57
|
show 2 more comments
4
4
4
You don't need to reduce anything. The headset is designed with a particular internal battery that has particular charging specifications, and its circuitry will not take more than whatever it is designed for (450mA in your case), no matter how capable the 5V source is, 1A or 10A or 30A.
FYI, USB cables do not "demand" anything. At most, if they have Type-C connector, they can inform the power provider and consumer about its wire gauge (current carrying capability). But you will be had pressed to find any of them yet.
answered Oct 18 '16 at 3:38
Ale..chenskiAle..chenski
8,54841731
You don't need to reduce anything. The headset is designed with a particular internal battery that has particular charging specifications, and its circuitry will not take more than whatever it is designed for (450mA in your case), no matter how capable the 5V source is, 1A or 10A or 30A.
FYI, USB cables do not "demand" anything. At most, if they have Type-C connector, they can inform the power provider and consumer about its wire gauge (current carrying capability). But you will be had pressed to find any of them yet.
answered Oct 18 '16 at 3:38
Ale..chenskiAle..chenski
8,54841731
answered Oct 18 '16 at 3:38
Ale..chenskiAle..chenski
8,54841731
answered Oct 18 '16 at 3:38
Ale..chenskiAle..chenski
8,54841731
answered Oct 18 '16 at 3:38
Ale..chenskiAle..chenski
8,54841731
8,54841731
I updated my question so I think the 2nd part of your answer is superluos. I suggested an edit and would then mark the answer with a green tick
– scjorge
Oct 18 '16 at 6:26
I am not sure I should do this. You have expressed a fairly common misconception about USB cables, I've seen this a half dozen times on stackexchange. Even in your edited form you still did not get it - a cable is a mere wire, it can't "block" 500mA.
– Ale..chenski
Oct 18 '16 at 6:37
by modifyed "cable" I mean: wire plus electrical component
– scjorge
Oct 18 '16 at 6:47
1
@scjorge I have rejected your edit. The second bullet could still useful for others who visit with a similar question to you
– Dave
Oct 18 '16 at 7:06
@AliChen... this issue of the mA has been going arround my mind these last days. What makes it possible that my phone and tablet charge faster with an AC-DC converter from HTC with 1.5A printed on than with their original (non-HTC) AC-DC converter with 1A printed on? should I open another question regarding this matter?
– scjorge
Oct 24 '16 at 5:57
|
show 2 more comments
I updated my question so I think the 2nd part of your answer is superluos. I suggested an edit and would then mark the answer with a green tick
– scjorge
Oct 18 '16 at 6:26
I am not sure I should do this. You have expressed a fairly common misconception about USB cables, I've seen this a half dozen times on stackexchange. Even in your edited form you still did not get it - a cable is a mere wire, it can't "block" 500mA.
– Ale..chenski
Oct 18 '16 at 6:37
by modifyed "cable" I mean: wire plus electrical component
– scjorge
Oct 18 '16 at 6:47
1
@scjorge I have rejected your edit. The second bullet could still useful for others who visit with a similar question to you
– Dave
Oct 18 '16 at 7:06
@AliChen... this issue of the mA has been going arround my mind these last days. What makes it possible that my phone and tablet charge faster with an AC-DC converter from HTC with 1.5A printed on than with their original (non-HTC) AC-DC converter with 1A printed on? should I open another question regarding this matter?
– scjorge
Oct 24 '16 at 5:57
I updated my question so I think the 2nd part of your answer is superluos. I suggested an edit and would then mark the answer with a green tick
– scjorge
Oct 18 '16 at 6:26
I updated my question so I think the 2nd part of your answer is superluos. I suggested an edit and would then mark the answer with a green tick
– scjorge
Oct 18 '16 at 6:26
I am not sure I should do this. You have expressed a fairly common misconception about USB cables, I've seen this a half dozen times on stackexchange. Even in your edited form you still did not get it - a cable is a mere wire, it can't "block" 500mA.
– Ale..chenski
Oct 18 '16 at 6:37
I am not sure I should do this. You have expressed a fairly common misconception about USB cables, I've seen this a half dozen times on stackexchange. Even in your edited form you still did not get it - a cable is a mere wire, it can't "block" 500mA.
– Ale..chenski
Oct 18 '16 at 6:37
by modifyed "cable" I mean: wire plus electrical component
– scjorge
Oct 18 '16 at 6:47
by modifyed "cable" I mean: wire plus electrical component
– scjorge
Oct 18 '16 at 6:47
1
1
@scjorge I have rejected your edit. The second bullet could still useful for others who visit with a similar question to you
– Dave
Oct 18 '16 at 7:06
@scjorge I have rejected your edit. The second bullet could still useful for others who visit with a similar question to you
– Dave
Oct 18 '16 at 7:06
@AliChen... this issue of the mA has been going arround my mind these last days. What makes it possible that my phone and tablet charge faster with an AC-DC converter from HTC with 1.5A printed on than with their original (non-HTC) AC-DC converter with 1A printed on? should I open another question regarding this matter?
– scjorge
Oct 24 '16 at 5:57
@AliChen... this issue of the mA has been going arround my mind these last days. What makes it possible that my phone and tablet charge faster with an AC-DC converter from HTC with 1.5A printed on than with their original (non-HTC) AC-DC converter with 1A printed on? should I open another question regarding this matter?
– scjorge
Oct 24 '16 at 5:57
|
show 2 more comments
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6
The device will only draw 500ma
– Ramhound
Oct 18 '16 at 2:14
1
This logic must be based on Grassman Algebra or something...
– Ale..chenski
Oct 18 '16 at 3:07
1
@DanielB, The simple Ohm or Kirchhoff formulas are applicable to circuit components with constant parameters. The AD-DC adapter is not a linear voltage source, and it will either abruptly reduce the voltage output when connected load exceeds its rated DC current capability (either along some artificially defined cut-off function, or just turn off). The device is also a non-linear load, starting with constant current, which will be reduced when the internal battery enters a constant voltage stage of charging. So these simple formulas do not apply.
– Ale..chenski
Oct 18 '16 at 6:55
1
@DanielB, I believe that the details are necessary because otherwise some curious people get confused with the discrepancy between their expectations and measured results.
– Ale..chenski
Oct 18 '16 at 7:20
1
@fixer1234, certainly the load is defined by Ohm's Law, if you mean that to consume 450mA from 5V it must have about 11 Ohms of effective resistance. However, there is a bunch of active elements that make this "impression", and this "resistor" will change in time as charging mode changes. So, given an unknown state of battery in the device when you plug it in first, you never know what its resistance is. How does the Ohm Law help here? So I am unsure what the objection is.
– Ale..chenski
Oct 18 '16 at 22:07