Charging phone battery with a lower voltage, coming from a bike charger?Risk of overloading a usb chargerHow...
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Charging phone battery with a lower voltage, coming from a bike charger?
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Charging phone battery with a lower voltage, coming from a bike charger?
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Since travel by bike a lot, and use my phone very often during those trips, I tried designing a USB phone bikecharger that would use my bike-dynamo that also powers my lamp.
Yesterday, I did my first tests. Without load, the circuit would give me 9,3V (even when the wheel was not spinning, which was a surprise to me :) ). With a 47 Ohm load, it would still give 9.3V. This is way too much for USB ofcourse, but 47Ohm is still quite big I think.
When a 10Ohm load was applied (I assumed 5V USB-voltage, 500mA charge current --> 10 ohm), the voltage dropped to 3,5V (it kept quite stable, it didn't fluctuate all that much. It seems the wheel speed doesn't influence the output voltage, except for when it spins too slow and the voltage is cut off entirely. I assume this is some kind of protection mechanism in the dynamo?).
So here's my question: what would happen is I were to attach a phone to this charger?
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery, so it will be blocked a few ms, but because of the open circuit, the voltage will rise to 5V and that will eventually be enough to charge. So in practice, it will just charge with a lower current (same as charging your phone on the USB-port of a netbook. Those often can't provide enough current, and the phone just charges slowly).
Is my guess right?
What are the risks of charging a phone-battery with too little voltage?
Should I worry about the charger outputting 9V when the phone battery is full? Because there won't be any load anymore then.
Thanks in advance!
ps. I also checked the signal with an oscilloscope to check the purity of the DC, and it was pretty clean. There were no big spikes and the level was quite stable. (I didn't try to pitch pennies on noise-supressing capacitors ;) )
Edit: added my schematic
EDIT2: J3 in the schematic is an LM2596 Buck module, using a MP1584 step-down regulator.
usb charger usb-device
asked 1 hour ago
OpifexOpifex
793
$endgroup$
add a comment |
$begingroup$
Since travel by bike a lot, and use my phone very often during those trips, I tried designing a USB phone bikecharger that would use my bike-dynamo that also powers my lamp.
Yesterday, I did my first tests. Without load, the circuit would give me 9,3V (even when the wheel was not spinning, which was a surprise to me :) ). With a 47 Ohm load, it would still give 9.3V. This is way too much for USB ofcourse, but 47Ohm is still quite big I think.
When a 10Ohm load was applied (I assumed 5V USB-voltage, 500mA charge current --> 10 ohm), the voltage dropped to 3,5V (it kept quite stable, it didn't fluctuate all that much. It seems the wheel speed doesn't influence the output voltage, except for when it spins too slow and the voltage is cut off entirely. I assume this is some kind of protection mechanism in the dynamo?).
So here's my question: what would happen is I were to attach a phone to this charger?
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery, so it will be blocked a few ms, but because of the open circuit, the voltage will rise to 5V and that will eventually be enough to charge. So in practice, it will just charge with a lower current (same as charging your phone on the USB-port of a netbook. Those often can't provide enough current, and the phone just charges slowly).
Is my guess right?
What are the risks of charging a phone-battery with too little voltage?
Should I worry about the charger outputting 9V when the phone battery is full? Because there won't be any load anymore then.
Thanks in advance!
ps. I also checked the signal with an oscilloscope to check the purity of the DC, and it was pretty clean. There were no big spikes and the level was quite stable. (I didn't try to pitch pennies on noise-supressing capacitors ;) )
Edit: added my schematic
EDIT2: J3 in the schematic is an LM2596 Buck module, using a MP1584 step-down regulator.
usb charger usb-device
asked 1 hour ago
OpifexOpifex
793
$endgroup$
$begingroup$
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery Why do you write this? How is the battery's internal resistance relevant? (Hint: it is not) You really need to learn how a battery in a phone is charged. Your contraption only needs to supply 5 V while being able to deliver enough current (like 1 A), then the charging circuit which is inside the phone will charge the battery. If your circuit cannot supply enough current then chances are that the phone will not charge the battery at all.
$endgroup$
– Bimpelrekkie
48 mins ago
$begingroup$
Bimpelrekkie: "internal resistance" is a wrong wording indeed, I meant that the phone will not charge because the voltage is too low --> the voltage will build up because of the open circuit --> 5V is eventually reached --> the battery will charge. This is what I meant.
$endgroup$
– Opifex
42 mins ago
add a comment |
$begingroup$
Since travel by bike a lot, and use my phone very often during those trips, I tried designing a USB phone bikecharger that would use my bike-dynamo that also powers my lamp.
Yesterday, I did my first tests. Without load, the circuit would give me 9,3V (even when the wheel was not spinning, which was a surprise to me :) ). With a 47 Ohm load, it would still give 9.3V. This is way too much for USB ofcourse, but 47Ohm is still quite big I think.
When a 10Ohm load was applied (I assumed 5V USB-voltage, 500mA charge current --> 10 ohm), the voltage dropped to 3,5V (it kept quite stable, it didn't fluctuate all that much. It seems the wheel speed doesn't influence the output voltage, except for when it spins too slow and the voltage is cut off entirely. I assume this is some kind of protection mechanism in the dynamo?).
So here's my question: what would happen is I were to attach a phone to this charger?
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery, so it will be blocked a few ms, but because of the open circuit, the voltage will rise to 5V and that will eventually be enough to charge. So in practice, it will just charge with a lower current (same as charging your phone on the USB-port of a netbook. Those often can't provide enough current, and the phone just charges slowly).
Is my guess right?
What are the risks of charging a phone-battery with too little voltage?
Should I worry about the charger outputting 9V when the phone battery is full? Because there won't be any load anymore then.
Thanks in advance!
ps. I also checked the signal with an oscilloscope to check the purity of the DC, and it was pretty clean. There were no big spikes and the level was quite stable. (I didn't try to pitch pennies on noise-supressing capacitors ;) )
Edit: added my schematic
EDIT2: J3 in the schematic is an LM2596 Buck module, using a MP1584 step-down regulator.
usb charger usb-device
asked 1 hour ago
OpifexOpifex
793
$endgroup$
Since travel by bike a lot, and use my phone very often during those trips, I tried designing a USB phone bikecharger that would use my bike-dynamo that also powers my lamp.
Yesterday, I did my first tests. Without load, the circuit would give me 9,3V (even when the wheel was not spinning, which was a surprise to me :) ). With a 47 Ohm load, it would still give 9.3V. This is way too much for USB ofcourse, but 47Ohm is still quite big I think.
When a 10Ohm load was applied (I assumed 5V USB-voltage, 500mA charge current --> 10 ohm), the voltage dropped to 3,5V (it kept quite stable, it didn't fluctuate all that much. It seems the wheel speed doesn't influence the output voltage, except for when it spins too slow and the voltage is cut off entirely. I assume this is some kind of protection mechanism in the dynamo?).
So here's my question: what would happen is I were to attach a phone to this charger?
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery, so it will be blocked a few ms, but because of the open circuit, the voltage will rise to 5V and that will eventually be enough to charge. So in practice, it will just charge with a lower current (same as charging your phone on the USB-port of a netbook. Those often can't provide enough current, and the phone just charges slowly).
Is my guess right?
What are the risks of charging a phone-battery with too little voltage?
Should I worry about the charger outputting 9V when the phone battery is full? Because there won't be any load anymore then.
Thanks in advance!
ps. I also checked the signal with an oscilloscope to check the purity of the DC, and it was pretty clean. There were no big spikes and the level was quite stable. (I didn't try to pitch pennies on noise-supressing capacitors ;) )
Edit: added my schematic
EDIT2: J3 in the schematic is an LM2596 Buck module, using a MP1584 step-down regulator.
usb charger usb-device
usb charger usb-device
asked 1 hour ago
OpifexOpifex
793
asked 1 hour ago
OpifexOpifex
793
edited 5 mins ago
Opifex
asked 1 hour ago
OpifexOpifex
793
asked 1 hour ago
OpifexOpifex
793
asked 1 hour ago
OpifexOpifex
793
793
$begingroup$
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery Why do you write this? How is the battery's internal resistance relevant? (Hint: it is not) You really need to learn how a battery in a phone is charged. Your contraption only needs to supply 5 V while being able to deliver enough current (like 1 A), then the charging circuit which is inside the phone will charge the battery. If your circuit cannot supply enough current then chances are that the phone will not charge the battery at all.
$endgroup$
– Bimpelrekkie
48 mins ago
$begingroup$
Bimpelrekkie: "internal resistance" is a wrong wording indeed, I meant that the phone will not charge because the voltage is too low --> the voltage will build up because of the open circuit --> 5V is eventually reached --> the battery will charge. This is what I meant.
$endgroup$
– Opifex
42 mins ago
add a comment |
$begingroup$
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery Why do you write this? How is the battery's internal resistance relevant? (Hint: it is not) You really need to learn how a battery in a phone is charged. Your contraption only needs to supply 5 V while being able to deliver enough current (like 1 A), then the charging circuit which is inside the phone will charge the battery. If your circuit cannot supply enough current then chances are that the phone will not charge the battery at all.
$endgroup$
– Bimpelrekkie
48 mins ago
$begingroup$
Bimpelrekkie: "internal resistance" is a wrong wording indeed, I meant that the phone will not charge because the voltage is too low --> the voltage will build up because of the open circuit --> 5V is eventually reached --> the battery will charge. This is what I meant.
$endgroup$
– Opifex
42 mins ago
$begingroup$
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery Why do you write this? How is the battery's internal resistance relevant? (Hint: it is not) You really need to learn how a battery in a phone is charged. Your contraption only needs to supply 5 V while being able to deliver enough current (like 1 A), then the charging circuit which is inside the phone will charge the battery. If your circuit cannot supply enough current then chances are that the phone will not charge the battery at all.
$endgroup$
– Bimpelrekkie
48 mins ago
$begingroup$
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery Why do you write this? How is the battery's internal resistance relevant? (Hint: it is not) You really need to learn how a battery in a phone is charged. Your contraption only needs to supply 5 V while being able to deliver enough current (like 1 A), then the charging circuit which is inside the phone will charge the battery. If your circuit cannot supply enough current then chances are that the phone will not charge the battery at all.
$endgroup$
– Bimpelrekkie
48 mins ago
$begingroup$
Bimpelrekkie: "internal resistance" is a wrong wording indeed, I meant that the phone will not charge because the voltage is too low --> the voltage will build up because of the open circuit --> 5V is eventually reached --> the battery will charge. This is what I meant.
$endgroup$
– Opifex
42 mins ago
$begingroup$
Bimpelrekkie: "internal resistance" is a wrong wording indeed, I meant that the phone will not charge because the voltage is too low --> the voltage will build up because of the open circuit --> 5V is eventually reached --> the battery will charge. This is what I meant.
$endgroup$
– Opifex
42 mins ago
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
First of all, the USB is not being fed directly to the phone’s battery. It’s actually being fed to the battery charger circuitry inside the phone. You cannot safely directly connect a normal regulated voltage source to a lithium ion battery without the distinct possibility of an explosion and/or fire! The charger circuitry adjusts the voltage and current fed to the battery depending on the state of charge.
Then, your question becomes what happens when you feed incorrect voltage into a phone’s charger circuitry. This would depend on the specific model, but we can generalize. With too low voltage, such as 3.5 volts, the phone will probably ignore it and refuse to charge. With too high voltage, such as 9.3 volts, the phone is likely to either shut down to protect itself or be damaged by the excessive voltage.
Your circuitry, which you have not really described, is totally unsuited to charging a mobile phone! You have a good chance of seriously damaging its circuitry. You should probably find a commercially produced buck/boost module, capable of taking in the maximum voltage from your dynamo and outputting a stable 5 volts at 500 ma, or whatever your phone needs.
answered 46 mins ago
DoxyLoverDoxyLover
5,51411221
$endgroup$
$begingroup$
You're right, I didn't describe the circuitry. I added my schematic to the original post now. A colleague of mine suggested a buck/boost module while I was designing it, but I assumed the boost wasn't necesarry because I would be able to achieve a stable 5V out of the dynamo. A buck/boost would have made the whole pcb a lot more bulky than I deemed acceptable on a bike.
$endgroup$
– Opifex
33 mins ago
$begingroup$
I am aware the USB is not fed directly into the battery. The voltage would be a multiple of 3.7V, so the 5V from the USB will be lowered internally. What I meant was, that the internal circuitry of the phone would cut everything down while it's lower than 5V, which would cause the voltage to rise because of the open contact. Once it reaches 5V, the circuit would close again. This is what I meant (but I admit it was very badly worded). Thing is that the 9.3V is only when it has no load, or a very high load. Wouldn't a normal USB charger also have a high voltage when it doesn't have a load?
$endgroup$
– Opifex
31 mins ago
add a comment |
$begingroup$
Obviously you have not any data of how your phone's USB connection is specified except it's USB. Don't risk your phone, the voltage must be 5V if you haven't better data. Too low voltage doesn't harm - except there's no charging, but 9V probably does. Have a voltage regulator, even 7805 with a couple of capacitors can be ok altough it wastes nearly half of the energy. A switching buck converter would be better in that sense. I do not recommend building them by yourself if you are not experienced electronics hobbyist who understands also the math of those circuits.
answered 45 mins ago
user287001user287001
9,4541517
$endgroup$
$begingroup$
7805 would be too bulky because it requires a heatsink, and also wastes a lot of the energy. I used an MP1584 step-down regulator instead.
$endgroup$
– Opifex
30 mins ago
$begingroup$
@Opifex MP1584 is fine. A roadworthy construction is your next challenge.
$endgroup$
– user287001
25 mins ago
$begingroup$
What do you mean with "construction"? A box? Or the circuit? was planning on 3D-printing something easily attachable to the frame.
$endgroup$
– Opifex
23 mins ago
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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$begingroup$
First of all, the USB is not being fed directly to the phone’s battery. It’s actually being fed to the battery charger circuitry inside the phone. You cannot safely directly connect a normal regulated voltage source to a lithium ion battery without the distinct possibility of an explosion and/or fire! The charger circuitry adjusts the voltage and current fed to the battery depending on the state of charge.
Then, your question becomes what happens when you feed incorrect voltage into a phone’s charger circuitry. This would depend on the specific model, but we can generalize. With too low voltage, such as 3.5 volts, the phone will probably ignore it and refuse to charge. With too high voltage, such as 9.3 volts, the phone is likely to either shut down to protect itself or be damaged by the excessive voltage.
Your circuitry, which you have not really described, is totally unsuited to charging a mobile phone! You have a good chance of seriously damaging its circuitry. You should probably find a commercially produced buck/boost module, capable of taking in the maximum voltage from your dynamo and outputting a stable 5 volts at 500 ma, or whatever your phone needs.
answered 46 mins ago
DoxyLoverDoxyLover
5,51411221
$endgroup$
$begingroup$
You're right, I didn't describe the circuitry. I added my schematic to the original post now. A colleague of mine suggested a buck/boost module while I was designing it, but I assumed the boost wasn't necesarry because I would be able to achieve a stable 5V out of the dynamo. A buck/boost would have made the whole pcb a lot more bulky than I deemed acceptable on a bike.
$endgroup$
– Opifex
33 mins ago
$begingroup$
I am aware the USB is not fed directly into the battery. The voltage would be a multiple of 3.7V, so the 5V from the USB will be lowered internally. What I meant was, that the internal circuitry of the phone would cut everything down while it's lower than 5V, which would cause the voltage to rise because of the open contact. Once it reaches 5V, the circuit would close again. This is what I meant (but I admit it was very badly worded). Thing is that the 9.3V is only when it has no load, or a very high load. Wouldn't a normal USB charger also have a high voltage when it doesn't have a load?
$endgroup$
– Opifex
31 mins ago
add a comment |
$begingroup$
First of all, the USB is not being fed directly to the phone’s battery. It’s actually being fed to the battery charger circuitry inside the phone. You cannot safely directly connect a normal regulated voltage source to a lithium ion battery without the distinct possibility of an explosion and/or fire! The charger circuitry adjusts the voltage and current fed to the battery depending on the state of charge.
Then, your question becomes what happens when you feed incorrect voltage into a phone’s charger circuitry. This would depend on the specific model, but we can generalize. With too low voltage, such as 3.5 volts, the phone will probably ignore it and refuse to charge. With too high voltage, such as 9.3 volts, the phone is likely to either shut down to protect itself or be damaged by the excessive voltage.
Your circuitry, which you have not really described, is totally unsuited to charging a mobile phone! You have a good chance of seriously damaging its circuitry. You should probably find a commercially produced buck/boost module, capable of taking in the maximum voltage from your dynamo and outputting a stable 5 volts at 500 ma, or whatever your phone needs.
answered 46 mins ago
DoxyLoverDoxyLover
5,51411221
$endgroup$
$begingroup$
You're right, I didn't describe the circuitry. I added my schematic to the original post now. A colleague of mine suggested a buck/boost module while I was designing it, but I assumed the boost wasn't necesarry because I would be able to achieve a stable 5V out of the dynamo. A buck/boost would have made the whole pcb a lot more bulky than I deemed acceptable on a bike.
$endgroup$
– Opifex
33 mins ago
$begingroup$
I am aware the USB is not fed directly into the battery. The voltage would be a multiple of 3.7V, so the 5V from the USB will be lowered internally. What I meant was, that the internal circuitry of the phone would cut everything down while it's lower than 5V, which would cause the voltage to rise because of the open contact. Once it reaches 5V, the circuit would close again. This is what I meant (but I admit it was very badly worded). Thing is that the 9.3V is only when it has no load, or a very high load. Wouldn't a normal USB charger also have a high voltage when it doesn't have a load?
$endgroup$
– Opifex
31 mins ago
add a comment |
$begingroup$
First of all, the USB is not being fed directly to the phone’s battery. It’s actually being fed to the battery charger circuitry inside the phone. You cannot safely directly connect a normal regulated voltage source to a lithium ion battery without the distinct possibility of an explosion and/or fire! The charger circuitry adjusts the voltage and current fed to the battery depending on the state of charge.
Then, your question becomes what happens when you feed incorrect voltage into a phone’s charger circuitry. This would depend on the specific model, but we can generalize. With too low voltage, such as 3.5 volts, the phone will probably ignore it and refuse to charge. With too high voltage, such as 9.3 volts, the phone is likely to either shut down to protect itself or be damaged by the excessive voltage.
Your circuitry, which you have not really described, is totally unsuited to charging a mobile phone! You have a good chance of seriously damaging its circuitry. You should probably find a commercially produced buck/boost module, capable of taking in the maximum voltage from your dynamo and outputting a stable 5 volts at 500 ma, or whatever your phone needs.
answered 46 mins ago
DoxyLoverDoxyLover
5,51411221
$endgroup$
First of all, the USB is not being fed directly to the phone’s battery. It’s actually being fed to the battery charger circuitry inside the phone. You cannot safely directly connect a normal regulated voltage source to a lithium ion battery without the distinct possibility of an explosion and/or fire! The charger circuitry adjusts the voltage and current fed to the battery depending on the state of charge.
Then, your question becomes what happens when you feed incorrect voltage into a phone’s charger circuitry. This would depend on the specific model, but we can generalize. With too low voltage, such as 3.5 volts, the phone will probably ignore it and refuse to charge. With too high voltage, such as 9.3 volts, the phone is likely to either shut down to protect itself or be damaged by the excessive voltage.
Your circuitry, which you have not really described, is totally unsuited to charging a mobile phone! You have a good chance of seriously damaging its circuitry. You should probably find a commercially produced buck/boost module, capable of taking in the maximum voltage from your dynamo and outputting a stable 5 volts at 500 ma, or whatever your phone needs.
answered 46 mins ago
DoxyLoverDoxyLover
5,51411221
answered 46 mins ago
DoxyLoverDoxyLover
5,51411221
answered 46 mins ago
DoxyLoverDoxyLover
5,51411221
answered 46 mins ago
DoxyLoverDoxyLover
5,51411221
5,51411221
$begingroup$
You're right, I didn't describe the circuitry. I added my schematic to the original post now. A colleague of mine suggested a buck/boost module while I was designing it, but I assumed the boost wasn't necesarry because I would be able to achieve a stable 5V out of the dynamo. A buck/boost would have made the whole pcb a lot more bulky than I deemed acceptable on a bike.
$endgroup$
– Opifex
33 mins ago
$begingroup$
I am aware the USB is not fed directly into the battery. The voltage would be a multiple of 3.7V, so the 5V from the USB will be lowered internally. What I meant was, that the internal circuitry of the phone would cut everything down while it's lower than 5V, which would cause the voltage to rise because of the open contact. Once it reaches 5V, the circuit would close again. This is what I meant (but I admit it was very badly worded). Thing is that the 9.3V is only when it has no load, or a very high load. Wouldn't a normal USB charger also have a high voltage when it doesn't have a load?
$endgroup$
– Opifex
31 mins ago
add a comment |
$begingroup$
You're right, I didn't describe the circuitry. I added my schematic to the original post now. A colleague of mine suggested a buck/boost module while I was designing it, but I assumed the boost wasn't necesarry because I would be able to achieve a stable 5V out of the dynamo. A buck/boost would have made the whole pcb a lot more bulky than I deemed acceptable on a bike.
$endgroup$
– Opifex
33 mins ago
$begingroup$
I am aware the USB is not fed directly into the battery. The voltage would be a multiple of 3.7V, so the 5V from the USB will be lowered internally. What I meant was, that the internal circuitry of the phone would cut everything down while it's lower than 5V, which would cause the voltage to rise because of the open contact. Once it reaches 5V, the circuit would close again. This is what I meant (but I admit it was very badly worded). Thing is that the 9.3V is only when it has no load, or a very high load. Wouldn't a normal USB charger also have a high voltage when it doesn't have a load?
$endgroup$
– Opifex
31 mins ago
$begingroup$
You're right, I didn't describe the circuitry. I added my schematic to the original post now. A colleague of mine suggested a buck/boost module while I was designing it, but I assumed the boost wasn't necesarry because I would be able to achieve a stable 5V out of the dynamo. A buck/boost would have made the whole pcb a lot more bulky than I deemed acceptable on a bike.
$endgroup$
– Opifex
33 mins ago
$begingroup$
You're right, I didn't describe the circuitry. I added my schematic to the original post now. A colleague of mine suggested a buck/boost module while I was designing it, but I assumed the boost wasn't necesarry because I would be able to achieve a stable 5V out of the dynamo. A buck/boost would have made the whole pcb a lot more bulky than I deemed acceptable on a bike.
$endgroup$
– Opifex
33 mins ago
$begingroup$
I am aware the USB is not fed directly into the battery. The voltage would be a multiple of 3.7V, so the 5V from the USB will be lowered internally. What I meant was, that the internal circuitry of the phone would cut everything down while it's lower than 5V, which would cause the voltage to rise because of the open contact. Once it reaches 5V, the circuit would close again. This is what I meant (but I admit it was very badly worded). Thing is that the 9.3V is only when it has no load, or a very high load. Wouldn't a normal USB charger also have a high voltage when it doesn't have a load?
$endgroup$
– Opifex
31 mins ago
$begingroup$
I am aware the USB is not fed directly into the battery. The voltage would be a multiple of 3.7V, so the 5V from the USB will be lowered internally. What I meant was, that the internal circuitry of the phone would cut everything down while it's lower than 5V, which would cause the voltage to rise because of the open contact. Once it reaches 5V, the circuit would close again. This is what I meant (but I admit it was very badly worded). Thing is that the 9.3V is only when it has no load, or a very high load. Wouldn't a normal USB charger also have a high voltage when it doesn't have a load?
$endgroup$
– Opifex
31 mins ago
add a comment |
$begingroup$
Obviously you have not any data of how your phone's USB connection is specified except it's USB. Don't risk your phone, the voltage must be 5V if you haven't better data. Too low voltage doesn't harm - except there's no charging, but 9V probably does. Have a voltage regulator, even 7805 with a couple of capacitors can be ok altough it wastes nearly half of the energy. A switching buck converter would be better in that sense. I do not recommend building them by yourself if you are not experienced electronics hobbyist who understands also the math of those circuits.
answered 45 mins ago
user287001user287001
9,4541517
$endgroup$
$begingroup$
7805 would be too bulky because it requires a heatsink, and also wastes a lot of the energy. I used an MP1584 step-down regulator instead.
$endgroup$
– Opifex
30 mins ago
$begingroup$
@Opifex MP1584 is fine. A roadworthy construction is your next challenge.
$endgroup$
– user287001
25 mins ago
$begingroup$
What do you mean with "construction"? A box? Or the circuit? was planning on 3D-printing something easily attachable to the frame.
$endgroup$
– Opifex
23 mins ago
add a comment |
$begingroup$
Obviously you have not any data of how your phone's USB connection is specified except it's USB. Don't risk your phone, the voltage must be 5V if you haven't better data. Too low voltage doesn't harm - except there's no charging, but 9V probably does. Have a voltage regulator, even 7805 with a couple of capacitors can be ok altough it wastes nearly half of the energy. A switching buck converter would be better in that sense. I do not recommend building them by yourself if you are not experienced electronics hobbyist who understands also the math of those circuits.
answered 45 mins ago
user287001user287001
9,4541517
$endgroup$
$begingroup$
7805 would be too bulky because it requires a heatsink, and also wastes a lot of the energy. I used an MP1584 step-down regulator instead.
$endgroup$
– Opifex
30 mins ago
$begingroup$
@Opifex MP1584 is fine. A roadworthy construction is your next challenge.
$endgroup$
– user287001
25 mins ago
$begingroup$
What do you mean with "construction"? A box? Or the circuit? was planning on 3D-printing something easily attachable to the frame.
$endgroup$
– Opifex
23 mins ago
add a comment |
$begingroup$
Obviously you have not any data of how your phone's USB connection is specified except it's USB. Don't risk your phone, the voltage must be 5V if you haven't better data. Too low voltage doesn't harm - except there's no charging, but 9V probably does. Have a voltage regulator, even 7805 with a couple of capacitors can be ok altough it wastes nearly half of the energy. A switching buck converter would be better in that sense. I do not recommend building them by yourself if you are not experienced electronics hobbyist who understands also the math of those circuits.
answered 45 mins ago
user287001user287001
9,4541517
$endgroup$
Obviously you have not any data of how your phone's USB connection is specified except it's USB. Don't risk your phone, the voltage must be 5V if you haven't better data. Too low voltage doesn't harm - except there's no charging, but 9V probably does. Have a voltage regulator, even 7805 with a couple of capacitors can be ok altough it wastes nearly half of the energy. A switching buck converter would be better in that sense. I do not recommend building them by yourself if you are not experienced electronics hobbyist who understands also the math of those circuits.
answered 45 mins ago
user287001user287001
9,4541517
edited 36 mins ago
answered 45 mins ago
user287001user287001
9,4541517
answered 45 mins ago
user287001user287001
9,4541517
answered 45 mins ago
user287001user287001
9,4541517
9,4541517
$begingroup$
7805 would be too bulky because it requires a heatsink, and also wastes a lot of the energy. I used an MP1584 step-down regulator instead.
$endgroup$
– Opifex
30 mins ago
$begingroup$
@Opifex MP1584 is fine. A roadworthy construction is your next challenge.
$endgroup$
– user287001
25 mins ago
$begingroup$
What do you mean with "construction"? A box? Or the circuit? was planning on 3D-printing something easily attachable to the frame.
$endgroup$
– Opifex
23 mins ago
add a comment |
$begingroup$
7805 would be too bulky because it requires a heatsink, and also wastes a lot of the energy. I used an MP1584 step-down regulator instead.
$endgroup$
– Opifex
30 mins ago
$begingroup$
@Opifex MP1584 is fine. A roadworthy construction is your next challenge.
$endgroup$
– user287001
25 mins ago
$begingroup$
What do you mean with "construction"? A box? Or the circuit? was planning on 3D-printing something easily attachable to the frame.
$endgroup$
– Opifex
23 mins ago
$begingroup$
7805 would be too bulky because it requires a heatsink, and also wastes a lot of the energy. I used an MP1584 step-down regulator instead.
$endgroup$
– Opifex
30 mins ago
$begingroup$
7805 would be too bulky because it requires a heatsink, and also wastes a lot of the energy. I used an MP1584 step-down regulator instead.
$endgroup$
– Opifex
30 mins ago
$begingroup$
@Opifex MP1584 is fine. A roadworthy construction is your next challenge.
$endgroup$
– user287001
25 mins ago
$begingroup$
@Opifex MP1584 is fine. A roadworthy construction is your next challenge.
$endgroup$
– user287001
25 mins ago
$begingroup$
What do you mean with "construction"? A box? Or the circuit? was planning on 3D-printing something easily attachable to the frame.
$endgroup$
– Opifex
23 mins ago
$begingroup$
What do you mean with "construction"? A box? Or the circuit? was planning on 3D-printing something easily attachable to the frame.
$endgroup$
– Opifex
23 mins ago
add a comment |
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$begingroup$
My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery Why do you write this? How is the battery's internal resistance relevant? (Hint: it is not) You really need to learn how a battery in a phone is charged. Your contraption only needs to supply 5 V while being able to deliver enough current (like 1 A), then the charging circuit which is inside the phone will charge the battery. If your circuit cannot supply enough current then chances are that the phone will not charge the battery at all.
$endgroup$
– Bimpelrekkie
48 mins ago
$begingroup$
Bimpelrekkie: "internal resistance" is a wrong wording indeed, I meant that the phone will not charge because the voltage is too low --> the voltage will build up because of the open circuit --> 5V is eventually reached --> the battery will charge. This is what I meant.
$endgroup$
– Opifex
42 mins ago