5 ms·

You could also charge at something higher than 5V, to keep the current manageable. Something like 1.8A/50V would deliver 90W, same as 18A/5V, still be within mo

by puetzk 10y ago

You could also charge at something higher than 5V, to keep the current manageable. Something like 1.8A/50V would deliver 90W, same as 18A/5V, still be within most of the safety guidelines, and allow a pretty similar connector bulk to microUSBr. Or looking at what's already popular, USB-C connectors already spec do 3A@5V while handling data, or in PD mode (re-purposing pins) can do 5A@20V (100W). So the first case of a charge 3000mAh in a minute doesn't seem completely ludicrous, though it's going to be challenging to make everything efficient enough the heat isn't prohibitive.

strainer 10y ago
Sorry the high voltage USB idea is no good: although 1.8A * 50V equals 18A * 5V in terms of power, 50V is no use to a 5V battery. You cant deliver power to a battery at a voltage higher than its charging voltage or it will burn. A 90W transformer will not fit into a phone for the foreseeable future.
- programmer_dude 10y ago
  If it's really a capacitor what you said does not apply. You can charge a capacitor at any voltage as long as you cut the current off when it acquires the desired charge / terminal voltage.
  - lutusp 10y ago
    > You can charge a capacitor at any voltage as long as you cut the current off when it acquires the desired charge / terminal voltage. Capacitors aren't charged with a voltage, they're charged with a current. The voltage is irrelevant until breakdown occurs. More current, faster charge, less current, slower charge. Also, there's the practical constraint that a capacitor's physical construction limits the amount of current that can be applied without overheating the capacitor's materials. The TL;DR: one does not apply a voltage to a capacitor, one applies a current. If you doubt this, try applying a constant-voltage source to a capacitor. See what happens.
    - programmer_dude 10y ago
      > one does not apply a voltage to a capacitor, one applies a current This is meaning less. You have to apply an EMF (https://en.wikipedia.org/wiki/Electromotive_force https://en.wikipedia.org/wiki/Electromotive_force) to get the current flowing in the first place. There is no current without EMF (measured in volts). When a capacitor is directly connected to a real world voltage source (with finite internal resistance) it shorts the power source and causes a massive amount of current to flow. I = VEmf/Rbatt (but only at t=0). This current causes charge to accumulate on the plates of the capacitor causing the voltage across the plates to rise while reducing the current flowing through it at the same time. The voltage across the capacitor rises (asymptotically) until it matches the potential of the power source. The voltage across the plates at any given time t is given by the equation: V = VEmf(1-e^(-t/(Rbatt * C))). See: https://en.wikipedia.org/wiki/RC_time_constant https://en.wikipedia.org/wiki/RC_time_constant Also the current at any given time is: I = VEmf/Rbatt * (e^(-t/(Rbatt * C))) Thus an ideal capacitor can be charged by a voltage source of any value (VEmf) to any voltage (less than or equal to VEmf) across its plates. There is no theoretical limit imposed by physics. Practical capacitors will experience a dielectric breakdown above their rated voltages: https://en.wikipedia.org/wiki/Electrical_breakdown https://en.wikipedia.org/wiki/Electrical_breakdown So all you have to do is remember to disconnect the power source before the voltage across the capacitor crosses its safe operating area. So if a capacitor is rated at 50V max terminal voltage then you need to disconnect the power source when you get 50v across the cap. You will reach 50v earlier if you use a 200v power source instead of a 50v source. The only thing that changes is the the time required to charge the cap (and the current like you said). A higher voltage power source or one with lower internal resistance can drive larger currents.
      - lutusp 10y ago
        Sorry for the late reply - I'm traveling. >> one does not apply a voltage to a capacitor, one applies a current > This is meaning less. A constant-current power supply is the preferred way to charge a capacitor wihtout destroying it. > You have to apply an EMF (https://en.wikipedia.org/wiki/Electromotive_force https://en.wikipedia.org/wiki/Electromotive_force) to get the current flowing in the first place. There is no current without EMF (measured in volts). In a circuit without resistance, current can flow with no potential difference, so the above claim is false. As to circuits other than ideal ones, I draw your attention to the behavior of superconductors, in which currents flow endlessly until interrupted. > The voltage across the capacitor rises (asymptotically) until it matches the potential of the power source. For a constant-current source, the rise is linear. For a constant-voltage source, the current at time zero is infinite, barring stray resistances. > You will reach 50v earlier if you use a 200v power source instead of a 50v source. False! The issue is current, not voltage. The voltage on a capacitor is the time integral of past applied currents. Capacitors have an upper limit to inrush current, and to charge one, you stay below that limit. The ideal charging source for a capacitor is a constant-current source, and the voltage driving the source is irrelevant until the capacitor's voltage limit is reached. Consider a capacitor that can tolerate 10 amperes of current and has a voltage limit of 50 volts. Let's say I want to charge it as quickly as possible. I will apply 10 amperes of constant current, and cut off the charge as the capacitor's voltage limit is approached. Will I be able to charge the capacitor faster from a source of 200 volts than from a source of 50 volts? Of course not. > Also the current at any given time is: I = VEmf/Rbatt * (e^(-t/(Rbatt * C))) I am mystified as to why you posted a battery equation, and/or reference to resistance, in this capacitor discussion, in particular when constant current sources are the current state of the art. Surely you realize one doesn't attach capacitors to voltage sources, or is this still not understood?
  - amluto 10y ago
    No. If you plug a capacitor charged to 1V into a voltage source at 100V, the 99V difference is going to be dropped somewhere and will dissipate 99% of your incoming power as heat. Averaged over a full charge (0-100V) you lose half your energy as heat. In a phone at these power levels, something will melt. But this is all moot. The phone contains a DC-DC converter that changes the voltage. Capacitors aren't really any different from batteries in this regard except that they have a very different discharge curve. P.S. 5V is already above your phone's battery voltage.
    - programmer_dude 10y ago
      1. The power is dissipated across the internal resistance of the power source. Just make sure you are using a power source which can handle the heat (and current) and you should be fine. 2. Charging circuits change all the time to accommodate new battery technology. It will change again for supercaps.
      - amluto 10y ago
        > 1. The power is dissipated across the internal resistance of the power source. Just make sure you are using a power source which can handle the heat (and current) and you should be fine. This is totally wrong. Suppose you approximate your power supply by its Norton equivalent circuit (the Norton equivalence is exact, but the power supply is probably nonlinear at some point). You have an open-circuit voltage V_s and internal resistance R_s. A good power supply will try to keep R_s small. Now you plug this power supply into a battery that currently has voltage V_b and internal resistance R_b. You seem to be assuming that R_b=0, which is odd. The current I = (V_s - V_b) / (R_s + R_b). For many power supplies, R_s and R_b are quite low, and if you have V_s = 20V and V_b = 3V (which would be the case for your average laptop power brick and a partly discharged single-cell or parallel-connected laptop battery), you get 17V / (R_s + R_b). For common power bricks and batteries, R_s + R_b is very small and I ends up being rather large. If you set R_b = 0 (as you did), then you're nearly short-circuiting your charger and expecting it to dissipate the resulting heat. If you change your scenario a bit and imagine a tiny USB-C phone charger at 9V or 12V and a phone battery at 3V, you're going to melt your charger. And your hypothetical design completely ignores the battery's preferred charging current, which I can pretty much guarantee you'll exceed. As a real-world example of how wrong you are, go get a 9V battery and a little 3V lightbulb. Connect them, cross your fingers, and imagine that light bulb will run correctly and all that excess voltage dissipates in the 9V battery. (And don't hold the light bulb with your bare hands while you do this, please.) Fortunately, chargers don't work like this. But please never design a power electronic circuit.
  - strainer 10y ago
    I cant figure out the best way to explain this, but the power - voltage - current relationship is not being grokked well in this thread. Mismatched voltages between sources and components in circuits generally lead to pyrotechnics, never magical charging schemes.
    - programmer_dude 10y ago
      I don't think you grok how a capacitor works. Please see my response for more information: https://news.ycombinator.com/item?id=13029901 https://news.ycombinator.com/item?id=13029901
      - strainer 10y ago
        I understand your conviction but sorry you will want to revise your outlook on this. Last attempt to explain - suppose you can charge a cap with a high voltage source, but stop charging when it reaches a lower voltage. The cap itself is never subjected to a high voltage in this situation (this is certain). It will receive current at the voltage it is at, and the power transferred to it will be the current * its voltage. No fast charging magic present. You cannot charge a battery or capacitor (directly) at voltages higher than their own voltage. If source voltage is suitable, resistive current limiting is plausible, but power is dissipated by resistance (within the cap and path to it) proportional to the square of the current. This is why efficient voltage transformation is required if source has a significantly higher potential.
        programmer_dude 10y ago
        Please see my comment here: https://news.ycombinator.com/item?id=13032719 https://news.ycombinator.com/item?id=13032719 to understand what I am trying to say.
        strainer 10y ago
        ok I see it, but this statement is just fundamentally wrong here: >capacitor can be charged at any voltage as long as you cut off the supply when the voltage across the capacitor reaches a certain threshold The cap in your circuit is not charging at any voltage, it is charging at Vn001 - its own voltage (the electrical potential created by virtue of its capacitance). In your circuit the only modelled resistance (and the drop in voltage between supply and the cap) is across R1. This is all to explain the point which you disagreed with but is iron cast - you cant run power into a phone at 50volts to fast charge its 4volt battery without a missing, quite impossibly cool and small voltage transformer inside the phone to make the idea work. Im not pointing this out to you to win an argument, Im explaining something which I didnt understand either when I began experimenting with ltspice and making charge circuitry. Take the advice or leave it now, but if you keep on with electronics you will find out sooner or later - "power stores charge and discharge > directly < at their own voltages" and if you dont have enough resistance between different electric potentials - take heed 'fun' will ensue of the dwarven kind ;)
B1FF_PSUVM 10y ago
> 1.8A/50V Then the capacitor would have to be able to withstand 50V. That usually means proportionally more spacing between electrodes. There's something called "breakdown voltage" (or more precisely dielectric strength), which in air is about 30 kV/cm, and is the reason for those ceramic spacers that keep high voltage lines apart from grounded metal. (If your electric field, measured in Volt/meter, exceeds the material's dielectric strength, you get ionised molecules and sparks. https://en.wikipedia.org/wiki/Dielectric_strength https://en.wikipedia.org/wiki/Dielectric_strength)

by puetzk 10y ago

strainer 10y ago

Sorry the high voltage USB idea is no good: although 1.8A * 50V equals 18A * 5V in terms of power, 50V is no use to a 5V battery. You cant deliver power to a battery at a voltage higher than its charging voltage or it will burn. A 90W transformer will not fit into a phone for the foreseeable future.

programmer_dude 10y ago
If it's really a capacitor what you said does not apply. You can charge a capacitor at any voltage as long as you cut the current off when it acquires the desired charge / terminal voltage.
- lutusp 10y ago
  > You can charge a capacitor at any voltage as long as you cut the current off when it acquires the desired charge / terminal voltage. Capacitors aren't charged with a voltage, they're charged with a current. The voltage is irrelevant until breakdown occurs. More current, faster charge, less current, slower charge. Also, there's the practical constraint that a capacitor's physical construction limits the amount of current that can be applied without overheating the capacitor's materials. The TL;DR: one does not apply a voltage to a capacitor, one applies a current. If you doubt this, try applying a constant-voltage source to a capacitor. See what happens.
  - programmer_dude 10y ago
    > one does not apply a voltage to a capacitor, one applies a current This is meaning less. You have to apply an EMF (https://en.wikipedia.org/wiki/Electromotive_force https://en.wikipedia.org/wiki/Electromotive_force) to get the current flowing in the first place. There is no current without EMF (measured in volts). When a capacitor is directly connected to a real world voltage source (with finite internal resistance) it shorts the power source and causes a massive amount of current to flow. I = VEmf/Rbatt (but only at t=0). This current causes charge to accumulate on the plates of the capacitor causing the voltage across the plates to rise while reducing the current flowing through it at the same time. The voltage across the capacitor rises (asymptotically) until it matches the potential of the power source. The voltage across the plates at any given time t is given by the equation: V = VEmf(1-e^(-t/(Rbatt * C))). See: https://en.wikipedia.org/wiki/RC_time_constant https://en.wikipedia.org/wiki/RC_time_constant Also the current at any given time is: I = VEmf/Rbatt * (e^(-t/(Rbatt * C))) Thus an ideal capacitor can be charged by a voltage source of any value (VEmf) to any voltage (less than or equal to VEmf) across its plates. There is no theoretical limit imposed by physics. Practical capacitors will experience a dielectric breakdown above their rated voltages: https://en.wikipedia.org/wiki/Electrical_breakdown https://en.wikipedia.org/wiki/Electrical_breakdown So all you have to do is remember to disconnect the power source before the voltage across the capacitor crosses its safe operating area. So if a capacitor is rated at 50V max terminal voltage then you need to disconnect the power source when you get 50v across the cap. You will reach 50v earlier if you use a 200v power source instead of a 50v source. The only thing that changes is the the time required to charge the cap (and the current like you said). A higher voltage power source or one with lower internal resistance can drive larger currents.
    - lutusp 10y ago
      Sorry for the late reply - I'm traveling. >> one does not apply a voltage to a capacitor, one applies a current > This is meaning less. A constant-current power supply is the preferred way to charge a capacitor wihtout destroying it. > You have to apply an EMF (https://en.wikipedia.org/wiki/Electromotive_force https://en.wikipedia.org/wiki/Electromotive_force) to get the current flowing in the first place. There is no current without EMF (measured in volts). In a circuit without resistance, current can flow with no potential difference, so the above claim is false. As to circuits other than ideal ones, I draw your attention to the behavior of superconductors, in which currents flow endlessly until interrupted. > The voltage across the capacitor rises (asymptotically) until it matches the potential of the power source. For a constant-current source, the rise is linear. For a constant-voltage source, the current at time zero is infinite, barring stray resistances. > You will reach 50v earlier if you use a 200v power source instead of a 50v source. False! The issue is current, not voltage. The voltage on a capacitor is the time integral of past applied currents. Capacitors have an upper limit to inrush current, and to charge one, you stay below that limit. The ideal charging source for a capacitor is a constant-current source, and the voltage driving the source is irrelevant until the capacitor's voltage limit is reached. Consider a capacitor that can tolerate 10 amperes of current and has a voltage limit of 50 volts. Let's say I want to charge it as quickly as possible. I will apply 10 amperes of constant current, and cut off the charge as the capacitor's voltage limit is approached. Will I be able to charge the capacitor faster from a source of 200 volts than from a source of 50 volts? Of course not. > Also the current at any given time is: I = VEmf/Rbatt * (e^(-t/(Rbatt * C))) I am mystified as to why you posted a battery equation, and/or reference to resistance, in this capacitor discussion, in particular when constant current sources are the current state of the art. Surely you realize one doesn't attach capacitors to voltage sources, or is this still not understood?
- amluto 10y ago
  No. If you plug a capacitor charged to 1V into a voltage source at 100V, the 99V difference is going to be dropped somewhere and will dissipate 99% of your incoming power as heat. Averaged over a full charge (0-100V) you lose half your energy as heat. In a phone at these power levels, something will melt. But this is all moot. The phone contains a DC-DC converter that changes the voltage. Capacitors aren't really any different from batteries in this regard except that they have a very different discharge curve. P.S. 5V is already above your phone's battery voltage.
  - programmer_dude 10y ago
    1. The power is dissipated across the internal resistance of the power source. Just make sure you are using a power source which can handle the heat (and current) and you should be fine. 2. Charging circuits change all the time to accommodate new battery technology. It will change again for supercaps.
    - amluto 10y ago
      > 1. The power is dissipated across the internal resistance of the power source. Just make sure you are using a power source which can handle the heat (and current) and you should be fine. This is totally wrong. Suppose you approximate your power supply by its Norton equivalent circuit (the Norton equivalence is exact, but the power supply is probably nonlinear at some point). You have an open-circuit voltage V_s and internal resistance R_s. A good power supply will try to keep R_s small. Now you plug this power supply into a battery that currently has voltage V_b and internal resistance R_b. You seem to be assuming that R_b=0, which is odd. The current I = (V_s - V_b) / (R_s + R_b). For many power supplies, R_s and R_b are quite low, and if you have V_s = 20V and V_b = 3V (which would be the case for your average laptop power brick and a partly discharged single-cell or parallel-connected laptop battery), you get 17V / (R_s + R_b). For common power bricks and batteries, R_s + R_b is very small and I ends up being rather large. If you set R_b = 0 (as you did), then you're nearly short-circuiting your charger and expecting it to dissipate the resulting heat. If you change your scenario a bit and imagine a tiny USB-C phone charger at 9V or 12V and a phone battery at 3V, you're going to melt your charger. And your hypothetical design completely ignores the battery's preferred charging current, which I can pretty much guarantee you'll exceed. As a real-world example of how wrong you are, go get a 9V battery and a little 3V lightbulb. Connect them, cross your fingers, and imagine that light bulb will run correctly and all that excess voltage dissipates in the 9V battery. (And don't hold the light bulb with your bare hands while you do this, please.) Fortunately, chargers don't work like this. But please never design a power electronic circuit.
- strainer 10y ago
  I cant figure out the best way to explain this, but the power - voltage - current relationship is not being grokked well in this thread. Mismatched voltages between sources and components in circuits generally lead to pyrotechnics, never magical charging schemes.
  - programmer_dude 10y ago
    I don't think you grok how a capacitor works. Please see my response for more information: https://news.ycombinator.com/item?id=13029901 https://news.ycombinator.com/item?id=13029901
    - strainer 10y ago
      I understand your conviction but sorry you will want to revise your outlook on this. Last attempt to explain - suppose you can charge a cap with a high voltage source, but stop charging when it reaches a lower voltage. The cap itself is never subjected to a high voltage in this situation (this is certain). It will receive current at the voltage it is at, and the power transferred to it will be the current * its voltage. No fast charging magic present. You cannot charge a battery or capacitor (directly) at voltages higher than their own voltage. If source voltage is suitable, resistive current limiting is plausible, but power is dissipated by resistance (within the cap and path to it) proportional to the square of the current. This is why efficient voltage transformation is required if source has a significantly higher potential.
      - programmer_dude 10y ago
        Please see my comment here: https://news.ycombinator.com/item?id=13032719 https://news.ycombinator.com/item?id=13032719 to understand what I am trying to say.
        strainer 10y ago
        ok I see it, but this statement is just fundamentally wrong here: >capacitor can be charged at any voltage as long as you cut off the supply when the voltage across the capacitor reaches a certain threshold The cap in your circuit is not charging at any voltage, it is charging at Vn001 - its own voltage (the electrical potential created by virtue of its capacitance). In your circuit the only modelled resistance (and the drop in voltage between supply and the cap) is across R1. This is all to explain the point which you disagreed with but is iron cast - you cant run power into a phone at 50volts to fast charge its 4volt battery without a missing, quite impossibly cool and small voltage transformer inside the phone to make the idea work. Im not pointing this out to you to win an argument, Im explaining something which I didnt understand either when I began experimenting with ltspice and making charge circuitry. Take the advice or leave it now, but if you keep on with electronics you will find out sooner or later - "power stores charge and discharge > directly < at their own voltages" and if you dont have enough resistance between different electric potentials - take heed 'fun' will ensue of the dwarven kind ;)

B1FF_PSUVM 10y ago

> 1.8A/50V Then the capacitor would have to be able to withstand 50V. That usually means proportionally more spacing between electrodes. There's something called "breakdown voltage" (or more precisely dielectric strength), which in air is about 30 kV/cm, and is the reason for those ceramic spacers that keep high voltage lines apart from grounded metal. (If your electric field, measured in Volt/meter, exceeds the material's dielectric strength, you get ionised molecules and sparks. https://en.wikipedia.org/wiki/Dielectric_strength https://en.wikipedia.org/wiki/Dielectric_strength)