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.
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.
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.
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.
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.
This is where you went wrong a capacitor is not a light bulb. Please take a look at this circuit: http://imgur.com/gallery/sOZ1F
The first graph shows the voltage and current across the capacitor when it is charged using a 10v source. The second shows the same parameters with a 5v source. The voltage source is cut-off by the zener diode at 4.7v in both cases.
The point I am trying to make is that the 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. Using a higher initial voltage will only charge your capacitor faster it will not necessarily destroy it.
Note1: Voltage across capacitor != supply voltage.
Note2: The above circuit was simulated using LTSpice.
I never said you couldn't charge a capacitor like this. I said (a) that it was a bad idea to charge a capacitor living in a portable device using a small power and (b) that it's extremely inefficient. (And (b) tends to imply (a) when the power brick is a tiny little thing with limited heat dissipation.) You also have the capacitor's leads to worry about if you scale to cell-phone-battery-sized capacitors.
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.
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) 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
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
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.
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?
A constant-current power supply is the preferred way to charge a capacitor wihtout destroying it.
Sure, I agree. I just wanted to use a simpler circuit to keep my explanation simple.
As to circuits other than ideal ones, I draw your attention to the behavior of superconductors, in which currents flow endlessly until interrupted.
I think you have cause and effect mixed up. How do you get this current started in the super conductor? What is the magnitude of this current? Infinite amps? You will still need an initial jolt (voltage source) to get the current started.
For a constant-current source, the rise is linear
You do realize there no such thing as a practical constant current source? A current source is just a voltage source which can adjust its terminal voltage to keep it's current output constant with changing load? Like I said you need to understand the cause and effect relationships here.
Capacitors have an upper limit to inrush current
Sure practical capacitors, conductors, resistors, ... everything has an Imax. I did say raise the voltage as long as you are within the safe operating area. SOAs are defined in terms of Imax, Vmax and Pmax.
I will apply 10 amperes of constant current
As I have already explained a constant current source is just a voltage source with current feedback. You are actually changing the terminal voltage to maintain your 10A. I agree charging at Imax is the fastest way of charging a capacitor but my circuit was simpler (constant voltage source with varying current vs your constant current source with varying terminal voltage).
Will I be able to charge the capacitor faster from a source of 200 volts than from a source of 50 volts?
This is a resounding yes. A constant voltage source rated at 200V will charge a capacitor faster than a source rated at 50V by virtue of the capacitor charging equation (assuming internal resistances are comparable).
Surely you realize one doesn't attach capacitors to voltage sources, or is this still not understood?
There is no reason why you cannot attach a voltage source to a capacitor. See: http://imgur.com/gallery/sOZ1F. Let me remind you a so called current source is also a voltage source. Current sources are a theoretical construct. Internal resistances have to be considered in a practical circuit otherwise you will have to deal with infinite currents at t=0.
You will still need an initial jolt (voltage source) to get the current started.
Yes, but that wasn't what I replied to. Your claim was this:
There is no current without EMF (measured in volts).
That's false. There can be massive currents without any potential difference. Again, superconductors show this behavior.
> For a constant-current source, the rise is linear
You do realize there no such thing as a practical constant current source?
Oh, really? So all those power supplies I designed for NASA were not really what they seemed? Constant-current sources are a normal part of electrical engineering practice and have been for decades.
Also, by using a switching inverter and manipulating the relationship between voltage and current in reactive elements, a constant-current source can be made very efficient. This is the ideal way to charge a supercapacitor, and it is standard practice. But I think I already said that.
I did say raise the voltage as long as you are within the safe operating area.
Yes, but that's not how you charge a capacitor -- you use a controlled current, not a voltage. The voltage follows the current, not the other way around.
> Will I be able to charge the capacitor faster from a source of 200 volts than from a source of 50 volts?
This is a resounding yes.
Quite false. For a given current that a capacitor can tolerate, the charging rate is the same regardless of the source voltage. The reason? You don't charge capacitors with voltage, you charge them with current.
The example I gave earlier was a capacitor rated at 50 volts and two current sources -- one that can deliver 200 volts and one that can deliver 50 volts. The charging rate is the same. The reason? You don't charge capacitors with voltage, you charge them with current.
There is no reason why you cannot attach a voltage source to a capacitor.
No reason at all. But don't be in the same room with a large capacitor and a voltage source that will supply substantial current to maintain a specified voltage. But, you know what? I already said that.
I've been designing power supply circuits for 40 years. My man-rated designs flew on the NASA Space Shuttle. I hold several patents. You're arguing with the wrong person.
Current sources are a theoretical construct.
Current sources are an everyday, trivial design task that all competent designers must learn to be regarded as employable. See the linked schematic above.
Internal resistances have to be considered in a practical circuit otherwise you will have to deal with infinite currents at t=0.
That is true for a voltage source. It is not true for a current source, and this is by design.
They're charged with power, which is current x voltage.
No, they are charged by currents. The normal power source for capacitor charging is a constant current source, one in which the voltage follows the desired current. The least desirable source is a constant-voltage supply, which (in the ideal case) will deliver infinite current at time zero.
The voltage on a capacitor is the time integral of past applied currents. The ideal charging scheme is a constant current source, which cuts off as the capacitor's voltage limit is reached.
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.
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.
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 ;)
To me the fundamentals have always been based on batteries being chemical reactions with their characteristic cell voltages. A cell definitely has its own discharge voltage, and can be used as a reference because the voltage remains relatively unchanged until the available chemicals are becoming effectively exhausted. It is an energy source based on the amount of chemicals available, plus a storage medium based on the reliable reversibility of the reaction.
In the case of rechargeable batteries, the desired reverse reaction can often be best obtained by carefully matching the recharge voltage & current dynamically, against the quantity of chemical you wish to react, including the amount undischarged and remaining to be reacted at any time. The recharge/maintenance voltages will need to be above the discharged/characterisitc voltages any time that cell recharging is expected to occur. The upper limit on charging voltage is based on the cell's characteristic voltage & the ability of the particular chemical system to withstand overvoltage. This is an electrochemical limit. The electrodes are separated by an electrolyte where having a low DC resistance is a desired property. As to the polarity of the DC voltage being provided or stored, this should always be respected, at your peril.
The working voltage of a battery cell can not be adjusted to match a particular circuit, the circuits must instead be closely designed to match the cell's characteristic voltage and chemical behavior. Increased energy storage requires more voluminous chemicals at the same voltage.
A capacitor whose plates can store charge across a dielectric has no characteristic voltage.
A capacitor can be a power store but there's no electrochemistry required.
Semantically, you can not "recharge" a capacitor since there is no original charge to begin with. You must first charge it with an energy source, which a battery by definition contains but a capacitor does not. A capacitor stores energy but it is not an actual energy source.
When charging a simple solid-state capacitor the upper limit on charging voltage is based on the electrical resistance of the particular dielectric & its ability to withstand overvoltage. This is a physical limit. The electrodes are separated by a dielectric where having a high DC resistance is a desired property. A simple solid-state capacitor functions equally well for operation at reversed polarity.
For an electrolytic capacitor, it is polarized because the chemicals it contains provide an effective increase in storage per volume, at the disadvantage of having high DC resistance in only one direction. An additional limit in maximum voltage is imposed by the electrochemical nature of these type capacitors, but they are still rated as against a physical limit. Plus the polarity of the DC voltage being stored in them should always be respected, your peril is once again at stake.
But no capacitor has its own ideal voltage, they are not at all like batteries in this regard.
A battery's maximum charging voltage is usually limited to a range close to its nominal voltage, imposed by the natural scientific ability of the chemicals to withstand overvoltage while retaining composure. There is very little leeway to work with here. To change the working voltage of a battery storage bank requires addressing the granularity of a different number of matched cells in series, plus circuitry carefully re-optimized for the incremented working battery voltage.
A capacitor's maximum charging voltage is only limited to a range within its rated voltage,
which is an engineering value assigned based on the natural scientific ability of the dielectric to withstand overvoltage while retaining composure. The order(s) of magnitude more leeway should be easily recognized. To change the working voltage of a capacitor storage bank requires only addressing the dielectric strength, which would not need to be changed if it was over-specified to begin with.
The working voltage of capacitor storage can be any arbitrary voltage selected for various engineering reasons, the working voltage only needs to be selected below the electrical rating of the components. Increased energy storage requires more voluminous dielectric, whether more capacitors having the same voltage rating, or bigger packages handling higher chosen voltages.
One of the differences between electronics and electrochemistry, capacitors are nominally marked with a usually conservative rating, but some parts may often be capable of truly handling twice the rated voltage reliably. Batteries not so much.
As a natural scientist designing circuits primarily using natural intelligence through experimentation & discovery, key efforts in computer science seem to be 100% helpful when invested in calculations which are too heavy or numerous otherwise.
As a computer scientist designing circuits primarily using software simulation, when key efforts in physical prototype building fail, this might reveal the investment in computer science itself to be less than 100% helpful.
Don't get me started on electrochemistry . . . the potential for reaction could be unlimited ;-)
Thanks, from reading your comment, I can see that I was a bit loose with the term 'charging voltage' - I was focusing on the internal electrical potential of the cap or battery(stabilised), but 'charging voltage' should pinpoint the external potential which is applied to the component.
In programdudes circuit, the internal and external voltage of the cap are the same, because the cap has no internal resistance modelled. But really the resistive and inductive functions of current stores like caps and batts are what moderates current flow , into or out of from them, and therefore they are what determines suitable charging voltages.
The resistive function of a cap is much simpler than a battery, but the capacity of each to store and release current makes their voltage (the ~pressure of their current) persistent in the respect that internal voltage will only change by releasing or storing current. Well there are caveats of course that chemical and other reactions may alter the stores potential over time, but current exchange is the fundamental cause of potential, and purpose of electrical storage.
Id like to edit this line but to late now:
> its own voltage (the electrical potential created by virtue of its capacitance)
It would be better to write "electric potential sustained by virtue of its capacity".
For caps vs batteries, the only modelling difference I see between them is their resistance functions - that function in batteries tends to be very complex, approximated very roughly by charging curves. Caps tend to be much simpler, but can still involve significant internal resistance and inductance especially to maximum performance applications.
In summary what ive struggled to explain for the record is - the charging(external) voltage of a battery or cap, is only separated by its internal voltage by its internal resistance. A store which will charge moderately at +0.5v above its internal potential, is liable to blow very quickly if 10 or 100 times that voltage is applied.
Comments
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.
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.
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.
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.
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.
This is where you went wrong a capacitor is not a light bulb. Please take a look at this circuit: http://imgur.com/gallery/sOZ1F
The first graph shows the voltage and current across the capacitor when it is charged using a 10v source. The second shows the same parameters with a 5v source. The voltage source is cut-off by the zener diode at 4.7v in both cases.
The point I am trying to make is that the 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. Using a higher initial voltage will only charge your capacitor faster it will not necessarily destroy it.
Note1: Voltage across capacitor != supply voltage.
Note2: The above circuit was simulated using LTSpice.
I never said you couldn't charge a capacitor like this. I said (a) that it was a bad idea to charge a capacitor living in a portable device using a small power and (b) that it's extremely inefficient. (And (b) tends to imply (a) when the power brick is a tiny little thing with limited heat dissipation.) You also have the capacitor's leads to worry about if you scale to cell-phone-battery-sized capacitors.
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.
This is meaning less. You have to apply an EMF (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
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
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.
Sorry for the late reply - I'm traveling.
A constant-current power supply is the preferred way to charge a capacitor wihtout destroying it.
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.
For a constant-current source, the rise is linear. For a constant-voltage source, the current at time zero is infinite, barring stray resistances.
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.
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?
No problem!
Sure, I agree. I just wanted to use a simpler circuit to keep my explanation simple.
I think you have cause and effect mixed up. How do you get this current started in the super conductor? What is the magnitude of this current? Infinite amps? You will still need an initial jolt (voltage source) to get the current started.
You do realize there no such thing as a practical constant current source? A current source is just a voltage source which can adjust its terminal voltage to keep it's current output constant with changing load? Like I said you need to understand the cause and effect relationships here.
Sure practical capacitors, conductors, resistors, ... everything has an Imax. I did say raise the voltage as long as you are within the safe operating area. SOAs are defined in terms of Imax, Vmax and Pmax.
As I have already explained a constant current source is just a voltage source with current feedback. You are actually changing the terminal voltage to maintain your 10A. I agree charging at Imax is the fastest way of charging a capacitor but my circuit was simpler (constant voltage source with varying current vs your constant current source with varying terminal voltage).
This is a resounding yes. A constant voltage source rated at 200V will charge a capacitor faster than a source rated at 50V by virtue of the capacitor charging equation (assuming internal resistances are comparable).
There is no reason why you cannot attach a voltage source to a capacitor. See: http://imgur.com/gallery/sOZ1F. Let me remind you a so called current source is also a voltage source. Current sources are a theoretical construct. Internal resistances have to be considered in a practical circuit otherwise you will have to deal with infinite currents at t=0.
Yes, but that wasn't what I replied to. Your claim was this:
That's false. There can be massive currents without any potential difference. Again, superconductors show this behavior.
Oh, really? So all those power supplies I designed for NASA were not really what they seemed? Constant-current sources are a normal part of electrical engineering practice and have been for decades.
https://en.wikipedia.org/wiki/Current_source
A typical, simple schematic of a constant current source:
http://www.allaboutcircuits.com/technical-articles/the-basic...
Also, by using a switching inverter and manipulating the relationship between voltage and current in reactive elements, a constant-current source can be made very efficient. This is the ideal way to charge a supercapacitor, and it is standard practice. But I think I already said that.
Yes, but that's not how you charge a capacitor -- you use a controlled current, not a voltage. The voltage follows the current, not the other way around.
Quite false. For a given current that a capacitor can tolerate, the charging rate is the same regardless of the source voltage. The reason? You don't charge capacitors with voltage, you charge them with current.
The example I gave earlier was a capacitor rated at 50 volts and two current sources -- one that can deliver 200 volts and one that can deliver 50 volts. The charging rate is the same. The reason? You don't charge capacitors with voltage, you charge them with current.
No reason at all. But don't be in the same room with a large capacitor and a voltage source that will supply substantial current to maintain a specified voltage. But, you know what? I already said that.
I've been designing power supply circuits for 40 years. My man-rated designs flew on the NASA Space Shuttle. I hold several patents. You're arguing with the wrong person.
Current sources are an everyday, trivial design task that all competent designers must learn to be regarded as employable. See the linked schematic above.
That is true for a voltage source. It is not true for a current source, and this is by design.
They're charged with power, which is current x voltage. If you have infinite current and zero voltage or vice versa, nothing happens.
No, they are charged by currents. The normal power source for capacitor charging is a constant current source, one in which the voltage follows the desired current. The least desirable source is a constant-voltage supply, which (in the ideal case) will deliver infinite current at time zero.
The voltage on a capacitor is the time integral of past applied currents. The ideal charging scheme is a constant current source, which cuts off as the capacitor's voltage limit is reached.
Sorry for the late reply, I am traveling.
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.
I don't think you grok how a capacitor works. Please see my response for more information: https://news.ycombinator.com/item?id=13029901
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.
Please see my comment here: https://news.ycombinator.com/item?id=13032719 to understand what I am trying to say.
ok I see it, but this statement is just fundamentally wrong here:
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 ;)
To me the fundamentals have always been based on batteries being chemical reactions with their characteristic cell voltages. A cell definitely has its own discharge voltage, and can be used as a reference because the voltage remains relatively unchanged until the available chemicals are becoming effectively exhausted. It is an energy source based on the amount of chemicals available, plus a storage medium based on the reliable reversibility of the reaction.
In the case of rechargeable batteries, the desired reverse reaction can often be best obtained by carefully matching the recharge voltage & current dynamically, against the quantity of chemical you wish to react, including the amount undischarged and remaining to be reacted at any time. The recharge/maintenance voltages will need to be above the discharged/characterisitc voltages any time that cell recharging is expected to occur. The upper limit on charging voltage is based on the cell's characteristic voltage & the ability of the particular chemical system to withstand overvoltage. This is an electrochemical limit. The electrodes are separated by an electrolyte where having a low DC resistance is a desired property. As to the polarity of the DC voltage being provided or stored, this should always be respected, at your peril.
The working voltage of a battery cell can not be adjusted to match a particular circuit, the circuits must instead be closely designed to match the cell's characteristic voltage and chemical behavior. Increased energy storage requires more voluminous chemicals at the same voltage.
A capacitor whose plates can store charge across a dielectric has no characteristic voltage.
A capacitor can be a power store but there's no electrochemistry required.
Semantically, you can not "recharge" a capacitor since there is no original charge to begin with. You must first charge it with an energy source, which a battery by definition contains but a capacitor does not. A capacitor stores energy but it is not an actual energy source.
When charging a simple solid-state capacitor the upper limit on charging voltage is based on the electrical resistance of the particular dielectric & its ability to withstand overvoltage. This is a physical limit. The electrodes are separated by a dielectric where having a high DC resistance is a desired property. A simple solid-state capacitor functions equally well for operation at reversed polarity.
For an electrolytic capacitor, it is polarized because the chemicals it contains provide an effective increase in storage per volume, at the disadvantage of having high DC resistance in only one direction. An additional limit in maximum voltage is imposed by the electrochemical nature of these type capacitors, but they are still rated as against a physical limit. Plus the polarity of the DC voltage being stored in them should always be respected, your peril is once again at stake.
But no capacitor has its own ideal voltage, they are not at all like batteries in this regard.
A battery's maximum charging voltage is usually limited to a range close to its nominal voltage, imposed by the natural scientific ability of the chemicals to withstand overvoltage while retaining composure. There is very little leeway to work with here. To change the working voltage of a battery storage bank requires addressing the granularity of a different number of matched cells in series, plus circuitry carefully re-optimized for the incremented working battery voltage.
A capacitor's maximum charging voltage is only limited to a range within its rated voltage, which is an engineering value assigned based on the natural scientific ability of the dielectric to withstand overvoltage while retaining composure. The order(s) of magnitude more leeway should be easily recognized. To change the working voltage of a capacitor storage bank requires only addressing the dielectric strength, which would not need to be changed if it was over-specified to begin with.
The working voltage of capacitor storage can be any arbitrary voltage selected for various engineering reasons, the working voltage only needs to be selected below the electrical rating of the components. Increased energy storage requires more voluminous dielectric, whether more capacitors having the same voltage rating, or bigger packages handling higher chosen voltages.
One of the differences between electronics and electrochemistry, capacitors are nominally marked with a usually conservative rating, but some parts may often be capable of truly handling twice the rated voltage reliably. Batteries not so much.
As a natural scientist designing circuits primarily using natural intelligence through experimentation & discovery, key efforts in computer science seem to be 100% helpful when invested in calculations which are too heavy or numerous otherwise.
As a computer scientist designing circuits primarily using software simulation, when key efforts in physical prototype building fail, this might reveal the investment in computer science itself to be less than 100% helpful.
Don't get me started on electrochemistry . . . the potential for reaction could be unlimited ;-)
Thanks, from reading your comment, I can see that I was a bit loose with the term 'charging voltage' - I was focusing on the internal electrical potential of the cap or battery(stabilised), but 'charging voltage' should pinpoint the external potential which is applied to the component.
In programdudes circuit, the internal and external voltage of the cap are the same, because the cap has no internal resistance modelled. But really the resistive and inductive functions of current stores like caps and batts are what moderates current flow , into or out of from them, and therefore they are what determines suitable charging voltages.
The resistive function of a cap is much simpler than a battery, but the capacity of each to store and release current makes their voltage (the ~pressure of their current) persistent in the respect that internal voltage will only change by releasing or storing current. Well there are caveats of course that chemical and other reactions may alter the stores potential over time, but current exchange is the fundamental cause of potential, and purpose of electrical storage.
Id like to edit this line but to late now: > its own voltage (the electrical potential created by virtue of its capacitance)
It would be better to write "electric potential sustained by virtue of its capacity".
For caps vs batteries, the only modelling difference I see between them is their resistance functions - that function in batteries tends to be very complex, approximated very roughly by charging curves. Caps tend to be much simpler, but can still involve significant internal resistance and inductance especially to maximum performance applications.
In summary what ive struggled to explain for the record is - the charging(external) voltage of a battery or cap, is only separated by its internal voltage by its internal resistance. A store which will charge moderately at +0.5v above its internal potential, is liable to blow very quickly if 10 or 100 times that voltage is applied.