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.
Comments
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.