Can a Hobby Charger charge the Lead-Acid battery that powers it?
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hellokitty[hk]
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Yeah but it wouldn't be charging it...
moderator007
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You would have the losses that the charger is using to up the voltage to try to charge it. The battery would just get weaker and weaker but it would take a really long time. You can't make more power than what is put in. The charger also checks the voltage as its charging thats how it knows when to stop. So in a sense you just have a closed loop from input of the charger to the output of the charger. I think the charger will recognize the higher voltage from itself and stop charging.
The answer would be no it would very slowly discharge it if it would even run at all.
It would be really great if what your where asking where true. We probably wouldn't need the power co. anymore.
The answer would be no it would very slowly discharge it if it would even run at all.
It would be really great if what your where asking where true. We probably wouldn't need the power co. anymore.
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I'm a little familiar about no such thing as 'perpetual power'. What makes me wonder (a little) is this: since the hobby charger has an input voltage of 10 to 18vdc, let's assume that it is powered by a DC-output AC power supply at 10VDC. This charger can then charge a lead-acid battery up to 14vdc. Now, let's change that PS into the lead-acid battery, the same battery to be charged, wouldn't it be possible, even just once? (I know power cannot perpetuate itself, btw) Just thinking out loud!
HKJ
Flashaholic
A hobby charge can both buck and boost the input voltage, i.e. the battery to charge can be below or above the supply voltage to the charger.
This makes it possible to charge a SLA battery with the charger (on a mains power supply) and later use the same SLA battery as power supply for the charger.
Most iChargers can also move power in reverse, i.e. when they discharge a battery they can move the power out to the supply (It is called regenerative discharge).
This makes it possible to charge a SLA battery with the charger (on a mains power supply) and later use the same SLA battery as power supply for the charger.
Most iChargers can also move power in reverse, i.e. when they discharge a battery they can move the power out to the supply (It is called regenerative discharge).
Remember that Voltage doesn't equal power, and that Voltage doesn't equal energy either.
Power is measured in Watts, and energy in Watthours.
You get the watts by multiplying Voltage with Current.
If the charger input voltage is 10V, and the input current is 1A, that's 10W. If the charger was 100% efficient, which is impossible, then it would be able to charge a 14V battery at about 0.714Amps, the same amount of watts as it was taking from its input.
Typical efficiencies, however, are around 70-85%. 15 - 30% is lost in the conversion.
Let's go back to your example, let's assume you're charging at 1A from your 12V battery. That's 12Watts. Let's assume your charger has a great efficiency of 85%, that means the output power will be 0.85 * 12Watts, or 10.2 Watts. Since the battery voltage starts at 12V, that's also the voltage at which the charge starts. But let's ignore that fact and just assume it will be outputting 14V as you said. Convert it back, 10.2W / 14Volts equals about 0.73 Amps. So, the charger is consuming 1A from the battery, while charging at 0.73Amps. A net loss of 0.27 Amps, no charging will take place at all.
But let's change the thought experiment, and assume we've got a cryogenic freezer for free, which along with some magic pixie dust enabled us to have a 100% efficient battery charger.. Even then, the charger would not be outputting more power than in it was taking in. As the output voltage rises, either the output current will drop, or the input current will rise, giving us a net energy of 0.
OK, now let's add a second, identical battery to the thought experiment. The second battery is full, and we will use our impossible 100% efficient charger to charge the first battery. When that battery is full, we reverse it, and move our energy from one battery to the other. With the 100% efficient charger, you'd think that you wont lose energy? Wrong again. The battery doesn't have 100% charge efficiency. What that means is, that if you put in 100 Watthours of energy into the battery, some of that energy will be lost in the form of the battery warming up both during charge and discharge. With lead-acid batteries, some is also lost when small amounts of the water in the electrolyte gets converted to hydrogen and oxygen. 100 Watthours in, less than 100 watthours out. (Probably something like 80 Watthours out)
Back in the real world, what this means is that with a realistic and achievable 85% efficient charger, and a lead-acid battery with 80% charge efficiency, giving us a combined efficiency of 0.85 * 0.80 = 0.68, or 68%. That means, that, for example, from 100 Watthours that the charger has consumed, ultimately only 68 Watthours would be available from the lead-acid battery that it charged!
Power is measured in Watts, and energy in Watthours.
You get the watts by multiplying Voltage with Current.
If the charger input voltage is 10V, and the input current is 1A, that's 10W. If the charger was 100% efficient, which is impossible, then it would be able to charge a 14V battery at about 0.714Amps, the same amount of watts as it was taking from its input.
Typical efficiencies, however, are around 70-85%. 15 - 30% is lost in the conversion.
Let's go back to your example, let's assume you're charging at 1A from your 12V battery. That's 12Watts. Let's assume your charger has a great efficiency of 85%, that means the output power will be 0.85 * 12Watts, or 10.2 Watts. Since the battery voltage starts at 12V, that's also the voltage at which the charge starts. But let's ignore that fact and just assume it will be outputting 14V as you said. Convert it back, 10.2W / 14Volts equals about 0.73 Amps. So, the charger is consuming 1A from the battery, while charging at 0.73Amps. A net loss of 0.27 Amps, no charging will take place at all.
But let's change the thought experiment, and assume we've got a cryogenic freezer for free, which along with some magic pixie dust enabled us to have a 100% efficient battery charger.. Even then, the charger would not be outputting more power than in it was taking in. As the output voltage rises, either the output current will drop, or the input current will rise, giving us a net energy of 0.
OK, now let's add a second, identical battery to the thought experiment. The second battery is full, and we will use our impossible 100% efficient charger to charge the first battery. When that battery is full, we reverse it, and move our energy from one battery to the other. With the 100% efficient charger, you'd think that you wont lose energy? Wrong again. The battery doesn't have 100% charge efficiency. What that means is, that if you put in 100 Watthours of energy into the battery, some of that energy will be lost in the form of the battery warming up both during charge and discharge. With lead-acid batteries, some is also lost when small amounts of the water in the electrolyte gets converted to hydrogen and oxygen. 100 Watthours in, less than 100 watthours out. (Probably something like 80 Watthours out)
Back in the real world, what this means is that with a realistic and achievable 85% efficient charger, and a lead-acid battery with 80% charge efficiency, giving us a combined efficiency of 0.85 * 0.80 = 0.68, or 68%. That means, that, for example, from 100 Watthours that the charger has consumed, ultimately only 68 Watthours would be available from the lead-acid battery that it charged!
PCC
Flashlight Enthusiast
Wouldn't the simple answer be: yes, just not at the same time.
recycledelectrons
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I've noticed that (with my cheap hoby charger) I put in fewer mAH than I pull out of batteries, almost like the battery wa giving something up while cycling. I'm sure I'm missing something, but my repeated observations have been bugging me for a few weeks now.
Do you mean the voltage?
No I meant the current. I haven't checked voltage, although, I have noticed that he voltage that the hobbycharger measures through the charge leads is often different than what it measures through the balance leads, but the difference approaches zero as the battery approaches full.
hellokitty[hk]
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What charger do you have?
http://www.candlepowerforums.com/vb...s-Check-your-hobby-chargers-current-accuracy!
HKJ
Flashaholic
The current is not always that precise on a hobby charger.
With the voltage you have the problem with voltage drop in the leads, when you are running current at the same time you are doing the measurement. This will often give a significant error when doing discharge curves at multiple amperes on a single cell.
The Voltage measured on one of the balance leads will sometimes be different to the to the main charge leads as the charger balances that particular cell.
Norm
