Sorry SeaMonkey, I have it wrong, my memory is dull. My circuit is more efficient with higher impedance load, less efficient without load.
"I always notice that my efficiency increase with 3V compared to 12V" is wrong.
"A circuit that its efficiency increase when the load impedance reduced" should be "A circuit that its efficiency increase with a load".
I don't measure resistance, is it enough to show with and without load?
Video on different experiment:
YouTube - COP = 69% battery charging battery
YouTube - Swap Charging front and back battery
With load my circuit consume less power.
YouTube - Input current reduce when there is load
Same source, same circuit different load. Sorry, I have it backward:
old circuit, single:
new circuit, single.
amp meter is in series with source battery. My circuit that do this is modified joule thief and stingo. My circuit is unpredictable, sometime I need to wake it up / power it up with a touch of hand.
"I always notice that my efficiency increase with 3V compared to 12V" is wrong.
"A circuit that its efficiency increase when the load impedance reduced" should be "A circuit that its efficiency increase with a load".
Originally posted by SeaMonkey
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Video on different experiment:
YouTube - COP = 69% battery charging battery
YouTube - Swap Charging front and back battery
Originally posted by SeaMonkey
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YouTube - Input current reduce when there is load
Originally posted by SeaMonkey
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old circuit, single:
Source voltage is 11.5V.
Input current without load = 1 Amp
Input current with load = 2 Amp
Current in the load (charging current) = 0.25 Amp
Voltage in the load part:
without load = 90V
with load = 9.4V
on battery = 2.72V
on total load:
Efficiency = (load voltage*load current)/(source voltage*source current)
Efficiency = (9.4V * 0.25)/(11.5V * 2)
Efficiency = (2.35 Watt)/(23 Watt)
Efficiency = 0.1 ~ 10%
on battery only:
Efficiency = (load voltage*load current)/(source voltage*source current)
Efficiency = (2.72V * 0.25)/(11.5V * 2)
Efficiency = (0.68 Watt)/(23 Watt)
Efficiency = 0.029 ~ 3%
Input current without load = 1 Amp
Input current with load = 2 Amp
Current in the load (charging current) = 0.25 Amp
Voltage in the load part:
without load = 90V
with load = 9.4V
on battery = 2.72V
on total load:
Efficiency = (load voltage*load current)/(source voltage*source current)
Efficiency = (9.4V * 0.25)/(11.5V * 2)
Efficiency = (2.35 Watt)/(23 Watt)
Efficiency = 0.1 ~ 10%
on battery only:
Efficiency = (load voltage*load current)/(source voltage*source current)
Efficiency = (2.72V * 0.25)/(11.5V * 2)
Efficiency = (0.68 Watt)/(23 Watt)
Efficiency = 0.029 ~ 3%
new circuit, single.
Performance data charging two 1.2V nicad (1000mAh):
Input: 0.7A @ 10.4V = 7.28watt
Output: 0.45A @ 3.2V = 1.44watt (at full battery)
Efficiency: 19%
Performance data charging one 12V gel SLA (7Ah/20hr):
Input: 0.62A @ 10.5V = 6.51watt
Output: 0.24A @ 13.1V = 3.14watt(initial voltage raising before going down)
Efficiency: 48%
Input: 0.7A @ 10.4V = 7.28watt
Output: 0.45A @ 3.2V = 1.44watt (at full battery)
Efficiency: 19%
Performance data charging one 12V gel SLA (7Ah/20hr):
Input: 0.62A @ 10.5V = 6.51watt
Output: 0.24A @ 13.1V = 3.14watt(initial voltage raising before going down)
Efficiency: 48%
Originally posted by SeaMonkey
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