John Urbanowski Amazing Radiant Battery Charger

[gotoluc]

Hi All,

the 2SC1768 is a discontinued type, though it is still available from some retailer. It is a 200V, 5A, high current gain NPN transistor

Here are some data on it from here http://www.datasheetarchive.com/data/2SC1768.html

2SC1768
  Si NPN Power BJT
   
V(BR)CEO (V)=150
V(BR)CBO (V)=200
I(C) Abs.(A) Collector Current=5.0
Absolute Max. Power Diss. (W)=50
I(CBO) Max. (A)=1.0m
h(FE) Min. Static Current Gain=400
h(FE) Max. Current gain.=3.0k
@I(C) (A) (Test Condition)=1.0
@V(CE) (V) (Test Condition)=4.0
f(T) Min. (Hz) Transition Freq=15M
Package=TO-3

A good substitue seems to be 2SD1090, see here http://www.alldatasheet.com/view.jsp?Searchword=2SD1090

its package which is different.

rgds,  Gyula

Thank you Gyula for the research on the transistor information and also finding a substitute.

Luc

[neptune]

When we talk about wire guage, do we mean SWG[ standard wire guage, used in the UK] or AWG { American wire guage] ? Swg has largely been replaced by wire diameter in millimetres. Note that in all systems, the diameter does not include the enamel insulation.

[gotoluc]

When we talk about wire guage, do we mean SWG[ standard wire guage, used in the UK] or AWG { American wire guage] ? Swg has largely been replaced by wire diameter in millimetres. Note that in all systems, the diameter does not include the enamel insulation.

Hi neptune,

it is AWG

Luc

[wattsup]

@gotoluc

Hi and good work with your friend. I do not want to put a damper on this idea but there is one main consideration of this whole circuit and that is he is using a 12vdc deep cycle marine battery in series with another 12vdc multi-celled battery. Total amperage availble is not known. But I would easily guesss it to be in the 600-800 ampere/hour range easy.

And from the consumption devices, from what I can tell, it is like hiring a fly to do a kamakasi flight onto the back of an elephant. It won't feel a thing.

What I am trying to say is the only way you can actually see something is to use smaller batteries so the effects of consumption can be measured in a more realistic time frame. The whole system would have to be scaled down to a more logical level.

Now if the circuit was drawing 40 amps at 24 volts, and the voltage off the batteries did not go down, this in my view would be something out of the ordinary. But so far,what I can understand is very normal for such high power batteries.

I have tried in the past running a 3hp motor in an RV loop with two 12 volts marine batteries and even then voltage went down about .1 volts every 10 minutes. The voltage draw was substantial enough to effect the batteries and show some loss in a normal time setting.

So what I am wondering is can this same experimentation be done with smaller 12 volts batteries, like in the 4 to 7 amp range so that if there are changes, they can be seen more quickly.

[gotoluc]

@gotoluc

Hi and good work with your friend. I do not want to put a damper on this idea but there is one main consideration of this whole circuit and that is he is using a 12vdc deep cycle marine battery in series with another 12vdc multi-celled battery. Total amperage availble is not known. But I would easily guesss it to be in the 600-800 ampere/hour range easy.

And from the consumption devices, from what I can tell, it is like hiring a fly to do a kamakasi flight onto the back of an elephant. It won't feel a thing.

What I am trying to say is the only way you can actually see something is to use smaller batteries so the effects of consumption can be measured in a more realistic time frame. The whole system would have to be scaled down to a more logical level.

Now if the circuit was drawing 40 amps at 24 volts, and the voltage off the batteries did not go down, this in my view would be something out of the ordinary. But so far,what I can understand is very normal for such high power batteries.

I have tried in the past running a 3hp motor in an RV loop with two 12 volts marine batteries and even then voltage went down about .1 volts every 10 minutes. The voltage draw was substantial enough to effect the batteries and show some loss in a normal time setting.

So what I am wondering is can this same experimentation be done with smaller 12 volts batteries, like in the 4 to 7 amp range so that if there are changes, they can be seen more quickly.

Hi  wattsup,  thanks for excellent and valid point!  You are on the right track. Below are a copy of posts from the Energetic Forum. I will ask John U. to do the test in purple below.

Luc


Here is a recomendation from Peter Lindemann from the Energetic Forum:

The proper procedure is to charge the batteries UP FULL and then let them settle to the 12.6 volt level.  Then discharge the battery with the headlamp to a LOADED voltage of 11.5 volts.  After turn off of the system, the battery should recover to above 12.1 volts.  Log both the length of time needed for the discharge and the current used by the lamp.  Then charge the battery back up to full and repeat the procedure of running the headlamp and the charger simultaneously, logging the data again.

My prediction is that the second procedure will drain the battery faster.  Unfortunately, there are many variables in this situation that are impossible to determine, such as the "exact charge state" of the batteries under test.  The size of the batteries he is using requires a testing procedure that could take MONTHS to determine the efficiency of the system properly.  One 6 hour run digging deep into the bottom reserves of a huge battery cannot produce any useful data.

Sorry,

Peter

Here is my reply:

Thank you again Peter for your reply and the test suggestion procedures needed here. This makes sense,  I will ask him to do to a fix load discharge test without the oscillator and calculate the time it takes for the loaded battery to get down to 11.5 volts. After a recharge he will retest with the same load but now with the oscillator working.

Something to note is that his load tests were at 12v so he was only loading one battery (marine) so one should not conclude that the power of 2 batteries are involve.

Luc