From: Patrick Jankowiak (eccm@swbell.net)
Date: Sat Sep 28 2002 - 16:05:08 PDT
I also forgot that the system charging voltage is probably more like 27.5VDC when the engine is running.
Well it is a start, and the idea was to do a slow charge based on the current going to zero at the desired voltage-limit. Maybe the resistance values and ratio could be shifted a bit? Hmm.
In any case, when the MV is running, you will get 28VDC at the switch, so the midpoint after the diode would be at 13.4V with no charging load (voltage limited charging). The problem is that without a
regulared something or another, or making the resistors lower in resistance (and thereby a bigger drain on the MV charging system) it would be likely to overcharge the 12V batery.
So go one step up to slightly higher tech:
Assuming 27.5VDC from the alternator, I was going to suggest a 13V 50 or 100 watt zener diode in place of the top resistor, to give 14.5V for charging.
(change the zener or add some silicon rectifiers at 0.6V each to reduce the voltage from 14.5 to whatever voltage you want the ultimate battery voltage to be when you stop the charging current)
Put a resistor in series with the zener, to limit the charge current. Assuming you want to charge with 14V, and the battery you use is fully charged at 12.7V:
zener diode rectifier 3 ohms resistor
__
|
| /| |\ |
27.5V----------------|< |---| >|___/\/\/\/\/\_____> 12V batt
| \| |/ |
|__ charging: 13.9V / 0A
| 12.7V / 0.4A
| | 11.9V / 0.6A
| 13.9V
14.5V
with a 1 ohm resistor,
13.9V / 0A
12.7V / 1.2A
11.9V / 2.0A
with no resistor? it would probably work fine, seeing as the wiring resitance is going to be a fraction of an ohm?
The considerations for the power ratings of the zener diode become considerable anyway. I am no expert on this, but I have done things like this by experiment.
Andreas Mehlhorn wrote:
>
> Patrick Jankowiak schrieb:
>
> > Low tech solution - wire two 15 ohm 15 watt resistors in series, bottom goes to ground. Top goes to MV powerswitch. Middle goes through a 100 volt 5 amp diode ($1.00) to feed the positive terminal of
> > the 12V battery. This is an old fashioned voltage splitter. The two resistors will get hot, but mount them under the dash somewhere. The regulation is not as good as the zener diode solution, but it
> > is alot cheaper and more 'period' as far as electronics goes.
> >
> > note: use fixed width font to see this properly, such as 'courier new'
> >
> > R1 R2
> > 24V from switch>__/\/\/\/\/\______/\/\/\/\/\____>ground
> > |
> > _|_
> > _\_/_
> > |
> > |
> > 12v battery
>
> Ok, very low tech. It works but remember, that the current you can get to charge your 12 V Battery
> is only around 0,2 A (200 milliamperes). At 12 Volts this is 2,4 Watts. Not very much!
>
> Why this: You have 2 x 15 Ohms on 24 Volts, this makes a current of 0,8 Amperes flow through
> both resistors. This means, that the two resistors make 19,8 Watts of heat all the time (0,8 A x 24 V)
>
> The current you may take off at the middle is normally 1/5th - 1/4th of the current going through the
> resistors. So you can take 0,2 x 0,8 A = 0,16 A or 0,25 x 0,8 A = 0,2 A.
>
> In reality you can take a little bit more but not much. If you take more, you will have less than 12 V
> on the middle connector and this means your 12 V battery won't charge.
>
> Best regards
> Andreas
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